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-rw-r--r--macros/luatex/latex/japanese-mathformulas/japanese-mathformulas.sty3779
1 files changed, 3779 insertions, 0 deletions
diff --git a/macros/luatex/latex/japanese-mathformulas/japanese-mathformulas.sty b/macros/luatex/latex/japanese-mathformulas/japanese-mathformulas.sty
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index 0000000000..201d70314e
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+++ b/macros/luatex/latex/japanese-mathformulas/japanese-mathformulas.sty
@@ -0,0 +1,3779 @@
+\NeedsTeXFormat{LaTeX2e}%
+
+\ProvidesPackage{mathformula}[2022/9/30,Version 1.0.0]%
+
+\RequirePackage{luatexja}%
+\RequirePackage{luatexja-fontspec}%
+\RequirePackage{luatexja-otf}%
+%\RequirePackage[hiragino-pron,deluxe,expert,bold]{luatexja-preset}%
+\RequirePackage{amsmath,amssymb,siunitx,ifthen,xparse,tikz,mathtools,graphics}%
+\usetikzlibrary{arrows,shapes,intersections,calc,angles,decorations.shapes,arrows.meta,quotes,through,decorations.text}%
+
+\newcommand{\空行}{\vskip\baselineskip}%
+\newcommand{\半空行}{\vskip.5\baselineskip}%
+\newcommand{\証明開始}{\noindent\textgt{【証明】}\par}%
+\newcommand{\証明終了}{\@rightalign{\ (Q.E.D.)}\par}%
+\newcommand{\数式カンマスペース}{,\ }%
+\newcommand*{\@rightalign}[1]%
+ {%
+ \hspace{\parfillskip}%
+ \mbox{}\linebreak[0]\mbox{}%
+ \nolinebreak[4]\hfill\mbox{#1}%
+ }%
+\newcommand{\Ttyuukakko}[1]{\left(#1\right)}%
+\newcommand{\Ttyuubracket}[1]{\left[#1\right]}%
+\newcommand{\Tdaikakko}[1]{\left\{#1\right\}}%
+\newcommand{\Tzettaiti}[1]{\left|#1\right|}%
+\def\shikimaru#1{\text{\quad$\cdots\cdots$\,\ajMaru{#1}}}
+\let\originalbigtriangleup\bigtriangleup
+\def\bigtriangleup#1{\originalbigtriangleup{\mathrm{#1}}}
+\DeclareRobustCommand\bunsuu{\@ifstar{\bunsuu@}{\@@bunsuu}}
+\def\@@bunsuu#1#2{%
+ \dfrac{\lower.44ex\hbox{$\,#1\,$}}{\lower-.1ex\hbox{$\,#2\,$}}}%
+\newlength{\@tempdima@math}
+\newlength{\@tempdimb@math}
+\newlength{\@tempdimd@math@math}
+\newlength{\@tempdimw@math}
+\DeclareRobustCommand\EMvphantom{\@ifstar{\EMvphantom@}{\EM@vphantom}}
+\def\EM@vphantom{\@ifnextchar[{\@EMvphantom}{\@EMvphantom[\apnd@ht]}}
+\def\@EMvphantom[#1]{\@ifnextchar[{\@@EMvphantom[#1]}{%
+ \@@EMvphantom[#1][#1]}}
+\def\@@EMvphantom[#1][#2]#3{{%
+% \edef\apnd@ht{#1}\edef\apnd@dp{#2}%
+ \@ifundefined{hakobanpush}{%
+ \@@@EMvphantom{#3}\ignorespaces
+ }{%
+ \hakobanpush\@@@EMvphantom{#3}\hakobanpop\ignorespaces
+ }%
+ \setlength{\@tempdima@math}{\ht0+#1}%
+ \setlength{\@tempdimb@math}{\dp0+#2}%
+ \vrule width \z@ height \@tempdima@math depth \@tempdimb@math
+}}
+\def\@@@EMvphantom#1{%
+ \ifmmode
+ \setbox0=\hbox{{$#1$}}%
+ \else
+ \setbox0=\hbox{{#1}}%
+ \fi
+}%
+\def\EMvphantom@{\@ifnextchar[{\@EMvphantom@}{\@EMvphantom@[\apnd@ht]}}%
+\def\@EMvphantom@[#1]{\@ifnextchar[{\@@EMvphantom@[#1]}{%
+ \@@EMvphantom@[#1][#1]}}%
+\def\@@EMvphantom@[#1][#2]#3{{\smash{#3}}%
+ \@@EMvphantom[#1][#2]{#3}\ignorespaces}%
+\newsavebox{\@tempboxd@math}
+\long\def\@colonfor#1:=#2\do#3{% % \@for の区切り記号を : に変えたバージョン
+ \expandafter\def\expandafter\@fortmp\expandafter{#2}%
+ \ifx\@fortmp\@empty \else
+ \expandafter\@colonforloop#2:{\@nil}:\@nil\@@#1{#3}\fi}
+\long\def\@colonforloop#1:#2:#3\@@#4#5{\def#4{#1}\ifx #4\@nnil \else
+ #5\def#4{#2}\ifx #4\@nnil \else#5\@icolonforloop #3\@@#4{#5}\fi\fi}
+\long\def\@icolonforloop#1:#2\@@#3#4{\def#3{#1}\ifx #3\@nnil
+ \expandafter\@fornoop \else
+ #4\relax\expandafter\@icolonforloop\fi#2\@@#3{#4}}%
+\def\phrases@math{\renewcommand{\arraystretch}{1}\@ifnextchar<{\@phrases@math}{\@phrases@math<lr>}}
+\def\@phrases@math<#1>{\@ifnextchar[{\@@phrases@math<#1>}{\@@phrases@math<#1>[l]}}
+\def\@@phrases@math<#1>[#2]{\@ifnextchar({\@@@phrases@math<#1>[#2]}{\@@@phrases@math<#1>[#2](c)}}
+\def\@@@phrases@math<#1>[#2](#3){\@ifnextchar|{\@@@@phrases@math<#1>[#2](#3)}{\@@@phrases@math<#1>[#2](#3)|0pt|}}
+\def\@@@@phrases@math<#1>[#2](#3)|#4|#5{%
+ \savebox{\@tempboxd@math}{%
+ \EMvphantom*[5pt]{%
+ \ensuremath{%
+ \ifthenelse{\equal{#1}{lr}\OR\equal{#1}{l}}{\left\{}{\left.}%
+ \kern-.5ex%
+ \setbox8\hbox{\begin{tabular}{#2}#5\end{tabular}}%
+ \setlength{\@tempdimw@math}{\dp8+#4}%
+ \setbox9 \vtop to \@tempdimw@math {\leftskip0pt\rightskip0pt\hsize\wd8\noindent\smash{\usebox8}}%
+ \usebox9%
+ \kern-.5ex%
+ \ifthenelse{\equal{#1}{lr}\OR\equal{#1}{r}}{\right\}\ }{\right.}%
+ }%
+ }%
+ }%
+ \if#3t%
+ \setlength{\@tempdimd@math@math}{-\ht\@tempboxd@math + \baselineskip}%
+ \else
+ \if#3b%
+ \setlength{\@tempdimd@math@math}{\ht\@tempboxd@math - \baselineskip + 1pt}%
+ \else
+ \setlength{\@tempdimd@math@math}{0pt}%
+ \fi
+ \fi
+ \raisebox{\@tempdimd@math@math}{\usebox{\@tempboxd@math}}%
+}
+
+\NewDocumentCommand{\二次式展開}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{a+b}^2=a^2+2ab+b^2$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {\[\Ttyuukakko{a+b}^2=a^2+2ab+b^2\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{a-b}^2=a^2-2ab+b^2$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {\[\Ttyuukakko{a-b}^2=a^2-2ab+b^2\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{x-a}\Ttyuukakko{x+a}=x^2-a^2$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}
+ {\[\Ttyuukakko{x-a}\Ttyuukakko{x+a}=x^2-a^2\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{x+a}\Ttyuukakko{x+b}=x+\Ttyuukakko{a+b}x+ab$}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{b}}%
+ {\[\Ttyuukakko{x+a}\Ttyuukakko{x+b}=x+\Ttyuukakko{a+b}x+ab\]}{\relax}%
+ }%
+
+
+\NewDocumentCommand{\二次式因数分解}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$a^2+2ab+b^2=\Ttyuukakko{a+b}^2$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {\[a^2+2ab+b^2=\Ttyuukakko{a+b}^2\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$a^2-2ab+b^2=\Ttyuukakko{a-b}^2$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {\[a^2-2ab+b^2=\Ttyuukakko{a-b}^2\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {$x^2-a^2=\Ttyuukakko{x-a}\Ttyuukakko{x+a}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}
+ {\[x^2-a^2=\Ttyuukakko{x-a}\Ttyuukakko{x+a}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{i}}%
+ {$x+\Ttyuukakko{a+b}x+ab=\Ttyuukakko{x+a}\Ttyuukakko{x+b}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{b}}%
+ {\[x+\Ttyuukakko{a+b}x+ab=\Ttyuukakko{x+a}\Ttyuukakko{x+b}\]}{\relax}%
+
+ }%
+
+
+\NewDocumentCommand{\平方根}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{i}}%
+ {$a$は実数として,$\sqrt{a^2}=\Tzettaiti{a}$}%
+ {\relax}%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{b}}%
+ {%
+ $a$は実数として,%
+ \[\sqrt{a^2}=\Tzettaiti{a}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質A}\AND\equal{#2}{i}}%
+ {%
+ $a\geqq0$のとき,%
+ $\Ttyuukakko{\sqrt{a}}^2=\Ttyuukakko{-\sqrt{a}}^2=a\数式カンマスペース\sqrt{a}\leqq0$%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質A}\AND\equal{#2}{b}}%
+ {%
+ $a\leqq0$のとき,%
+ \[\Ttyuukakko{\sqrt{a}}^2=\Ttyuukakko{-\sqrt{a}}^2=a\数式カンマスペース\sqrt{a}\leqq0\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質B}\AND\equal{#2}{i}}%
+ {$\sqrt{a}=\Tzettaiti{a}$}%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質B}\AND\equal{#2}{b}}%
+ {\[\sqrt{a}=\Tzettaiti{a}\]}%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質C}\AND\equal{#2}{i}}%
+ {%
+ $a>0\数式カンマスペース b>0\数式カンマスペース a\neq b$のとき,%
+ $\sqrt{a}\sqrt{b}=\sqrt{ab}$%
+ }%
+ {\relax}
+ \ifthenelse{\equal{#1}{性質C}\AND\equal{#2}{b}}%
+ {%
+ $a>0\数式カンマスペース b>0\数式カンマスペース a\neq b$のとき,%
+ \[\sqrt{a}\sqrt{b}=\sqrt{ab}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質D}\AND\equal{#2}{i}}%
+ {$\bunsuu{\sqrt{a}}{\sqrt{b}}=\sqrt{\bunsuu{a}{b}}$}%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質D}\AND\equal{#2}{b}}%
+ {\[\bunsuu{\sqrt{a}}{\sqrt{b}}=\sqrt{\bunsuu{a}{b}}\]}%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質E}\AND\equal{#2}{i}}%
+ {$\sqrt{k^2a}=k\sqrt{a}$}%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質E}\AND\equal{#2}{b}}%
+ {\[\sqrt{k^2a}=k\sqrt{a}\]}%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\一次不等式}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{性質A}\AND\equal{#2}{i}}%
+ {$a<b$のとき\数式カンマスペース $a+c<b+c$}{\relax}%
+ \ifthenelse{\equal{#1}{性質A}\AND\equal{#2}{b}}%
+ {%
+ $a<b$のとき\数式カンマスペース %
+ \[a+c<b+c\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質B}\AND\equal{#2}{i}}%
+ {$c>0$のとき,$ac<bc$}{\relax}%
+ \ifthenelse{\equal{#1}{性質B}\AND\equal{#2}{b}}%
+ {%
+ $c>0$のとき,%
+ \[ac<bc\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質C}\AND\equal{#2}{i}}%
+ {$c<0$のとき,$ac>bc$}{\relax}%
+ \ifthenelse{\equal{#1}{性質C}\AND\equal{#2}{b}}%
+ {%
+ $c<0$のとき,%
+ \[ac>bc\]%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\集合}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{積集合}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{A\cap B}$}{\relax}%
+ \ifthenelse{\equal{#1}{積集合}\AND\equal{#2}{b}}%
+ {\[\Ttyuukakko{A\cap B}\]}{\relax}%
+ \ifthenelse{\equal{#1}{和集合}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{A\cup B}$}{\relax}%
+ \ifthenelse{\equal{#1}{和集合}\AND\equal{#2}{b}}%
+ {\[\Ttyuukakko{A\cup B}\]}{\relax}%
+ \ifthenelse{\equal{#1}{補集合}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{\overline{A}}$}{\relax}%
+ \ifthenelse{\equal{#1}{補集合}\AND\equal{#2}{b}}%
+ {\[\Ttyuukakko{\overline{A}}\]}{\relax}%
+ }%
+
+
+\NewDocumentCommand{\対偶}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定理}\AND\equal{#2}{i}}%
+ {%
+ $P$ならば$Q$の命題において,\par%
+ 逆は$Q$ならば$P$\par%
+ 裏は$P$でないならば$Q$でない\par%
+ 対偶は$Q$でないならば$P$でない\par%
+ 対偶と元の命題の真偽は一致する。%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{定理}\AND\equal{#2}{b}}%
+ {%
+ $P$ならば$Q$の命題において,\par%
+ 逆は$Q$ならば$P$\par%
+ 裏は$P$でないならば$Q$でない\par%
+ 対偶は$Q$でないならば$P$でない\par%
+ 対偶と元の命題の真偽は一致する。%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ 命題を「$p$ならば$q$」とし$p$の真理集合を$P$\数式カンマスペース $q$の真理集合を$Q$とする。\par%
+ 「$p$ならば$q$」が真のとき,$Q\subset P\Leftrightarrow\overline{P}\subset\overline{Q}$より対偶命題「$q$でないならば$p$でない」は真。\par%
+ 「$p$ならば$q$」が偽のとき,$Q\not\subset P\Leftrightarrow\overline{P}\not\subset\overline{Q}$より対偶命題「$q$でないならば$p$でない」は偽。\par
+ 従って,対偶命題と元の命題の真偽は一致する。%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\newcommand{\背理法}{命題$P$ならば$Q$に対して$P$でないならば$Q$と仮定して矛盾を示す。}%
+
+
+\NewDocumentCommand{\二次関数}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{標準形}\AND\equal{#2}{i}}%
+ {$y=a\Ttyuukakko{x-p}^2+q$}{\relax}%
+ \ifthenelse{\equal{#1}{標準形}\AND\equal{#2}{b}}%
+ {\[y=a\Ttyuukakko{x-p}^2+q\]}{\relax}%
+ \ifthenelse{\equal{#1}{一般形}\AND\equal{#2}{i}}%
+ {$y=ax^2+bx+c$}{\relax}%
+ \ifthenelse{\equal{#1}{一般形}\AND\equal{#2}{b}}%
+ {\[y=ax^2+bx+c\]}{\relax}%
+ \ifthenelse{\equal{#1}{切片形}\AND\equal{#2}{i}}%
+ {$y=a\Ttyuukakko{x-\alpha}\Ttyuukakko{x-\beta}$}{\relax}%
+ \ifthenelse{\equal{#1}{切片形}\AND\equal{#2}{b}}%
+ {\[y=a\Ttyuukakko{x-\alpha}\Ttyuukakko{x-\beta}\]}{\relax}%
+ \ifthenelse{\equal{#1}{平方完成}\AND\equal{#2}{i}}%
+ {$y=ax^2+bx+c$に対して,$y=a\Ttyuukakko{x+\bunsuu{b}{2a}}-\bunsuu{b^2-4ac}{4a}$}{\relax}%
+ \ifthenelse{\equal{#1}{平方完成}\AND\equal{#2}{b}}%
+ {%
+ $y=ax^2+bx+c$に対して,%
+ \[y=a\Ttyuukakko{x+\bunsuu{b}{2a}}-\bunsuu{b^2-4ac}{4a}\]%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\二次方程式の解の公式}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{i}}%
+ {$ax^2+bx+c=0 \Ttyuukakko{a\neq0}$に対して,$x=\bunsuu{-b\pm\sqrt{b^2-4ac}}{2a}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{b}}%
+ {%
+ $ax^2+bx+c=0 \Ttyuukakko{a\neq0}$に対して,%
+ \[x=\bunsuu{-b\pm\sqrt{b^2-4ac}}{2a}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{証明A}}%
+ {%
+ \証明開始%
+ \vspace{-1\zw}%
+ \begin{align*}%
+ ax^2+bx+c&=0&\\%
+ a\Ttyuukakko{x^2+\bunsuu{b}{a}x}+c&=0&\\%
+ a\Tdaikakko{\Ttyuukakko{x+\bunsuu{b}{2a}}^2-\bunsuu{b^2}{4a^2}}+c&=0&\\%
+ a\Ttyuukakko{x+\bunsuu{b}{2a}}^2-\bunsuu{b^2}{4a}+c&=0&\\%
+ \Ttyuukakko{x+\bunsuu{b}{2a}}^2&=\bunsuu{b^2-4ac}{4a^2}&\\%
+ x&=\bunsuu{-b\pm\sqrt{b^2-4ac}}{2a}%
+ \end{align*}%
+ \証明終了%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{証明B}}%
+ {%
+ \証明開始%
+ \vspace{-1\zw}%
+ \begin{align*}%
+ ax^2+bx+c&=0&\\%
+ 4a^2x^2+4abx+4ac&=0&\\%
+ \Ttyuukakko{2ax+b}^2-b^2+4ac&=0&\\%
+ 2ax+b&=\pm\sqrt{b^2-4ac}&\\%
+ x&=\bunsuu{-b\pm\sqrt{b^2-4ac}}{2a}%
+ \end{align*}%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\三角比の定義}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定義A}\AND\equal{#2}{i}}%
+ {%
+ \begin{tikzpicture}%
+ \draw(0,0)--(1.5,0)--(1.5,2)--cycle;%
+ \draw(0,0)node[below]{A};%
+ \draw(1.5,0)node[below]{B};%
+ \draw(1.5,2)node[above]{C};%
+ \draw(0,0)coordinate(A);%
+ \draw(1.5,0)coordinate(B);%
+ \draw(1.5,2)coordinate(C);%
+ \draw pic[draw,black,thin,angle radius=0.3cm]{right angle=A--B--C};%
+ \draw pic["$\theta$",draw=black,->,thin,angle eccentricity=1.4,angle radius=0.4cm]{angle=B--A--C};%
+ \end{tikzpicture}%
+ \空行%
+ 図の様な直角三角形ABCにおいて$\angle\mathrm{CAB}=\theta$のとき,%
+ \[%
+ \sin\theta=\bunsuu{\text{BC}}{\text{AC}}\数式カンマスペース%
+ \cos\theta=\bunsuu{\text{AB}}{\text{AC}}\数式カンマスペース%
+ \tan\theta=\bunsuu{\text{BC}}{\text{AB}}%
+ \]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{定義B}\AND\equal{#2}{i}}%
+ {%
+ \begin{tikzpicture}%
+ \draw(0,0)--(1.05,1.4);%
+ \draw[dashed](0.75,1)--(0,1);%
+ \draw(0,0)node[below right]{O};%
+ \draw(0.75,0)node[below]{$x$};%
+ \draw(0,1)node[left]{$y$};%
+ \draw(0.8,1)node[right]{P$\Ttyuukakko{x,y}$};%
+ \draw(0,0)circle[radius=1.25];%
+ \draw(0,-1.25)node[below left]{$-r$};%
+ \draw(-1.25,0)node[below left]{$-r$};%
+ \draw(0,1.25)node[above left]{$r$};%
+ \draw(1.25,0)node[below right]{$r$};%
+ \draw[->,>=stealth,semithick](-1.5,0)--(1.5,0)node[right]{$x$};%
+ \draw[->,>=stealth,semithick](0,-1.5)--(0,1.5)node[above]{$y$};%
+ \draw[dashed](0,0)coordinate(O)-- (0.75,0)coordinate(Q)-- (0.75,1)coordinate(P);%
+ \draw pic["$\theta$",draw=black,->,thin,angle eccentricity=1.4,angle radius=0.4cm] {angle=Q--O--P};%
+ \end{tikzpicture}%
+ \空行%
+ 図において%
+ \[\sin\theta=\bunsuu{y}{r}\数式カンマスペース\cos\theta=\bunsuu{x}{r}\数式カンマスペース\tan\theta=\bunsuu{y}{x}\]%
+ このとき,$r=1$にしても一般性を失わない。%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\三角比の相互関係}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$\sin^2\theta+\cos^2\theta=1$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {\[\sin^2\theta+\cos^2\theta=1\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$\tan\theta=\bunsuu{\sin\theta}{\cos\theta}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {\[\tan\theta=\bunsuu{\sin\theta}{\cos\theta}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {$1+\tan^2\theta=\bunsuu{1}{\cos^2\theta}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}%
+ {\[1+\tan^2\theta=\bunsuu{1}{\cos^2\theta}\]}{\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ \begin{tikzpicture}%
+ \draw(0,0)--(1.05,1.4);%
+ \draw[dashed](0.75,1)--(0,1);%
+ \draw(0,0)node[below right]{O};%
+ \draw(0.75,0)node[below]{$x$};%
+ \draw(0,1)node[left]{$y$};%
+ \draw(0.8,1)node[right]{P$\Ttyuukakko{x,y}$};%
+ \draw(0,0)circle[radius=1.25];%
+ \draw(0,-1.25)node[below left]{$-r$};%
+ \draw(-1.25,0)node[below left]{$-r$};%
+ \draw(0,1.25)node[above left]{$r$};%
+ \draw(1.25,0)node[below right]{$r$};%
+ \draw[->,>=stealth,semithick](-1.5,0)--(1.5,0)node[right]{$x$};%
+ \draw[->,>=stealth,semithick](0,-1.5)--(0,1.5)node[above]{$y$};%
+ \draw[dashed](0,0)coordinate(O)-- (0.75,0)coordinate(Q)-- (0.75,1)coordinate(P);%
+ \draw pic["$\theta$",draw=black, ->,thin,angle eccentricity=1.4,angle radius=0.4cm]{angle=Q--O--P};%
+ \end{tikzpicture}%
+ \空行%
+ 図において,$\sin\theta=\bunsuu{y}{r}\数式カンマスペース\quad\cos\theta=\bunsuu{x}{r}$より%
+ \[\sin^2\theta+\cos^2\theta=\bunsuu{y^2+x^2}{r^2}\]%
+ ここで,三平方の定理より$x^2+y^2=r^2$なので%
+ \[\sin^2\theta+\cos^2\theta=\bunsuu{r^2}{r^2}=1\]%
+ \空行%
+ $\sin\theta=\bunsuu{y}{r}\数式カンマスペース\quad\cos\theta=\bunsuu{x}{r}\quad\tan\theta=\bunsuu{y}{x}$より%
+ \[\bunsuu{\sin\theta}{\cos\theta}=\bunsuu{y}{x}=\tan\theta\]%
+ \空行%
+ $\sin^2\theta+\cos^2\theta=1$の両辺を$\cos^2\theta$で割ることで,%
+ \[\bunsuu{\sin^2\theta}{\cos^2\theta}+1=\bunsuu{1}{\cos^2\theta}\]%
+ ここで,$\bunsuu{\sin\theta}{\cos\theta}=\tan\theta$なので%
+ \[\tan^2\theta+1=\bunsuu{1}{\cos^2\theta}\]%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\ユークリッド幾何の公理}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公理A}\AND\equal{#2}{i}}%
+ {二つの異なる二点を与えることで,それを通る直線が一意的に決定する。}{\relax}%
+ \ifthenelse{\equal{#1}{公理B}\AND\equal{#2}{i}}%
+ {一つの直線$l$と$l$上にない一つの点が与えられたとき,与えられた点を通り,$l$に平行な直線をただ一つ引くことができる。}{\relax}%
+ }%
+
+
+\newcommand{\直線}{両方向に限りなく伸びたまっすぐな線。}%
+
+
+\newcommand{\線分}{直線ABのうち,二点A\数式カンマスペース Bを端とする部分。}%
+
+
+\newcommand{\半直線}{直線ABのうち,一方の点を端とし,もう一方に限りなく伸びた部分。}%
+
+
+\newcommand{\距離}
+ {%
+ 空でない集合Xの元$x\数式カンマスペース y$にたいして,実数値$d(x\数式カンマスペース y)$が定義され,%
+ \[d(x\数式カンマスペース y)=0\Leftrightarrow x=y\数式カンマスペース\quad(x\数式カンマスペース y)=d(y\数式カンマスペース x)\数式カンマスペース\quad(x\数式カンマスペース y)\leqq d(x\数式カンマスペース y)+d(y\数式カンマスペース x)\]%
+ の三つの性質を満たす$d$をX上の距離といい,$(\text{X}\数式カンマスペース d)$を距離空間という。 %
+ }%
+
+
+\newcommand{\円}{平面上の一点から等しい距離にある点の集合。}%
+
+
+\newcommand{\弧}{円周上の二点A\数式カンマスペース Bに対して,A\数式カンマスペース Bによって分けられた円周の各々の部分を弧ABといい,$\overarc{AB}$と表す。}%
+
+
+\newcommand{\弦}{弧の両端を結んだ線分。}%
+
+
+\newcommand{\中心角}{円の中心を頂点として,2辺が弧の両端を通る角を,その弧に対する中心角という。}%
+
+
+\NewDocumentCommand{\対頂角}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{i}}%
+ {%
+ \begin{tikzpicture}%
+ \draw(0,0)--(2,2);%
+ \draw(2,0)--(0,2);%
+ \draw(0,0)coordinate(O);%
+ \draw(2,2)coordinate(A);%
+ \draw(2,0)coordinate(B);%
+ \draw(0,2)coordinate(C); %
+ \draw(1,1)coordinate(D);%
+ \draw pic["A",draw=black,-,thin,angle eccentricity=1.4,angle radius=0.4cm] {angle=A--D--C};%
+ \draw pic["B",draw=black,-,thin,angle eccentricity=1.4,angle radius=0.4cm] {angle=O--D--B};%
+ \end{tikzpicture}%
+ \空行%
+ 図において,$\angle\mathrm{A}$と$\angle\mathrm{B}$を対頂角という。%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質}\AND\equal{#2}{i}}%
+ {対頂角は等しい。}{\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ \begin{tikzpicture}%
+ \draw(0,0)--(2,2);%
+ \draw(2,0)--(0,2);%
+ \draw(0,0)coordinate(O);%
+ \draw(2,2)coordinate(A);%
+ \draw(2,0)coordinate(B);%
+ \draw(0,2)coordinate(C);%
+ \draw(1,1)coordinate(D);%
+ \draw pic["A",draw=black,-,thin,angle eccentricity=1.4,angle radius=0.4cm] {angle=A--D--C};%
+ \draw pic["\,C",draw=black,-,thin,angle eccentricity=1.4,angle radius=0.3cm] {angle=B--D--A};%
+ \draw pic["B",draw=black,-,thin,angle eccentricity=1.4,angle radius=0.4cm] {angle=O--D--B};%
+ \end{tikzpicture}%
+ \空行%
+ \[180^\circ =\angle\mathrm{A}+\angle\mathrm{C}\]%
+ \[180^\circ=\angle\mathrm{B}+\angle\mathrm{C}\]%
+ \[\Leftrightarrow\angle\mathrm{A}=\angle\mathrm{B}\]%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\錯角}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{i}}%
+ {%
+ \begin{tikzpicture}%
+ \draw(-1,-0.5)--(2,1);%
+ \draw(-1,-1)--(2,-1);%
+ \draw(0,-2)--(2,2);%
+ \draw(2,2)coordinate(A);%
+ \draw(1.3333,0.66666)coordinate(B);%
+ \draw(2,1)coordinate(C);%
+ \draw(2,-1)coordinate(D);%
+ \draw(0.5,-1)coordinate(E);%
+ \draw(0,-2)coordinate(F);%
+ \draw(-1,-1)coordinate(G);%
+ \draw(-1,-0.5)coordinate(H);%
+ \draw pic["\,A",draw=black,-,thin,angle eccentricity=1.4,angle radius=0.3cm] {angle=E--B--C};%
+ \draw pic["B\,\,\,",draw=black,-,thin,angle eccentricity=1.4,angle radius=0.3cm] {angle=B--E--G};%
+ \end{tikzpicture}
+ \空行%
+ 図において,$\angle\mathrm{A}$と$\angle\mathrm{B}$を錯角という。%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質}\AND\equal{#2}{i}}%
+ {直線$l\数式カンマスペース m$において,錯角が等しい$\Leftrightarrow$直線$l,m$は平行。}{\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ \begin{tikzpicture}%
+ \draw(-1,1)--(2,1);%
+ \draw(-1,-1)--(2,-1);%
+ \draw(0,-2)--(2,2);%
+ \draw(2,2)coordinate(A);%
+ \draw(1.5,1)coordinate(B);%
+ \draw(2,1)coordinate(C);%
+ \draw(2,-1)coordinate(D);%
+ \draw(0.5,-1)coordinate(E);%
+ \draw(0,-2)coordinate(F);%
+ \draw(-1,-1)coordinate(G);%
+ \draw(-1,1)coordinate(H);%
+ \draw pic["\,\,\,A",draw=black,-,thin,angle eccentricity=1.4,angle radius=0.3cm] {angle=E--B--C};%
+ \draw pic["B\,\,\,",draw=black,-,thin,angle eccentricity=1.4,angle radius=0.3cm] {angle=A--E--G};%
+ \draw pic["C\,\,\,",draw=black,-,thin,angle eccentricity=1.4,angle radius=0.3cm] {angle=A--B--H};%
+ \end{tikzpicture}
+ \空行%
+ \begin{enumerate}%
+ \item 「平行ならば錯角が等しい」の証明。%
+ \空行%
+ 対頂角は等しいので,%
+ \[\angle\mathrm{A}=\angle\mathrm{C}\]%
+ ここで,$\angle\mathrm{B}$と$\angle\mathrm{C}$は同位角なので等しいので,%
+ \[\angle\mathrm{A}=\angle\mathrm{B}\]%
+ \item 「錯角が等しいならば平行」の証明。%
+ \空行%
+ 錯角が等しいので,%
+ \[\angle\mathrm{A}=\angle\mathrm{B}\]%
+ 対頂角は等しいので,%
+ \[\angle\mathrm{A}=\angle\mathrm{C}\]%
+ \[\Leftrightarrow\angle\mathrm{C}=\angle\mathrm{B}\]%
+ 即ち,同位角が等しいので二直線は平行。%
+ \end{enumerate}%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+\NewDocumentCommand{\同位角}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{i}}%
+ {%
+ \begin{tikzpicture}%
+ \draw(-1,-0.5)--(2,1);%
+ \draw(-1,-1)--(2,-1);%
+ \draw(0,-2)--(2,2);%
+ \draw(2,2)coordinate(A);%
+ \draw(1.3333,0.66666)coordinate(B);%
+ \draw(2,1)coordinate(C);%
+ \draw(2,-1)coordinate(D);%
+ \draw(0.5,-1)coordinate(E);%
+ \draw(0,-2)coordinate(F);%
+ \draw(-1,-1)coordinate(G);%
+ \draw(-1,-0.5)coordinate(H);%
+ \draw pic["\,\,A",draw=black,-,thin,angle eccentricity=1.4,angle radius=0.4cm] {angle=C--B--A};%
+ \draw pic["\,\,B",draw=black,-,thin,angle eccentricity=1.4,angle radius=0.4cm] {angle=D--E--B};%
+ \end{tikzpicture}
+ \空行%
+ 図において,$\angle\mathrm{A}$と$\angle\mathrm{B}$を同位角という。
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公理}\AND\equal{#2}{i}}%
+ {直線$l,m$において,同位角が等しい$\Leftrightarrow$直線$l\数式カンマスペース m$は平行。}{\relax}%
+ }%
+
+
+\NewDocumentCommand{\正弦定理}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{i}}%
+ {三角形ABCの外接円の半径をRとして,$\bunsuu{a}{\sin\text{A}}=2\text{R}\text{\ (}b\数式カンマスペース\text{B
+}\数式カンマスペース c\数式カンマスペース\text{Cについても同様に成立})$}{\relax}%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{b}}%
+ {%
+ 三角形ABCの外接円の半径をRとして,%
+ \[\bunsuu{a}{sin\text{A}}=2\text{R}\]%
+ ($b,B,c,C$についても同様に成立)%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ \空行%
+ \begin{tikzpicture}%
+ \draw(0,1.25)coordinate(A)-- (1,-0.75)coordinate(C)-- (-1,-0.75)coordinate(B);%
+ \draw pic["A",draw=black,-,thin,angle eccentricity=1.4,angle radius=0.4cm] {angle=B--A--C};%
+ \draw(-1,0.75)coordinate(D);%
+ \draw pic["D",draw=black,-,thin,angle eccentricity=1.4,angle radius=0.4cm] {angle=B--D--C};%
+ \draw(0,1.25)--(1,-0.75)--(-1,-0.75)--cycle;%
+ \draw(0,1.25)node[above]{A};%
+ \draw(1,-0.75)node[below]{C};%
+ \draw(-1,-0.75)node[left]{B};%
+ \draw(-1,0.75)node[left]{D};%
+ \draw(-1,0.75)--(1,-0.75)--(-1,-0.75)--cycle;%
+ \draw [line width=0.2pt] (B) to [bend right=27] node [fill=white,midway] { $a$ }(C);%
+ \draw(0,0)circle[radius=1.25];%
+ \draw pic[draw,black,thin,angle radius=0.3cm] {right angle=D--B--C};%
+ \draw(0,0)node[above]{O};%
+ \draw(0,0)coordinate(O);%
+ \fill[black](O)circle(0.03);%
+ \end{tikzpicture}%
+ \空行%
+ 図において円周角の定理より,%
+ \[\angle\mathrm{A}=\angle\mathrm{D}\]%
+ なので,円Oの半径をRとして$\sin\text{A}=\sin\text{D}=\bunsuu{a}{2\text{R}}$より,%
+ \[\bunsuu{a}{\sin\text{A}}=2\text{R}\]%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\余弦定理}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{i}}%
+ {三角形ABCにおいて,$a^2=b^2+c^2-2bc\cos\text{A}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{b}}%
+ {%
+ 三角形ABCにおいて,%
+ \[a^2=b^2+c^2-2bc\cos\text{A}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ \空行%
+ \begin{tikzpicture}%
+ \draw(0,0)--(2.5,0)--(1.5,2)--cycle;%
+ \draw(0,0)node[below]{A};%
+ \draw(1.5,2)node[above]{B};%
+ \draw(2.5,0)node[below]{C};%
+ \draw(1.5,0)node[below]{H};%
+ \draw(0,0)coordinate(A);%
+ \draw(1.5,0)coordinate(H);%
+ \draw(1.5,2)coordinate(B);%
+ \draw(1.5,0)--(1.5,2);%
+ \draw pic[draw,black,thin,angle radius=0.3cm] {right angle=A--H--B};%
+ \draw pic["$A$",draw=black,-,thin,angle eccentricity=1.4,angle radius=0.4cm] {angle=H--A--B};%
+ \end{tikzpicture}%
+ \空行%
+ 図において$\text{BC}=a,\text{CA}=b,\text{AC}=c$として,%
+ \[\text{BH}=c\sin\text{A},\quad\text{AH}=c\cos\text{A}\]%
+ また,三角形BHCに三平方の定理を用いることにより%
+ \[\text{CB}^2=\text{BH}^2+\text{HC}^2\]%
+ ここで,$\text{HC}=\text{AC}-\text{AH}=b-c\cos\text{A},\quad\text{BH}=c\sin\text{A}$より%
+ \begin{align*}%
+ a^2&=\Ttyuukakko{c\sin\text{A}}^2+\Ttyuukakko{b-c\cos\text{A}}^2&\\%
+ &=c^2\sin^2\text{A}+b^2-2bc\cos\text{A}+c^2\cos^2\text{A}&\\%
+ &=c^2\Ttyuukakko{1-\cos^2\text{A}}+b^2-2bc\cos\text{A}+c^2\cos^2\text{A}&\\%
+ &=b^2+c^2-2bc\cos\text{A}%
+ \end{align*}%
+ よって,%
+ \[a^2=b^2+c^2-2bc\cos\text{A}\]%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\三角形の面積}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{i}}%
+ {三角形ABCの面積を$S$として,$S=\bunsuu{1}{2}bc\sin\text{A}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{b}}%
+ {%
+ 三角形ABCの面積を$S$として,%
+ \[S=\bunsuu{1}{2}bc\sin\text{A}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ \begin{tikzpicture}%
+ \draw(0,0)--(2.5,0)--(1.5,2)--cycle;%
+ \draw(0,0)node[below]{A};%
+ \draw(1.5,2)node[above]{B};%
+ \draw(2.5,0)node[below]{C};%
+ \draw(1.5,0)node[below]{H};%
+ \draw(0,0)coordinate(A);%
+ \draw(1.5,0)coordinate(H);%
+ \draw(1.5,2)coordinate(B);%
+ \draw(1.5,0)--(1.5,2);%
+ \draw pic[draw,black,thin,angle radius=0.3cm] {right angle=A--H--B};%
+ \draw pic["$A$",draw=black,-,thin,angle eccentricity=1.4,angle radius=0.4cm] {angle=H--A--B};%
+ \end{tikzpicture}%
+ \空行%
+ 図において%
+ \[\text{BC}=a\数式カンマスペース\text{CA}=B\数式カンマスペース\text{AC}=c\]%
+ また,三角形ABCの面積を$S$として$S=\bunsuu{1}{2}\text{AC}\times\text{BH}$と,$AB\sin\text{A}=\text{BH}$から,%
+ \[S=\bunsuu{1}{2}bc\sin\text{A}\]%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\場合の数と確率}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{和集合の要素の個数}\AND\equal{#2}{i}}%
+ {$n\Ttyuukakko{A\cup B}=n\Ttyuukakko{A}+n\Ttyuukakko{B}-n\Ttyuukakko{A\cap B}$}{\relax}%
+ \ifthenelse{\equal{#1}{和集合の要素の個数}\AND\equal{#2}{b}}%
+ {\[n\Ttyuukakko{A\cup B}=n\Ttyuukakko{A}+n\Ttyuukakko{B}-n\Ttyuukakko{A\cap B}\]}{\relax}%
+ \ifthenelse{\equal{#1}{補集合の要素の個数}\AND\equal{#2}{i}}%
+ {全体集合を$U$として,$n\Ttyuukakko{\overline{A}}=n\Ttyuukakko{U}-n\Ttyuukakko{A}$}{\relax}%
+ \ifthenelse{\equal{#1}{補集合の要素の個数}\AND\equal{#2}{b}}%
+ {全体集合を$U$として,\[n\Ttyuukakko{\overline{A}}=n\Ttyuukakko{U}-n\Ttyuukakko{A}\]}{\relax}%
+ \ifthenelse{\equal{#1}{和の法則}\AND\equal{#2}{i}}%
+ {二つの事象A\数式カンマスペース Bにたいして,Aの起こりかたが$a$通り,Bの起こりかたが$b$通りのとき,AまたはBの起こる場合の数は$a+b$通り}{\relax}%
+ \ifthenelse{\equal{#1}{和の法則}\AND\equal{#2}{b}}%
+ {二つの事象A\数式カンマスペース Bにたいして,Aの起こりかたが$a$通り,Bの起こりかたが$b$通りのとき,AまたはBの起こる場合の数は$a+b$通り}{\relax}%
+ \ifthenelse{\equal{#1}{積の法則}\AND\equal{#2}{i}}%
+ {二つの事象A\数式カンマスペース Bにたいして,Aの起こりかたが$a$通り,Bの起こりかたが$b$通りのとき,AかつBの起こる場合の数は$ab$通り}{\relax}%
+ \ifthenelse{\equal{#1}{積の法則}\AND\equal{#2}{b}}%
+ {二つの事象A\数式カンマスペース Bにたいして,Aの起こりかたが$a$通り,Bの起こりかたが$b$通りのとき,AかつBの起こる場合の数は$ab$通り}{\relax}%
+ \ifthenelse{\equal{#1}{順列}\AND\equal{#2}{i}}%
+ {異なる$n$個のものから$r$個選んで並べる場合の数は${}_{n}P_{r}=\bunsuu{n!}{\Ttyuukakko{n-r}!}$}{\relax}%
+ \ifthenelse{\equal{#1}{順列}\AND\equal{#2}{b}}%
+ {%
+ 異なる$n$個のものから$r$個選んで並べる場合の数は%
+ \[{}_{n}P_{r}=\bunsuu{n!}{\Ttyuukakko{n-r}!}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{順列の証明}\AND\equal{#2}{i}}%
+ {%
+ \証明開始%
+ 異なる$n$個のものから$r$個選んで並べる場合の数は,%
+ \[n\times\Ttyuukakko{n-1}\times\Ttyuukakko{n-2}\times\cdots\Ttyuukakko{n-r+1}=\bunsuu{n!}{\Ttyuukakko{n-r}!}\]%
+ ここで,$\bunsuu{n!}{\Ttyuukakko{n-r}!}$を${}_{n} P_{r}$と表す。%
+ \証明終了%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{円順列}\AND\equal{#2}{i}}%
+ {異なる$n$個のものを円に並べる場合の数は$\Ttyuukakko{n-1}!$}{\relax}%
+ \ifthenelse{\equal{#1}{円順列}\AND\equal{#2}{b}}%
+ {異なる$n$個のものを円に並べる場合の数は\[\Ttyuukakko{n-1}!\]}{\relax}%
+ \ifthenelse{\equal{#1}{円順列の証明}\AND\equal{#2}{i}}%
+ {%
+ \証明開始%
+ $n$個のものを円形に並べるとき,1つを固定して考えると,残り$n-1$個を並べる順列の個数に等しい。よって$\Ttyuukakko{n-1}!$通りとなる。%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{重複順列}\AND\equal{#2}{i}}%
+ {$n$個から$r$個,重複を許して並べる場合の数は$n^r$}{\relax}%
+ \ifthenelse{\equal{#1}{重複順列}\AND\equal{#2}{b}}%
+ {$n$個から$r$個,重複を許して並べる場合の数は\[n^r\]}{\relax}%
+ \ifthenelse{\equal{#1}{組み合わせ}\AND\equal{#2}{i}}%
+ {異なる$n$個のものから$r$個選ぶ場合の数は,${}_{n}C_{r}=\bunsuu{n!}{r!\Ttyuukakko{n-r}!}$}{\relax}%
+ \ifthenelse{\equal{#1}{組み合わせ}\AND\equal{#2}{b}}%
+ {%
+ 異なる$n$個のものから$r$個選ぶ場合の数は,%
+ \[{}_{n}C_{r}=\bunsuu{n!}{r!\Ttyuukakko{n-r}!}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{組み合わせの証明}\AND\equal{#2}{i}}%
+ {%
+ \証明開始%
+ 異なる$n$個のものから$r$個選ぶ場合の数は,順列を重複度で割ったものなので%
+ \[\bunsuu{{}_{n} P_{r}}{r!}=\bunsuu{n!}{r!\Ttyuukakko{n-r}!}\]%
+ ここで,$\bunsuu{n!}{r!\Ttyuukakko{n-r}!}$を${}_{n}C_{r}$と表す。
+ \証明終了%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{同じものを含む順列}\AND\equal{#2}{i}}%
+ {aが$p$個,bが$q$個,cが$r$個,とある時,それら全部を並べる場合の数は,$\bunsuu{n!}{p!q!r!}$(ただし,$p+q+r=n$)}{\relax}%
+ \ifthenelse{\equal{#1}{同じものを含む順列}\AND\equal{#2}{b}}%
+ {%
+ aが$p$個,bが$q$個,cが$r$個,とある時,それら全部を並べる場合の数は,%
+ \[\bunsuu{n!}{p!q!r!}\text{\ (ただし,$p+q+r=n$)}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{同じものを含む順列の証明}\AND\equal{#2}{i}}%
+ {%
+ \証明開始%
+ $n$個のものを並べる場合の数は$n!$通りだが,$n$個の中に同じものが含まれているので,重複度で割ることで$\bunsuu{n!}{p!q!r!}$を得る。%
+ \証明終了%
+ }%
+ {\relax}%
+
+ \ifthenelse{\equal{#1}{確率の定義}\AND\equal{#2}{i}}%
+ {全事象$U$のどの根元事象も同様に確からしいとき,事象Aの起こる確率は,$P\Ttyuukakko{A}=\bunsuu{n\Ttyuukakko{A}}{n\Ttyuukakko{U}}$}{\relax}%
+ \ifthenelse{\equal{#1}{確率の定義}\AND\equal{#2}{b}}%
+ {%
+ 全事象$U$のどの根元事象も同様に確からしいとき,事象Aの起こる確率は,%
+ \[P\Ttyuukakko{A}=\bunsuu{n\Ttyuukakko{A}}{n\Ttyuukakko{U}}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{排反の定義}\AND\equal{#2}{i}}%
+ {事象A\数式カンマスペース Bが同時に起こりえないとき}{\relax}%
+ \ifthenelse{\equal{#1}{排反の定義}\AND\equal{#2}{b}}%
+ {事象A\数式カンマスペース Bが同時に起こりえないとき}{\relax}%
+ \ifthenelse{\equal{#1}{確率の性質A}\AND\equal{#2}{i}}%
+ {任意の事象Aに対して,$0\leqq A\leqq1$}{\relax}%
+ \ifthenelse{\equal{#1}{確率の性質A}\AND\equal{#2}{b}}%
+ {%
+ 任意の事象Aに対して,%
+ \[0\leqq A\leqq1\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{確率の性質B}\AND\equal{#2}{i}}%
+ {全事象Uの確率$P\Ttyuukakko{U}=1$}{\relax}%
+ \ifthenelse{\equal{#1}{確率の性質B}\AND\equal{#2}{b}}%
+ {%
+ 全事象Uの確率%
+ \[P\Ttyuukakko{U}=1\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{和事象の確率}\AND\equal{#2}{i}}%
+ {$P\Ttyuukakko{A\cup B}=P\Ttyuukakko{A}+P\Ttyuukakko{B}-P\Ttyuukakko{A\cap B}$}{\relax}%
+ \ifthenelse{\equal{#1}{和事象の確率}\AND\equal{#2}{b}}%
+ {\[P\Ttyuukakko{A\cup B}=P\Ttyuukakko{A}+P\Ttyuukakko{B}-P\Ttyuukakko{A\cap B}\]}{\relax}%
+ \ifthenelse{\equal{#1}{余事象の確率}\AND\equal{#2}{i}}%
+ {$P\Ttyuukakko{\overline{A}}=1-P\Ttyuukakko{A}$}{\relax}%
+ \ifthenelse{\equal{#1}{余事象の確率}\AND\equal{#2}{b}}%
+ {\[P\Ttyuukakko{\overline{A}}=1-P\Ttyuukakko{A}\]}{\relax}%
+ \ifthenelse{\equal{#1}{独立な事象の確率}\AND\equal{#2}{i}}%
+ {事象AとBが独立のとき,事象Aが起こりかつ事象Bが起こる確率$p$は,$p=P\Ttyuukakko{A}P\Ttyuukakko{B}$}{\relax}%
+ \ifthenelse{\equal{#1}{独立な事象の確率}\AND\equal{#2}{b}}%
+ {%
+ 事象AとBが独立のとき,事象Aが起こりかつ事象Bが起こる確率$p$は,%
+ \[p=P\Ttyuukakko{A}P\Ttyuukakko{B}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{反復試行の確率}\AND\equal{#2}{i}}%
+ {一回の試行で事象Aの起こる確率を$p$として,この試行を$n$回行う反復試行でAが$r$回起こる確率は,${}_{n}C_{r}\Ttyuukakko{p}^r\Ttyuukakko{1-p}^{n-r}$}{\relax}%
+ \ifthenelse{\equal{#1}{反復試行の確率}\AND\equal{#2}{b}}%
+ {%
+ 一回の試行で事象Aの起こる確率を$p$として,この試行を$n$回行う反復試行でAが$r$回起こる確率は,%
+ \[{}_{n}C_{r}\Ttyuukakko{p}^r\Ttyuukakko{1-p}^{n-r}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{反復試行の確率の証明}\AND\equal{#2}{i}}%
+ {%
+ \証明開始%
+ $n$回の試行のうち事象Aが$r$回起こる順番の場合の数は${}_{n} C_{r}$通り。さらに,Aが起こる確率は$p$で$r$回起こり,Aの余事象が起こる確率は$p-1$で$n-r$回起こるので,%
+ \[{}_{n}C_{r}\Ttyuukakko{p}^r\Ttyuukakko{1-p}^{n-r}\]%
+ となる。
+ \証明終了%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{条件付き確率}\AND\equal{#2}{i}}%
+ {事象Aが起こったときの事象Bの起こる確率は,$P_{A}\Ttyuukakko{B}=\bunsuu{P\Ttyuukakko{A\cap B}}{P\Ttyuukakko{A}}$}{\relax}%
+ \ifthenelse{\equal{#1}{条件付き確率}\AND\equal{#2}{b}}%
+ {%
+ 事象Aが起こったときの事象Bの起こる確率は,%
+ \[P_{A}\Ttyuukakko{B}=\bunsuu{P\Ttyuukakko{A\cap B}}{P\Ttyuukakko{A}}\]%
+ }%
+ {\relax}%
+ }%
+
+
+\newcommand{\図形の性質}[1]%
+ {%
+ \ifthenelse{\equal{#1}{内心}}%
+ {%
+ \begin{tikzpicture}%
+ \draw(-2.4,0.2)--(0,2)--(0,-3)--cycle;%
+ \draw(-1,0)circle[radius=1];%
+ \draw(-2.4,0.2)node[left]{A};%
+ \draw(0,2)node[above]{B};%
+ \draw(0,-3)node[below]{C};%
+ \draw(-1,0)node[below]{O};%
+ \end{tikzpicture}%
+ \空行%
+ 図においてOが内心%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{外心}}%
+ {%
+ \begin{tikzpicture}%
+ \draw(-2,0)--(1.6,1.2)--(1.2,-1.6)--cycle;%
+ \draw(0,0)circle[radius=2];%
+ \draw(0,0)node[below]{O};%
+ \draw(-2,0)node[left]{A};%
+ \draw(1.6,1.2)node[right]{B};%
+ \draw(1.2,-1.6)node[below]{C};%
+ \end{tikzpicture}%
+ \空行%
+ 図においてOが外心%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{垂心}}%
+ {%
+ \begin{tikzpicture}%
+ \draw(0,0)--(3,0)--(2,3)--cycle;%
+ \draw(0,0)--(2.7,0.9);%
+ \draw(2,3)--(2,0);%
+ \draw(3,0)--(0.92,1.4);%
+ \draw(0,0)node[below]{A};%
+ \draw(3,0)node[below]{C};%
+ \draw(2,3)node[above]{B};%
+ \draw(2.7,0.9)node[right]{P};%
+ \draw(2,0)node[below]{Q};%
+ \draw(0.92309,1.46154)node[left]{R};%
+ \draw(2,1)node[right]{H};%
+ \end{tikzpicture}%
+ \空行%
+ 図においてHが垂心%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{重心}}%
+ {%
+ \begin{tikzpicture}%
+ \draw(0,0)--(3,0)--(2,3)--cycle;%
+ \draw(0,0)--(2.5,1.5);%
+ \draw(3,0)--(1,1.5);%
+ \draw(2,3)--(1.5,0);%
+ \draw(0,0)node[below]{A};%
+ \draw(3,0)node[below]{B};%
+ \draw(2,3)node[above]{C};%
+ \draw(1.5,0)node[below]{D};%
+ \draw(2.5,1.5)node[right]{E};%
+ \draw(1,1.5)node[left]{F};%
+ \draw(1.75,1)node[right]{G};%
+ \end{tikzpicture}%
+ \空行%
+ 図においてGが重心%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{傍心}}%
+ {%
+ \begin{tikzpicture}
+ \draw(-0.4,2.8)--(-1,2)--(0.666666,2)--cycle;%
+ \draw(-0.4,2.8)node[above]{A};%
+ \draw(-1,2)node[left]{B};%
+ \draw(0.666666,2)node[right]{C};%
+ \draw(-0.4,2.8)--(-2.8,-0.4);%
+ \draw(-2.8,-0.4)node[below]{H};%
+ \draw(-0.4,2.8)--(2.8,0.4);%
+ \draw(2.8,0.4)node[right]{G};%
+ \draw(-0.4,2.8)--(0,0);%
+ \draw(0,0)node[below]{I};%
+ \draw(-1,2)--(0,0);%
+ \draw(0.666666,2)--(0,0);%
+ \draw(0,2)node[above]{D};%
+ \draw(-1.6,1.2)node[left]{E};%
+ \draw(1.2,1.6)node[right]{F};%
+ \draw(0,0)circle[radius=2];%
+ \end{tikzpicture}%
+ \空行%
+ 図においてIが傍心%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{チェバの定理}}%
+ {%
+ \begin{tikzpicture}
+ \draw(2,3)--(0,0)--(3,0)--cycle;%
+ \draw(2,3)node[above]{A};%
+ \draw(2,3)--(1.5,0);%
+ \draw(0,0)node[below]{B};%
+ \draw(0,0)--(2.75,0.75);%
+ \draw(3,0)node[below]{C};%
+ \draw(3,0)--(0.5,0.75);%
+ \draw(1.5,0)node[below]{P};%
+ \draw(2.75,0.75)node[right]{Q};%
+ \draw(0.5,0.75)node[left]{R};%
+ \draw(1.62,0.8)node[right]{O};%
+ \end{tikzpicture}
+ \空行%
+ $\bunsuu{\text{BP}}{\text{PC}}\cdot\bunsuu{\text{CQ}}{\text{QA}}\cdot\bunsuu{\text{AR}}{\text{RB}}=1$%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{チェバの定理の証明}}%
+ {%
+ \証明開始%
+ \begin{tikzpicture}%
+ \draw(2,3)--(0,0)--(3,0)--cycle;%
+ \draw(2,3)node[above]{A};%
+ \draw(2,3)--(1.5,0);%
+ \draw(0,0)node[below]{B};%
+ \draw(0,0)--(2.75,0.75);%
+ \draw(3,0)node[below]{C};%
+ \draw(3,0)--(0.5,0.75);%
+ \draw(1.5,0)node[below]{P};%
+ \draw(2.75,0.75)node[right]{Q};%
+ \draw(0.5,0.75)node[left]{R};%
+ \draw(1.62,0.8)node[right]{O};%
+ \end{tikzpicture}%
+ \空行%
+ 図において三角形の面積比を考えると,%
+ \[\bigtriangleup{ABO}:\bigtriangleup{ACO}=\mathrm{BP}:\mathrm{CP}\]%
+ \[\Leftrightarrow\bunsuu{\bigtriangleup{ABO}}{\bigtriangleup{ACO}}=\bunsuu{BP}{PC}\]%
+ 同様にして,%
+ \[\bunsuu{\bigtriangleup{BCO}}{\bigtriangleup{BAO}}=\bunsuu{CQ}{QA}\]%
+ \[\bunsuu{\bigtriangleup{CAO}}{\bigtriangleup{CBO}}=\bunsuu{AR}{RB}\]%
+ ここで,%
+ \[\bunsuu{\bigtriangleup{ABO}}{\bigtriangleup{ACO}}\cdot\bunsuu{\bigtriangleup{BCO}}{\bigtriangleup{BAO}}\cdot\bunsuu{\bigtriangleup{CAO}}{\bigtriangleup{CBO}}=1\]%
+ \[\Leftrightarrow\bunsuu{\text{BP}}{\text{PC}}\cdot\bunsuu{\text{CQ}}{\text{QA}}\cdot\bunsuu{\text{AR}}{\text{RB}}=1\]%
+ \証明終了%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{メネラウスの定理}}%
+ {%
+ \begin{tikzpicture}%
+ \draw(2,3)--(0,0)--(3,0)--cycle;%
+ \draw(2,3)node[above]{A};%
+ \draw(0,0)node[below]{B};%
+ \draw(3,0)node[right]{C};%
+ \draw(3,0)--(3.11111,-0.33333);%
+ \draw(2.25,0)node[below]{P};%
+ \draw(3.11111,-0.33333)node[below]{Q};%
+ \draw(3.11111,-0.33333)--(0.5,0.75);%
+ \draw(0.5,0.75)node[left]{R};%
+ \end{tikzpicture}%
+ \空行%
+ $\bunsuu{\text{BP}}{\text{PC}}\cdot\bunsuu{\text{CQ}}{\text{QA}}\cdot\bunsuu{\text{AR}}{\text{RB}}=1$%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{メネラウスの定理の証明}}%
+ {%
+ \証明開始%
+ \begin{tikzpicture}%
+ \draw(2,3)--(0,0)--(3,0)--cycle;%
+ \draw(2,3)node[above]{A};%
+ \draw(0,0)node[below]{B};%
+ \draw(3,0)node[right]{C};%
+ \draw(3,0)--(3.11111,-0.33333);%
+ \draw(2.25,0)node[below]{P};%
+ \draw(3.11111,-0.33333)node[below]{Q};%
+ \draw(3.11111,-0.33333)--(0.5,0.75);%
+ \draw(0.5,0.75)node[left]{R};%
+ \draw(0.65,0.975)node[above left]{S};%
+ \draw(0.65,0.975)--(3,0);%
+ \end{tikzpicture}%
+ \空行%
+ $\text{SC}/ \!/ \text{RP}$より,%
+ \[\text{RA}:\text{SR}=\text{QA}:\text{CQ},\text{BR}:\text{RS}=\text{BP}:\text{PC}\]%
+ \[\Leftrightarrow\bunsuu{\text{CQ}}{\text{QA}}=\bunsuu{\text{SR}}{\text{AR}},\bunsuu{\text{BP}}{\text{PC}}=\bunsuu{\text{BR}}{\text{RS}}\]%
+ \[\bunsuu{\text{BP}}{\text{PC}}\cdot\bunsuu{\text{CQ}}{\text{QA}}\cdot\bunsuu{\text{AR}}{\text{RB}}=\bunsuu{\text{BR}}{\text{RS}}\cdot\bunsuu{\text{SR}}{\text{AR}}
+ \cdot\bunsuu{\text{AR}}{\text{RB}}=1\]%
+ \証明終了%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{円周角の定理}}%
+ {%
+ \begin{tikzpicture}%
+ \draw(-1.6,-1.2)--(1.2,-1.6)--(1.2,1.6)--cycle;%
+ \draw(-1.6,-1.2)--(1.2,-1.6)--(-2,0)--cycle;%
+ \draw(-1.6,-1.2)node[left]{A};%
+ \draw(1.2,-1.6)node[right]{B};%
+ \draw(1.2,1.6)node[above]{P};%
+ \draw(-2,0)node[left]{Q};%
+ \draw(0,0)circle[radius=2];%
+ \end{tikzpicture}%
+ \空行%
+ $\angle\mathrm{APB}=\angle\mathrm{AQB}$%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{円周角の定理の証明}}%
+ {%
+ \証明開始%
+ \begin{tikzpicture}%%
+ \draw(-1.6,-1.2)--(1.2,-1.6)--(1.2,1.6)--cycle;%
+ \draw(-1.6,-1.2)--(1.2,-1.6)--(0,0)--cycle;%
+ \draw(-1.6,-1.2)node[left]{A};%
+ \draw(1.2,-1.6)node[right]{B};%
+ \draw(1.2,1.6)node[above]{P};%
+ \draw(0,0)node[right]{O};%
+ \draw[dashed](-1.2,-1.6)--(1.2,1.6);%
+ \draw(-0.6,-1.4)node[above]{D};%
+ \draw(0,0)circle[radius=2];%
+ \end{tikzpicture}%
+ \空行%
+ 三角形AOP,BOPは二等辺三角形なので,%
+ \[\angle\mathrm{APO}=\angle\mathrm{OAP}\数式カンマスペース\angle\mathrm{BPO}=\angle\mathrm{OBP}\]%
+ 三角形の外角より,%
+ \[\angle\mathrm{AOD}=2\angle\mathrm{APO}\数式カンマスペース\angle\mathrm{BOD}=2\angle\mathrm{BPO}\]%
+ \[\Leftrightarrow\angle\mathrm{AOB}=2\angle\mathrm{APB}\]%
+ \空行%
+ \begin{tikzpicture}%
+ \draw(-1.6,-1.2)--(1.6,-1.2)--(1.6,1.2)--cycle;%
+ \draw(-1.6,-1.2)--(1.6,-1.2)--(0,0)--cycle;%
+ \draw(-1.6,-1.2)node[left]{A};%
+ \draw(1.6,-1.2)node[right]{B};%
+ \draw(1.6,1.2)node[above]{P};%
+ \draw(0,0)node[above]{O};%
+ \draw(0,0)circle[radius=2];%
+ \end{tikzpicture}%
+ \空行%
+ 三角形OPBは二等辺三角形なので,%
+ \[\angle\mathrm{OPB}=\angle\mathrm{OBP}\]%
+ 三角形の外角より%
+ \[\angle\mathrm{AOB}=2\angle\mathrm{OPB}\]%
+ \空行%
+ \begin{tikzpicture}%
+ \draw(-1.6,-1.2)--(1.2,-1.6)--(0,0)--cycle;%
+ \draw(-1.6,-1.2)--(1.2,-1.6)--(-2,0)--cycle;%
+ \draw[dashed](-2,0)--(2,0);%
+ \draw(-1.6,-1.2)node[left]{A};%
+ \draw(1.2,-1.6)node[right]{B};%
+ \draw(0,0)node[above]{O};%
+ \draw(2,0)node[right]{D};%
+ \draw(-2,0)node[left]{Q};%
+ \draw(0,0)circle[radius=2];%
+ \end{tikzpicture}%
+ \空行%
+ 三角形QOA,OQBは二等辺三角形なので,%
+ \[\angle\mathrm{OQA}=\angle\mathrm{OAQ}\数式カンマスペース\angle\mathrm{OQB}=\angle\mathrm{OBQ}\]%
+ となる,\par%
+ 三角形の外角より,%
+ \[\angle\mathrm{OQA}+\angle\mathrm{OAQ}=\angle\mathrm{DOA}\数式カンマスペース\angle\mathrm{OQB}+\angle\mathrm{OBQ}=\angle\mathrm{DOB}\]%
+ \[\Leftrightarrow\angle\mathrm{DOA}-\angle\mathrm{DOB}=2\Ttyuukakko{\angle\mathrm{OQA}-\angle\mathrm{BQO}}\]%
+ \[\Leftrightarrow\angle\mathrm{AOB}=2\angle\mathrm{AQB}\]%
+ \空行%
+ 従って,円に内接する三角形について,円周角の$2$倍が中心角である。\par%
+ \空行%
+ \begin{tikzpicture}%
+ \draw(-1.6,-1.2)--(1.2,-1.6)--(1.2,1.6)--cycle;%
+ \draw(-1.6,-1.2)--(1.2,-1.6)--(-2,0)--cycle;%
+ \draw(0,0)--(-1.6,-1.2)--(1.2,-1.6)--cycle;%
+ \draw(-1.6,-1.2)node[left]{A};%
+ \draw(1.2,-1.6)node[right]{B};%
+ \draw(1.2,1.6)node[above]{P};%
+ \draw(-2,0)node[left]{Q};%
+ \draw(0,0)node[above]{O};%
+ \draw(0,0)circle[radius=2];%
+ \end{tikzpicture}%
+ \空行%
+ \[\angle\mathrm{APB}=2\angle\mathrm{AOB},\angle\mathrm{AQB}=2\angle\mathrm{AOB}\]%
+ \[\Leftrightarrow\angle\mathrm{AQB}=\angle\mathrm{APB}\]が成立。
+ \証明終了%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{内接四角形の定理}}%
+ {%
+ \begin{tikzpicture}%
+ \draw(-1.6,-1.2)--(1.6,-1.2)--(1.2,1.6)--(0,2);%
+ \draw(-1.6,-1.2)--(0,2);%
+ \draw(-1.6,-1.2)--(3,-1.2);%
+ \draw(-1.6,-1.2)node[left]{A};%
+ \draw(1.6,-1.2)node[below]{B};%
+ \draw(1.2,1.6)node[above]{C};%
+ \draw(0,2)node[above]{D};%
+ \draw(3,-1.2)node[below]{T};%
+ \draw(0,0)circle[radius=2];%
+ \end{tikzpicture}%
+ \空行%
+ $\angle\mathrm{ADC}=\angle\mathrm{CBT}$%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{内接四角形の定理の証明}}%
+ {%
+ \証明開始%
+ \begin{tikzpicture}%
+ \draw(-1.6,-1.2)--(1.6,-1.2)--(1.2,1.6)--(0,2);%
+ \draw(-1.6,-1.2)--(0,2);%
+ \draw(-1.6,-1.2)--(3,-1.2);%
+ \draw(-1.6,-1.2)node[left]{A};%
+ \draw(1.6,-1.2)node[below]{B};%
+ \draw(1.2,1.6)node[above]{C};%
+ \draw(0,2)node[above]{D};%
+ \draw(3,-1.2)node[below]{T};%
+ \draw(0,0)circle[radius=2];%
+ \draw(-1.6,-1.2)--(0,0);%
+ \draw(1.2,1.6)--(0,0);%
+ \draw(0,0)coordinate(O);%
+ \fill[black](O)circle(0.03);%
+ \end{tikzpicture}%
+ \空行%
+ \[\angle\mathrm{AOC}=2\angle\mathrm{ABC}\]%
+ \[\angle\mathrm{AOC}=2\angle\mathrm{ADC}\]%
+ ここで,$\angle\mathrm{ABC}+\angle\mathrm{ADC}=180^\circ$%
+ \[\Leftrightarrow\angle\mathrm{AOC}+\angle\mathrm{AOC}=180^\circ\]%
+ \証明終了%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{接弦定理}}%
+ {%
+ \begin{tikzpicture}%
+ \draw(0,-2)--(2,0)--(-1.2,1.6)--cycle;%
+ \draw(0,-2)node[below]{A};%
+ \draw(2,0)node[right]{B};%
+ \draw(-1.2,1.6)node[above]{C};%
+ \draw(3,-2)--(-3,-2);%
+ \draw(3,-2)node[below]{T};%
+ \draw(-3,-2)node[below]{S};%
+ \draw(0,0)circle[radius=2];%
+ \end{tikzpicture}
+ \空行%
+ $\angle\mathrm{BAT}=\angle\mathrm{ACB}$%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{接弦定理の証明}}%
+ {%
+ \証明開始%
+ \begin{enumerate}%
+ \item 鋭角のとき%
+ \空行%
+ \begin{tikzpicture}%
+ \draw(0,-2)--(2,0)--(-1.2,1.6)--cycle;%
+ \draw[dashed](0,-2)--(2,0)--(0,2)--cycle;%
+ \draw(0,-2)node[below]{A};%
+ \draw(2,0)node[right]{B};%
+ \draw(-1.2,1.6)node[above]{C};%
+ \draw(3,-2)--(-3,-2);%
+ \draw(3,-2)node[below]{T};%
+ \draw(-3,-2)node[below]{S};%
+ \draw(0,2)node[above]{E};%
+ \draw(0,0)circle[radius=2];%
+ \draw(0,2)coordinate(E);%
+ \draw(2,0)coordinate(B);%
+ \draw(0,-2)coordinate(A);%
+ \draw pic[draw,black,thin,angle radius=0.3cm] {right angle=E--B--A};%
+ \end{tikzpicture}%
+ \空行%
+ 三角形ACBとABEについて円周角の定理より,%
+ \[\angle\mathrm{ACB}=\angle\mathrm{AEB}\]%
+ ここで,三角形ABEについて%
+ \[\angle\mathrm{BEA}+\angle\mathrm{BAE}=90^\circ\]%
+ また,ATが円の接線なので$\angle\mathrm{BAE}+\angle\mathrm{BAT}=90^\circ$から,%
+ \[\angle\mathrm{BAT}=\angle\mathrm{AEB}\]%
+ \[\Leftrightarrow\angle\mathrm{ACB}=\angle\mathrm{BAT}\]%
+ \空行%
+ \item 直角のとき%
+ \空行
+ \begin{tikzpicture}%
+ \draw(0,-2)--(2,0)--(0,2)--cycle;%
+ \draw(0,-2)node[below]{A};%
+ \draw(2,0)node[right]{B};%
+ \draw(3,-2)--(-3,-2);%
+ \draw(3,-2)node[below]{T};%
+ \draw(-3,-2)node[below]{S};%
+ \draw(0,2)node[above]{E};%
+ \draw(0,0)circle[radius=2];%
+ \draw(0,2)coordinate(E);%
+ \draw(2,0)coordinate(B);%
+ \draw(0,-2)coordinate(A);%
+ \draw pic[draw,black,thin,angle radius=0.3cm] {right angle=E--B--A};%
+ \end{tikzpicture}%
+ \空行%
+ ATが円の接線なので,%
+ \[\angle\mathrm{EAS}=90^\circ\]%
+ \[\Leftrightarrow\angle\mathrm{EBA}=\angle\mathrm{EAS}\]%
+ \空行%
+ \item 鈍角のとき%
+ \空行%
+ \begin{tikzpicture}%
+ \draw(0,-2)--(-2,0)--(-1.2,1.6)--cycle;
+ \draw(0,-2)node[below]{A};%
+ \draw(-2,0)node[left]{B};%
+ \draw(-1.2,1.6)node[above]{C};%
+ \draw(3,-2)--(-3,-2);%
+ \draw(3,-2)node[below]{T};%
+ \draw(-3,-2)node[below]{S};%
+ \draw(0,0)circle[radius=2];%
+ \end{tikzpicture}%
+ \空行%
+ 鋭角のときの接弦定理より,%
+ \[\angle\mathrm{BCA}=\angle\mathrm{BAS}\]%
+ また,三角形ABCにおいて%
+ \[\angle\mathrm{ABC}=\angle\mathrm{ACB}+\angle\mathrm{BAC}\]%
+ \[\Leftrightarrow\angle\mathrm{ABC}=\angle\mathrm{CAT}\]%
+ \空行%
+ \end{enumerate}%
+ 従って円に内接する三角形について成り立つことが証明された。%
+ \証明終了%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{内角と外角の二等分線}}%
+ {%
+ \begin{tikzpicture}%
+ \draw(5.3,0.8)--(0,0)--(4.8,0)--cycle;%
+ \draw(5.3,0.8)--(0,0)--(4,0)--cycle;%
+ \draw(5.3,0.8)--(0,0)--(6,0)--cycle;%
+ \draw(5.3,0.8)node[above]{A};%
+ \draw(0,0)--(6.3,0.950943395);%
+ \draw(0,0)node[below]{B};%
+ \draw(4.8,0)node[below]{C};%
+ \draw(4,0)node[below]{P};%
+ \draw(6,0)node[below]{Q};%
+ \draw(6.3,0.950943395)node[above]{R};%
+ \end{tikzpicture}%
+ \空行%
+ $\angle\mathrm{BAP}=\angle\mathrm{PAC},\angle\mathrm{CAQ}=\angle\mathrm{QAR}$のとき,\par%
+ $\text{BP}:\text{PC}=\text{BQ}:\text{QC}=\text{AB}:\text{AC}$%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{方べきの定理A}}%
+ {%
+ \begin{tikzpicture}%
+ \draw(0,2)--(1.6,-1.2);%
+ \draw(0,2)node[above]{A};%
+ \draw(1.6,-1.2)node[below]{B};%
+ \draw(-1.6,-1.2)--(1.2,1.6);%
+ \draw(-1.6,-1.2)node[below]{C};%
+ \draw(1.2,1.6)node[above]{D};%
+ \draw(0.7,0.9)node[right]{P};%
+ \draw(0,0)circle[radius=2];%
+ \draw(0,0)node[below]{O};%
+ \end{tikzpicture}%
+ \空行%
+ $\text{PA}\cdot\text{PB}=\text{PC}\cdot\text{PD}$%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{方べきの定理Aの証明}}%
+ {%
+ \証明開始%
+ \begin{tikzpicture}%
+ \draw(0,2)--(1.6,-1.2);%
+ \draw(0,2)node[above]{A};%
+ \draw(1.6,-1.2)node[below]{B};%
+ \draw(-1.6,-1.2)--(1.2,1.6);%
+ \draw(-1.6,-1.2)node[below]{C};%
+ \draw(1.2,1.6)node[above]{D};%
+ \draw(0.7,0.9)node[right]{P};%
+ \draw(0,0)circle[radius=2];%
+ \draw(0,0)node[below]{O};%
+ \end{tikzpicture}%
+ \空行%
+ 円周角の定理より,%
+ \[\angle\mathrm{CAP}=\angle\mathrm{BDP},\quad\angle\mathrm{ACP}=\angle\mathrm{DBP}\]%
+ 三角形ACPと三角形DBPは相似なので,\par%
+ \[\text{PA}\cdot\text{PB}=\text{PC}\cdot\text{PD}\]%
+ \証明終了%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{方べきの定理B}}%
+ {%
+ \begin{tikzpicture}%
+ \draw(-1.2,-1.6)node[below]{A};%
+ \draw(1.2,-1.6)--(-3.6,-1.6);%
+ \draw(1.2,-1.6)node[below]{B};%
+ \draw(-2,0)node[left]{C};%
+ \draw(0,2)--(-3.6,-1.6);%
+ \draw(0,2)node[above]{D};%
+ \draw(-3.6,-1.6)node[below]{P};%
+ \draw(0,0)circle[radius=2];%
+ \end{tikzpicture}%
+ \空行%
+ $\text{PA}\cdot\text{PB}=\text{PC}\cdot\text{PD}$%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{方べきの定理Bの証明}}%
+ {%
+ \証明開始%
+ \begin{tikzpicture}%
+ \draw(-1.2,-1.6)node[below]{A};%
+ \draw(1.2,-1.6)--(-3.6,-1.6);%
+ \draw(1.2,-1.6)node[below]{B};%
+ \draw(-2,0)node[left]{C};%
+ \draw(0,2)--(-3.6,-1.6);%
+ \draw(0,2)node[above]{D};%
+ \draw(-3.6,-1.6)node[below]{P};%
+ \draw(0,0)circle[radius=2];%
+ \draw(-1.2,-1.6)--(-2,0);%
+ \end{tikzpicture}%
+ \空行%
+ 内接四角形の証明より,%
+ \[\angle\mathrm{CDB}=\angle\mathrm{CAP}\数式カンマスペース\angle\mathrm{DBA}=\angle\mathrm{PCA}\]%
+ 三角形ACPと三角形DPBは相似なので,%
+ \[\text{PA}\cdot\text{PB}=\text{PC}\cdot\text{PD}\]%
+ \証明終了%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{方べきの定理C}}%
+ {%
+ \begin{tikzpicture}%
+ \draw(-1.6,-1.2)node[below]{A};%
+ \draw(1.6,-1.2)--(-4.93,-1.2);%
+ \draw(1.6,-1.2)node[right]{B};%
+ \draw(-1.2,1.6)--(-4.93,-1.2);%
+ \draw(-1.2,1.6)node[above]{T};%
+ \draw(-4.93,-1.2)node[below]{P};%
+ \draw(0,0)circle[radius=2];%
+ \end{tikzpicture}%
+ \空行%
+ $\text{PA}\cdot\text{PB}=\text{PT}^2$%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{方べきの定理Cの証明}}%
+ {%
+ \証明開始%
+ \begin{tikzpicture}%
+ \draw(-1.6,-1.2)node[below]{A};%
+ \draw(1.6,-1.2)--(-4.93,-1.2);%
+ \draw(1.6,-1.2)node[right]{B};%
+ \draw(-1.2,1.6)--(-4.93,-1.2);%
+ \draw(-1.2,1.6)node[above]{T};%
+ \draw(-4.93,-1.2)node[below]{P};%
+ \draw(0,0)circle[radius=2];%
+ \draw(-1.6,-1.2)--(-1.2,1.6);%
+ \draw(-1.2,1.6)--(1.6,-1.2);%
+ \end{tikzpicture}%
+ \空行%
+ 接弦定理より,%
+ \[\angle\mathrm{TBA}=\angle\mathrm{PTA}\]%
+ これと,$\angle\mathrm{P}$共通なので三角形PTAと三角形PBTは相似より,%
+ \[\text{PA}\cdot\text{PB}=\text{PT}^2\]%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+%%%%%%%%%%%%%%%%%%%%ここから数\UTF{2161}B%%%%%%%%%%%%%%%%%%%%
+\NewDocumentCommand{\三次式展開}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{a+b}^{3}=a^{3}+3a^2b+3ab^2+b^{3}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {\[\Ttyuukakko{a+b}^{3}=a^{3}+3a^2b+3ab^2+b^{3}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{a-b}^{3}=a^{3}-3a^2b+3ab^2-b^{3}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {\[\Ttyuukakko{a-b}^{3}=a^{3}-3a^2b+3ab^2-b^{3}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{a+b}\Ttyuukakko{a^2-ab+b^2}=a^{3}+b^{3}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}%
+ {\[\Ttyuukakko{a+b}\Ttyuukakko{a^2-ab+b^2}=a^{3}+b^{3}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{a-b}\Ttyuukakko{a^2+ab+b^2}=a^{3}-b^{3}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{b}}%
+ {\[\Ttyuukakko{a-b}\Ttyuukakko{a^2+ab+b^2}=a^{3}-b^{3}\]}{\relax}%
+ }%
+
+
+\NewDocumentCommand{\三次式因数分解}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$a^{3}+b^{3}=\Ttyuukakko{a+b}\Ttyuukakko{a^2-ab+b^2}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {\[a^{3}+b^{3}=\Ttyuukakko{a+b}\Ttyuukakko{a^2-ab+b^2}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$a^{3}-b^{3}=\Ttyuukakko{a-b}\Ttyuukakko{a^2+ab+b^2}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {\[a^{3}-b^{3}=\Ttyuukakko{a-b}\Ttyuukakko{a^2+ab+b^2}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {$a^{3}+3a^2b+3ab^2+b^{3}=\Ttyuukakko{a+b}^{3}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}%
+ {\[a^{3}+3a^2b+3ab^2+b^{3}=\Ttyuukakko{a+b}^{3}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{i}}%
+ {$a^{3}-3a^2b+3ab^2-b^{3}=\Ttyuukakko{a-b}^{3}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{b}}%
+ {\[a^{3}-3a^2b+3ab^2-b^{3}=\Ttyuukakko{a-b}^{3}\]}{\relax}%
+ }%
+
+
+\NewDocumentCommand{\二項定理}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{a+b}^{n}={}_{n}C_{0} a^{n}+{}_{n}C_{1} a^{n-1}b+{}_{n}C_{2} a^{n-2}b^2+....{}_{n}C_{n-1} ab^{n-1}+{}_{n}C_{n} b^{n}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{b}}%
+ {\[\Ttyuukakko{a+b}^{n}={}_{n}C_{0} a^{n}+{}_{n}C_{1} a^{n-1}b+{}_{n}C_{2} a^{n-2}b^2+....{}_{n}C_{n-1} ab^{n-1}+{}_{n}C_{n} b^{n}\]}{\relax}%
+ \ifthenelse{\equal{#1}{一般項}\AND\equal{#2}{i}}%
+ {${}_{n}C_{r}a^{n-r}b^{r}$}{\relax}%
+ \ifthenelse{\equal{#1}{一般項}\AND\equal{#2}{b}}%
+ {\[{}_{n}C_{r}a^{n-r}b^{r}\]}{\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ $\Ttyuukakko{a+b}^{n}$を展開すると,$a^{r}b^{n-r}$の項の係数は$n$個の$a$から$r$個$a$を選ぶ場合の数に等しいので係数は${}_{n} C_{r}$よって,一般項は%
+ \[{}_{n}C_{r}a^{n-r}b^{r}\]%
+ この$r$に$1$から順番に自然数を代入したものが二項定理となる。%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\分数式}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$\bunsuu{A}{B}\times\bunsuu{C}{D}=\bunsuu{AC}{BD}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {\[\bunsuu{A}{B}\times\bunsuu{C}{D}=\bunsuu{AC}{BD}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$\bunsuu{A}{B}\div \bunsuu{C}{D}=\bunsuu{AD}{BC}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {\[\bunsuu{A}{B}\div \bunsuu{C}{D}=\bunsuu{AD}{BC}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {$\bunsuu{A}{C}+\bunsuu{B}{C}=\bunsuu{A+B}{C}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}%
+ {\[\bunsuu{A}{C}+\bunsuu{B}{C}=\bunsuu{A+B}{C}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{i}}%
+ {$\bunsuu{A}{C}-\bunsuu{B}{C}=\bunsuu{A-B}{C}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{b}}%
+ {\[\bunsuu{A}{C}-\bunsuu{B}{C}=\bunsuu{A-B}{C}\]}{\relax}%
+ }%
+
+
+\NewDocumentCommand{\相加相乗平均}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{i}}%
+ {$a>0\数式カンマスペース b>0$のとき,$\bunsuu{a+b}{2}\geqq\sqrt{ab}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{b}}%
+ {%
+ $a>0\数式カンマスペース b>0$のとき,%
+ \[\bunsuu{a+b}{2}\geqq\sqrt{ab}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ $a+b-2\sqrt{ab}\geqq0$を示す。%
+ \[a+b-2\sqrt{ab}=\Ttyuukakko{\sqrt{a}-\sqrt{b}}^2\]%
+ より,$\sqrt{a}-\sqrt{b}$は実数なので,%
+ \[\Ttyuukakko{\sqrt{a}-\sqrt{b}}^2\geqq0\]%
+ よって,$a>0\数式カンマスペース b>0$のとき,%
+ \[\bunsuu{a+b}{2}\geqq\sqrt{ab}\text{\ (等号成立条件は$a=b$)}\]%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\虚数の定義}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{i}}%
+ {$i=\sqrt{-1}$}{\relax}%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{b}}%
+ {\[i=\sqrt{-1}\]}{\relax}%
+ }%
+
+
+\NewDocumentCommand{\複素数の定義}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{i}}%
+ {実数$a\数式カンマスペース b$を用いて,$a+bi$}{\relax}%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{b}}%
+ {%
+ 実数$a\数式カンマスペース b$を用いて,%
+ \[a+bi\]%
+ }%
+ {\relax}%
+ }%
+
+
+\newcommand{\二次方程式の解の判別}%
+ {%
+ $ax^2+bx+c=0\数式カンマスペース\Ttyuukakko{a\neq0}$の判別式を$D=b^2-4ac$とすると,%
+ \phrases@math[l]%
+ {%
+ $D>0$のとき,異なる二つの実数解\\%
+ $D=0$のとき,重解\\%
+ $D<0$のとき,異なる二つの虚数解%
+ }%
+ を持つ。
+ }%
+
+
+\NewDocumentCommand{\解と係数の関係}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{二次方程式の解と係数の関係A}\AND\equal{#2}{i}}%
+ {$ax^2+bx+c=0 \Ttyuukakko{a\neq0}$の解を$\alpha\数式カンマスペース\beta$として,$\alpha+\beta=-\bunsuu{b}{a}$}{\relax}%
+ \ifthenelse{\equal{#1}{二次方程式の解と係数の関係A}\AND\equal{#2}{b}}%
+ {%
+ $ax^2+bx+c=0 \Ttyuukakko{a\neq0}$の解を$\alpha\数式カンマスペース\beta$として,%
+ \[\alpha+\beta=-\bunsuu{b}{a}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{二次方程式の解と係数の関係B}\AND\equal{#2}{i}}%
+ {$ax^2+bx+c=0 \Ttyuukakko{a\neq0}$の解を$\alpha\数式カンマスペース\beta$として,$\alpha\beta=\bunsuu{c}{a}$}{\relax}%
+ \ifthenelse{\equal{#1}{二次方程式の解と係数の関係B}\AND\equal{#2}{b}}%
+ {%
+ $ax^2+bx+c=0 \Ttyuukakko{a\neq0}$の解を$\alpha\数式カンマスペース\beta$として,%
+ \[\alpha\beta=\bunsuu{c}{a}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{二次方程式の解と係数の関係の証明}\AND\equal{#2}{i}}%
+ {%
+ \証明開始%
+ \[ax^2+bx+c=a\Ttyuukakko{x-\alpha}\Ttyuukakko{x-\beta}=a\Tdaikakko{x^2-\Ttyuukakko{\alpha+\beta}x+\alpha\beta}\]%
+ \[\Leftrightarrow ax^2+bx+c=a\Ttyuukakko{x^2+\bunsuu{b}{a}x+\bunsuu{c}{a}}\]%
+ 係数比較することで,%
+ \[\alpha+\beta=-\bunsuu{b}{a}\数式カンマスペース\alpha\beta=\bunsuu{c}{a}\]%
+ \証明終了%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{三次方程式の解と係数の関係A}\AND\equal{#2}{i}}%
+ {$ax^{3}+bx^2+cx+d=0\Ttyuukakko{a\neq0}$の解を$\alpha\数式カンマスペース\beta\数式カンマスペース\gamma$として,$\alpha+\beta+\gamma=-\bunsuu{b}{a}$}{\relax}%
+ \ifthenelse{\equal{#1}{三次方程式の解と係数の関係A}\AND\equal{#2}{b}}%
+ {%
+ $ax^{3}+bx^2+cx+d=0\Ttyuukakko{a\neq0}$の解を$\alpha\数式カンマスペース\beta\数式カンマスペース\gamma$として,%
+ \[\alpha+\beta+\gamma=-\bunsuu{b}{a}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{三次方程式の解と係数の関係B}\AND\equal{#2}{i}}%
+ {$ax^{3}+bx^2+cx+d=0\Ttyuukakko{a\neq0}$の解を$\alpha\数式カンマスペース\beta\数式カンマスペース\gamma$として,$\alpha\beta+\beta\gamma+\gamma\alpha=\bunsuu{c}{a}$}{\relax}%
+ \ifthenelse{\equal{#1}{三次方程式の解と係数の関係B}\AND\equal{#2}{b}}%
+ {%
+ $ax^{3}+bx^2+cx+d=0\Ttyuukakko{a\neq0}$の解を$\alpha\数式カンマスペース\beta\数式カンマスペース\gamma$として,%
+ \[\alpha\beta+\beta\gamma+\gamma\alpha=\bunsuu{c}{a}\]
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{三次方程式の解と係数の関係C}\AND\equal{#2}{i}}%
+ {$ax^{3}+bx^2+cx+d=0\Ttyuukakko{a\neq0}$の解を$\alpha\数式カンマスペース\beta\数式カンマスペース\gamma$として,$\alpha\beta\gamma=-\bunsuu{d}{a}$}{\relax}%
+ \ifthenelse{\equal{#1}{三次方程式の解と係数の関係C}\AND\equal{#2}{b}}%
+ {%
+ $ax^{3}+bx^2+cx+d=0\Ttyuukakko{a\neq0}$の解を$\alpha\数式カンマスペース\beta\数式カンマスペース\gamma$として,%
+ \[\alpha\beta\gamma=-\bunsuu{d}{a}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{三次方程式の解と係数の関係の証明}\AND\equal{#2}{i}}%
+ {%
+ \証明開始%
+ \vspace{-1\zw}
+ \[ax^{3}+bx^2+cx+d=a\Ttyuukakko{x-\alpha}\Ttyuukakko{x-\beta}\Ttyuukakko{x-\gamma}=a\Tdaikakko{x^3-\Ttyuukakko{\alpha+\beta+\gamma}x^2+\Ttyuukakko{\alpha\beta+\beta\gamma+\gamma\alpha}x-\alpha\beta\gamma}\]%
+ \[\Leftrightarrow ax^{3}+bx^2+cx+d=a\Ttyuukakko{x^3+\bunsuu{b}{a}x^2+\bunsuu{c}{a}x+\bunsuu{d}{a}}\]%
+ 係数比較することで,\par%
+ \[\alpha+\beta+\gamma=-\bunsuu{b}{a}\数式カンマスペース\alpha\beta+\beta\gamma+\gamma\alpha=\bunsuu{c}{a}\数式カンマスペース\alpha\beta\gamma=-\bunsuu{d}{a}\]%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\剰余定理}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定理A}\AND\equal{#2}{i}}%
+ {整式 $P\Ttyuukakko{x}$を$x-k$で割った余りは$P\Ttyuukakko{k}$}{\relax}%
+ \ifthenelse{\equal{#1}{定理A}\AND\equal{#2}{b}}%
+ {%
+ 整式 $P\Ttyuukakko{x}$を$x-k$で割った余りは%
+ \[P\Ttyuukakko{k}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{定理B}\AND\equal{#2}{i}}%
+ {整式$P\Ttyuukakko{x}$を$ax-b$で割った余りは$P\Ttyuukakko{\bunsuu{b}{a}}$}{\relax}%
+ \ifthenelse{\equal{#1}{定理B}\AND\equal{#2}{b}}%
+ {%
+ 整式$P\Ttyuukakko{x}$を$ax-b$で割った余りは%
+ \[P\Ttyuukakko{\bunsuu{b}{a}}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ $P\Ttyuukakko{x}$を$\Ttyuukakko{x-k}$で割った商を$Q\Ttyuukakko{x}$あまりを$R$として,%
+ \[P\Ttyuukakko{x}=\Ttyuukakko{x-k}Q\Ttyuukakko{x}+R\]%
+ $x=k$のとき,%
+ \[P\Ttyuukakko{k}=R\]%
+ よって,余りは%
+ \[P\Ttyuukakko{k}\]%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\因数定理}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定理}\AND\equal{#2}{i}}%
+ {%
+ 整式$P\Ttyuukakko{x}$が$x-k$を因数に持つ$\Leftrightarrow P\Ttyuukakko{k}=0$%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{定理}\AND\equal{#2}{b}}%
+ {%
+ 整式$P\Ttyuukakko{x}$が$x-k$を因数に持つ%
+ \[\Leftrightarrow P\Ttyuukakko{k}=0\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ 剰余の定理より,$x-k$で割った余りが$0$なので,%
+ \[P\Ttyuukakko{k}=0\]%
+ 剰余の定理より,$P\Ttyuukakko{k}=0$ということは$P\Ttyuukakko{x}$を$x-k$で割った余りが$0$ということなので,$P\Ttyuukakko{x}$は$x-k$を因数に持つ。%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\点の座標}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{二点間の距離}\AND\equal{#2}{i}}%
+ {$A\Ttyuukakko{x_{1}\数式カンマスペース y_{1}}\数式カンマスペース B\Ttyuukakko{x_{2}\数式カンマスペース y_{2}}$として,線分$AB$間の距離は,$\sqrt{\Ttyuukakko{x_{2}-x_{1}}^2-\Ttyuukakko{y_{2}-y_{1}}^2}$}{\relax}%
+ \ifthenelse{\equal{#1}{二点間の距離}\AND\equal{#2}{b}}%
+ {%
+ $A\Ttyuukakko{x_{1}\数式カンマスペース y_{1}}\数式カンマスペース B\Ttyuukakko{x_{2}\数式カンマスペース y_{2}}$として,線分$AB$間の距離は,%
+ \[\sqrt{\Ttyuukakko{x_{2}-x_{1}}^2-\Ttyuukakko{y_{2}-y_{1}}^2}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{内分点の座標}\AND\equal{#2}{i}}%
+ {$A\Ttyuukakko{x_{1}\数式カンマスペース y_{1}}\数式カンマスペース B\Ttyuukakko{x_{2}\数式カンマスペース y_{2}}$として,線分$AB$を$m:n$に内分する点の座標は,$\Ttyuukakko{\bunsuu{nx_{1}+mx_{2}}{n+m}\数式カンマスペース\bunsuu{ny_{1}+my_{2}}{n+m}}$}{\relax}%
+ \ifthenelse{\equal{#1}{内分点の座標}\AND\equal{#2}{b}}%
+ {%
+ $A\Ttyuukakko{x_{1}\数式カンマスペース y_{1}}\数式カンマスペース B\Ttyuukakko{x_{2}\数式カンマスペース y_{2}}$として,線分$AB$を$m:n$に内分する点の座標は,%
+ \[\Ttyuukakko{\bunsuu{nx_{1}+mx_{2}}{n+m}\数式カンマスペース\bunsuu{ny_{1}+my_{2}}{n+m}}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{内分点の座標の証明}\AND\equal{#2}{i}}%
+ {%
+ \証明開始%
+ $m:n$に内分する点の座標を$P\Ttyuukakko{x\数式カンマスペース y}$として,%
+ \[m:n=x-x_{1}:x_{2}-x\]%
+ \[\Leftrightarrow\Ttyuukakko{\bunsuu{nx_{1}+mx_{2}}{n+m}\数式カンマスペース\bunsuu{ny_{1}+my_{2}}{n+m}}\]%
+ \証明終了%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{外分点の座標}\AND\equal{#2}{i}}%
+ {$A\Ttyuukakko{x_{1}\数式カンマスペース y_{1}}\数式カンマスペース B\Ttyuukakko{x_{2}\数式カンマスペース y_{2}}$として,線分$AB$を$m:n$に外分する点の座標は,$\Ttyuukakko{\bunsuu{-nx_{1}+mx_{2}}{m-n}\数式カンマスペース \bunsuu{-ny_{1}+my_{2}}{m-n}}$}{\relax}%
+ \ifthenelse{\equal{#1}{外分点の座標}\AND\equal{#2}{b}}%
+ {%
+ $A\Ttyuukakko{x_{1}\数式カンマスペース y_{1}}\数式カンマスペース B\Ttyuukakko{x_{2}\数式カンマスペース y_{2}}$として,線分$AB$を$m:n$に外分する点の座標は,%
+ \[\Ttyuukakko{\bunsuu{-nx_{1}+mx_{2}}{m-n}\数式カンマスペース \bunsuu{-ny_{1}+my_{2}}{m-n}}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{外分点の座標の証明}\AND\equal{#2}{b}}%
+ {%
+ \証明開始%
+ \begin{enumerate}%
+ \item $m>n$のとき\par%
+ $n:m$に外分する点の座標を$P\Ttyuukakko{x\数式カンマスペース y}$として,%
+ \[m:n=x-x_{1}:x-x_{2}\]%
+ \[\Leftrightarrow\Ttyuukakko{\bunsuu{-nx_{1}+mx_{2}}{m-n}\数式カンマスペース \bunsuu{-ny_{1}+my_{2}}{m-n}}\]%
+ \item $m<n$のとき\par%
+ $n:m$に外分する点の座標を$P\Ttyuukakko{x\数式カンマスペース y}$として,%
+ \[m:n=x-x_{2}:x-x_{1}\]%
+ \[\Leftrightarrow\Ttyuukakko{\bunsuu{-nx_{1}+mx_{2}}{m-n}\数式カンマスペース \bunsuu{-ny_{1}+my_{2}}{m-n}}\]%
+ \end{enumerate}%
+ よって$m\数式カンマスペース n$の大小に依らず%
+ \[\Ttyuukakko{\bunsuu{-nx_{1}+mx_{2}}{m-n}\数式カンマスペース \bunsuu{-ny_{1}+my_{2}}{m-n}}\]%
+ となる。%
+ \証明終了%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{中点の座標}\AND\equal{#2}{i}}%
+ {$A\Ttyuukakko{x_{1}\数式カンマスペース y_{1}}\数式カンマスペース B\Ttyuukakko{x_{2}\数式カンマスペース y_{2}}$として,線分$AB$の中点は,$\Ttyuukakko{\bunsuu{x_{1}+x_{2}}{2}\数式カンマスペース\bunsuu{y_{1}+y_{2}}{2}}$}{\relax}%
+ \ifthenelse{\equal{#1}{中点の座標}\AND\equal{#2}{b}}%
+ {%
+ $A\Ttyuukakko{x_{1}\数式カンマスペース y_{1}}\数式カンマスペース B\Ttyuukakko{x_{2}\数式カンマスペース y_{2}}$として,線分$AB$の中点は,%
+ \[\Ttyuukakko{\bunsuu{x_{1}+x_{2}}{2}\数式カンマスペース\bunsuu{y_{1}+y_{2}}{2}}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{中点の座標の証明}\AND\equal{#2}{i}}%
+ {%
+ \証明開始%
+ 内分点の公式において$m=n$のとき,%
+ \[\Ttyuukakko{\bunsuu{x_{1}+x_{2}}{2}\数式カンマスペース\bunsuu{y_{1}+y_{2}}{2}}\]%
+ \証明終了%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{重心の座標}\AND\equal{#2}{i}}%
+ {$A\Ttyuukakko{x_{1}\数式カンマスペース y_{1}}\数式カンマスペース B\Ttyuukakko{x_{2}\数式カンマスペース y_{2}}\数式カンマスペース C\Ttyuukakko{x_{3}\数式カンマスペース y_{3}}$として,三角形$ABC$の重心の座標は,$\Ttyuukakko{\bunsuu{x_{1}+x_{2}+x_{3}}{3}\数式カンマスペース\bunsuu{y_{1}+y_{2}+y_{3}}{3}}$}{\relax}%
+ \ifthenelse{\equal{#1}{重心の座標}\AND\equal{#2}{b}}%
+ {%
+ $A\Ttyuukakko{x_{1}\数式カンマスペース y_{1}}\数式カンマスペース B\Ttyuukakko{x_{2}\数式カンマスペース y_{2}}\数式カンマスペース C\Ttyuukakko{x_{3}\数式カンマスペース y_{3}}$として,三角形$ABC$の重心の座標は,%
+ \[\Ttyuukakko{\bunsuu{x_{1}+x_{2}+x_{3}}{3}\数式カンマスペース\bunsuu{y_{1}+y_{2}+y_{3}}{3}}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{重心の座標の証明}\AND\equal{#2}{i}}%
+ {%
+ \証明開始%
+ $A$と$B$の中点$M$の座標は$\Ttyuukakko{\bunsuu{x_{1}+x_{2}}{2}\数式カンマスペース\bunsuu{y_{1}+y_{2}}{2}}$\par%
+ 重心は$CM$を$2:1$に内分するので,重心の座標は内分点の公式より,%
+ \[\Ttyuukakko{\bunsuu{x_{1}+x_{2}+x_{3}}{3}\数式カンマスペース\bunsuu{y_{1}+y_{2}+y_{3}}{3}}\]%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\直線の方程式}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$ax+by+c=0$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {\[ax+by+c=0\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {点$\Ttyuukakko{x_{1}\数式カンマスペース y_{1}}$を通り傾きが$m$の直線は,$y-y_{1}=m\Ttyuukakko{x-x_{1}}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {%
+ 点$\Ttyuukakko{x_{1}\数式カンマスペース y_{1}}$を通り傾きが$m$の直線は,%
+ \[y-y_{1}=m\Ttyuukakko{x-x_{1}}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {異なる二点$\Ttyuukakko{x_{1}\数式カンマスペース y_{1}}\数式カンマスペース\Ttyuukakko{x_{2}\数式カンマスペース y_{2}}$を通る直線$\Ttyuukakko{x_{1}\neq x_{2}}$は,$y-y_{1}=\bunsuu{y_{2}-y_{1}}{x_{2}-x_{1}}\Ttyuukakko{x-x_{1}}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}%
+ {%
+ 異なる二点$\Ttyuukakko{x_{1}\数式カンマスペース y_{1}}\数式カンマスペース\Ttyuukakko{x_{2}\数式カンマスペース y_{2}}$を通る直線,$\Ttyuukakko{x_{1}\neq x_{2}}$は,%
+ \[y-y_{1}=\bunsuu{y_{2}-y_{1}}{x_{2}-x_{1}}\Ttyuukakko{x-x_{1}}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式Bの証明}}%
+ {%
+ \証明開始%
+ 傾き$m$なので,$y=mx+a$と置ける($a$は切片)。\par%
+ ここで,$\Ttyuukakko{x_{1\数式カンマスペース x_{2}}}$を通るので,$y_{1}=mx_{1}+a$となり,連立することで%
+ \[y-y_{1}=m\Ttyuukakko{x-x_{1}}\]%
+ を得る。%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\二直線の関係}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {二直線$y=m_{1}x+n_{1}\数式カンマスペース y=m_{2}x+n_{2}$が平行$\Leftrightarrow m_{1}=m_{2}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {%
+ 二直線$y=m_{1}x+n_{1}\数式カンマスペース y=m_{2}x+n_{2}$が平行%
+ \[\Leftrightarrow m_{1}=m_{2}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {二直線$y=m_{1}x+n_{1}\数式カンマスペース y=m_{2}x+n_{2}$が垂直$\Leftrightarrow m_{1}m_{2}=-1$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {%
+ 二直線$y=m_{1}x+n_{1}\数式カンマスペース y=m_{2}x+n_{2}$が垂直%
+ \[\Leftrightarrow m_{1}m_{2}=-1\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式Bの証明}\AND\equal{#2}{i}}%
+ {%
+ \証明開始%
+ $y=mx_{1}$上に点A$\Ttyuukakko{1\数式カンマスペース m_{1}}$\数式カンマスペース $y=mx_{2}$上にB$\Ttyuukakko{-m_{1}\数式カンマスペース 1}$をとる。\par%
+ H$\Ttyuukakko{1\数式カンマスペース 0}$\数式カンマスペース I$\Ttyuukakko{0\数式カンマスペース 1}$として,$\bigtriangleup{OAH}$と$\bigtriangleup{OBI}$は合同。よって,%
+ \[m_{1}m_{2}=-1\]%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\点と直線の距離}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{i}}%
+ {%
+ 点$\Ttyuukakko{x_{1}\数式カンマスペース y_{2}}$と直線$ax+bx+c=0$の距離は,$\bunsuu{\Tzettaiti{ax_{1}+by_{2}+c}}{\sqrt{a^2+b^2}}$%
+ }{\relax}%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{b}}%
+ {%
+ 点$\Ttyuukakko{x_{1}\数式カンマスペース y_{2}}$と直線$ax+bx+c=0$の距離は,%
+ \[\bunsuu{\Tzettaiti{ax_{1}+by_{2}+c}}{\sqrt{a^2+b^2}}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ 全体を$x$軸方向に$-x_{1}$\数式カンマスペース $y$軸方向に$-y_{1}$平行移動するとき,直線$l$は$a\Ttyuukakko{x+x_{1}}+b\Ttyuukakko{y+y_{1}}+c=0$となる。\par%
+ また,直線$l$に原点Oからおろした垂線との交点をHとする。ここでOH間の距離を$d$と置くと,%
+ \begin{enumerate}%
+ \item $a\neq0$のとき\par%
+ 直線$l$の垂線の傾きは$b$の値に依らず,$y=\bunsuu{b}{a}$となる。\par%
+ よって,Hの座標は二式を連立することで得られ,%
+ \[\Ttyuukakko{\bunsuu{-a\Ttyuukakko{ax_{1}+by_{1}+c}}{a^2+b^2}\数式カンマスペース\bunsuu{-b\Ttyuukakko{ax_{1}+by_{1}+c}}{a^2+b^2}}\]%
+ \begin{align*}%
+ \Leftrightarrow d&=\sqrt{\Tdaikakko{\Ttyuukakko{\bunsuu{-a\Ttyuukakko{ax_{1}+by_{1}+c}}{a^2+b^2}}^2+\Tdaikakko{\bunsuu{-b\Ttyuukakko{ax_{1}+by_{1}+c}}{a^2+b^2}}}^2}&\\%
+ &=\bunsuu{\Tzettaiti{ax_{1}+by_{2}+c}}{\sqrt{a^2+b^2}} %
+ \end{align*}%
+ \item $a=0$のとき\par%
+ 直線$l$は$y=-\bunsuu{by_{1}+c}{b}$となるので,%
+ \begin{align*}%
+ d&=\Tzettaiti{-\bunsuu{by_{1}+c}{b}}&\\%
+ &=\bunsuu{\Tzettaiti{by_{1}+c}}{\Tzettaiti{b}}&\\%
+ \end{align*}%
+ これは,$\bunsuu{\Tzettaiti{ax_{1}+by_{2}+c}}{\sqrt{a^2+b^2}}$に$a=0$を代入したものである。
+ \end{enumerate}%
+ よって,いずれの場合も%
+ \[\bunsuu{\Tzettaiti{ax_{1}+by_{2}+c}}{\sqrt{a^2+b^2}}\]%
+ を得る。%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\円の方程式}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{i}}%
+ {中心$\Ttyuukakko{a\数式カンマスペース b}$で半径$r$の円は,$\Ttyuukakko{x-a}^2+\Ttyuukakko{y-b}^2=r^2$と表す($x^2+y^2+Ax+By+C=0\Ttyuukakko{l^2+m^2-4n>0}$の形でもよい)。}{\relax}%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{i}}%
+ {%
+ 中心$\Ttyuukakko{a\数式カンマスペース b}$で半径$r$の円は,%
+ \[\Ttyuukakko{x-a}^2+\Ttyuukakko{y-b}^2=r^2\]%
+ また,円は%
+ \[x^2+y^2+Ax+By+C=0\Ttyuukakko{A^2+B^2-4C>0}\]%
+ とも表せられる。%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ 円の中心をO\数式カンマスペース 円周上の任意の点を$P\Ttyuukakko{x\数式カンマスペース y}$として,三平方の定理より%
+ \[\Ttyuukakko{x-a}^2+\Ttyuukakko{y-b}^2=r^2\]%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\円と直線}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{i}}%
+ {円$x^2+y^2=r^2$上の点 $\Ttyuukakko{x_{1}\数式カンマスペース y_{1}}$における接線の方程式は,$xx_{1}+yy_{1}=r^2$}{\relax}%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{b}}%
+ {%
+ 円$x^2+y^2=r^2$上の点 $\Ttyuukakko{x_{1}\数式カンマスペース y_{1}}$における接線の方程式は,%
+ \[xx_{1}+yy_{1}=r^2\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ \begin{enumerate}%
+ \item $x_{0}\neq0\数式カンマスペース y_{0}\neq0$のとき\par%
+ $A\Ttyuukakko{x_{0}\数式カンマスペース y_{0}}$と置いて,OAの傾きは$\bunsuu{y_{0}}{x_{0}}$となる。接線の傾きはこれに垂直なので,$-\bunsuu{x_{0}}{y_{0}}$また接線は点$\Ttyuukakko{x_{0}\数式カンマスペース y_{0}}$を通るので%
+ \[y=-\bunsuu{x_{0}}{y_{0}}\Ttyuukakko{x-x_{0}}+y_{0}\]%
+ より,$\Ttyuukakko{x_{0}\数式カンマスペース y_{0}}$が$x^2+y^2=r^2$上に存在することに留意して,$x_{0}x+y_{0}y=r^2$となる。\par%
+ \item $x_{0}\neq0$のとき\par%
+ $y_{0}=\pm r$より接線は$y=\pm r\text{\ (複合同順)}$\par%
+ \item $y_{0}=0$のとき\par%
+ $x_{0}=\pm r$より接線は$x=\pm r\text{\ (複合同順)}$%
+ \end{enumerate}%
+ よって,接線の方程式は%
+ \[xx_{1}+yy_{1}=r^2\]%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\三角関数の相互関係}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$\sin^2 \theta+\cos^2 \theta=1$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {\[\sin^2 \theta+\cos^2 \theta=1\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$\tan\theta =\bunsuu{\sin\theta}{\cos\theta}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {\[\tan\theta =\bunsuu{\sin\theta}{\cos\theta}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {$1+\tan^2 \theta=\bunsuu{1}{\cos^2 \theta}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}%
+ {\[1+\tan^2 \theta=\bunsuu{1}{\cos^2 \theta}\]}{\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ \begin{tikzpicture}%
+ \draw(0,0)--(1.05,1.4);%
+ \draw[dashed](0.75,1)--(0,1);%
+ \draw(0,0)node[below right]{O};%
+ \draw(0.75,0)node[below]{$x$};%
+ \draw(0,1)node[left]{$y$};%
+ \draw(0.8,1)node[right]{P$\Ttyuukakko{x,y}$};%
+ \draw(0,0)circle[radius=1.25];%
+ \draw(0,-1.25)node[below left]{$-r$};%
+ \draw(-1.25,0)node[below left]{$-r$};%
+ \draw(0,1.25)node[above left]{$r$};%
+ \draw(1.25,0)node[below right]{$r$};%
+ \draw[->,>=stealth,semithick](-1.5,0)--(1.5,0)node[right]{$x$};%
+ \draw[->,>=stealth,semithick](0,-1.5)--(0,1.5)node[above]{$y$};%
+ \draw[dashed](0,0)coordinate(O)-- (0.75,0)coordinate(Q)-- (0.75,1)coordinate(P);%
+ \draw pic["$\theta$",draw=black,->,thin,angle eccentricity=1.4,angle radius=0.4cm] {angle=Q--O--P};%
+ \end{tikzpicture}%
+ \空行%
+ 図において,$\sin\theta=\bunsuu{y}{r}\数式カンマスペース\quad\cos\theta=\bunsuu{x}{r}$より%
+ \[\sin^2\theta+\cos^2\theta=\bunsuu{y^2+x^2}{r^2}\]%
+ ここで,三平方の定理より$x^2+y^2=r^2$なので\par%
+ $\sin^2\theta+\cos^2\theta=\bunsuu{r^2}{r^2}=1$%
+ \空行%
+ $\sin\theta=\bunsuu{y}{r}\数式カンマスペース\quad\cos\theta=\bunsuu{x}{r}\quad\tan\theta=\bunsuu{y}{x}$より\par%
+ $\bunsuu{\sin\theta}{\cos\theta}=\bunsuu{y}{x}=\tan\theta$%
+ \空行%
+ $\sin^2\theta+\cos^2\theta=1$の両辺を$\cos^2\theta$で割ることで,\par%
+ \[\bunsuu{\sin^2\theta}{\cos^2\theta}+1=\bunsuu{1}{\cos^2\theta}\]%
+ ここで,$\bunsuu{\sin\theta}{\cos\theta}=\tan\theta$なので\par%
+ $\tan^2\theta+1=\bunsuu{1}{\cos^2\theta}$%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\三角関数の性質}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{性質A}\AND\equal{#2}{i}}%
+ {$\sin\Ttyuukakko{-\theta}=-\sin\theta$}{\relax}%
+ \ifthenelse{\equal{#1}{性質A}\AND\equal{#2}{b}}%
+ {\[\sin\Ttyuukakko{-\theta}=-\sin\theta\]}{\relax}%
+ \ifthenelse{\equal{#1}{性質B}\AND\equal{#2}{i}}%
+ {$\cos\Ttyuukakko{-\theta}=\cos\theta$}{\relax}%
+ \ifthenelse{\equal{#1}{性質B}\AND\equal{#2}{b}}%
+ {\[\cos\Ttyuukakko{-\theta}=\cos\theta\]}{\relax}%
+ \ifthenelse{\equal{#1}{性質C}\AND\equal{#2}{i}}%
+ {$\tan\Ttyuukakko{-\theta}=-\tan\theta$}{\relax}%
+ \ifthenelse{\equal{#1}{性質C}\AND\equal{#2}{b}}%
+ {\[\tan\Ttyuukakko{-\theta}=-\tan\theta\]}{\relax}%
+ \ifthenelse{\equal{#1}{性質D}\AND\equal{#2}{i}}%
+ {$\sin\Ttyuukakko{\theta+\pi}=-\sin\theta$}{\relax}%
+ \ifthenelse{\equal{#1}{性質D}\AND\equal{#2}{b}}%
+ {\[\sin\Ttyuukakko{\theta+\pi}=-\sin\theta\]}{\relax}%
+ \ifthenelse{\equal{#1}{性質E}\AND\equal{#2}{i}}%
+ {$\cos\Ttyuukakko{\theta+\pi}=-\cos\theta$}{\relax}%
+ \ifthenelse{\equal{#1}{性質E}\AND\equal{#2}{b}}%
+ {\[\cos\Ttyuukakko{\theta+\pi}=-\cos\theta\]}{\relax}%
+ \ifthenelse{\equal{#1}{性質F}\AND\equal{#2}{i}}%
+ {$\tan\Ttyuukakko{\theta+\pi}=\tan\theta$}{\relax}%
+ \ifthenelse{\equal{#1}{性質F}\AND\equal{#2}{b}}%
+ {\[\tan\Ttyuukakko{\theta+\pi}=\tan\theta\]}{\relax}%
+ \ifthenelse{\equal{#1}{性質G}\AND\equal{#2}{i}}%
+ {$\sin\Ttyuukakko{\pi-\theta}=\sin\theta$}{\relax}%
+ \ifthenelse{\equal{#1}{性質G}\AND\equal{#2}{b}}%
+ {\[\sin\Ttyuukakko{\pi-\theta}=\sin\theta\]}{\relax}%
+ \ifthenelse{\equal{#1}{性質H}\AND\equal{#2}{i}}%
+ {$\cos\Ttyuukakko{\pi-\theta}=-\cos\theta$}{\relax}%
+ \ifthenelse{\equal{#1}{性質H}\AND\equal{#2}{b}}%
+ {\[\cos\Ttyuukakko{\pi-\theta}=-\cos\theta\]}{\relax}%
+ \ifthenelse{\equal{#1}{性質I}\AND\equal{#2}{i}}%
+ {$\tan\Ttyuukakko{\pi-\theta}=-\tan\theta$}{\relax}%
+ \ifthenelse{\equal{#1}{性質I}\AND\equal{#2}{b}}%
+ {\[\tan\Ttyuukakko{\pi-\theta}=-\tan\theta\]}{\relax}%
+ \ifthenelse{\equal{#1}{性質J}\AND\equal{#2}{i}}%
+ {$\sin\Ttyuukakko{\bunsuu{\pi}{2}-\theta}=\cos\theta$}{\relax}%
+ \ifthenelse{\equal{#1}{性質J}\AND\equal{#2}{b}}%
+ {\[\sin\Ttyuukakko{\bunsuu{\pi}{2}-\theta}=\cos\theta\]}{\relax}%
+ \ifthenelse{\equal{#1}{性質K}\AND\equal{#2}{i}}%
+ {$\cos\Ttyuukakko{\bunsuu{\pi}{2}-\theta}=\sin\theta$}{\relax}%
+ \ifthenelse{\equal{#1}{性質K}\AND\equal{#2}{b}}%
+ {\[\cos\Ttyuukakko{\bunsuu{\pi}{2}-\theta}=\sin\theta\]}{\relax}%
+ \ifthenelse{\equal{#1}{性質L}\AND\equal{#2}{i}}%
+ {$\tan\Ttyuukakko{\bunsuu{\pi}{2}-\theta}=\bunsuu{1}{\tan\theta}$}{\relax}%
+ \ifthenelse{\equal{#1}{性質L}\AND\equal{#2}{b}}%
+ {\[\tan\Ttyuukakko{\bunsuu{\pi}{2}-\theta}=\bunsuu{1}{\tan\theta}\]}{\relax}%
+ \ifthenelse{\equal{#1}{性質M}\AND\equal{#2}{i}}%
+ {$\sin\Ttyuukakko{\theta+\bunsuu{\pi}{2}}=\cos\theta$}{\relax}%
+ \ifthenelse{\equal{#1}{性質M}\AND\equal{#2}{b}}%
+ {\[\sin\Ttyuukakko{\theta+\bunsuu{\pi}{2}}=\cos\theta\]}{\relax}%
+ \ifthenelse{\equal{#1}{性質N}\AND\equal{#2}{i}}%
+ {$\cos\Ttyuukakko{\theta+\bunsuu{\pi}{2}}=-\sin\theta$}{\relax}%
+ \ifthenelse{\equal{#1}{性質N}\AND\equal{#2}{b}}%
+ {\[\cos\Ttyuukakko{\theta+\bunsuu{\pi}{2}}=-\sin\theta\]}{\relax}%
+ \ifthenelse{\equal{#1}{性質O}\AND\equal{#2}{i}}%
+ {$\tan\Ttyuukakko{\theta+\bunsuu{\pi}{2}}=-\bunsuu{1}{\tan\theta}$}{\relax}%
+ \ifthenelse{\equal{#1}{性質O}\AND\equal{#2}{b}}%
+ {\[\tan\Ttyuukakko{\theta+\bunsuu{\pi}{2}}=-\bunsuu{1}{\tan\theta}\]}{\relax}%
+ }%
+
+
+\NewDocumentCommand{\三角関数の加法定理}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$\sin\Ttyuukakko{\alpha\pm\beta}=\sin\alpha \cos\beta\pm \cos\alpha \sin\beta$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {\[\sin\Ttyuukakko{\alpha\pm\beta}=\sin\alpha \cos\beta\pm \cos\alpha \sin\beta\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$\cos\Ttyuukakko{\alpha\pm\beta}=\cos\alpha \cos\beta\mp \sin\alpha \sin\beta$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {\[\cos\Ttyuukakko{\alpha\pm\beta}=\cos\alpha \cos\beta\mp \sin\alpha \sin\beta\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {$\tan\Ttyuukakko{\alpha\pm\beta}=\bunsuu{\tan\alpha \pm \tan\beta}{1\mp \tan\alpha \tan\beta}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}%
+ {\[\tan\Ttyuukakko{\alpha\pm\beta}=\bunsuu{\tan\alpha \pm \tan\beta}{1\mp \tan\alpha \tan\beta}\]}{\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ \begin{tikzpicture}%
+ \draw(0,0)--(0.75,1);%
+ \draw(0,0)node[below right]{O};%
+ \draw(0.8,1)node[right]{P$\Ttyuukakko{\cos\alpha,\sin\alpha}$};%
+ \draw(0,0)circle[radius=1.25];%
+ \draw(0,-1.25)node[below left]{$-1$};%
+ \draw(-1.25,0)node[below left]{$-1$};%
+ \draw(0,1.25)node[above left]{$1$};%
+ \draw(1.25,0)node[below right]{$1$};%
+ \draw[->,>=stealth,semithick](-1.5,0)--(1.5,0)node[right]{$x$};%
+ \draw[->,>=stealth,semithick](0,-1.5)--(0,1.5)node[above]{$y$};%
+ \draw(-1,0.75)node[left]{Q$\Ttyuukakko{\cos\beta,\sin\beta}$};%
+ \draw(-1,0.75)coordinate(Q);%
+ \draw(-1,0.75)--(0.75,1);%
+ \draw(-1,0.75)--(0,0);%
+ \draw(0,0)coordinate(O)-- (0.75,0)coordinate(R)-- (0.75,1)coordinate(P);%
+ \draw pic["$\alpha$",draw=black,-,thin,angle eccentricity=1.4,angle radius=0.5cm] {angle=R--O--P};%
+ \draw pic["$\beta$",draw=black,-,thin,angle eccentricity=1.4,angle radius=0.4cm] {angle=R--O--Q};%
+ \end{tikzpicture}%
+ \空行%
+ 図において,三角関数の性質より$\cos\Ttyuukakko{\beta-\alpha}=\cos\Ttyuukakko{\alpha-\beta}$なので,三角形QOPについて余弦定理より%
+ \[\mathrm{QP}^2=1^2+1^2-2\cdot1\cdot1\cdot\cos\Ttyuukakko{\alpha-\beta}\]%
+ また,QP間の距離について三平方の定理を用いて%
+ \[\mathrm{QP}^2=\Ttyuukakko{\cos\beta-\cos\alpha}^2+\Ttyuukakko{\sin\alpha-\sin\beta}^2\]%
+ \[\Leftrightarrow 2-2\cos\Ttyuukakko{\alpha-\beta}=\Ttyuukakko{\cos\beta-\cos\alpha}^2+\Ttyuukakko{\sin\alpha-\sin\beta}^2\]%
+ 両辺整理して,%
+ \[\cos\Ttyuukakko{\alpha-\beta}=\cos\alpha\cos\beta+\sin\alpha\sin\beta\]%
+ を得る。\par%
+ また,$\sin-\theta=-\sin\theta$より,%
+ \[\cos\Ttyuukakko{\alpha+\beta}=\cos\alpha\cos\beta-\sin\alpha\sin\beta\]%
+ \空行%
+ \[\cos\Ttyuukakko{\alpha-\beta}=\cos\alpha\cos\beta+\sin\alpha\sin\beta\]%
+ において,$\alpha$を$\bunsuu{\pi}{2}-\alpha$にすることで,%
+ \[\sin\Ttyuukakko{\alpha+\beta}=\sin\alpha \cos\beta+ \cos\alpha \sin\beta\]%
+ ここで,$\beta$を$-\beta$にすることで,%
+ \[\sin\Ttyuukakko{\alpha-\beta}=\sin\alpha \cos\beta-\cos\alpha \sin\beta\]%
+ \空行%
+ $\tan\theta=\bunsuu{\sin\theta}{\cos\theta}$より,%
+ \[\tan\Ttyuukakko{\alpha\pm\beta}=\bunsuu{\sin\alpha \cos\beta\pm \cos\alpha \sin\beta}{\cos\alpha \cos\beta\mp \sin\alpha \sin\beta}\]%
+ 両辺を$\cos\alpha\cos\beta$でわることで,%
+ \[\tan\Ttyuukakko{\alpha\pm\beta}=\bunsuu{\tan\alpha \pm \tan\beta}{1\mp \tan\alpha \tan\beta}\]%
+ を得る。%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\三角関数の二倍角の公式}{ m O{i} }%%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$\sin2\alpha=2\sin\alpha\cos\beta$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {\[\sin2\alpha=2\sin\alpha\cos\beta\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$\cos2\alpha=\cos^{2}\alpha-\sin^{2}\alpha$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {\[\cos2\alpha=\cos^{2}\alpha-\sin^{2}\alpha\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {$\cos2\alpha=2\cos^{2}\alpha-1$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}%
+ {\[\cos2\alpha=2\cos^{2}\alpha-1\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{i}}%
+ {$\cos2\alpha=1-2\sin^{2}\alpha$}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{b}}%
+ {\[\cos2\alpha=1-2\sin^{2}\alpha\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式E}\AND\equal{#2}{i}}%
+ {$\tan2\alpha=\frac{2\tan\alpha}{1-\tan^{2}\alpha}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式E}\AND\equal{#2}{b}}%
+ {\[\tan2\alpha=\frac{2\tan\alpha}{1-\tan^{2}\alpha}\]}{\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ 三角関数の加法定理\par%
+ \[\sin\Ttyuukakko{\alpha+\beta}=\sin\alpha \cos\beta+ \cos\alpha \sin\beta\]%
+ \[\cos\Ttyuukakko{\alpha+\beta}=\cos\alpha \cos\beta- \sin\alpha \sin\beta\]%
+ \[\tan\Ttyuukakko{\alpha+\beta}=\bunsuu{\tan\alpha + \tan\beta}{1- \tan\alpha \tan\beta}\]%
+ において,$\alpha=\beta=\theta$として,%
+ \[\sin2\theta=2\sin\theta\cos\theta\]%
+ \[\cos2\theta=\cos^{2}\theta-\sin^{2}\theta\]%
+ \[\tan2\theta=\bunsuu{2\tan\theta}{1-\tan^{2}\theta}\]%
+ を得る。\par%
+ また,$\cos2\theta=\cos^{2}\theta-\sin^{2}\theta$において,三角関数の相互関係$\sin^2\theta+\cos^2\theta=1$を用いて,%
+ \[\cos2\theta=2\cos^{2}\theta-1\]%
+ \[\Leftrightarrow\cos2\theta=1-2\sin^{2}\theta\]%
+ を得る。%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\三角関数の三倍角の公式}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$\sin3\alpha=-4\sin^{3}\alpha+3\sin\alpha$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {\[\sin3\alpha=-4\sin^{3}\alpha+3\sin\alpha\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$\cos3\alpha=4\cos^{3}\alpha-3\cos\alpha$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {\[\cos3\alpha=4\cos^{3}\alpha-3\cos\alpha\]}{\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ 三角関数の加法定理%
+ \[\sin\Ttyuukakko{\alpha+\beta}=\sin\alpha \cos\beta+ \cos\alpha \sin\beta\]%
+ \[\cos\Ttyuukakko{\alpha+\beta}=\cos\alpha \cos\beta- \sin\alpha \sin\beta\]%
+ において,$\alpha=\theta\数式カンマスペース\beta=2\theta$のとき,%
+ \[\sin3\theta=\sin\theta \cos2\theta+ \cos\theta \sin2\theta\]%
+ \[\cos3\theta=\cos\theta\cos2\theta-\sin\theta\sin2\theta\]%
+ 二倍角の公式と三角関数の相互関係より,%
+ \begin{align*}%
+ \sin3\theta&=\sin\theta\Ttyuukakko{1-2\sin^{2}\theta}+2\sin\theta\cos^2\theta&\\%
+ &=\sin\theta-2\sin^{3}\theta+2\sin\theta\Ttyuukakko{1-\sin^2\theta}&\\%
+ &=-4\sin^{3}\theta+3\sin\theta&\\%
+ \cos3\theta&=\cos\theta\Ttyuukakko{2\cos^2\theta-1}-2\sin^2\theta\cos\theta&\\%
+ &=2\cos^{3}\theta-\cos\theta-2\Ttyuukakko{1-\cos^2\theta}\cos\theta &\\
+ &=4\cos^{3}\theta-3\cos\theta%
+ \end{align*}%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\三角関数の積和公式}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$\sin\alpha\cos\beta=\bunsuu{\sin\Ttyuukakko{\alpha+\beta}+\sin\Ttyuukakko{\alpha-\beta}}{2}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {\[\sin\alpha\cos\beta=\bunsuu{\sin\Ttyuukakko{\alpha+\beta}+\sin\Ttyuukakko{\alpha-\beta}}{2}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$\cos\alpha\cos\beta=\bunsuu{\cos\Ttyuukakko{\alpha+\beta}+\cos\Ttyuukakko{\alpha-\beta}}{2}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {\[\cos\alpha\cos\beta=\bunsuu{\cos\Ttyuukakko{\alpha+\beta}+\cos\Ttyuukakko{\alpha-\beta}}{2}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {$\sin\alpha\sin\beta=\bunsuu{\cos\Ttyuukakko{\alpha+\beta}-\cos\Ttyuukakko{\alpha-\beta}}{2}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}%
+ {\[\sin\alpha\sin\beta=\bunsuu{\cos\Ttyuukakko{\alpha+\beta}-\cos\Ttyuukakko{\alpha-\beta}}{2}\]}{\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ 三角関数の加法定理%
+ \begin{align*}
+ \sin\Ttyuukakko{\alpha+\beta}=\sin\alpha\cos\beta+\cos\alpha\sin\beta\shikimaru{1}\\%
+ \sin\Ttyuukakko{\alpha-\beta}=\sin\alpha\cos\beta-\cos\alpha\sin\beta\shikimaru{2}\\%
+ \cos\Ttyuukakko{\alpha+\beta}=\cos\alpha\cos\beta-\sin\alpha\sin\beta\shikimaru{3}\\%
+ \cos\Ttyuukakko{\alpha-\beta}=\cos\alpha\cos\beta+\sin\alpha\sin\beta\shikimaru{4}%
+ \end{align*}
+ より,$\text{\ajMaru{1}}+\text{\ajMaru{2}}$から%
+ \[\sin\alpha\cos\beta=\bunsuu{1}{2}\Tdaikakko{\sin\Ttyuukakko{\alpha+\beta}+\sin\Ttyuukakko{\alpha-\beta}}\]%
+ \半空行%
+ $\text{\ajMaru{3}}+\text{\ajMaru{4}}$から%
+ \[\cos\alpha\cos\beta=\bunsuu{1}{2}\Tdaikakko{\cos\Ttyuukakko{\alpha+\beta}+\cos\Ttyuukakko{\alpha-\beta}}\]%
+ \半空行%
+ $\text{\ajMaru{4}}-\text{\ajMaru{3}}$から%
+ \[\sin\alpha\sin\beta=\bunsuu{1}{2}\Tdaikakko{\cos\Ttyuukakko{\alpha-\beta}-\cos\Ttyuukakko{\alpha+\beta}}\]%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\三角関数の和積公式}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$\sin{A}+\sin{B}=2 \sin\bunsuu{A+B}{2}\cos\bunsuu{A-B}{2}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {\[\sin{A}+\sin{B}=2 \sin\bunsuu{A+B}{2}\cos\bunsuu{A-B}{2}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$\sin{A}-\sin{B}=2 \cos\bunsuu{A+B}{2}\sin\bunsuu{A-B}{2}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {\[\sin{A}-\sin{B}=2 \cos\bunsuu{A+B}{2}\sin\bunsuu{A-B}{2}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {$\cos{A}+\cos{B}=2 \cos\bunsuu{A+B}{2}\cos\bunsuu{A-B}{2}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}%
+ {\[\cos{A}+\cos{B}=2 \cos\bunsuu{A+B}{2}\cos\bunsuu{A-B}{2}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{i}}%
+ {$\cos{A}-\cos{B}=-2 \sin\bunsuu{A+B}{2}\sin\bunsuu{A-B}{2}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{b}}%
+ {\[\cos{A}-\cos{B}=-2 \sin\bunsuu{A+B}{2}\sin\bunsuu{A-B}{2}\]}{\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ 三角関数の積和の公式%
+ \[\sin\alpha\cos\beta=\bunsuu{1}{2}\Tdaikakko{\sin\Ttyuukakko{\alpha+\beta}+\sin\Ttyuukakko{\alpha-\beta}}\]%
+ \[\cos\alpha\cos\beta=\bunsuu{1}{2}\Tdaikakko{\cos\Ttyuukakko{\alpha+\beta}+\cos\Ttyuukakko{\alpha-\beta}}\]%
+ \[\sin\alpha\sin\beta=\bunsuu{1}{2}\Tdaikakko{\cos\Ttyuukakko{\alpha-\beta}-\cos\Ttyuukakko{\alpha+\beta}}\]%
+ において,$\alpha+\beta=A\数式カンマスペース\alpha-\beta=B$と置くことで,$\alpha=\bunsuu{A+B}{2}\数式カンマスペース\beta=\bunsuu{A-B}{2}$となるので,%
+ \[\sin{A}+\sin{B}=2 \sin\bunsuu{A+B}{2}\cos\bunsuu{A-B}{2}\]%
+ \[\sin{A}-\sin{B}=2 \cos\bunsuu{A+B}{2}\sin\bunsuu{A-B}{2}\]%
+ \[\cos{A}+\cos{B}=2 \cos\bunsuu{A+B}{2}\cos\bunsuu{A-B}{2}\]%
+ \[\cos{A}-\cos{B}=-2 \sin\bunsuu{A+B}{2}\sin\bunsuu{A-B}{2}\]%
+ \証明終了%
+ となる。%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\三角関数の合成}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{i}}%
+ {$a\sin\theta+b\cos\theta=\sqrt{a^2+b^2}\sin\Ttyuukakko{\theta+\alpha}\text{\ (ただし,$\sin\alpha=\bunsuu{b}{\sqrt{a^2+b^2}}\数式カンマスペース\cos\alpha=\bunsuu{a}{\sqrt{a^2+b^2}}$)}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{b}}%
+ {%
+ \[a\sin\theta+b\cos\theta=\sqrt{a^2+b^2}\sin\Ttyuukakko{\theta+\alpha}\text{\ (ただし,$\sin\alpha=\bunsuu{b}{\sqrt{a^2+b^2}}\数式カンマスペース\cos\alpha=\bunsuu{a}{\sqrt{a^2+b^2}}$)}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ 三角関数の加法定理\par%
+ $\sin\Ttyuukakko{\alpha+\beta}=\sin\alpha \cos\beta+ \cos\alpha \sin\beta$について,%
+ \[\bunsuu{a}{\sqrt{a^2+b^2}}=\cos\alpha\数式カンマスペース\bunsuu{b}{\sqrt{a^2+b^2}}=\sin\alpha\]%
+ とすることで,\par%
+ \[a\sin\theta+b\cos\theta=\sqrt{a^2+b^2}\sin\Ttyuukakko{\theta+\alpha}\]%
+ となる。%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\有理数の指数}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$a>0$また$m\数式カンマスペース n$が正の整数,$r$が正の有理数のとき,$a^{\bunsuu{m}{n}}=\sqrt[n]{a^{m}}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {%
+ $a>0$また$m\数式カンマスペース n$が正の整数,$r$が正の有理数のとき,%
+ \[a^{\bunsuu{m}{n}}=\sqrt[n]{a^{m}}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$a>0$また$n$が正の整数のとき,$a^{\bunsuu{1}{n}}=\sqrt[n]{a}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {%
+ $a>0$また$n$が正の整数のとき,%
+ \[a^{\bunsuu{1}{n}}=\sqrt[n]{a}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {$a>0$,$r$が正の有理数のとき,$a^{-r}=\bunsuu{1}{a^{r}}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}%
+ {%
+ $a>0$,$r$が正の有理数のとき,%
+ \[a^{-r}=\bunsuu{1}{a^{r}}\]%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\指数法則}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$a>0$また,$r\数式カンマスペース s$は有理数のとき,$a^{r}a^{s}=a^{r+s}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {%
+ $a>0$また,$r\数式カンマスペース s$は有理数のとき,%
+ \[a^{r}a^{s}=a^{r+s}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$a>0$また,$r\数式カンマスペース s$は有理数のとき,$\Ttyuukakko{a^{r}}^{s}=a^{rs}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {%
+ $a>0$また,$r\数式カンマスペース s$は有理数のとき,%
+ \[\Ttyuukakko{a^{r}}^{s}=a^{rs}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {$a>0\数式カンマスペース b>0$また,$r$は有理数のとき,$\Ttyuukakko{ab}^{r}=a^{r}b^{r}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}%
+ {%
+ $a>0\数式カンマスペース b>0$また,$r$は有理数のとき,%
+ \[\Ttyuukakko{ab}^{r}=a^{r}b^{r}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{i}}%
+ {$a>0$また,$r\数式カンマスペース s$は有理数のとき,$\bunsuu{a^{r}}{a^{s}}=a^{r-s}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{b}}%
+ {%
+ $a>0$また,$r\数式カンマスペース s$は有理数のとき,%
+ \[\bunsuu{a^{r}}{a^{s}}=a^{r-s}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式E}\AND\equal{#2}{i}}%
+ {$a>0\数式カンマスペース b>0$また,$r$は有理数のとき,$\Ttyuukakko{\bunsuu{a}{b}}^{r}=\bunsuu{a^{r}}{b^{r}}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式E}\AND\equal{#2}{b}}%
+ {%
+ $a>0\数式カンマスペース b>0$また,$r$は有理数のとき,%
+ \[\Ttyuukakko{\bunsuu{a}{b}}^{r}=\bunsuu{a^{r}}{b^{r}}\]%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\対数の定義}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{i}}%
+ {%
+ $a>0\数式カンマスペース b>0$また,$r\数式カンマスペース s$は有理数とする。\par%
+ $a^{p}=M$ならば,$\log_{a}M$,$\log_{a}M \log_{a}M$ならば,$a^{p}=M$%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{b}}%
+ {%
+ $a>0\数式カンマスペース b>0$また,$r\数式カンマスペース s$は有理数とする。\par%
+ $a^{p}=M$ならば,$\log_{a}M$\par%
+ $\log_{a}M$ならば,$a^{p}=M$%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\対数の性質}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$a>0\数式カンマスペース a\neq1$とするとき,$\log_{a}a=1$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {%
+ $a>0\数式カンマスペース a\neq1$とするとき,%
+ \[\log_{a}a=1\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$a>0\数式カンマスペース a\neq1$とするとき,$\log_{a}1=0$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {%
+ $a>0\数式カンマスペース a\neq1$とするとき,%
+ \[\log_{a}1=0\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {$a>0\数式カンマスペース a\neq1$とするとき,$\log_{a}\bunsuu{1}{a}=-1$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}%
+ {%
+ $a>0\数式カンマスペース a\neq1$とするとき,%
+ \[\log_{a}\bunsuu{1}{a}=-1\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{i}}%
+ {$a>0\数式カンマスペース a\neq1\数式カンマスペース M>0\数式カンマスペース N>0$とするとき,$\log_{a}MN=\log_{a}M+\log_{a}N$}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{b}}%
+ {%
+ $a>0\数式カンマスペース a\neq1\数式カンマスペース M>0\数式カンマスペース N>0$とするとき,%
+ \[\log_{a}MN=\log_{a}M+\log_{a}N\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式E}\AND\equal{#2}{i}}%
+ {$a>0\数式カンマスペース a\neq1\数式カンマスペース M>0\数式カンマスペース N>0$とするとき,$\log_{a}\bunsuu{M}{N}=\log_{a}M-\log_{a}N$}{\relax}%
+ \ifthenelse{\equal{#1}{公式E}\AND\equal{#2}{b}}%
+ {%
+ $a>0\数式カンマスペース a\neq1\数式カンマスペース M>0\数式カンマスペース N>0$とするとき,%
+ \[\log_{a}\bunsuu{M}{N}=\log_{a}M-\log_{a}N\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式F}\AND\equal{#2}{i}}%
+ {$a>0\数式カンマスペース a\neq1\数式カンマスペース M>0\数式カンマスペース N>0$とするとき,$\log_{a}M^{k}=k\log_{a}M$}{\relax}%
+ \ifthenelse{\equal{#1}{公式F}\AND\equal{#2}{b}}%
+ {%
+ $a>0\数式カンマスペース a\neq1\数式カンマスペース M>0\数式カンマスペース N>0$とするとき,%
+ \[\log_{a}M^{k}=k\log_{a}M\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ $p=\log_{a}M\数式カンマスペース q=\log_{a}N$として,\par%
+ $MN=a^{p}a^{q}$指数法則より%
+ \[MN=a^{p+q}\]%
+ ここで,対数の定義より%
+ \[\log_{a}MN=p+q\]%
+    \[\Leftrightarrow\log_{a}MN=\log_{a}M+\log_{a}N\]%
+ \空行%
+ $p=\log_{a}M\数式カンマスペース q=\log_{a}N$として,%
+ \[\bunsuu{M}{N}=\bunsuu{a^{p}}{a^{q}}\]%
+ 指数法則より%
+ \[\bunsuu{M}{N}=a^{p-q}\]%
+ ここで,対数の定義より%
+ \[\log_{a}\bunsuu{M}{N}=p-q\]%
+ \[\Leftrightarrow\log_{a}\bunsuu{M}{N}=\log_{a}M-\log_{a}N\]%
+ \空行%
+ $p=\log_{a}M$として,$a^{p}=M$より両辺$k$乗して%
+ \[a^{pk}=M^{k}\]%
+ 対数を取ると%
+ \[pk=\log_{a}M^{k}\]%
+ $p=\log_{a}M$より,%
+ \[\log_{a}M^{k}=k\log_{a}M\]%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\底の変換公式}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{i}}%
+ {$a\数式カンマスペース b\数式カンマスペース c$は正の実数で,$a\neq1\数式カンマスペース b\neq1\数式カンマスペース c\neq1$のとき,$\log_{a}b=\bunsuu{\log_{c}b}{\log_{c}a}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{b}}%
+ {%
+ $a\数式カンマスペース b\数式カンマスペース c$は正の実数で,$a\neq1\数式カンマスペース b\neq1\数式カンマスペース c\neq1$のとき,%
+ \[\log_{a}b=\bunsuu{\log_{c}b}{\log_{c}a}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ 対数の定義より$a^{\log_{a}b}=b$が成立。\par%
+ 底が$c$の対数を取ると,%
+ \[\log_{c}a^{\log_{a}b}=\log_{c}b\]%
+ 対数の性質より,%
+ \[\log_{a}b\log_{c}a=\log_{c}b\]%
+ よって,%
+ \[\log_{a}b=\bunsuu{\log_{c}b}{\log_{c}a}\]%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\導関数の定義}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{i}}%
+ {$f'\Ttyuukakko{x}=\displaystyle\lim_{h \to 0}\bunsuu{f\Ttyuukakko{x+h}-f\Ttyuukakko{x}}{h}$}{\relax}%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{b}}%
+ {\[f'\Ttyuukakko{x}=\displaystyle\lim_{h \to 0}\bunsuu{f\Ttyuukakko{x+h}-f\Ttyuukakko{x}}{h}\]}{\relax}%
+ }%
+
+
+\NewDocumentCommand{\べき乗関数と定数関数の導関数}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{x^{n}}'=nx^{n-1}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {\[\Ttyuukakko{x^{n}}'=nx^{n-1}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{c}'=0$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {\[\Ttyuukakko{c}'=0\]}{\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ 導関数の定義より,\par%
+ \[\Ttyuukakko{x^{n}}'=\displaystyle\lim_{h \to 0}\bunsuu{\Ttyuukakko{x+h}^{n}-x^{n}}{h}\]%
+ 二項定理より,\par%
+ \begin{align*}%
+ \Ttyuukakko{x^{n}}'&=\displaystyle\lim_{h \to 0} \bunsuu{\Ttyuukakko{x+h}^{n}-x^{n}}{h}&\\%
+ &=\displaystyle\lim_{h \to 0}\bunsuu{{}_{n}C_{0} x^{n}+{}_{n}C_{1} x^{n-1}h+{}_{n}C_{2}x^{n-2}h^2+\cdots\cdot{}_{n}C_{n-1} xh^{n-1}+{}_{n}C_{n} h^{n}-x^{n}}{h}&\\%
+ &=\displaystyle\lim_{h \to 0}\Ttyuukakko{{}_{n}C_{1} x^{n-1}+{}_{n}C_{2}x^{n-2}h+\cdots+{}_{n}C_{n-1} xh^{n-2}+{}_{n}C_{n} h^{n-1}}&\\%
+ &=\displaystyle\lim_{h \to 0}\Tdaikakko{{}_{n}C_{1} x^{n-1}+\Ttyuukakko{{}_{n}C_{2}x^{n-2}+\cdots\cdot{}_{n}C_{n-1} xh^{n-3}+{}_{n}C_{n} h^{n-2}}h}&\\%
+ &={}_{n}C_{1} x^{n-1}&\\%
+ &=nx^{n-1}%
+ \end{align*}%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\導関数の性質}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {${kf\Ttyuukakko{x}}'=kf'\Ttyuukakko{x}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {\[{kf\Ttyuukakko{x}}'=kf'\Ttyuukakko{x}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {${f\Ttyuukakko{x}\pm g\Ttyuukakko{x}}'=f'\Ttyuukakko{x}\pm g'\Ttyuukakko{x}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {\[{f\Ttyuukakko{x}\pm g\Ttyuukakko{x}}'=f'\Ttyuukakko{x}\pm g'\Ttyuukakko{x}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {${kf\Ttyuukakko{x}+lg\Ttyuukakko{x}}'=kf'\Ttyuukakko{x}+lg'\Ttyuukakko{x}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}%
+ {\[{kf\Ttyuukakko{x}+lg\Ttyuukakko{x}}'=kf'\Ttyuukakko{x}+lg'\Ttyuukakko{x}\]}{\relax}%
+ }%
+
+
+\NewDocumentCommand{\接線の方程式}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{i}}%
+ {曲線$y=f\Ttyuukakko{x}$上の点$\Ttyuukakko{a\数式カンマスペース f\Ttyuukakko{a}}$における曲線の接線の方程式は,$y-f\Ttyuukakko{a}=f'\Ttyuukakko{x}\Ttyuukakko{x-a}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{b}}%
+ {%
+ 曲線$y=f\Ttyuukakko{x}$上の点$\Ttyuukakko{a\数式カンマスペース f\Ttyuukakko{a}}$における曲線の接線の方程式は,%
+ \[y-f\Ttyuukakko{a}=f'\Ttyuukakko{x}\Ttyuukakko{x-a}\]%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\不定積分の定義}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{i}}%
+ {$F'\Ttyuukakko{x}=f\Ttyuukakko{x}$のとき,$\int_{}^{} f\Ttyuukakko{x}dx=F\Ttyuukakko{x}+C$($C$は積分定数)}{\relax}%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{b}}%
+ {%
+ $F'\Ttyuukakko{x}=f\Ttyuukakko{x}$のとき,%
+ \[\int_{}^{} f\Ttyuukakko{x}dx=F\Ttyuukakko{x}+C\]%
+ ($C$は積分定数)%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\べき乗関数の不定積分}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{i}}%
+ {$\int_{}^{} x^{n}dx=\bunsuu{1}{n+1}x^{n+1}+C\text{\ ($C$は積分定数)}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{b}}%
+ {%
+ \[\int_{}^{} x^{n}dx=\bunsuu{1}{n+1}x^{n+1}+C\text{\ ($C$は積分定数)}\]%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\不定積分の性質}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$\displaystyle \int_{}^{} kf\Ttyuukakko{x}dx=k\displaystyle \int_{}^{} f\Ttyuukakko{x}dx$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {\[\int_{}^{} kf\Ttyuukakko{x}dx=k\int_{}^{} f\Ttyuukakko{x}dx\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$\displaystyle \int_{}^{} {f\Ttyuukakko{x}\pm g\Ttyuukakko{x}}dx=\displaystyle \int_{}^{} f\Ttyuukakko{x}dx\pm\displaystyle \int_{}^{} g\Ttyuukakko{x}dx$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {\[\int_{}^{} {f\Ttyuukakko{x}\pm g\Ttyuukakko{x}}dx=\int_{}^{} f\Ttyuukakko{x}dx\pm\int_{}^{} g\Ttyuukakko{x}dx\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {$\displaystyle \int_{}^{} {kf\Ttyuukakko{x}+lg\Ttyuukakko{x}}dx=k\displaystyle \int_{}^{} f\Ttyuukakko{x}dx+l\displaystyle \int_{}^{} g\Ttyuukakko{x}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}%
+ {\[\int_{}^{} {kf\Ttyuukakko{x}+lg\Ttyuukakko{x}}dx=k\int_{}^{} f\Ttyuukakko{x}dx+l\int_{}^{} g\Ttyuukakko{x}\]}{\relax}%
+ }%
+
+
+\NewDocumentCommand{\定積分の定義}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{i}}%
+ {曲線$y=f\Ttyuukakko{x}$と$x$軸(区間は$a$から$b$)に囲まれた部分の面積$S$について,$F'\Ttyuukakko{x}=f\Ttyuukakko{x}$のとき,$S=\int_{b}^{a} f\Ttyuukakko{x}dx=[F\Ttyuukakko{x}]^{b}_{a}=F\Ttyuukakko{b}-F\Ttyuukakko{a}$}{\relax}%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{b}}%
+ {%
+ 曲線$y=f\Ttyuukakko{x}$と$x$軸(区間は$a$から$b$)に囲まれた部分の面積$S$について,$F'\Ttyuukakko{x}=f\Ttyuukakko{x}$のとき,%
+ \[S=\int_{b}^{a} f\Ttyuukakko{x}dx=[F\Ttyuukakko{x}]^{b}_{a}=F\Ttyuukakko{b}-F\Ttyuukakko{a}\]%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\定積分の性質}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$\int_{b}^{a} kf\Ttyuukakko{x}dx=k\int_{b}^{a} f\Ttyuukakko{x}dx$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {\[\int_{b}^{a} kf\Ttyuukakko{x}dx=k\int_{b}^{a} f\Ttyuukakko{x}dx\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$\int_{b}^{a} {f\Ttyuukakko{x}\pm g\Ttyuukakko{x}}dx=\int_{b}^{a} f\Ttyuukakko{x}dx\pm\int_{b}^{a} g\Ttyuukakko{x}dx$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {\[\int_{b}^{a} {f\Ttyuukakko{x}\pm g\Ttyuukakko{x}}dx=\int_{b}^{a} f\Ttyuukakko{x}dx\pm\int_{b}^{a} g\Ttyuukakko{x}dx\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {$\int_{a}^{a} f\Ttyuukakko{x}dx=0$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}%
+ {\[\int_{a}^{a} f\Ttyuukakko{x}dx=0\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{i}}%
+ {$\int_{b}^{a} f\Ttyuukakko{x}dx=-\int_{a}^{b} f\Ttyuukakko{x}dx$}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{b}}%
+ {\[\int_{b}^{a} f\Ttyuukakko{x}dx=-\int_{a}^{b} f\Ttyuukakko{x}dx\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式E}\AND\equal{#2}{i}}%
+ {$\int_{b}^{a} f\Ttyuukakko{x}dx=\int_{a}^{c} f\Ttyuukakko{x}dx+\int_{c}^{b} f\Ttyuukakko{x}dx$}{\relax}%
+ \ifthenelse{\equal{#1}{公式E}\AND\equal{#2}{b}}%
+ {\[\int_{b}^{a} f\Ttyuukakko{x}dx=\int_{a}^{c} f\Ttyuukakko{x}dx+\int_{c}^{b} f\Ttyuukakko{x}dx\]}{\relax}%
+ }%
+
+
+\NewDocumentCommand{\ベクトルの演算}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$k\数式カンマスペース l$が実数のとき,$\vec{a}+\vec{b}=\vec{b}+\vec{a}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {%
+ $k\数式カンマスペース l$が実数のとき%
+ \[\vec{a}+\vec{b}=\vec{b}+\vec{a}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$k\数式カンマスペース l$が実数のとき,$\Ttyuukakko{\vec{a}+\vec{b}}+\vec{c}=\vec{a}+\Ttyuukakko{\vec{b}+\vec{c}}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {%
+ $k\数式カンマスペース l$が実数のとき%
+ \[\Ttyuukakko{\vec{a}+\vec{b}}+\vec{c}=\vec{a}+\Ttyuukakko{\vec{b}+\vec{c}}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {$\vec{a}+\Ttyuukakko{a\vec{a}}=\vec{0}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}%
+ {\[\vec{a}+\Ttyuukakko{a\vec{a}}=\vec{0}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{i}}%
+ {$\vec{a}+\vec{0}=\vec{a}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{b}}%
+ {\[\vec{a}+\vec{0}=\vec{a}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式E}\AND\equal{#2}{i}}%
+ {$\vec{a}-\vec{b}=\vec{a}+\Ttyuukakko{-\vec{b}}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式E}\AND\equal{#2}{b}}%
+ {\[\vec{a}-\vec{b}=\vec{a}+\Ttyuukakko{-\vec{b}}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式F}\AND\equal{#2}{i}}%
+ {$k\数式カンマスペース l$が実数のとき,$k\Ttyuukakko{l\vec{a}}=l\Ttyuukakko{k\vec{b}}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式F}\AND\equal{#2}{b}}%
+ {%
+ $k\数式カンマスペース l$が実数のとき%
+ \[k\Ttyuukakko{l\vec{a}}=l\Ttyuukakko{k\vec{b}}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式G}\AND\equal{#2}{i}}%
+ {$k\数式カンマスペース l$が実数のとき,$\Ttyuukakko{k+l}\vec{a}=k\vec{a}+l\vec{a}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式G}\AND\equal{#2}{b}}%
+ {%
+ $k\数式カンマスペース l$が実数のとき%
+ \[\Ttyuukakko{k+l}\vec{a}=k\vec{a}+l\vec{a}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式H}\AND\equal{#2}{i}}%
+ {$k$が実数のとき,$k\Ttyuukakko{\vec{a}+\vec{b}}=k\vec{a}+k\vec{b}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式H}\AND\equal{#2}{b}}%
+ {%
+ $k$が実数のとき%
+ \[k\Ttyuukakko{\vec{a}+\vec{b}}=k\vec{a}+k\vec{b}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式I}\AND\equal{#2}{i}}%
+ {$\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式I}\AND\equal{#2}{b}}%
+ {\[\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式J}\AND\equal{#2}{i}}%
+ {$\overrightarrow{OA}-\overrightarrow{OB}=\overrightarrow{BA}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式J}\AND\equal{#2}{b}}%
+ {\[\overrightarrow{OA}-\overrightarrow{OB}=\overrightarrow{BA}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式K}\AND\equal{#2}{i}}%
+ {$\overrightarrow{AA}=\vec{0}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式K}\AND\equal{#2}{b}}%
+ {\[\overrightarrow{AA}=\vec{0}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式L}\AND\equal{#2}{i}}%
+ {$\overrightarrow{BA}=\overrightarrow{AB}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式L}\AND\equal{#2}{b}}%
+ {\[\overrightarrow{BA}=\overrightarrow{AB}\]}{\relax}%
+ }%
+
+
+\NewDocumentCommand{\平面ベクトルの分解}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{i}}%
+ {$\vec{a}\neq0\数式カンマスペース\vec{b}\neq0$で,$\vec{a}$と$\vec{b}$が平行でないとき,任意の$\vec{p}$はただ一通りに,$\vec{p}=s\vec{a}+t\vec{b}$の形に表せられる。}{\relax}%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{b}}%
+ {%
+ $\vec{a}\neq0\数式カンマスペース\vec{b}\neq0$で,$\vec{a}$と$\vec{b}$が平行でないとき,任意の$\vec{p}$はただ一通りに,%
+ \[\vec{p}=s\vec{a}+t\vec{b}\]%
+ の形に表せられる。%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\平面ベクトルの成分}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$\vec{a}=\Ttyuukakko{a_{1}\数式カンマスペース a_{2}}\数式カンマスペース\vec{b}=\Ttyuukakko{b_{1}\数式カンマスペース b_{2}}$とすると,$\vec{a}=\vec{b}\Leftrightarrow a_{1}=b_{1}\数式カンマスペース a_{2}=b_{2}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {%
+ $\vec{a}=\Ttyuukakko{a_{1}\数式カンマスペース a_{2}}\数式カンマスペース\vec{b}=\Ttyuukakko{b_{1}\数式カンマスペース b_{2}}$とすると,%
+ $\vec{a}=\vec{b}$%
+ \[\Leftrightarrow a_{1}=b_{1}\数式カンマスペース a_{2}=b_{2}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$\vec{a}=\Ttyuukakko{a_{1}\数式カンマスペース a_{2}}\数式カンマスペース\vec{b}=\Ttyuukakko{b_{1}\数式カンマスペース b_{2}}$とすると,$a_{1}=b_{1}\数式カンマスペース a_{2}=b_{2}\Leftrightarrow\vec{a}=\vec{b}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {%
+ $\vec{a}=\Ttyuukakko{a_{1}\数式カンマスペース a_{2}}\数式カンマスペース\vec{b}=\Ttyuukakko{b_{1}\数式カンマスペース b_{2}}$とすると,%
+ $a_{1}=b_{1}\数式カンマスペース a_{2}=b_{2}$%
+ \[\Leftrightarrow\vec{a}=\vec{b}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {$\vec{a}=\Ttyuukakko{a_{1}\数式カンマスペース a_{2}}$とすると,$\Tzettaiti{\vec{a}}=\sqrt{a_{1}^2+a_{2}^2}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}%
+ {%
+ $\vec{a}=\Ttyuukakko{a_{1}\数式カンマスペース a_{2}}$とすると,%
+ \[\Tzettaiti{\vec{a}}=\sqrt{a_{1}^2+a_{2}^2}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{i}}%
+ {$\vec{a}=\Ttyuukakko{a_{1}\数式カンマスペース a_{2}}\数式カンマスペース\vec{b}=\Ttyuukakko{b_{1}\数式カンマスペース b_{2}}$また,$k\数式カンマスペース l$を実数として,$k\vec{a}+l\vec{b}=k\Ttyuukakko{a_{1}\数式カンマスペース a_{2}}+l\Ttyuukakko{b_{1}\数式カンマスペース b_{2}}=\Ttyuukakko{ka_{1}+lb_{1}\数式カンマスペース ka_{2}+lb_{2}}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{b}}%
+ {%
+ $\vec{a}=\Ttyuukakko{a_{1}\数式カンマスペース a_{2}}\数式カンマスペース\vec{b}=\Ttyuukakko{b_{1}\数式カンマスペース b_{2}}$また,$k\数式カンマスペース l$を実数として,%
+ \[k\vec{a}+l\vec{b}=k\Ttyuukakko{a_{1}\数式カンマスペース a_{2}}+l\Ttyuukakko{b_{1}\数式カンマスペース b_{2}}=\Ttyuukakko{ka_{1}+lb_{1}\数式カンマスペース ka_{2}+lb_{2}}\]%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\ベクトルの成分と大きさ}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$A\Ttyuukakko{a_{1}\数式カンマスペース a_{2}}\数式カンマスペース B\Ttyuukakko{b_{1}\数式カンマスペース b_{2}}$とすると,$\overrightarrow{AB}=\Ttyuukakko{b_{1}-a_{1}\数式カンマスペース b_{2}-a_{2}}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {%
+ $A\Ttyuukakko{a_{1}\数式カンマスペース a_{2}}\数式カンマスペース B\Ttyuukakko{b_{1}\数式カンマスペース b_{2}}$とすると,%
+ \[\overrightarrow{AB}=\Ttyuukakko{b_{1}-a_{1}\数式カンマスペース b_{2}-a_{2}}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$A\Ttyuukakko{a_{1}\数式カンマスペース a_{2}}\数式カンマスペース B\Ttyuukakko{b_{1}\数式カンマスペース b_{2}}$とすると,$\Tzettaiti{\overrightarrow{AB}}=\sqrt{\Ttyuukakko{b_{1}-a_{1}}^2+\Ttyuukakko{b_{2}-a_{2}}^2}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {%
+ $A\Ttyuukakko{a_{1}\数式カンマスペース a_{2}}\数式カンマスペース B\Ttyuukakko{b_{1}\数式カンマスペース b_{2}}$とすると,%
+ \[\Tzettaiti{\overrightarrow{AB}}=\sqrt{\Ttyuukakko{b_{1}-a_{1}}^2+\Ttyuukakko{b_{2}-a_{2}}^2}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ 三平方の定理より,%
+ \[\Tzettaiti{\overrightarrow{AB}}=\sqrt{\Ttyuukakko{b_{1}-a_{1}}^2+\Ttyuukakko{b_{2}-a_{2}}^2}\]%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\平面ベクトルの内積}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{i}}%
+ {ベクトルの内積は,$\vec{a} \cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta \Ttyuukakko{0^{\circ} \leqq \theta \leqq 180^{\circ}}\text{\ (ただし,$\theta$は$\vec{a}$と$\vec{b}$のなす角}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{b}}%
+ {%
+ ベクトルの内積は,%
+ \[\vec{a} \cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta \Ttyuukakko{0^{\circ} \leqq \theta \leqq 180^{\circ}}\text{\ (ただし,$\theta$は$\vec{a}$と$\vec{b}$のなす角}\]%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\内積の性質}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$\vec{a} \cdot\vec{b}=\vec{b} \cdot\vec{a}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {\[\vec{a} \cdot\vec{b}=\vec{b} \cdot\vec{a}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{\vec{a}+\vec{b}} \cdot\vec{c}=\vec{a} \cdot\vec{c}+\vec{b} \cdot\vec{c}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {\[\Ttyuukakko{\vec{a}+\vec{b}} \cdot\vec{c}=\vec{a} \cdot\vec{c}+\vec{b} \cdot\vec{c}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {$\vec{c} \cdot\Ttyuukakko{\vec{b}+\vec{c}}=\vec{a} \cdot\vec{c}+\vec{b} \cdot\vec{c}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}%
+ {\[\vec{c} \cdot\Ttyuukakko{\vec{b}+\vec{c}}=\vec{a} \cdot\vec{c}+\vec{b} \cdot\vec{c}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{i}}%
+ {$k$が実数のとき,$\Ttyuukakko{k\vec{a}} \cdot\vec{b}=\vec{a} \cdot\Ttyuukakko{k\vec{b}}=k\Ttyuukakko{\vec{a} \cdot\vec{b}}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{b}}%
+ {%
+ $k$が実数のとき,%
+ \[\Ttyuukakko{k\vec{a}} \cdot\vec{b}=\vec{a} \cdot\Ttyuukakko{k\vec{b}}=k\Ttyuukakko{\vec{a} \cdot\vec{b}}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式E}\AND\equal{#2}{i}}%
+ {$\vec{a} \cdot\vec{a}=\Tzettaiti{\vec{a}}^2$}{\relax}%
+ \ifthenelse{\equal{#1}{公式E}\AND\equal{#2}{b}}%
+ {\[\vec{a} \cdot\vec{a}=\Tzettaiti{\vec{a}}^2\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式F}\AND\equal{#2}{i}}%
+ {$\Tzettaiti{\vec{a}}=\sqrt{\vec{a} \cdot\vec{a}}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式F}\AND\equal{#2}{b}}%
+ {\[\Tzettaiti{\vec{a}}=\sqrt{\vec{a} \cdot\vec{a}}\]}{\relax}%
+ }%
+
+
+\NewDocumentCommand{\平面ベクトルの平行条件}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{条件}\AND\equal{#2}{i}}%
+ {%
+ $\vec{a}\neq\vec{0}\数式カンマスペース\vec{b}\neq\vec{0}$また,$k$は実数とする,\par%
+ $\vec{a}/ \!/ \vec{b}\Leftrightarrow\vec{b}=k\vec{a}$,$\vec{b}=k\vec{a}$%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{条件}\AND\equal{#2}{b}}%
+ {%
+ $\vec{a}\neq\vec{0}\数式カンマスペース\vec{b}\neq\vec{0}$また,$k$は実数とする,\par%
+ $\vec{a}/ \!/ \vec{b}$%
+ \[\Leftrightarrow\vec{b}=k\vec{a}\]%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\平面ベクトルの垂直条件}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{条件}\AND\equal{#2}{i}}%
+ {%
+ $\vec{a}\neq\vec{0}\数式カンマスペース\vec{b}\neq\vec{0}$また,$k$は実数とする,\par%
+ $\vec{a} \perp \vec{b}\Leftrightarrow\vec{a} \cdot\vec{b}=0$%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{条件}\AND\equal{#2}{b}}%
+ {%
+ $\vec{a}\neq\vec{0}\数式カンマスペース\vec{b}\neq\vec{0}$また,$k$は実数とする,\par%
+ $\vec{a} \perp \vec{b}$%
+ \[\Leftrightarrow\vec{a} \cdot\vec{b}=0\]%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\位置ベクトル}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$A\Ttyuukakko{\vec{a}}\数式カンマスペース B\Ttyuukakko{\vec{b}}$とする,線分$AB$を$m:n$に内分する点は,$\bunsuu{n\vec{a}+m\vec{b}}{m+n}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {%
+ $A\Ttyuukakko{\vec{a}}\数式カンマスペース B\Ttyuukakko{\vec{b}}$とする,線分$AB$を$m:n$に内分する点は,%
+ \[\bunsuu{n\vec{a}+m\vec{b}}{m+n}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{内分点の位置ベクトルの証明}\AND\equal{#2}{i}}%
+ {%
+ \証明開始%
+ $P\Ttyuukakko{\vec{p}}$が$A\Ttyuukakko{\vec{a}}\数式カンマスペース B\Ttyuukakko{\vec{b}}$を$m:n$に内分するとき,%
+ \begin{align*}%
+ \vec{p}&=\vec{a}+\bunsuu{m}{m+n}\Ttyuukakko{\vec{b}-\vec{a}}&\\%
+ &=\bunsuu{n\vec{a}+m\vec{b}}{m+n}&\\%
+ \end{align*}%
+ \証明終了%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$A\Ttyuukakko{\vec{a}}\数式カンマスペース B\Ttyuukakko{\vec{b}}$とする,線分$AB$を$m:n$に外分する点は,$\bunsuu{-n\vec{a}+m\vec{b}}{m-n}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {%
+ $A\Ttyuukakko{\vec{a}}\数式カンマスペース B\Ttyuukakko{\vec{b}}$とする,線分$AB$を$m:n$に外分する点は,%
+ \[\bunsuu{-n\vec{a}+m\vec{b}}{m-n}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{外分点の位置ベクトルの証明}\AND\equal{#2}{i}}%
+ {%
+ \証明開始%
+ $m:n$に外分ということは$m:-n$に内分ということなので,$\bunsuu{-n\vec{a}+m\vec{b}}{m-n}$%
+ \証明終了%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {$A\Ttyuukakko{\vec{a}}\数式カンマスペース B\Ttyuukakko{\vec{b}}$とする,線分$AB$の中点は,$\bunsuu{\vec{a}+\vec{b}}{2}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}%
+ {%
+ $A\Ttyuukakko{\vec{a}}\数式カンマスペース B\Ttyuukakko{\vec{b}}$とする,線分$AB$の中点は,%
+ \[\bunsuu{\vec{a}+\vec{b}}{2}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{i}}%
+ {$A\Ttyuukakko{\vec{a}}\数式カンマスペース B\Ttyuukakko{\vec{b}}\数式カンマスペース C\Ttyuukakko{\vec{c}}$とする,三角形$ABC$の重心は,$\bunsuu{\vec{a}+\vec{b}+\vec{c}}{3}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{b}}%
+ {%
+ $A\Ttyuukakko{\vec{a}}\数式カンマスペース B\Ttyuukakko{\vec{b}}\数式カンマスペース C\Ttyuukakko{\vec{c}}$とする,三角形$ABC$の重心は,%
+ \[\bunsuu{\vec{a}+\vec{b}+\vec{c}}{3}\]%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\ベクトル方程式}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$s\数式カンマスペース t$を実数とする。点$A\Ttyuukakko{\vec{a}}$をとおり,$\vec{d}$に平行な直線は,$\vec{p}=\vec{a}+t\vec{b}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {%
+ $s\数式カンマスペース t$を実数とする。点$A\Ttyuukakko{\vec{a}}$をとおり,$\vec{d}$に平行な直線は,%
+ \[\vec{p}=\vec{a}+t\vec{b}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$s\数式カンマスペース t$を実数とする。二点$A\Ttyuukakko{\vec{a}}\数式カンマスペース B\Ttyuukakko{\vec{b}}$を通る直線は,$\vec{p}=\Ttyuukakko{1-t}\vec{a}+t\vec{b}\数式カンマスペース\vec{p}=a\vec{a}+t\vec{b}\text{\ (ただし,$s+t=1$)}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {%
+ $s\数式カンマスペース t$を実数とする。二点$A\Ttyuukakko{\vec{a}}\数式カンマスペース B\Ttyuukakko{\vec{b}}$を通る直線は,%
+ \[\vec{p}=\Ttyuukakko{1-t}\vec{a}+t\vec{b}\数式カンマスペース\vec{p}=a\vec{a}+t\vec{b}\text{\ (ただし,$s+t=1$)}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {点$A\Ttyuukakko{\vec{a}}$を通り,$\vec{n}$に垂直な直線$\vec{p}$について,$\vec{n}\cdot\Ttyuukakko{\vec{p}-\vec{a}}=0$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}%
+ {%
+ 点$A\Ttyuukakko{\vec{a}}$を通り,$\vec{n}$に垂直な直線$\vec{p}$について,%
+ \[\vec{n}\cdot\Ttyuukakko{\vec{p}-\vec{a}}=0\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{i}}%
+ {中心$C\Ttyuukakko{\vec{c}}$,半径$r$の円は,$\Tzettaiti{\vec{p}-\vec{c}}=r\数式カンマスペース\Ttyuukakko{\vec{p}-\vec{c}}\cdot\Ttyuukakko{\vec{p}-\vec{c}}=r^2$}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{b}}%
+ {%
+ 中心$C\Ttyuukakko{\vec{c}}$,半径$r$の円は,%
+ \[\Tzettaiti{\vec{p}-\vec{c}}=r\]%
+ \[\Ttyuukakko{\vec{p}-\vec{c}}\cdot\Ttyuukakko{\vec{p}-\vec{c}}=r^2\]%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\等差数列}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{一般項}\AND\equal{#2}{i}}%
+ {初項$a_{1}$,公差$d$のとき,$a_{n}=a_{1}+\Ttyuukakko{n-1}d$}{\relax}%
+ \ifthenelse{\equal{#1}{一般項}\AND\equal{#2}{b}}%
+ {%
+ 初項$a_{1}$,公差$d$のとき,%
+ \[a_{n}=a_{1}+\Ttyuukakko{n-1}d\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{総和}\AND\equal{#2}{i}}%
+ {$S_{n}=\bunsuu{n\Ttyuukakko{a_{1}+a_{n}}}{2}$}{\relax}%
+ \ifthenelse{\equal{#1}{総和}\AND\equal{#2}{b}}%
+ {\[S_{n}=\bunsuu{n\Ttyuukakko{a_{1}+a_{n}}}{2}\]}{\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ \[S_{n}=a_{1}+\Ttyuukakko{a_{1}+d}+\Ttyuukakko{a_{1}+2d}+\cdots+\Tdaikakko{a_{1}+\Ttyuukakko{n-1}d}\]%
+ \[S_{n}=\Tdaikakko{a_{1}+\Ttyuukakko{n-1}d}+\cdots+a_{1}+\Ttyuukakko{a_{1}+d}+\Ttyuukakko{a_{1}+2d}\]%
+ 連立して,$2S=\Ttyuukakko{a_{1}+a_{n}}n$より,\par%
+ $S_{n}=\bunsuu{n\Ttyuukakko{a_{1}+a_{n}}}{2}$%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\等比数列}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{一般項}\AND\equal{#2}{i}}%
+ {$a_{n}=ar^{n-1}$}{\relax}%
+ \ifthenelse{\equal{#1}{一般項}\AND\equal{#2}{b}}%
+ {\[a_{n}=ar^{n-1}\]}{\relax}%
+ \ifthenelse{\equal{#1}{総和}\AND\equal{#2}{i}}%
+ {%
+ $r\neq1$のとき,$S_{n}=\bunsuu{a_{1}\Ttyuukakko{1-r^{n}}}{1-r}$もしくは,$\bunsuu{a_{1}\Ttyuukakko{r^{n}-1}}{r-1}$\par%
+ $r=1$のとき,$S_{n}=na_{1}$%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{総和}\AND\equal{#2}{b}}%
+ {%
+ $r\neq1$のとき,%
+ \[S_{n}=\bunsuu{a_{1}\Ttyuukakko{1-r^{n}}}{1-r}\]%
+ もしくは,%
+ \[S_{n}=\bunsuu{a_{1}\Ttyuukakko{r^{n}-1}}{r-1}\]%
+ $r=1$のとき,%
+ \[S_{n}=na_{1}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ \[S_{n}=a_{1}+ra_{1}+r^2a_{1}+\cdots+r^{n-1}a_{1}\]%
+ \[S_{n}r=ra_{1}+r^2a_{2}+r^{3}a_{1}+\cdots+r^{n}\]%
+ 連立することで,$S\Ttyuukakko{1-r}=a_{1}-r^{n}a_{1}$となる。\par%
+ よって,%
+ \[S=\bunsuu{a_{1}\Ttyuukakko{1-r^{n}}}{1-r}\]%
+ また,$\bunsuu{-1}{-1}$をかけることで,%
+ \[S=\bunsuu{a_{1}\Ttyuukakko{r^{n}-1}}{r-1}\]%
+ 以上より,%
+ \[S=\bunsuu{a_{1}\Ttyuukakko{1-r^{n}}}{1-r}=\bunsuu{a_{1}\Ttyuukakko{r^{n}-1}}{r-1}\]%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\シグマの公式}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{i}}%
+ {$c$は$k$に無関係なとき,$\displaystyle \sum_{k=1}^{n} c=nc$}{\relax}%
+ \ifthenelse{\equal{#1}{公式A}\AND\equal{#2}{b}}%
+ {%
+ $c$は$k$に無関係なとき,%
+ \[\displaystyle \sum_{k=1}^{n} c=nc\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{i}}%
+ {$\displaystyle \sum_{k=1}^{n} k=\bunsuu{1}{2}n\Ttyuukakko{n+1}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式B}\AND\equal{#2}{b}}%
+ {\[\displaystyle \sum_{k=1}^{n} k=\bunsuu{1}{2}n\Ttyuukakko{n+1}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{i}}%
+ {$\displaystyle \sum_{k=1}^{n} k^2=\bunsuu{1}{6}n\Ttyuukakko{n+1}\Ttyuukakko{2n+1}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式C}\AND\equal{#2}{b}}%
+ {\[\displaystyle \sum_{k=1}^{n} k^2=\bunsuu{1}{6}n\Ttyuukakko{n+1}\Ttyuukakko{2n+1}\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{i}}%
+ {$\displaystyle \sum_{k=1}^{n} k^{3}=\Tdaikakko{\bunsuu{1}{2}n\Ttyuukakko{n+1}}^2$}{\relax}%
+ \ifthenelse{\equal{#1}{公式D}\AND\equal{#2}{b}}%
+ {\[\displaystyle \sum_{k=1}^{n} k^{3}=\Tdaikakko{\bunsuu{1}{2}n\Ttyuukakko{n+1}}^2\]}{\relax}%
+ \ifthenelse{\equal{#1}{公式E}\AND\equal{#2}{i}}%
+ {$\displaystyle \sum_{k=1}^{n} r^{k-1}=\bunsuu{\Ttyuukakko{1-r^{n}}}{1-r}=\bunsuu{r^{n}-1}{r-1}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式E}\AND\equal{#2}{b}}%
+ {\[\displaystyle \sum_{k=1}^{n} r^{k-1}=\bunsuu{\Ttyuukakko{1-r^{n}}}{1-r}=\bunsuu{r^{n}-1}{r-1}\]}{\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ $\Ttyuukakko{k+1}^{3}=k^{3}+3k^2+3k+1$を用いる。\par%
+ $\Ttyuukakko{k+1}^{3}-k^{3}=3k^2+3k+1$の$k$に$1$から$n$までの自然数を代入したものを足したものは,%
+ \[\Ttyuukakko{n+1}^{3}-1=3\displaystyle \sum_{k=1}^{n} k^2+3\displaystyle \sum_{k=1}^{n} k=\bunsuu{1}{2}n\Ttyuukakko{n+1}+n\]%
+ \[\Leftrightarrow\displaystyle \sum_{k=1}^{n} k^2=\bunsuu{1}{6}n\Ttyuukakko{n+1}\Ttyuukakko{2n+1}\]
+ となる。%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\シグマの性質}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{性質}\AND\equal{#2}{i}}%
+ {$p\数式カンマスペース q$が$k$に無関係な定数のとき,$\displaystyle\sum_{k=1}^{n}\Ttyuukakko{pa_{k}+qb_{k}}=p\displaystyle\sum_{k=1}^{n}a_{k}+q\displaystyle\sum_{k=1}^{n}a_{k}$}{\relax}%
+ \ifthenelse{\equal{#1}{性質}\AND\equal{#2}{b}}%
+ {%
+ $p\数式カンマスペース q$が$k$に無関係な定数のとき,%
+ \[\displaystyle\sum_{k=1}^{n}\Ttyuukakko{pa_{k}+qb_{k}}=p\displaystyle\sum_{k=1}^{n}a_{k}+q\displaystyle\sum_{k=1}^{n}a_{k}\]%
+ }%
+ {\relax}%
+ }%
+
+\NewDocumentCommand{\階差数列}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{一般項}\AND\equal{#2}{i}}%
+ {数列${a_{n}}$の階差数列を${b_{n}}$とすると,$2\leqq n$のとき,$a_{n}=a_{1}+\displaystyle\sum_{k=1}^{n-1}b_{k}$}{\relax}%
+ \ifthenelse{\equal{#1}{一般項}\AND\equal{#2}{b}}%
+ {%
+ 数列${a_{n}}$の階差数列を${b_{n}}$とすると,$2\leqq n$のとき,%
+ \[a_{n}=a_{1}+\displaystyle\sum_{k=1}^{n-1}b_{k}\]%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\漸化式}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{等差型}\AND\equal{#2}{i}}%
+ {$a_{n+1}=a_{n}+d$のとき,$a_{n}=a_{1}+\Ttyuukakko{n-1}d$}{\relax}%
+ \ifthenelse{\equal{#1}{等差型}\AND\equal{#2}{b}}%
+ {%
+ $a_{n+1}=a_{n}+d$のとき,%
+ \[a_{n}=a_{1}+\Ttyuukakko{n-1}d\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{等比型}\AND\equal{#2}{i}}%
+ {$a_{n+1}=ra_{n}$のとき,$a_{n}=a_{1}r^{n-1}$}{\relax}%
+ \ifthenelse{\equal{#1}{等比型}\AND\equal{#2}{b}}%
+ {%
+ $a_{n+1}=ra_{n}$のとき,%
+ \[a_{n}=a_{1}r^{n-1}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{階差型}\AND\equal{#2}{i}}%
+ {$a_{n+1}-a_{n}=f\Ttyuukakko{n}$のとき,$a_{1}+\displaystyle \sum_{k=1}^{n-1}f\Ttyuukakko{k}$ただし,$2\leqq n$}{\relax}%
+ \ifthenelse{\equal{#1}{階差型}\AND\equal{#2}{b}}%
+ {%
+ $a_{n+1}-a_{n}=f\Ttyuukakko{n}$のとき,%
+ \[a_{1}+\displaystyle \sum_{k=1}^{n-1}f\Ttyuukakko{k}\text{\ (ただし,$2\leqq n$)}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{特性方程式}\AND\equal{#2}{i}}%
+ {$a_{n+1}=pa_{n}+q \Ttyuukakko{p\neq0\数式カンマスペース q\neq0}$のとき,$a_{n+1}-c=p\Ttyuukakko{a_{n}-c}$と変形して等差型に(ただし,$c=pc+q$を満たす)。}{\relax}%
+ \ifthenelse{\equal{#1}{特性方程式}\AND\equal{#2}{b}}%
+ {%
+ $a_{n+1}=pa_{n}+q \Ttyuukakko{p\neq0\数式カンマスペース q\neq0}$のとき,%
+ \[a_{n+1}-c=p\Ttyuukakko{a_{n}-c}\]%
+ と変形して等差型に(ただし,$c=pc+q$を満たす)。%
+ }%
+ {\relax}%
+ }%
+
+
+\newcommand{\数学的帰納法}{自然数$n$に関する命題$P$が全ての自然数$n$について成立することを証明するには,$n=1$のときに$P$が成立することと,$n=k$のときに$P$が成立するという仮定のもと,$n=k+1$が成立することを証明する。}%
+
+
+%%%%%%%%%%%%%%%%%%%%ここから数\UTF{2162}%%%%%%%%%%%%%%%%%%%%
+\NewDocumentCommand{\共役複素数}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{i}}%
+ {$\alpha=a+bi$のとき,共役な複素数$\overline{\alpha}$は$a-bi$}{\relax}%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{b}}%
+ {%
+ $\alpha=a+bi$のとき,共役な複素数$\overline{\alpha}$は%
+ \[a-bi\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質A}\AND\equal{#2}{i}}%
+ {$z$が実数かつ,$\overline{z}=z$ならば,$z$が実数。}{\relax}%
+ \ifthenelse{\equal{#1}{性質A}\AND\equal{#2}{b}}%
+ {$z$が実数かつ,$\overline{z}=z$ならば,$z$が実数。}{\relax}%
+ \ifthenelse{\equal{#1}{性質B}\AND\equal{#2}{i}}%
+ {$z$が純虚数ならば,$\overline{z}=-z\数式カンマスペース z\neq0$}{\relax}%
+ \ifthenelse{\equal{#1}{性質B}\AND\equal{#2}{b}}%
+ {%
+ $z$が純虚数ならば,%
+ \[\overline{z}=-z\数式カンマスペース z\neq0\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質C}\AND\equal{#2}{i}}%
+ {$\overline{z}=-z\数式カンマスペース z\neq0$ならば,$z$が純虚数。 }{\relax}%
+ \ifthenelse{\equal{#1}{性質C}\AND\equal{#2}{b}}%
+ {%
+ \[\overline{z}=-z\数式カンマスペース z\neq0\]%
+ ならば,$z$が純虚数。 %
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質D}\AND\equal{#2}{i}}%
+ {$\overline{\alpha+\beta}=\overline{\alpha}+\overline{\beta}$}{\relax}%
+ \ifthenelse{\equal{#1}{性質D}\AND\equal{#2}{b}}%
+ {\[\overline{\alpha+\beta}=\overline{\alpha}+\overline{\beta}\]}{\relax}%
+ \ifthenelse{\equal{#1}{性質E}\AND\equal{#2}{i}}%
+ {$\overline{\alpha-\beta}=\overline{\alpha}-\overline{\beta}$}{\relax}%
+ \ifthenelse{\equal{#1}{性質E}\AND\equal{#2}{b}}%
+ {\[\overline{\alpha-\beta}=\overline{\alpha}-\overline{\beta}\]}{\relax}%
+ \ifthenelse{\equal{#1}{性質F}\AND\equal{#2}{i}}%
+ {$\overline{\alpha\beta}=\overline{\alpha}\overline{\beta}$}{\relax}%
+ \ifthenelse{\equal{#1}{性質F}\AND\equal{#2}{b}}%
+ {\[\overline{\alpha\beta}=\overline{\alpha}\overline{\beta}\]}{\relax}%
+ \ifthenelse{\equal{#1}{性質G}\AND\equal{#2}{i}}%
+ {$\overline{\Ttyuukakko{\bunsuu{\alpha}{\beta}}}=\bunsuu{\overline{\alpha}}{\overline{\beta}}$}{\relax}%
+ \ifthenelse{\equal{#1}{性質G}\AND\equal{#2}{b}}%
+ {\[\overline{\Ttyuukakko{\bunsuu{\alpha}{\beta}}}=\bunsuu{\overline{\alpha}}{\overline{\beta}}\]}{\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ $\alpha=a+bi\数式カンマスペース\beta=c+di$\quad($a\数式カンマスペース b c\数式カンマスペース d$は実数かつ$a\neq0\数式カンマスペース c\neq0$)として,%
+ \begin{align*}%
+ \overline{\alpha+\beta}&=\overline{\Ttyuukakko{a+c}+\Ttyuukakko{b+d}i}&\\%
+ &=\Ttyuukakko{a+c}-\Ttyuukakko{b+d}i&\\%
+ &=\Ttyuukakko{a-ci}+\Ttyuukakko{b-di}&\\%
+ &=\overline{\alpha}+\overline{\beta}%
+ \end{align*}%
+ \begin{align*}%
+ \overline{\alpha\beta}&=\overline{\Ttyuukakko{a+bi}\Ttyuukakko{c+di}}&\\%
+ &=\overline{\Ttyuukakko{ac-bd}+\Ttyuukakko{ad+bc}i}&\\
+ &=\Ttyuukakko{ac-bd}-\Ttyuukakko{ad+bc}i&\\%
+ &=\Ttyuukakko{a-bi}\Ttyuukakko{c-di}&\\%
+ &=\overline{\alpha}\overline{\beta}%
+ \end{align*}%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\複素数の絶対値}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{i}}%
+ {複素数$z=a+bi$に対して,$\Tzettaiti{z}=\Tzettaiti{a+bi}=\sqrt{a^2+b^2}$}{\relax}%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{b}}%
+ {%
+ 複素数$z=a+bi$に対して,%
+ \[\Tzettaiti{z}=\Tzettaiti{a+bi}=\sqrt{a^2+b^2}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質A}\AND\equal{#2}{i}}%
+ {$\Tzettaiti{z}=\Tzettaiti{\overline{z}}=\Tzettaiti{-z}$}{\relax}%
+ \ifthenelse{\equal{#1}{性質A}\AND\equal{#2}{b}}%
+ {\[\Tzettaiti{z}=\Tzettaiti{\overline{z}}=\Tzettaiti{-z}\]}{\relax}%
+ \ifthenelse{\equal{#1}{性質B}\AND\equal{#2}{i}}%
+ {$z\overline{z}=\Tzettaiti{z^2}$}{\relax}%
+ \ifthenelse{\equal{#1}{性質B}\AND\equal{#2}{b}}%
+ {\[z\overline{z}=\Tzettaiti{z^2}\]}{\relax}%
+ }%
+
+
+\NewDocumentCommand{\極形式}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{i}}%
+ {複素数$\alpha=a+bi$について,$\alpha=r\Ttyuukakko{\cos\theta+i\sin\theta}\text{\ (ただし$z>0$)}$また,$r=\Tzettaiti{\alpha}=\sqrt{a^2+b^2}\数式カンマスペース\cos\theta=\bunsuu{a}{r}\数式カンマスペース\sin\theta=\bunsuu{b}{r}$を極形式という。}{\relax}%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{b}}%
+ {%
+ 複素数$\alpha=a+bi$について,%
+ \[\alpha=r\Ttyuukakko{\cos\theta+i\sin\theta}\text{\ (ただし$z>0$)}\]%
+ また,$r=\Tzettaiti{\alpha}=\sqrt{a^2+b^2}\数式カンマスペース\cos\theta=\bunsuu{a}{r}\数式カンマスペース\sin\theta=\bunsuu{b}{r}$を極形式という。%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質A}\AND\equal{#2}{i}}%
+ {%
+ $\alpha\数式カンマスペース\beta\数式カンマスペース\gamma$を複素数とする。$\alpha=r_{1}\Ttyuukakko{\cos\theta_{1}+i\sin\theta_{1}}\数式カンマスペース\beta=r_{2}\Ttyuukakko{\cos\theta_{2}+i\sin\theta_{2}}$のとき,\par%
+ $\alpha\beta=r_{1}r_{2}\Tdaikakko{\cos\Ttyuukakko{\theta_{1}+\theta_{2}}+i\sin\Ttyuukakko{\theta_{1}+\theta_{2}}}$}{\relax}%
+ \ifthenelse{\equal{#1}{性質A}\AND\equal{#2}{b}}%
+ {%
+ $\alpha\数式カンマスペース\beta\数式カンマスペース\gamma$を複素数とする。$\alpha=r_{1}\Ttyuukakko{\cos\theta_{1}+i\sin\theta_{1}}\数式カンマスペース\beta=r_{2}\Ttyuukakko{\cos\theta_{2}+i\sin\theta_{2}}$のとき,%
+ \[\alpha\beta=r_{1}r_{2}\Tdaikakko{\cos\Ttyuukakko{\theta_{1}+\theta_{2}}+i\sin\Ttyuukakko{\theta_{1}+\theta_{2}}}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質B}\AND\equal{#2}{i}}%
+ {%
+ $\alpha\数式カンマスペース\beta\数式カンマスペース\gamma$を複素数とする。$\alpha=r_{1}\Ttyuukakko{\cos\theta_{1}+i\sin\theta_{1}}\数式カンマスペース\beta=r_{2}\Ttyuukakko{\cos\theta_{2}+i\sin\theta_{2}}$のとき,\par%
+ $\bunsuu{\alpha}{\beta}=\bunsuu{r_{1}}{r_{2}}\Tdaikakko{\cos\Ttyuukakko{\theta_{1}+\theta_{2}}+i\sin\Ttyuukakko{\theta_{1}+\theta_{2}}}$}{\relax}%
+ \ifthenelse{\equal{#1}{性質B}\AND\equal{#2}{b}}%
+ {%
+ $\alpha\数式カンマスペース\beta\数式カンマスペース\gamma$を複素数とする。$\alpha=r_{1}\Ttyuukakko{\cos\theta_{1}+i\sin\theta_{1}}\数式カンマスペース\beta=r_{2}\Ttyuukakko{\cos\theta_{2}+i\sin\theta_{2}}$のとき,%
+ \[\bunsuu{\alpha}{\beta}=\bunsuu{r_{1}}{r_{2}}\Tdaikakko{\cos\Ttyuukakko{\theta_{1}+\theta_{2}}+i\sin\Ttyuukakko{\theta_{1}+\theta_{2}}}\]%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\偏角}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{i}}%
+ {%
+ 複素数$\alpha=a+bi$について,$\alpha=r\Ttyuukakko{\cos\theta+i\sin\theta}$\par
+ ただし$z>0$のとき$\theta$を偏角といい,$\mathrm{aug}\alpha$で表す。%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{b}}%
+ {%
+ 複素数$\alpha=a+bi$について,%
+ \[\alpha=r\Ttyuukakko{\cos\theta+i\sin\theta}\]%
+ ただし$z>0$のとき$\theta$を偏角といい,%
+ \[\mathrm{aug}\alpha\]%
+ で表す。%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質A}\AND\equal{#2}{i}}%
+ {$\alpha\数式カンマスペース\beta\数式カンマスペース\gamma$を複素数とする。$\alpha=r_{1}\Ttyuukakko{\cos\theta_{1}+i\sin\theta_{1}}\数式カンマスペース\beta=r_{2}\Ttyuukakko{\cos\theta_{2}+i\sin\theta_{2}}$のとき,$\theta_{1}=\mathrm{arg}\alpha$また,$\mathrm{arg}\alpha=\theta_{1}+2n\pi$ ($n$は整数)}{\relax}%
+ \ifthenelse{\equal{#1}{性質A}\AND\equal{#2}{b}}%
+ {%
+ $\alpha\数式カンマスペース\beta\数式カンマスペース\gamma$を複素数とする。$\alpha=r_{1}\Ttyuukakko{\cos\theta_{1}+i\sin\theta_{1}}\数式カンマスペース\beta=r_{2}\Ttyuukakko{\cos\theta_{2}+i\sin\theta_{2}}$のとき,%
+ \[\theta_{1}=\theta_{1}+2n\pi=\mathrm{arg}\alpha\]%
+ ($n$は整数)%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質B}\AND\equal{#2}{i}}%
+ {$\alpha\数式カンマスペース\beta\数式カンマスペース\gamma$を複素数とする。$\alpha=r_{1}\Ttyuukakko{\cos\theta_{1}+i\sin\theta_{1}}\数式カンマスペース\beta=r_{2}\Ttyuukakko{\cos\theta_{2}+i\sin\theta_{2}}$のとき,$\mathrm{arg}z_{1}z_{2}=\mathrm{arg}z_{1}+\mathrm{arg}z_{2}$}{\relax}%
+ \ifthenelse{\equal{#1}{性質B}\AND\equal{#2}{b}}%
+ {%
+ $\alpha\数式カンマスペース\beta\数式カンマスペース\gamma$を複素数とする。$\alpha=r_{1}\Ttyuukakko{\cos\theta_{1}+i\sin\theta_{1}}\数式カンマスペース\beta=r_{2}\Ttyuukakko{\cos\theta_{2}+i\sin\theta_{2}}$のとき,%
+ \[\mathrm{arg}z_{1}z_{2}=\mathrm{arg}z_{1}+\mathrm{arg}z_{2}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質C}\AND\equal{#2}{i}}%
+ {$\alpha\数式カンマスペース\beta\数式カンマスペース\gamma$を複素数とする。$\alpha=r_{1}\Ttyuukakko{\cos\theta_{1}+i\sin\theta_{1}}\数式カンマスペース\beta=r_{2}\Ttyuukakko{\cos\theta_{2}+i\sin\theta_{2}}$のとき,$\mathrm{arg}\bunsuu{z_{1}}{z_{2}}=\mathrm{arg}z_{1}-\mathrm{arg}z_{2}$}{\relax}%
+ \ifthenelse{\equal{#1}{性質C}\AND\equal{#2}{b}}%
+ {%
+ $\alpha\数式カンマスペース\beta\数式カンマスペース\gamma$を複素数とする。$\alpha=r_{1}\Ttyuukakko{\cos\theta_{1}+i\sin\theta_{1}}\数式カンマスペース\beta=r_{2}\Ttyuukakko{\cos\theta_{2}+i\sin\theta_{2}}$のとき,%
+ \[\mathrm{arg}\bunsuu{z_{1}}{z_{2}}=\mathrm{arg}z_{1}-\mathrm{arg}z_{2}\]%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\ドモアブルの定理}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{i}}%
+ {$n$が整数のとき,$\Ttyuukakko{\cos\theta+i\sin\theta}^{n}=\cos n\theta+i\sin n\theta$}{\relax}%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{b}}%
+ {%
+ $n$が整数のとき,%
+ \[\Ttyuukakko{\cos\theta+i\sin\theta}^{n}=\cos n\theta+i\sin n\theta\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{証明}}%
+ {%
+ \証明開始%
+ 複素数%
+ \[\alpha_{1}=r_{1}\Ttyuukakko{\cos\theta_{1}+i\sin\theta_{1}}\数式カンマスペース\alpha_{2}=r_{2}\Ttyuukakko{\cos\theta_{2}+i\sin\theta_{2}}\ldots\alpha_{n}=r_{n}\Ttyuukakko{\cos\theta_{n}+i\sin\theta_{n}}\]
+ に対して,$\alpha_{1}\alpha_{2}\cdots\alpha_{n}$を考えると,三角関数の積和の公式から%
+ \[\alpha_{1}\alpha_{2}\cdots\alpha_{n}=r_{1}r_{2}\cdots r_{n}\Tdaikakko{\cos\Ttyuukakko{\theta_{1}+\theta_{2}+\cdots+\theta_{n}}+i\sin\Ttyuukakko{\theta_{1}+\theta_{2}+\cdots+\theta_{n}}}\]%
+ となる。ここで,$\alpha_{1}=\alpha_{2}=\cdots=\alpha_{n}$のとき,%
+ \[\alpha^{n}=r^{n}\Ttyuukakko{\cos\theta+i\sin\theta}^{n}=r^{n}\Ttyuukakko{\cos n\theta+i\sin n\theta}\]%
+ \[\Leftrightarrow\Ttyuukakko{\cos\theta+i\sin\theta}^{n}=\cos n\theta+i\sin n\theta\]%
+ を得る。%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\放物線}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{i}}%
+ {定点$F$ (焦点)と$F$を通らない直線$l$ (準線)があるとき,焦点と準線からの距離の和が一定な点の軌跡。}{\relax}%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{b}}%
+ {定点$F$ (焦点)と$F$を通らない直線$l$ (準線)があるとき,焦点と準線からの距離の和が一定な点の軌跡。}{\relax}%
+ \ifthenelse{\equal{#1}{性質A}\AND\equal{#2}{i}}%
+ {放物線は$y^2=4px$と表せられる。}{\relax}%
+ \ifthenelse{\equal{#1}{性質A}\AND\equal{#2}{b}}%
+ {%
+ 放物線は%
+ \[y^2=4px\]%
+ と表せられる。%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質B}\AND\equal{#2}{i}}%
+ {放物線の焦点は$F\Ttyuukakko{p\数式カンマスペース 0}$}{\relax}%
+ \ifthenelse{\equal{#1}{性質B}\AND\equal{#2}{b}}%
+ {%
+ 放物線の焦点は%
+ \[F\Ttyuukakko{p\数式カンマスペース 0}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質C}\AND\equal{#2}{i}}%
+ {放物線の準線は$x=-p$}{\relax}%
+ \ifthenelse{\equal{#1}{性質C}\AND\equal{#2}{b}}%
+ {%
+ 放物線の準線は%
+ \[x=-p\]%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\楕円}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{i}}%
+ {二つの焦点$F$と$F'$からの距離の和が一定な点の軌跡。}{\relax}%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{b}}%
+ {二つの焦点$F$と$F'$からの距離の和が一定な点の軌跡。}{\relax}%
+ \ifthenelse{\equal{#1}{性質A}\AND\equal{#2}{i}}%
+ {楕円は$\bunsuu{x^2}{a^2}+\bunsuu{y^2}{b^2}=1$と表せられる。}{\relax}%
+ \ifthenelse{\equal{#1}{性質A}\AND\equal{#2}{b}}%
+ {%
+ 楕円は%
+ \[\bunsuu{x^2}{a^2}+\bunsuu{y^2}{b^2}=1\]%
+ と表せられる。%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質B}\AND\equal{#2}{i}}%
+ {楕円の焦点は$F\Ttyuukakko{\sqrt{a^2-b^2}\数式カンマスペース 0}$と,$F'\Ttyuukakko{\sqrt{a^2-b^2}\数式カンマスペース 0}$}{\relax}%
+ \ifthenelse{\equal{#1}{性質B}\AND\equal{#2}{b}}%
+ {%
+ 楕円の焦点は%
+ \[F\Ttyuukakko{\sqrt{a^2-b^2}\数式カンマスペース 0} F'\Ttyuukakko{\sqrt{a^2-b^2}\数式カンマスペース 0}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質C}\AND\equal{#2}{i}}%
+ {楕円の二つの焦点からの距離の和は$2a$である。}{\relax}%
+ \ifthenelse{\equal{#1}{性質C}\AND\equal{#2}{b}}%
+ {%
+ 楕円の二つの焦点からの距離の和は%
+ \[2a\]%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\双曲線}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{i}}%
+ {二つの焦点$F$と$F'$からの距離の差が$0$でなく一定な点の軌跡。}{\relax}%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{b}}%
+ {二つの焦点$F$と$F'$からの距離の差が$0$でなく一定な点の軌跡。}{\relax}%
+ \ifthenelse{\equal{#1}{性質A}\AND\equal{#2}{i}}%
+ {双曲線は$\bunsuu{x^2}{a^2}-\bunsuu{y^2}{b^2}=1$と表せられる。}{\relax}%
+ \ifthenelse{\equal{#1}{性質A}\AND\equal{#2}{b}}%
+ {%
+ 双曲線は%
+ \[\bunsuu{x^2}{a^2}-\bunsuu{y^2}{b^2}=1\]%
+ と表せられる。%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質B}\AND\equal{#2}{i}}%
+ {双曲線の焦点は$F\Ttyuukakko{\sqrt{a^2+b^2}\数式カンマスペース 0}$と,$F'\Ttyuukakko{\sqrt{a^2+b^2}\数式カンマスペース 0}$}{\relax}%
+ \ifthenelse{\equal{#1}{性質B}\AND\equal{#2}{b}}%
+ {%
+ 双曲線の焦点は%
+ \[F\Ttyuukakko{\sqrt{a^2+b^2}\数式カンマスペース 0} F'\Ttyuukakko{\sqrt{a^2+b^2}\数式カンマスペース 0}\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質C}\AND\equal{#2}{i}}%
+ {双曲線の二つの焦点からの距離の差は$2a$ }{\relax}%
+ \ifthenelse{\equal{#1}{性質C}\AND\equal{#2}{b}}%
+ {%
+ 双曲線の二つの焦点からの距離の差は%
+ \[2a\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{性質D}\AND\equal{#2}{i}}%
+ {双曲線の漸近線は$\bunsuu{x}{a}-\bunsuu{y}{b}=0\数式カンマスペース\bunsuu{x}{a}+\bunsuu{y}{b}=0$}{\relax}%
+ \ifthenelse{\equal{#1}{性質D}\AND\equal{#2}{b}}%
+ {%
+ 双曲線の漸近線は%
+ \[\bunsuu{x}{a}-\bunsuu{y}{b}=0\数式カンマスペース\bunsuu{x}{a}+\bunsuu{y}{b}=0\]%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\連続な関数}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{i}}%
+ {定義域の$x$の値$a$に関して,$\displaystyle \lim_{x \to a}f\Ttyuukakko{x}=f\Ttyuukakko{a}$のとき,$f\Ttyuukakko{x}$は$x=a$で連続。}{\relax}%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{b}}%
+ {%
+ 定義域の$x$の値$a$に関して,%
+ \[\displaystyle \lim_{x \to a}f\Ttyuukakko{x}=f\Ttyuukakko{a}\]%
+ のとき,$f\Ttyuukakko{x}$は$x=a$で連続。%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\中間値の定理}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{i}}%
+ {閉区間$[a\数式カンマスペース b]$で連続な関数$f\Ttyuukakko{x}$について,$f\Ttyuukakko{a}\neq f\Ttyuukakko{b}$のとき,$f\Ttyuukakko{a}$と$f\Ttyuukakko{b}$の間の任意の実数$k$について,$f\Ttyuukakko{c}=k$となる$c$が少なからず一つ存在する。}{\relax}%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{b}}%
+ {%
+ 閉区間$[a\数式カンマスペース b]$で連続な関数$f\Ttyuukakko{x}$について,$f\Ttyuukakko{a}\neq f\Ttyuukakko{b}$のとき,$f\Ttyuukakko{a}$と$f\Ttyuukakko{b}$の間の任意の実数$k$について,%
+ \[f\Ttyuukakko{c}=k\]%
+ となる$c$が少なからず一つ存在する。%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\平均値の定理}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{i}}%
+ {関数$f\Ttyuukakko{x}$が閉区間$[a\数式カンマスペース b]$で連続,開区間$\Ttyuukakko{a\数式カンマスペース b}$で微分可能ならば,$\bunsuu{f\Ttyuukakko{b}-f\Ttyuukakko{a}}{b-a}=f'\Ttyuukakko{c} \Ttyuukakko{a<c<b}$を満たす$c$が存在する。}{\relax}%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{b}}%
+ {%
+ 関数$f\Ttyuukakko{x}$が閉区間$[a\数式カンマスペース b]$で連続,開区間$\Ttyuukakko{a\数式カンマスペース b}$で微分可能ならば,%
+ \[\bunsuu{f\Ttyuukakko{b}-f\Ttyuukakko{a}}{b-a}=f'\Ttyuukakko{c} \Ttyuukakko{a<c<b}\]%
+ を満たす$c$が存在する。%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\微分}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{i}}%
+ {$f'\Ttyuukakko{x}=\displaystyle\lim_{h \to 0}\bunsuu{f\Ttyuukakko{x+h}-f\Ttyuukakko{x}}{h}$}{\relax}%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{b}}%
+ {\[f'\Ttyuukakko{x}=\displaystyle\lim_{h \to 0}\bunsuu{f\Ttyuukakko{x+h}-f\Ttyuukakko{x}}{h}\]}{\relax}%
+ \ifthenelse{\equal{#1}{積の微分公式}\AND\equal{#2}{i}}%
+ {$\Tdaikakko{f\Ttyuukakko{x}g\Ttyuukakko{x}}'=f'\Ttyuukakko{x}g\Ttyuukakko{x}+f\Ttyuukakko{x}g'\Ttyuukakko{x}$}{\relax}%
+ \ifthenelse{\equal{#1}{積の微分公式}\AND\equal{#2}{b}}%
+ {\[\Tdaikakko{f\Ttyuukakko{x}g\Ttyuukakko{x}}'=f'\Ttyuukakko{x}g\Ttyuukakko{x}+f\Ttyuukakko{x}g'\Ttyuukakko{x}\]}{\relax}%
+ \ifthenelse{\equal{#1}{商の微分公式}\AND\equal{#2}{i}}%
+ {$\Tdaikakko{\bunsuu{f\Ttyuukakko{x}}{g\Ttyuukakko{x}}}'=\bunsuu{f'\Ttyuukakko{x}g\Ttyuukakko{x}-f\Ttyuukakko{x}g'\Ttyuukakko{x}}{\Tdaikakko{g\Ttyuukakko{x}}^2}$}{\relax}%
+ \ifthenelse{\equal{#1}{商の微分公式}\AND\equal{#2}{b}}%
+ {\[\Tdaikakko{\bunsuu{f\Ttyuukakko{x}}{g\Ttyuukakko{x}}}'=\bunsuu{f'\Ttyuukakko{x}g\Ttyuukakko{x}-f\Ttyuukakko{x}g'\Ttyuukakko{x}}{\Tdaikakko{g\Ttyuukakko{x}}^2}\]}{\relax}%
+ \ifthenelse{\equal{#1}{合成関数の微分}\AND\equal{#2}{i}}%
+ {$\Tdaikakko{f\Ttyuukakko{g\Ttyuukakko{x}}}'=f'\Ttyuukakko{g\Ttyuukakko{x}}g'\Ttyuukakko{x}$}{\relax}%
+ \ifthenelse{\equal{#1}{合成関数の微分}\AND\equal{#2}{b}}%
+ {\[\Tdaikakko{f\Ttyuukakko{g\Ttyuukakko{x}}}'=f'\Ttyuukakko{g\Ttyuukakko{x}}g'\Ttyuukakko{x}\]}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の微分公式A}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{c}'=0$}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の微分公式A}\AND\equal{#2}{b}}%
+ {\[\Ttyuukakko{c}'=0\]}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の微分公式B}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{x^{\alpha}}'=\alpha x^{\alpha-1}$ ($\alpha$は実数)}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の微分公式B}\AND\equal{#2}{b}}%
+ {%
+ \[\Ttyuukakko{x^{\alpha}}'=\alpha x^{\alpha-1}\]%
+ $\alpha$は実数%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{初等関数の微分公式C}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{\sin x}'=\cos x$}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の微分公式C}\AND\equal{#2}{b}}%
+ {\[\Ttyuukakko{\sin x}'=\cos x\]}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の微分公式D}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{\cos x}'=-\sin x$}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の微分公式D}\AND\equal{#2}{b}}%
+ {\[\Ttyuukakko{\cos x}'=-\sin x\]}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の微分公式E}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{\tan x}'=\bunsuu{1}{\cos^2x}$}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の微分公式E}\AND\equal{#2}{b}}%
+ {\[\Ttyuukakko{\tan x}'=\bunsuu{1}{\cos^2x}\]}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の微分公式F}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{\log\Tzettaiti{x}}'=\bunsuu{1}{x}$}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の微分公式F}\AND\equal{#2}{b}}%
+ {\[\Ttyuukakko{\log\Tzettaiti{x}}'=\bunsuu{1}{x}\]}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の微分公式G}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{\log_{a}\Tzettaiti{x}}'=\bunsuu{1}{x\log a}$}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の微分公式G}\AND\equal{#2}{b}}%
+ {\[\Ttyuukakko{\log_{a}\Tzettaiti{x}}'=\bunsuu{1}{x\log a}\]}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の微分公式H}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{e^{x}}'=e^{x}$}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の微分公式H}\AND\equal{#2}{b}}%
+ {\[\Ttyuukakko{e^{x}}'=e^{x}\]}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の微分公式I}\AND\equal{#2}{i}}%
+ {$\Ttyuukakko{a^{x}}'=a^{x}\log a$}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の微分公式I}\AND\equal{#2}{b}}%
+ {\[\Ttyuukakko{a^{x}}'=a^{x}\log a\]}{\relax}%
+ \ifthenelse{\equal{#1}{三角関数の微分公式の証明}\AND\equal{#2}{i}}%
+ {%
+ \証明開始%
+ \begin{align*}%
+ \Ttyuukakko{\sin x}'&=\displaystyle\lim_{h \to 0} \bunsuu{\sin\Ttyuukakko{x+h}-\sin x}{h}&\\%
+ &=\displaystyle\lim_{h \to 0} \bunsuu{\sin x\cos h+\cos x\sin x-\sin x}{h}&\\%
+ &=\displaystyle\lim_{h \to 0} \Ttyuukakko{\cos x\bunsuu{\sin h}{h}-\sin x\bunsuu{1-\cos x}{h}\bunsuu{1+\cos h}{1+\cos h}}&\\%
+ &=\displaystyle\lim_{h \to 0} \Ttyuukakko{\cos x\bunsuu{\sin h}{h}-\sin x\bunsuu{1}{1+\cos h}\bunsuu{\sin^2h}{h^2}h}&\\%
+ &=\cos x\cdot1-\sin x\cdot\bunsuu{1}{2}1^2\cdot0&\\%
+ &=\cos x&\\%
+ \Ttyuukakko{\cos x}'&=\Tdaikakko{\sin\Ttyuukakko{\bunsuu{\pi}{2}-x}}'&\\%
+ &=\cos\Ttyuukakko{\bunsuu{\pi}{2}-x}\cdot\Ttyuukakko{-1}&\\%
+ &=-\sin x%
+ \end{align*}%
+ \証明終了%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{対数関数の微分公式の証明}\AND\equal{#2}{i}}%
+ {%
+ \証明開始%
+ \begin{align*}%
+ \Ttyuukakko{\log x}'&=\displaystyle\lim_{h \to 0} \bunsuu{\log\Ttyuukakko{x+h}-\log x}{h}&\\%
+ &=\displaystyle\lim_{h \to 0} \bunsuu{\log\Ttyuukakko{1+\bunsuu{x}{h}}}{h}&\\%
+ \end{align*}%
+ ここで$\bunsuu{h}{x}=t$とおくと,$h=tx$となり$\displaystyle\lim_{h \to 0} t=0$なので,%
+ \begin{align*}%
+ \Ttyuukakko{\log x}'&=\displaystyle\lim_{h \to 0} \bunsuu{\log\Ttyuukakko{1+t}}{xt}&\\%
+ &=\displaystyle\lim_{h \to 0} \Tdaikakko{\bunsuu{\log\Ttyuukakko{1+t}}{t}\cdot\bunsuu{1}{x}}&\\%
+ &=\displaystyle\lim_{h \to 0} \log\Ttyuukakko{1+t}^{\bunsuu{1}{t}}\cdot\bunsuu{1}{x}&\\%
+ &=\log e\cdot\bunsuu{1}{x}&\\%
+ &=\bunsuu{1}{x}
+ \end{align*}%
+ $f\Ttyuukakko{x}=e^{x}$とおく。\par%
+ \begin{align*}%
+ \Ttyuukakko{e^{x}}'&=\displaystyle\lim_{h \to 0} \bunsuu{e^{x+h}-e^{x}}{h}&\\%
+ &=e^{x}\displaystyle\lim_{h \to 0} \bunsuu{e^{h}-1}{h}%
+ \end{align*}%
+ ここで$\Ttyuukakko{\log x}'=\bunsuu{1}{x}$より,$y=\log x$の$x=1$においての接線の傾きは$1$であり,$y=\log x$と$y=e^{x}$は$y=x$において対称であるので$y=e^{x}$の$x=0$においての接線の傾きも$1$なので,
+ \[f'\Ttyuukakko{0}\displaystyle\lim_{h \to 0} \bunsuu{e^{h}-1}{h}=1\]%
+ よって,%
+ \begin{align*}%
+ \Ttyuukakko{e^{x}}'&=e^{x}\displaystyle\lim_{h \to 0} \bunsuu{e^{h}-1}{h}&\\%
+ &=e^{x}\cdot1&\\%
+ &=e^{x}
+ \end{align*}%
+ \証明終了%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\法線の方程式}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{i}}%
+ {曲線$f\Ttyuukakko{x}$上の点$A\Ttyuukakko{a\数式カンマスペース f\Ttyuukakko{a}}$における法線の方程式は,$y-f\Ttyuukakko{a}=-\bunsuu{1}{f'\Ttyuukakko{a}}\Ttyuukakko{x-a}$}{\relax}%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{b}}%
+ {%
+ 曲線$f\Ttyuukakko{x}$上の点$A\Ttyuukakko{a\数式カンマスペース f\Ttyuukakko{a}}$における法線の方程式は,%
+ \[y-f\Ttyuukakko{a}=-\bunsuu{1}{f'\Ttyuukakko{a}}\Ttyuukakko{x-a}\]%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\不定積分}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{i}}%
+ {$F'\Ttyuukakko{x}=f\Ttyuukakko{x}$とすると,$\displaystyle \int_{}^{} f\Ttyuukakko{x}dx=F\Ttyuukakko{x}+C$ ($C$は積分定数)}{\relax}%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{b}}%
+ {%
+ $F'\Ttyuukakko{x}=f\Ttyuukakko{x}$とすると,%
+ \[\displaystyle \int_{}^{} f\Ttyuukakko{x}dx=F\Ttyuukakko{x}+C\]%
+ ($C$は積分定数)%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{置換積分}\AND\equal{#2}{i}}%
+ {$\displaystyle \int_{}^{} f\Ttyuukakko{x}dx=\displaystyle \int_{}^{}f\Ttyuukakko{g\Ttyuukakko{t}}g'\Ttyuukakko{t}dt$ ($x=g\Ttyuukakko{t}$に置換)}{\relax}%
+ \ifthenelse{\equal{#1}{置換積分}\AND\equal{#2}{b}}%
+ {%
+ \[\displaystyle \int_{}^{} f\Ttyuukakko{x}dx=\displaystyle \int_{}^{}f\Ttyuukakko{g\Ttyuukakko{t}}g'\Ttyuukakko{t}dt\]%
+ ($x=g\Ttyuukakko{t}$に置換)%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{部分積分}\AND\equal{#2}{i}}%
+ {$\displaystyle \int_{}^{} f\Ttyuukakko{x}g'\Ttyuukakko{x}dx=f\Ttyuukakko{x}g\Ttyuukakko{x}-\displaystyle \int_{}^{}f'\Ttyuukakko{x}g\Ttyuukakko{x}$}{\relax}%
+ \ifthenelse{\equal{#1}{部分積分}\AND\equal{#2}{b}}%
+ {\[\displaystyle \int_{}^{} f\Ttyuukakko{x}g'\Ttyuukakko{x}dx=f\Ttyuukakko{x}g\Ttyuukakko{x}-\displaystyle \int_{}^{}f'\Ttyuukakko{x}g\Ttyuukakko{x}\]}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の積分公式A}\AND\equal{#2}{i}}%
+ {$C$は積分定数とする。$\displaystyle \int_{}^{} x^{\alpha}dx=\bunsuu{1}{\alpha+1}x^{\alpha+1}+C$}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の積分公式A}\AND\equal{#2}{b}}%
+ {%
+ $C$は積分定数とする。%
+ \[\displaystyle \int_{}^{} x^{\alpha}dx=\bunsuu{1}{\alpha+1}x^{\alpha+1}+C\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{初等関数の積分公式B}\AND\equal{#2}{i}}%
+ {$C$は積分定数とする。$\displaystyle \int_{}^{} \bunsuu{1}{x}dx=\log\Tzettaiti{x}+C$}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の積分公式B}\AND\equal{#2}{b}}%
+ {%
+ $C$は積分定数とする。%
+ \[\displaystyle \int_{}^{} \bunsuu{1}{x}dx=\log\Tzettaiti{x}+C\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{初等関数の積分公式C}\AND\equal{#2}{i}}%
+ {$C$は積分定数とする。$\displaystyle \int_{}^{} \sin xdx=-\cos x+C$}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の積分公式C}\AND\equal{#2}{b}}%
+ {%
+ $C$は積分定数とする。%
+ \[\displaystyle \int_{}^{} \sin xdx=-\cos x+C\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{初等関数の積分公式D}\AND\equal{#2}{i}}%
+ {$C$は積分定数とする。$\displaystyle \int_{}^{} \cos xdx=\sin x+C$}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の積分公式D}\AND\equal{#2}{b}}%
+ {%
+ $C$は積分定数とする。%
+ \[\displaystyle \int_{}^{} \cos xdx=\sin x+C\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{初等関数の積分公式E}\AND\equal{#2}{i}}%
+ {$C$は積分定数とする。$\displaystyle \int_{}^{} e^{x}dx=e^{x}+C$}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の積分公式E}\AND\equal{#2}{b}}%
+ {%
+ $C$は積分定数とする。%
+ \[\displaystyle \int_{}^{} e^{x}dx=e^{x}+C\]%
+ }%
+ {\relax}%
+ \ifthenelse{\equal{#1}{初等関数の積分公式F}\AND\equal{#2}{i}}%
+ {$C$は積分定数とする。$\displaystyle \int_{}^{} a^{x}dx=\bunsuu{a^{x}}{\log a}+C$}{\relax}%
+ \ifthenelse{\equal{#1}{初等関数の積分公式F}\AND\equal{#2}{b}}%
+ {%
+ $C$は積分定数とする。%
+ \[\displaystyle \int_{}^{} a^{x}dx=\bunsuu{a^{x}}{\log a}+C\]%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\定積分}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{i}}%
+ {曲線$y=f\Ttyuukakko{x}$と$x$軸(間は$a$から$b$)に囲まれた部分の面積$S$について,$F'\Ttyuukakko{x}=f\Ttyuukakko{x}$のとき,$S=\displaystyle \int_{b}^{a} f\Ttyuukakko{x}dx=[F\Ttyuukakko{x}]^{b}_{a}=F\Ttyuukakko{b}-F\Ttyuukakko{a}$}{\relax}%
+
+ \ifthenelse{\equal{#1}{定義}\AND\equal{#2}{b}}%
+ {%
+ 曲線$y=f\Ttyuukakko{x}$と$x$軸(間は$a$から$b$)に囲まれた部分の面積$S$について,$F'\Ttyuukakko{x}=f\Ttyuukakko{x}$のとき,%
+ \[S=\displaystyle \int_{b}^{a} f\Ttyuukakko{x}dx=[F\Ttyuukakko{x}]^{b}_{a}=F\Ttyuukakko{b}-F\Ttyuukakko{a}\]%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\区分求積法}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{i}}%
+ {$\displaystyle \lim_{n \to \infty} \sum_{k=0}^{n-1}f\Ttyuukakko{x_{k}}\mathit{\Delta}x=\displaystyle \lim_{n \to \infty} \sum_{k=1}^{n}f\Ttyuukakko{x_{k}}\mathit{\Delta}x$ここで,$\mathit{\Delta}x=\bunsuu{b-a}{n}\数式カンマスペース x_{k}=a+k\mathit{\Delta}x$}{\relax}%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{b}}%
+ {%
+ \[\displaystyle \lim_{n \to \infty} \sum_{k=0}^{n-1}f\Ttyuukakko{x_{k}}\mathit{\Delta}x=\displaystyle \lim_{n \to \infty} \sum_{k=1}^{n}f\Ttyuukakko{x_{k}}\mathit{\Delta}x\]%
+ ここで,%
+ \[\mathit{\Delta}x=\bunsuu{b-a}{n}\数式カンマスペース x_{k}=a+k\mathit{\Delta}x\]%
+ }%
+ {\relax}%
+ }%
+
+
+\NewDocumentCommand{\体積の積分}{ m O{i} }%
+ {%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{i}}%
+ {曲線$y=f\Ttyuukakko{x}$と$x$軸の間の部分($a\leqq x\leqq b$)を$x$軸の周りに一回転させてできる回転体の体積は,$V=\pi\displaystyle \int_{a}^{b} \Tdaikakko{f\Ttyuukakko{x}}^2dx$}{\relax}%
+ \ifthenelse{\equal{#1}{公式}\AND\equal{#2}{b}}%
+ {%
+ 曲線$y=f\Ttyuukakko{x}$と$x$軸の間の部分($a\leqq x\leqq b$)を$x$軸の周りに一回転させてできる回転体の体積は,%
+ \[V=\pi\displaystyle \int_{a}^{b} \Tdaikakko{f\Ttyuukakko{x}}^2dx\]%
+ }%
+ {\relax}%
+ }% \ No newline at end of file