%
%	gbb60166@gmail.com	https://unilab.gbb60166.jp/prekou/prekou.htm
%
% aspectratio= は 1610, 169, 149, 54, 32 の中から選べる（省略時は 43）
% C:\texlive\2024\texmf-dist\tex\latex\beamer\beamer.cls
%\documentclass[20pt,dvipdfmx,usepdftitle=false,hyperref={pdfpagemode=FullScreen},aspectratio=169]{beamer}
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% pdfの栞の字化けを防ぐ
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% テーマ
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% navi. symbolsは目立たないが，dvipdfmxを使うと機能しないので非表示に
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\usepackage{bxdpx-beamer,pxjahyper,minijs}
%\usepackage{graphicx}
%\usepackage{tkokugo,furikana,tsayusen,shiika,sfkanbun,jdkintou,plext}
\usepackage{type1cm}
\usepackage[deluxe]{otf}
%\usepackage{tikzsymbols}
%\usepackage[dvipdfmx]{graphicx}%
\def\pgfsysdriver{pgfsys-dvipdfmx.def}%（graphicxパッケージを使用しない場合はこの行を有効に）
%\def\pgfsysdriver{pgfsys-dvips.def}%デフォルト
%\usepackage[e]{esvect}
\usepackage{tikz}%（これで、pgfとpgfforが読み込まれます。）
%\usepackage{tikz-3dplot}
%\usetikzlibrary{calc}
%\usetikzlibrary{arrows.meta,patterns,angles,intersections,backgrounds}
%\usepackage{animate}
%\usepackage{pgfplots}
%\pgfplotsset{compat=newest}
%\pgfplotsset{compat=newest, every axis/.append style={line width=1pt}}
%\PassOptionsToPackage{dvipdfmx}{graphicx}
\nofiles

\hypersetup{% Jul 18 2023
	pdftitle={プレ高数学科（LaTeX Beamer で汎用PDFプレゼンを作る）},
	pdfauthor={gbb60166@gmail.com},
	pdfsubject={https://unilab.gbb60166.jp/prekou/prekou.htm}
	}

% フォントはお好みで
%\usepackage{txfonts}

\mathversion{bold}
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\setbeamerfont{frametitle}{size=\normalsize,series=\bfseries}
\setbeamertemplate{frametitle}[default][center]
\usefonttheme{professionalfonts}

% 参考にしたURL
% http://windom.phys.hirosaki-u.ac.jp/fswiki/wiki.cgi?page=LaTeX+Beamer%A4%C7%A5%D7%A5%EC%A5%BC%A5%F3%A5%C6%A1%BC%A5%B7%A5%E7%A5%F3

\newcommand{\Slash}[1]{\ooalign{\hfil\kern-3pt／\hfil\crcr$#1$}}
\everymath{\displaystyle}

\newcommand{\maruwaku}[1]{\begin{tikzpicture}[scale=0.7, baseline={([yshift=-22pt] current bounding box.north)}]
\filldraw[color=CUDBlue, line width=1pt, rounded corners=2pt] (-0.1,0)--(2.1,0)--(2.1,1.1)--(-0.1,1.1)--cycle;
\draw(1,0.5) node[white]{#1};
\end{tikzpicture}
}

\setbeamersize{text margin left=5mm,text margin right=5mm}
\fboxrule=0.8pt

\makeatletter
\newcommand{\hooklen}[2]{\settowidth{\@tempdima}{\(#1\)}
%\advance\@tempdima by.3ex
% ↑ 数式モードで式の前後に入るスペースを制御したかったが、
%    難しいのでやめた。段々難解なコードになっているのでやめた方がよい？
\hbox to\@tempdima{\hfil \(#2\)\hfil}}
%■等号付き不等号 by 美文書作成入門
\newcommand{\myleqq}{\mathrel{\mathpalette\gl@align<}}
\newcommand{\mygeqq}{\mathrel{\mathpalette\gl@align>}}
\newcommand{\gl@align}[2]{\lower.6ex\vbox{\baselineskip\z@skip\lineskip\z@
	\ialign{$\m@th#1\hfil##\hfil$\crcr#2\crcr=\crcr}}}
\makeatother

% カラーユニバーサルデザインを調べたつもりだがあまり自信がありません
% http://www.fukushihoken.metro.tokyo.jp/kiban/machizukuri/kanren/color.files/colorudguideline.pdf
% http://jfly.iam.u-tokyo.ac.jp/colorset/

% ■ アクセントカラー小面積を目立たせる高彩度色
\definecolor{CUDRed}{RGB}{255,75,0}
\definecolor{CUDGreen}{RGB}{3,175,122}
\definecolor{CUDBlue}{RGB}{0,90,255}
\definecolor{CUDCyan}{RGB}{77,196,255}
\definecolor{CUDMagenta}{RGB}{153,0,153}
\definecolor{CUDYellow}{RGB}{255,241,0}
\definecolor{CUDBrown}{RGB}{128,64,0}

\definecolor{CUDOrange}{RGB}{246,170,0}

% ■ ベースカラー広い面積の塗り分けに用いる低・中彩度色
\definecolor{CUDPink}{RGB}{255,202,191}
\definecolor{CUDBrightGreen}{RGB}{119,217,168}
\definecolor{CUDLime}{RGB}{216,242,85}
\definecolor{CUDCream}{RGB}{255,255,128}
\definecolor{CUDBrightCyan}{RGB}{191,228,255}
%\definecolor{CUD}{RGB}{}%

\setbeamercolor{CUDBrightGreen}{fg=black,bg=CUDBrightGreen!50}
\setbeamercolor{CUDBrightCyan}{fg=black,bg=CUDBrightCyan}
\setbeamercolor{CUDCream}{fg=black,bg=CUDCream!75}

%\setbeamercolor{frametitle}{use=structure,fg=white,bg=structure.fg}
%\setbeamercolor{palette primary}{use=structure,fg=white,bg=structure.fg}
%\setbeamercolor{frametitle}{bg=CUDOrange}
%\setbeamercolor{palette primary}{bg=CUDOrange}
%\setbeamercolor{frametitle}{bg=CUDCyan!25,fg=black}
%\setbeamercolor{palette primary}{bg=CUDCyan!25,fg=black}
%\setbeamercolor{frametitle}{bg=CUDCream!75,fg=black!80}
%\setbeamercolor{palette primary}{bg=CUDCream!75,fg=black!80}

\newcommand{\myfrac}[2]{\mathchoice%■ 修正 Oct 8,2023
	{%\fontsize{20}{20}\selectfont%	\displaystyle
		\frac{\raisebox{-.6ex}{\(#1\)}}{\raisebox{.4ex}{\(#2\)}}}%
	{\textstyle{\frac{#1}{#2}}}%	\textstyle
	{%\fontsize{13}{13}\selectfont%	\scriptstyle
		\frac{\raisebox{0ex}{\(#1\)}}{\raisebox{-.4ex}{\(#2\)}}}%
	{%\fontsize{11}{11}\selectfont%	\scriptscriptstyle
		\frac{\raisebox{0ex}{\(#1\)}}{\raisebox{-.4ex}{\(#2\)}}}%
	}


\begin{document}

\title{プレ高数学科}\author{gbb60166}

% ■■ 変数 yurika を定義する	2021 Feb 19
% sin, cos,tan の値で有理化をした数値を掲載している教科書がありました
\newif\ifyurika
% \yurikatrue は有理化された答えが表示（\yurikafalse は有理化されない答え）
%\yurikatrue
\yurikafalse

\newcommand{\ittan}{\begin{tikzpicture}[scale=0.7,
	baseline={([yshift=-22pt] current bounding box.north)}]
\filldraw[very thick,draw=CUDBlue,fill=CUDYellow,rounded corners=1pt]
	(0,0)--++(0.8,0)--++(0,0.8)--++(-0.8,0)--cycle;
\draw(0.4,0.4) node[CUDBlue]
	{\fontsize{7pt}{7pt}\selectfont
		\shortstack{一旦\\[-2pt]停止}};
\end{tikzpicture}
}

%■ 新しいプラスマイナス記号を自作（TikZを利用）
\newcommand{\newpm}{\,\tikz[baseline=-0.1ex]
	\draw[line width=0.1ex,line cap=round](0,0)--++(0.6em,0)
					(0,0.8ex)--++(0.6em,0)
					(0.3em,0.25ex)--++(0,1.1ex);
	\,}


%■■■■■■■■■■■■■　テスト領域　■■■■■■■■■■■■■■



%\end{document}
%■■■■■■■■■■■■■　完成品　■■■■■■■■■■■■■■

\begin{frame}[t]
\frametitle{三角比の相互関係}

\maruwaku{公式}\ 
\begin{minipage}[t]{0.82\textwidth}
\ajKakko{1}\quad
	\( \tan A=\myfrac{\ \sin A\ }{\cos A} \)

\bigskip
\ajKakko{2}\quad\( \sin^2 A +\cos^2 A=1 \)

\quad{\scriptsize\( \Bigl[\quad 
(\sin A)^2 +(\cos A)^2=1 \)の
意味です \quad \(\Bigl]\)}

\bigskip
\ajKakko{3}\quad\( 1+\tan^2 A =\myfrac{1}{\cos^2 A} \)

\quad{\scriptsize\( \Bigl[\quad 
	1 +(\tan A)^2=\myfrac{1}{(\cos A)^2} \)の
	意味です \quad \(\Bigl]\)}
\end{minipage}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{1}\ \( \sin A=\tfrac{\,1\,}{4} \)
	%分母の数字が切れるので見えない支柱を入れる
	\rule[-12pt]{0pt}{3pt}のとき\( \cos A,~\tan A \)を求めよ}

数学が苦手な人は図を描いて求めた方がよいだろう。\pause

\( \sin A=\myfrac{\alt<4->{\textcolor{CUDBlue}{\text{縦}}}{\text{縦}} }{
\alt<4->{\textcolor{CUDRed}{\text{斜め}}}{\text{斜め}}} \)だから\pause
\( \sin A=\myfrac{\,\alt<4->{\textcolor{CUDBlue}{1}}{1}\, }{
\alt<4->{\textcolor{CUDRed}{4}}{4}} \)となるには、\pause
\begin{tikzpicture}[rotate=0, very thick, scale=1,
	baseline={( [yshift=-36pt] current bounding box.north)}]
\footnotesize
\coordinate [label=left:{$A$}] (A) at (0,0);
\coordinate (B) at (3,2);
\coordinate (C) at (3,0);
\fill[CUDPink!50](A)--(B)--(C)--cycle;
%\draw[line join=round] (A)--(B) node[midway,above]{$4$}--(C) node[midway,right]{$1$}--cycle ;
\draw[CUDRed,line cap=round] (A)--(B) node[midway,above]{$4$};
\draw[CUDBlue,line cap=round] (B)--(C) node[midway,right]{$1$};
\draw[line cap=round](A)--(C);
\draw (C)++(-0.3,0)--++(0,0.3)--++(0.3,0);
\end{tikzpicture}
という三角形を考えればよい。
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{1}\ \( \sin A=\tfrac{\,1\,}{4} \)
	\rule[-12pt]{0pt}{3pt}のとき\( \cos A,~\tan A \)を求めよ}
\begin{tikzpicture}[rotate=0, very thick, scale=1,
	baseline={( [yshift=-36pt] current bounding box.north)}]
\footnotesize
\coordinate [label=left:{$A$}] (A) at (0,0);
\coordinate (B) at (3,2);
\coordinate (C) at (3,0);
\fill[CUDPink!50](A)--(B)--(C)--cycle;
\draw[CUDRed,line cap=round] (A)--(B) node[midway,above]{$4$};
\draw[CUDBlue,line cap=round] (B)--(C) node[midway,right]{$1$};
\draw[line cap=round] (A)--(C);
\draw (C)++(-0.3,0)--++(0,0.3)--++(0.3,0);
%\draw[CUDGreen,dashed] (A) to[out=-30,in=210]%■ 好みの曲がり具合にするには
%	node[inner sep=2pt,fill=white]{$x$} (C);% to[out=  がいいかも
%■ \draw[bend] で描いてみた	May 10 2023
%\draw[bend right,CUDGreen,dashed,distance=5ex]
\draw[bend right=30,CUDGreen,dashed]
	(A) to node[inner sep=2pt,fill=white,circle]{$x$} (C);
\end{tikzpicture}
\( \cos A,~\tan A \)を求めるには\textcolor{CUDGreen}{$x$}を
計算しなければならない。\pause
三平方の定理より

\medskip
\centerline{\colorbox{CUDCream}{\(\text{\LARGE
	\textcolor{CUDRed}{斜め}}^2=\text{○}^2+\text{△}^2  \)}}
\medskip

である。\pause
今の場合\( \textcolor{CUDRed}{4}^2=
	\textcolor{CUDGreen}{x}^2+\textcolor{CUDBlue}{1}^2 \)となる。
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{1}\ \( \sin A=\tfrac{\,1\,}{4} \)
	\rule[-12pt]{0pt}{3pt}のとき\( \cos A,~\tan A \)を求めよ}
\vspace*{-44pt}
\begin{eqnarray*}
\textcolor{CUDRed}{4}^2 &=&
	\textcolor{CUDGreen}{x}^2+\textcolor{CUDBlue}{1}^2 \\ \pause
16 &=& x^2+1 \\ \pause
16-1 &=& x^2 \\ \pause
\hooklen{16-1}{15} &=& x^2 \\ \pause
\newpm\sqrt{15} &=& \sqrt{x^2} \\
\text{\footnotesize\( x>0 \)なので}\quad & & \\
\ittan\ \sqrt{15} &=& x
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{1}\ \( \sin A=\tfrac{\,1\,}{4} \)
	\rule[-12pt]{0pt}{3pt}のとき\( \cos A,~\tan A \)を求めよ}
\begin{tikzpicture}[rotate=0, very thick, scale=1,
	baseline={( [yshift=-36pt] current bounding box.north)}]
\footnotesize
\coordinate [label=left:{$A$}] (A) at (0,0);
\coordinate (B) at (3,2);
\coordinate (C) at (3,0);
%\fill[CUDPink!50](A)--(B)--(C)--cycle;
%\draw[line cap=round](A)-- node[above]{$4$}(B);
%\draw[line cap=round](B)--(C) node[right]{$1$};
%\draw[line cap=round](A)--(C) node[below,fill=CUDCream]{$\sqrt{15}$};
\filldraw[fill=CUDPink!50,draw=black](A)
	-- node[above]{$4$}(B)
	-- node[right]{$1$}(C)
	-- node[yshift=-1.6ex,fill=CUDCream,inner sep=1pt]{$\sqrt{15}$} cycle;
\draw (C)++(-0.3,0)--++(0,0.3)--++(0.3,0);
\end{tikzpicture}
となるので\pause

\vspace*{-36pt}
\hspace*{5em}\( \cos A=\myfrac{\text{横}}{\text{斜め}}\pause =
\myfrac{\sqrt{15}}{4} \)\ \□答\pause 

\bigskip
\hspace*{5em}\( \tan A=\myfrac{\,\text{縦}\,}{\text{横}}\pause
	=\myfrac{1}{\sqrt{15}}
\ifyurika
	=\myfrac{\sqrt{15}}{15}
\fi \)\ \□答
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{1}\ \( \sin A=\tfrac{\,1\,}{4} \)
	\rule[-12pt]{0pt}{3pt}のとき\( \cos A,~\tan A \)を求めよ}
公式を使って、計算だけで解くなら

\bigskip
\maruwaku{公式} \( \sin^2 A +\cos^2 A=1 \)

\bigskip
\hspace*{7ex}に\( \sin A = \myfrac{\,1\,}{4} \)を代入して

\[ \Bigl(\myfrac{\,1\,}{4}\Bigr)^2 +\cos^2 A=1 \]
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{1}\ \( \sin A=\tfrac{\,1\,}{4} \)
	\rule[-12pt]{0pt}{3pt}のとき\( \cos A,~\tan A \)を求めよ}
\vspace*{-4ex}
\begin{eqnarray*}
\Bigl(\myfrac{\,1\,}{4}\Bigr)^2 +\cos^2 A &=& 1 \\
\hooklen{\Bigl(\myfrac{\,1\,}{4}\Bigr)^2}{\myfrac{1}{\,16\,}}
	+\cos^2 A &=& 1 \\
\cos^2 A &=& 1-\myfrac{1}{\,16\,} \\
\cos^2 A &=& \myfrac{16}{\,16\,}-\myfrac{1}{\,16\,}
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{1}\ \( \sin A=\tfrac{\,1\,}{4} \)
	\rule[-12pt]{0pt}{3pt}のとき\( \cos A,~\tan A \)を求めよ}
\vspace*{-4ex}
\begin{eqnarray*}
\cos^2 A &=& \myfrac{16}{\,16\,}-\myfrac{1}{\,16\,} \\
\cos^2 A &=& \hooklen{\myfrac{16}{\,16\,}-\myfrac{1}{\,16\,}}
		{\myfrac{15}{\,16\,}}\\
\sqrt{\cos^2 A} &=& \newpm\sqrt{\myfrac{15}{\,16\,}}\\
\cos A &=& \myfrac{\newpm\sqrt{15}}{\sqrt{16}}
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{1}\ \( \sin A=\tfrac{\,1\,}{4} \)
	\rule[-12pt]{0pt}{3pt}のとき\( \cos A,~\tan A \)を求めよ}
\vspace*{-4ex}
\begin{eqnarray*}
\cos A &=& \myfrac{\newpm\sqrt{15}}{\sqrt{16}}\\
\cos A &=& \myfrac{\newpm\sqrt{15}}{4}
\end{eqnarray*}

\( \cos A>0 \)なので\( \cos A=\myfrac{\sqrt{15}}{4} \)
\ \□答\quad となる。
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{1}\ \( \sin A=\tfrac{\,1\,}{4} \)
	\rule[-12pt]{0pt}{3pt}のとき\( \cos A,~\tan A \)を求めよ}
次に

\bigskip
\maruwaku{公式} \( \tan A=\myfrac{\ \sin A \ }{\cos A} \)

\bigskip
\centerline{に\( \sin A=\myfrac{\,1\,}{4},\ 
\cos A=\myfrac{\sqrt{15}}{4} \)を代入して}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{1}\ \( \sin A=\tfrac{\,1\,}{4} \)
	\rule[-12pt]{0pt}{3pt}のとき\( \cos A,~\tan A \)を求めよ}
\vspace*{-4ex}
\begin{eqnarray*}
\tan A &=& \myfrac{\ \sin A \ }{\cos A}\\
	&=& \myfrac{\ \myfrac{\,1\,}{4}\ }{\ \myfrac{\sqrt{15}}{4}\ } \pause
	\quad = \myfrac{\ \myfrac{\,1\,}{4}\textcolor{CUDRed}{\times4}\ }{
		\myfrac{\sqrt{15}}{4}\textcolor{CUDRed}{\times4}}\\\pause
	&=& \myfrac{1}{\sqrt{15}}
\ifyurika
		\quad =\myfrac{\sqrt{15}}{15}
\fi
\quad\text{\□答}
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\begin{frame}[t]
\frametitle{\ajKakko{2}\ \( \tan A=\tfrac{\sqrt{5}}{3} \)
	\rule[-12pt]{0pt}{3pt}のとき\( \sin A,~\cos A \)を求めよ}
\pause

\( \tan A=\myfrac{\alt<4->{\textcolor{CUDBlue}{\,\text{縦}\,}}{\,\text{縦}\,} }{
\alt<4->{\textcolor{CUDRed}{\text{横}}}{\text{横}}} \)だから\pause
\( \tan A=\myfrac{\alt<4->{\textcolor{CUDBlue}{\sqrt{5}}}{\sqrt{5}} }{
\alt<4->{\textcolor{CUDRed}{3}}{3}} \)となるには、\pause
\begin{tikzpicture}[rotate=0, very thick, scale=1,
	baseline={( [yshift=-36pt] current bounding box.north)}]
\footnotesize
\coordinate [label=left:{$A$}] (A) at (0,0);
\coordinate (B) at (3,2.23);
\coordinate (C) at (3,0);
\fill[CUDPink!50](A)--(B)--(C)--cycle;
\draw[CUDRed, line cap=round](A)--(C) node[midway,below]{$3$};
\draw[CUDBlue, line cap=round](B)--(C) node[midway,right]{$\sqrt{5}$};
\draw[line cap=round] (A)--(B);
\draw (C)++(-0.3,0)--++(0,0.3)--++(0.3,0);
\end{tikzpicture}
という三角形を考えればよい。
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{2}\ \( \tan A=\tfrac{\sqrt{5}}{3} \)
	\rule[-12pt]{0pt}{3pt}のとき\( \sin A,~\cos A \)を求めよ}
\begin{tikzpicture}[rotate=0, very thick, scale=1,
	baseline={( [yshift=-36pt] current bounding box.north)}]
\footnotesize
\coordinate [label=left:{$A$}] (A) at (0,0);
\coordinate (B) at (3,2.23);
\coordinate (C) at (3,0);
\fill[CUDPink!50](A)--(B)--(C)--cycle;
\draw[CUDRed, line cap=round](A)--(C) node[midway,below]{$3$};
\draw[CUDBlue, line cap=round](B)--(C) node[midway,right]{$\sqrt{5}$};
\draw[line cap=round](A)--(B);
\draw (C)++(-0.3,0)--++(0,0.3)--++(0.3,0);
%\draw[CUDGreen,dashed](A) to[out=60,in=190] 
%	node[inner sep=3pt,fill=white]{$x$} (B);
%■ \draw[bend] で描いてみた	May 10 2023
%\draw[bend left,CUDGreen,dashed,distance=5ex]
%■ 改良案
\draw[bend left=30,CUDGreen,dashed]
	(A) to node[inner sep=3pt,fill=white,circle]{$x$} (B);
\end{tikzpicture}
\( \sin A,~\cos A \)を求めるには\textcolor{CUDGreen}{$x$}を
計算しなければならない。\pause
三平方の定理は

\medskip
\centerline{\colorbox{CUDCream}{\(\text{\LARGE
	\textcolor{CUDGreen}{斜め}}^2=\text{○}^2+\text{△}^2  \)}}
\medskip

だ。\pause
今の場合\quad\( \textcolor{CUDGreen}{x}^2=
\textcolor{CUDRed}{3}^2+\textcolor{CUDBlue}{\sqrt{5}\kern4pt}^2 \)となる。
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{2}\ \( \tan A=\tfrac{\sqrt{5}}{3} \)
	\rule[-12pt]{0pt}{3pt}のとき\( \sin A,~\cos A \)を求めよ}
\vspace*{-48pt}
\begin{eqnarray*}
\textcolor{CUDGreen}{x}^2 &=&
	\textcolor{CUDRed}{3}^2
	+ \textcolor{CUDBlue}{\sqrt{5}\kern4pt}^2 \\ \pause
x^2 &=& \hooklen{3^2}{9}+\hooklen{{\sqrt{5}\kern4pt}^2}{5} \\ \pause
x^2 &=& \quad 14 \\ \pause
\sqrt{x^2} &=& \newpm\sqrt{14} \\
\text{\footnotesize\( x>0 \)なので}\quad & & \\
x &=& \sqrt{14}\ \ittan
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{2}\ \( \tan A=\tfrac{\sqrt{5}}{3} \)
	\rule[-12pt]{0pt}{3pt}のとき\( \sin A,~\cos A \)を求めよ}
\begin{tikzpicture}[rotate=0, very thick, scale=1,
	baseline={( [yshift=-36pt] current bounding box.north)}]
\footnotesize
\coordinate [label=left:{$A$}] (A) at (0,0);
\coordinate (B) at (3,2.23);
\coordinate (C) at (3,0);
\filldraw[fill=CUDPink!50,draw=black](A)
	-- node[above left,fill=CUDCream,inner sep=1pt]{$\sqrt{14}$}(B)
	-- node[right]{$\sqrt{5}$}(C)
	-- node[below]{$3$} cycle;
\draw (C)++(-0.3,0)--++(0,0.3)--++(0.3,0);
\end{tikzpicture}
となるので\pause

\vspace*{-2.2ex}
\hspace*{4em}\( \sin A=\myfrac{\text{縦}}{\text{斜め}}\pause =
	\myfrac{\sqrt{5}}{\sqrt{14}}
\ifyurika
	=\myfrac{\sqrt{70}}{14}
\fi \)\ \□答\pause 

\bigskip
\hspace*{4em}\( \cos A=\myfrac{\text{横}}{\text{斜め}}\pause
	=\myfrac{3}{\sqrt{14}}
\ifyurika
	=\myfrac{3\sqrt{14}}{14}
\fi \)\ \□答
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{3}\ \( \cos A=\tfrac{\,2\,}{\sqrt{13}} \)
	\rule[-12pt]{0pt}{3pt}のとき\( \sin A,~\tan A \)を求めよ}
\pause
\( \cos A=\myfrac{\alt<4->{\textcolor{CUDBlue}{\text{横}}}{\text{横}} }{
	\alt<4->{\textcolor{CUDRed}{\text{斜め}}}{\text{斜め}}} \)だから\pause
\( \cos A=\myfrac{\,\alt<4->{\textcolor{CUDBlue}{2}}{2}\, }{
	\alt<4->{\textcolor{CUDRed}{\sqrt{13}}}{\sqrt{13}}} \)となるには、
		\pause
\begin{tikzpicture}[rotate=0, very thick, scale=1,
	baseline={( [yshift=-36pt] current bounding box.north)}]
\footnotesize
\coordinate [label=left:{$A$}] (A) at (0,0);
\coordinate (B) at (3,2);
\coordinate (C) at (3,0);
\fill[CUDPink!50](A)--(B)--(C)--cycle;
\draw[CUDRed,line cap=round](A)--(B) node[midway,above]{$\sqrt{13}$\kern12pt};
\draw[line cap=round] (B)--(C);
\draw[CUDBlue, line cap=round] (A)--(C) node[midway,below]{$2$};
\draw (C)++(-0.3,0)--++(0,0.3)--++(0.3,0);
\end{tikzpicture}
という三角形を考えればよい。
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{3}\ \( \cos A=\tfrac{\,2\,}{\sqrt{13}} \)
	\rule[-12pt]{0pt}{3pt}のとき\( \sin A,~\tan A \)を求めよ}
\vspace*{-1ex}
\begin{tikzpicture}[rotate=0, very thick, scale=1,
	baseline={( [yshift=-36pt] current bounding box.north)}]
\footnotesize
\coordinate [label=left:{$A$}] (A) at (0,0);
\coordinate (B) at (3,2);
\coordinate (C) at (3,0);
\fill[CUDPink!50](A)--(B)--(C)--cycle;
\draw[CUDRed,line cap=round](A)--(B) node[midway,above]{$\sqrt{13}$\kern12pt};
\draw[line cap=round] (B)--(C);
\draw[CUDBlue, line cap=round] (A)--(C) node[midway,below]{$2$};
\draw (C)++(-0.3,0)--++(0,0.3)--++(0.3,0);
%\draw[CUDGreen,dashed] (B) to[out=-45,in=45]%■ 好みの曲がり具合にするには
%	node[inner sep=2pt,fill=white]{$x$} (C);% to[out=  がいいかも
%■ \draw[bend] で描いてみた	May 10 2023
% \draw[bend left,CUDGreen,dashed,distance=4ex]
%■ 改良案
\draw[bend left=45,CUDGreen,dashed]
	(B) to node[inner sep=1pt,fill=white,circle]{$x$} (C);
\end{tikzpicture}
\( \sin A,~\tan A \)を求めるには\textcolor{CUDGreen}{$x$}を
計算しなければならない。\pause
三平方の定理より

\medskip
\centerline{\colorbox{CUDCream}{\(\text{\LARGE
	\textcolor{CUDRed}{斜め}}^2=\text{○}^2+\text{△}^2  \)}}
\medskip

である。\pause
今の場合\( {\textcolor{CUDRed}{\sqrt{13}}\kern4pt}^2=
\textcolor{CUDGreen}{x}^2+\textcolor{CUDBlue}{2}^2 \)となる。
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{3}\ \( \cos A=\tfrac{\,2\,}{\sqrt{13}} \)
	\rule[-12pt]{0pt}{3pt}のとき\( \sin A,~\tan A \)を求めよ}
\vspace*{-40pt}
\begin{eqnarray*}
{\textcolor{CUDRed}{\sqrt{13}}\kern4pt}^2 &=&
	\textcolor{CUDGreen}{x}^2+\textcolor{CUDBlue}{2}^2 \\ \pause
\hooklen{13-4}{13} &=& x^2+4 \\ \pause
13-4 &=& x^2 \\ \pause
\hooklen{13-4}{9} &=& x^2 \\ \pause
\newpm\sqrt{9} &=& \sqrt{x^2} \\
\newpm 3 \ &=& x
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{3}\ \( \cos A=\tfrac{\,2\,}{\sqrt{13}} \)
	\rule[-12pt]{0pt}{3pt}のとき\( \sin A,~\tan A \)を求めよ}
\vspace*{-40pt}
\begin{eqnarray*}
\newpm 3 &=& x \\
\text{\footnotesize\( x>0 \)なので}\quad & & \\
\ittan\ 3 &=& x
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}[t]
\frametitle{\ajKakko{3}\ \( \cos A=\tfrac{\,2\,}{\sqrt{13}} \)
	\rule[-12pt]{0pt}{3pt}のとき\( \sin A,~\tan A \)を求めよ}
\begin{tikzpicture}[rotate=0, very thick, scale=1,
	baseline={( [yshift=-36pt] current bounding box.north)}]
\footnotesize
\coordinate [label=left:{$A$}] (A) at (0,0);
\coordinate (B) at (3,2);
\coordinate (C) at (3,0);
\filldraw[fill=CUDPink!50,draw=black](A)
	-- node[above left]{$\sqrt{13}$}(B)
	-- node[xshift=0.6em,fill=CUDCream,inner sep=3pt]{$3$}(C)
	-- node[below]{$2$}cycle;
\draw (C)++(-0.3,0)--++(0,0.3)--++(0.3,0);
\end{tikzpicture}
となるので\pause

\vspace*{-2.2ex}
\hspace*{4em}\( \sin A=\myfrac{\text{縦}}{\text{斜め}}\pause
=\myfrac{3}{\sqrt{13}}
\ifyurika
	=\myfrac{3\sqrt{13}}{13}
\fi \)\ \□答\pause 

\bigskip
\hspace*{4em}\( \tan A=\myfrac{\,\text{縦}\,}{\text{横}}\pause
=\myfrac{\,3\,}{2} \)\ \□答
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\end{document}


