【数学I】例題4.1.5：余角・補角の三角比（One More）★★

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題I4.1.5：余角・補角の三角比（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(1) $\sin 50^{\circ},\cos 130^{\circ}$を$45^{\circ}$以下の三角比で表せ．また，$\sin 40^{\circ}\cos 130^{\circ}+\sin 50^{\circ}\cos 140^{\circ}$を簡単にせよ．

(2) $\triangle\mathrm{ABC}$の3つの内角$\angle\mathrm{A},\angle\mathrm{B},\angle\mathrm{C}$の大きさを，それぞれ$A,B,C$とするとき，等式$\sin\frac{C}{2}=\cos\frac{A+B}{2}$が成り立つことを証明せよ．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題I4.1.5の解答（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(1)

\[
\sin 50^{\circ}=\sin(90^{\circ}-40^{\circ})=\cos 40^{\circ}\cdots(\mathrm{i}),
\]

\[
\cos 130^{\circ}=\cos(180^{\circ}-50^{\circ})=-\cos 50^{\circ}=-\cos(90^{\circ}-40^{\circ})=-\sin 40^{\circ}\cdots(\mathrm{ii})
\]

$\cos 140^{\circ}=\cos(180^{\circ}-40^{\circ})=-\cos 40^{\circ}$であるから，これと(i)，(ii)より，

\begin{align*}
&\sin 40^{\circ}\cos 130^{\circ}+\sin 50^{\circ}\cos 140^{\circ}\\
&=\sin 40^{\circ}(-\sin 40^{\circ})+\cos 40^{\circ}(-\cos 40^{\circ})\\
&=-\sin^240^{\circ}-\cos^240^{\circ}\\
&=-(\sin^240^{\circ}+\cos^240^{\circ})\\
&=-1
\end{align*}

(2) $A+B+C=180^{\circ}$であるから，$A+B=180^{\circ}-C$

したがって，$\frac{A+B}{2}=\frac{180^{\circ}-C}{2}=90^{\circ}-\frac{C}{2}$

ゆえに，$\cos\frac{A+B}{2}=\cos(90^{\circ}-\frac{C}{2})=\sin\frac{C}{2}$

よって，等式は成り立つ．$\blacksquare$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題I4.1.5：余角・補角の三角比（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(1) $\sin 55^{\circ},\cos 125^{\circ}$を$45^{\circ}$以下の三角比で表せ．また，$\sin 35^{\circ}\cos 125^{\circ}+\sin 55^{\circ}\cos 145^{\circ}$を簡単にせよ．

(2) $\triangle\mathrm{ABC}$の3つの内角$\angle\mathrm{A},\angle\mathrm{B},\angle\mathrm{C}$の大きさを，それぞれ$A,B,C$とするとき，等式$\tan\frac{A}{2}\tan\frac{B+C}{2}=1$が成り立つことを証明せよ．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題I4.1.5の解答（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(1)

\[
\sin 55^{\circ}=\sin(90^{\circ}-35^{\circ})=\cos 35^{\circ}\cdots(\mathrm{i}),
\]

\[
\cos 125^{\circ}=\cos(180^{\circ}-55^{\circ})=-\cos 55^{\circ}=-\cos(90^{\circ}-35^{\circ})=-\sin 35^{\circ}\cdots(\mathrm{ii})
\]

$\cos 145^{\circ}=\cos(180^{\circ}-35^{\circ})=-\cos 35^{\circ}$であるから，これと(i)，(ii)より，

\begin{align*}
&\sin 35^{\circ}\cos 125^{\circ}+\sin 55^{\circ}\cos 145^{\circ}\\
&=\sin 35^{\circ}(-\sin 35^{\circ})+\cos 35^{\circ}(-\cos 35^{\circ})\\
&=-\sin^235^{\circ}-\cos^235^{\circ}\\
&=-(\sin^235^{\circ}+\cos^235^{\circ})\\
&=-1
\end{align*}

(2) $A+B+C=180^{\circ}$であるから，$B+C=180^{\circ}-A$

したがって，$\frac{B+C}{2}=\frac{180^{\circ}-A}{2}=90^{\circ}-\frac{A}{2}$

ゆえに，$\tan\frac{A}{2}\tan\frac{B+C}{2}=\tan\frac{A}{2}\cdot\frac{1}{\tan\frac{A}{2}}=1$

よって，等式は成り立つ．$\blacksquare$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題I4.1.5の側注：余角・補角の三角比（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
$\sin(90^{\circ}-\theta)=\cos\theta$

% 側注2
$\cos(180^{\circ}-\theta)=-\cos\theta,\cos(90^{\circ}-\theta)=\sin\theta$

% 側注3
$\cos(180^{\circ}-\theta)=-\cos\theta$

% 側注4
すべて$40^\circ$の三角比に式変形する．

% 側注5
$\sin^2\theta+\cos^2\theta=1$

% 側注6
$\cos(90^{\circ}-\theta)=\sin\theta$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題I4.1.5の側注：余角・補角の三角比（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% 側注1
$\sin(90^{\circ}-\theta)=\cos\theta$

% 側注2
$\cos(180^{\circ}-\theta)=-\cos\theta,\cos(90^{\circ}-\theta)=\sin\theta$

% 側注3
$\cos(180^{\circ}-\theta)=-\cos\theta$

% 側注4
すべて$35^\circ$の三角比に式変形する．

% 側注5
$\sin^2\theta+\cos^2\theta=1$

% 側注6
$\tan(90^{\circ}-\theta)=\frac{1}{\tan\theta}$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題I4.1.5の考え方・補足など：余角・補角の三角比（One More）★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 考え方
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(1) $90^{\circ}-\theta$（余角）と$180^{\circ}-\theta$（補角）の三角比を用いると，すべて$40^\circ$の三角比に式変形することができる．

(2) 1つの辺を式変形して（ここでは右辺），もう一方の辺と一致することを示す．$A,B,C$は$\triangle\mathrm{ABC}$の3つの内角であるから，$A+B+C=180^{\circ}$であることを利用すればよい．

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 余角・補角の三角比
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(1) 余角の三角比$(0^{\circ}\leqq\theta\leqq 90^{\circ})$

\[
\sin(90^{\circ}-\theta)=\cos\theta,\cos(90^{\circ}-\theta)=\sin\theta,\tan(90^{\circ}-\theta)=\frac{1}{\tan\theta}(\theta\neq 0^{\circ},\theta\neq 90^{\circ})
\]

(2) 補角の三角比$(0^{\circ}\leqq\theta\leqq 180^{\circ})$

\[
\sin(180^{\circ}-\theta)=\sin\theta,\cos(180^{\circ}-\theta)=-\cos\theta,\tan(180^{\circ}-\theta)=-\tan\theta(\theta\neq 90^{\circ})
\]

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 余談
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

$\sin(90^{\circ}+\theta)=\cos\theta,\cos(90^{\circ}+\theta)=-\sin\theta,\tan(90^{\circ}+\theta)=-\frac{1}{\tan\theta}$が成り立つ．