t=\tan\bunsuu{\theta}{2}\ (t\neqq\pm1)とするとき,\ \sin\theta,\ \cos\theta,\ \tan\theta\ をtで表せ.$ \\[.5zh] 三角関数の媒介変数表示（有理関数表示）}$}}}} \\\\[.5zh] $\bm{\textcolor{red}{t=\tan\bunsuu{\theta}{2}\ とおくと,\ 三角関数\,\sin\theta,\ \cos\theta,\ \tan\theta\,をtのみの式に変換できる.}}$ \\[.5zh] この三角関数の媒介変数表示は,\ ある種の問題で利用することになる. \\\\ 変換は,\ $\bm{\textcolor{red}{\theta=2\cdot\bunsuu{\theta}{2}}}\ と考えて,\ \bm{\textcolor{cyan}{2倍角の公式}}や\bm{\textcolor{magenta}{三角関数の相互関係}}を利用する.$ \\\\\\ $\bm{\tan\theta}=\tan\hspace{-.2zw}\left(\textcolor{red}{2\cdot\bunsuu{\theta}{2}}\right)=\textcolor{cyan}{\bunsuu{2\tan\bunsuu{\theta}{2}}{1-\tan^2\bunsuu{\theta}{2}}}=\bm{\bunsuu{2t}{1-t^2}}$ \\\\[1zh] $\bm{\cos\theta}=\cos\hspace{-.2zw}\left(\textcolor{red}{2\cdot\bunsuu{\theta}{2}}\right)=\textcolor{cyan}{2\cos^2\bunsuu{\theta}{2}-1}=2\cdot\textcolor{magenta}{\bunsuu{1}{1+\tan^2\bunsuu{\theta}{2}}}-1=\bunsuu{2}{1+t^2}-1=\bm{\bunsuu{1-t^2}{1+t^2}}$ \\\\[1zh] $\bm{\sin\theta}=\textcolor{magenta}{\tan\theta\cos\theta}=\bunsuu{2t}{1-t^2}\cdot\bunsuu{1-t^2}{1+t^2}=\bm{\bunsuu{2t}{1+t^2}}$ \\\\\\ \centerline{{\small $\left[\textcolor{brown}{\begin{array}{l} \sin\theta\,を直接求めるのは厄介なので,\ 先に\,\bm{\tan\theta,\ \cos\theta\,を求め,\ その2つから\ \sin\theta\,を求める}とよい. \\[1zh] 2倍角の公式 \tan2\theta=\bunsuu{2\tan\theta}{1-\tan^2\theta} \cos2\theta=2\cos^2\theta-1 \\[1zh] 1+\tan^2\theta=\bunsuu{1}{\cos^2\theta}\ より \cos^2\theta=\bunsuu{1}{1+\tan^2\theta} \\\\ \tan\theta=\bunsuu{\sin\theta}{\cos\theta}\ より \sin\theta=\tan\theta\cos\theta $\sin\theta$\ を直接求める方法も2つ示しておく. \\\\ $\bm{\sin\theta}=\sin\hspace{-.2zw}\left(\textcolor{red}{2\cdot\bunsuu{\theta}{2}}\right)=\textcolor{cyan}{2\sin\bunsuu{\theta}{2}\cos\bunsuu{\theta}{2}}$ \\[.5zh] $\phantom{\bm{\sin\theta}}=\bunsuu{2\,\sin\bunsuu{\theta}{2}\cos\bunsuu{\theta}{2}}{\textcolor{magenta}{\sin^2\bunsuu{\theta}{2}+\cos^2\bunsuu{\theta}{2}}}$ $\left[\textcolor{brown}{\,\sin^2\bunsuu{\theta}{2}+\cos^2\bunsuu{\theta}{2}=1で割った}\,\right]$ \\[.5zh] $\phantom{\bm{\sin\theta}}=\bunsuu{2\,\textcolor{magenta}{\tan\bunsuu{\theta}{2}}}{\textcolor{magenta}{1+\tan^2\bunsuu{\theta}{2}}}$ \ \,\hspace{.2zw}$\left[\,\textcolor{brown}{分母・分子を \cos^2\bunsuu{\theta}{2}\ で割り,\ \bunsuu{\sin\bunsuu{\theta}{2}}{\cos\bunsuu{\theta}{2}}=\tan\bunsuu{\theta}{2}\,とした}\,\right]$ \\[.5zh] $\phantom{\bm{\sin\theta}}=\bm{\bunsuu{2t}{1+t^2}}$ \\\\\\ $\bm{\sin\theta}=\sin\hspace{-.2zw}\left(\textcolor{red}{2\cdot\bunsuu{\theta}{2}}\right)=\textcolor{cyan}{2\sin\bunsuu{\theta}{2}\cos\bunsuu{\theta}{2}}$ \\[.5zh] $\phantom{\bm{\sin\theta}}=2\cdot\bunsuu{\sin\bunsuu{\theta}{2}}{\cos\bunsuu{\theta}{2}}\cdot\cos^2\bunsuu{\theta}{2}$ \ \ $\left[\,\textcolor{brown}{\bunsuu{\cos\bunsuu{\theta}{2}}{\cos\bunsuu{\theta}{2}}\ を掛けた}\,\right]$ \\[.5zh] $\phantom{\bm{\sin\theta}}=2\,\textcolor{magenta}{\tan\bunsuu{\theta}{2}}\cdot\textcolor{magenta}{\bunsuu{1}{1+\tan^2\bunsuu{\theta}{2}}}$ $\left[\,\textcolor{brown}{\cos^2\theta=\bunsuu{1}{1+\tan^2\theta}\,を適用した}\,\right]$ \\[.5zh] $\phantom{\bm{\sin\theta}}=\bm{\bunsuu{2t}{1+t^2}}$ \\\\\\ 上級者は結果を暗記しておくことが望ましい. \\[1zh] $\bm{\textcolor{cyan}{t=\tan\bunsuu{\theta}{2}}}\ のとき$ $\bm{\sin\theta=\textcolor{red}{\bunsuu{2t}{1+t^2}},\ \ \cos\theta=\textcolor{red}{\bunsuu{1-t^2}{1+t^2}},\ \ \tan\theta=\textcolor{red}{\bunsuu{2t}{1-t^2}}}$
