$\displaystyle t=\tan\frac{x}{2}$ とおくと, $$ \sin x = \frac{2t}{1+t^2},\quad \cos x = \frac{1-t^2}{1+t^2} $$ が成り立つことを証明せよ.
解答例 1
\begin{align*} \sin x &= 2\sin\frac{x}{2}\cos\frac{x}{2} = 2\tan\frac{x}{2}\cos^2\frac{x}{2} \\ &= \frac{\displaystyle\biggl. 2\tan\frac{x}{2}}{\displaystyle\biggl.1+\tan^2\frac{x}{2}} = \frac{2t}{1+t^2}, \\ \cos x &= 2\cos^2\frac{x}{2}-1 = \frac{2}{\displaystyle\biggl. 1+\tan^2\frac{x}{2}} - 1 \\ &= \frac{\displaystyle\biggl. 1-\tan^2\frac{x}{2}}{\displaystyle\biggl. 1+\tan^2\frac{x}{2}} = \frac{1-t^2}{1+t^2}. \end{align*}
最終更新日：2011年11月02日