tan(x/2)=t のとき sinx, cosx, tanxをtを使って表す

【方法１による導出】

${\tan}\,x\\ ={\tan}\,(\dfrac{x}{2}+\dfrac{x}{2})\\ =\dfrac{{\tan}\,\dfrac{x}{2}+\tan\,\dfrac{x}{2}}{1-\tan\dfrac{x}{2}\tan\dfrac{x}{2}}\\ =\dfrac{2t}{1-t^2}$

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$\cos\,x=\cos\,(2\cdot\dfrac{x}{2})=2{\cos}^2\,\dfrac{x}{2}-1$

${\cos}^2\,\dfrac{x}{2}=\dfrac{1}{1+\tan^2\,{\dfrac{x}{2}}}=\dfrac{1}{1+t^2}$
だから
$\cos\,x=\dfrac{2}{1+t^2}-1=$$\dfrac{1-t^2}{1+t^2}$

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$\sin\,x$
$=\tan\,x\cdot\cos\,x$
$=\dfrac{2t}{1-t^2}\cdot\dfrac{1-t^2}{1+t^2}$
$=\dfrac{2t}{1+t^2}$

【方法２による導出】

$\sin x$
$=\dfrac{2\sin \dfrac{x}{2} \cos \dfrac{x}{2}}{\sin^2 \dfrac{x}{2}+\cos^2 \dfrac{x}{2}}$
$=\dfrac{2\tan \dfrac{x}{2}}{\tan^2 \dfrac{x}{2}+1}$
$=\dfrac{2t}{1+t^2}$

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$\cos x$
$=\dfrac{\cos^2\dfrac{x}{2}-\sin^2 \dfrac{x}{2}}{\cos^2 \dfrac{x}{2}+\sin^2 \dfrac{x}{2}}$
$=\dfrac{1-\tan^2 \dfrac{x}{2}}{1+\tan^2 \dfrac{x}{2}}$
$=\dfrac{1-t^2}{1+t^2}$

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$\tan x$
$=\dfrac{\sin x}{\cos x}$

$=\dfrac{2t}{1-t^2}$