正切半角公式

1. 将角统一为$\frac{\alpha }{2}$
2. 将函数名称统一为$\tan$
3. 任意实数都可以表示为$\tan \frac{\alpha }{2}$的形式，可以用正切函数换元
4. 在某些积分中,可以将含有三角函数的积分变为有理分式的积分.

$\sin \alpha = \frac{{2\tan \frac{\alpha }{2}}}{{1 + \tan ^2 \frac{\alpha }{2}}}$
$\cos \alpha = \frac{{1 - \tan ^2 \frac{\alpha }{2}}}{{1 + \tan ^2 \frac{\alpha }{2}}}$
$\tan \alpha = \frac{{2\tan \frac{\alpha }{2}}}{{1 - \tan ^2 \frac{\alpha }{2}}}$

\begin{align} \tan\left(\frac{\eta}{2} \pm \frac{\theta}{2}\right) & = \frac{\sin\eta \pm \sin\theta}{\cos\eta + \cos\theta} = -\frac{\cos\eta - \cos\theta}{\sin\eta \mp \sin\theta}, \\[10pt] \tan\left(\pm\frac{\theta}{2}\right) & = \frac{\pm\sin\theta}{1 + \cos\theta} = \frac{\pm\tan\theta}{\sec\theta + 1} = \frac{\pm 1}{\csc\theta + \cot\theta}, ~~~~(\eta = 0) \\[10pt] \tan\left(\pm\frac{\theta}{2}\right) & = \frac{1-\cos\theta}{\pm\sin\theta} = \frac{\sec\theta-1}{\pm\tan\theta} = \pm(\csc\theta-\cot\theta), ~~~~(\eta=0) \\[10pt] \tan\left(\frac{\pi}{4} \pm \frac{\theta}{2} \right) & = \frac{1 \pm \sin\theta}{\cos\theta} = \sec\theta \pm \tan\theta = \frac{\csc\theta \pm 1}{\cot\theta}, ~~~~(\eta=\frac{\pi}{2}) \\[10pt] \tan\left(\frac{\pi}{4} \pm \frac{\theta}{2} \right) & = \frac{\cos\theta}{1 \mp \sin\theta} = \frac{1}{\sec\theta \mp \tan\theta} = \frac{\cot\theta}{\csc\theta \mp 1}, ~~~~(\eta=\frac{\pi}{2}) \\[10pt] \frac{1 - \tan\frac{\theta}{2}}{1 + \tan\frac{\theta}{2}} & = \sqrt{\frac{1 - \sin\theta}{1 + \sin\theta}}. \end{align}

万能公式的证明

$\sin \alpha = 2\sin \frac{\alpha}{2}\cos \frac{\alpha}{2} =\frac{2\sin \frac{\alpha}{2}\cos \frac{\alpha}{2}}{\cos^2 \frac{\alpha}{2}+\sin^2\frac{\alpha}{2}} = \frac{2\sin \frac{\alpha}{2}\cos \frac{\alpha}{2} \div \cos^2 \frac{\alpha}{2}}{(\cos^2 \frac{\alpha}{2}+\sin^2\frac{\alpha}{2})\div \cos^2 \frac{\alpha}{2}} = \frac{2\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}}{1+\frac{\sin^2\frac{\alpha}{2}}{\cos^2 \frac{\alpha}{2}}} =\frac{2\tan\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}$

$\tan \alpha = \frac{{2\tan \frac{\alpha }{2}}}{{1 - \tan ^2 \frac{\alpha }{2}}}$

$\cos \alpha = \frac{\sin\alpha}{\tan\alpha}=\frac{\frac{{2\tan \frac{\alpha }{2}}}{{1 + \tan ^2 \frac{\alpha }{2}}}}{\frac{{2\tan \frac{\alpha }{2}}}{{1 - \tan ^2 \frac{\alpha }{2}}}} = \frac{{1 - \tan ^2 \frac{\alpha }{2}}}{{1 + \tan ^2 \frac{\alpha }{2}}}$

双曲函数

$t = \tanh\tfrac{1}{2}\theta = \frac{\sinh\theta}{\cosh\theta+1} = \frac{\cosh\theta-1}{\sinh\theta}$

 $\cosh\theta = \frac{1 + t^2}{1 - t^2},$ $\sinh\theta = \frac{2t}{1 - t^2},$ $\tanh\theta = \frac{2t}{1 + t^2},$ $\coth\theta = \frac{1 + t^2}{2t},$ $\mathrm{sech}\,\theta = \frac{1 - t^2}{1 + t^2},$ $\mathrm{csch}\,\theta = \frac{1 - t^2}{2t},$

 $e^{\theta} = \frac{1 + t}{1 - t},$ $e^{-\theta} = \frac{1 - t}{1 + t}.$

θT而得出下面的双曲反正切和自然对数之间的关系：

$\operatorname{artanh} t = \frac{1}{2}\ln\frac{1+t}{1-t}.$