# 正切半角公式

1. 统一为${\displaystyle {\frac {\alpha }{2}}}$
2. 函数名称统一为${\displaystyle \tan }$
3. 任意实数都可以${\displaystyle \tan {\frac {\alpha }{2}}}$形式表達，可用正切函数换元
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。

${\displaystyle \sin \alpha ={\frac {2\tan {\frac {\alpha }{2}}}{1+\tan ^{2}{\frac {\alpha }{2}}}}}$
${\displaystyle \cos \alpha ={\frac {1-\tan ^{2}{\frac {\alpha }{2}}}{1+\tan ^{2}{\frac {\alpha }{2}}}}}$
${\displaystyle \tan \alpha ={\frac {2\tan {\frac {\alpha }{2}}}{1-\tan ^{2}{\frac {\alpha }{2}}}}}$
${\displaystyle \cot \alpha ={\frac {1-\tan ^{2}{\frac {\alpha }{2}}}{2\tan {\frac {\alpha }{2}}}}}$
${\displaystyle \sec \alpha ={\frac {1+\tan ^{2}{\frac {\alpha }{2}}}{1-\tan ^{2}{\frac {\alpha }{2}}}}}$
${\displaystyle \csc \alpha ={\frac {1+\tan ^{2}{\frac {\alpha }{2}}}{2\tan {\frac {\alpha }{2}}}}}$

{\displaystyle {\begin{aligned}\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{aligned}}}

## 万能公式的证明

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

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

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

### 几何证明

${\displaystyle t={\frac {\sin \phi }{1+\cos \phi }}={\frac {\sin \phi (1-\cos \phi )}{(1+\cos \phi )(1-\cos \phi )}}={\frac {1-\cos \phi }{\sin \phi }}}$

## 双曲函数

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

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

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

θT而得出下面的双曲反正切函數自然对数之间的关系：

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