正切半角公式

正切半角公式又称万能公式，这一组公式有四个功能：

1. 将角统一为${\displaystyle {\frac {\alpha }{2}}}$[1]；
2. 将函数名称统一为${\displaystyle \tan }$；
3. 任意实数都可以${\displaystyle \tan {\frac {\alpha }{2}}}$的形式表達，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。

因此，这组公式被称为以切表弦公式，简称以切表弦。它们是由二倍角公式求得的。

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

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

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

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

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

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

而被称为萬能公式的原因是利用${\displaystyle \tan {\frac {\alpha }{2}}}$的代換可以解決一些有關三角函数的積分。参见三角换元法。

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