u = tan x 2 sin x = 2 u 1 + u 2 cos x = 1 − u 2 1 + u 2 d x = 2 1 + u 2 d u tan x = 2 u 1 − u 2 tan ( α ± β ) = tan α ± tan β 1 ∓ tan α tan β {\displaystyle {\begin{aligned}u=\tan {\frac {x}{2}}\qquad \sin x={\frac {2u}{1+u^{2}}}\qquad \cos x={\frac {1-u^{2}}{1+u^{2}}}\\dx={\frac {2}{1+u^{2}}}du\qquad \tan x={\frac {2u}{1-u^{2}}}\\\tan(\alpha \pm \beta )={\frac {\tan \alpha \pm \tan \beta }{1\mp \tan \alpha \tan \beta }}\end{aligned}}}