正接定理
a − b a + b = 2 sin  1 2 ( α − β ) cos  1 2 ( α + β ) 2 sin  1 2 ( α + β ) cos  1 2 ( α − β ) = tan  [ 1 2 ( α − β ) ] tan  [ 1 2 ( α + β ) ] . ◼ {\displaystyle {\frac {a-b}{a+b}}={\frac {2\sin {\tfrac {1}{2}}\left(\alpha -\beta \right)\cos {\tfrac {1}{2}}\left(\alpha +\beta \right)}{2\sin {\tfrac {1}{2}}\left(\alpha +\beta \right)\cos {\tfrac {1}{2}}\left(\alpha -\beta \right)}}={\frac {\tan[{\frac {1}{2}}(\alpha -\beta )]}{\tan[{\frac {1}{2}}(\alpha +\beta )]}}.\qquad \blacksquare }
この証明を変形して以下の式を導くことができる。
tan  α ± β 2 = sin  α ± sin  β cos  α + cos  β {\displaystyle \tan {\frac {\alpha \pm \beta }{2}}={\frac {\sin \alpha \pm \sin \beta }{\cos \alpha +\cos \beta }}}