Tangenttilause

Sinilauseen mukaan

${\displaystyle {\frac {a}{\sin \alpha }}={\frac {b}{\sin \beta }}.}$

Olkoon

${\displaystyle d={\frac {a}{\sin \alpha }}={\frac {b}{\sin \beta }},}$

jolloin

${\displaystyle a=d\sin \alpha {\text{ ja }}b=d\sin \beta .\,}$

Tällöin

${\displaystyle {\frac {a-b}{a+b}}={\frac {d\sin \alpha -d\sin \beta }{d\sin \alpha +d\sin \beta }}={\frac {\sin \alpha -\sin \beta }{\sin \alpha +\sin \beta }}.}$

Kun käytetään identiteettiä

${\displaystyle \sin(\alpha )\pm \sin(\beta )=2\sin \left({\frac {\alpha \pm \beta }{2}}\right)\cos \left({\frac {\alpha \mp \beta }{2}}\right),\;}$

saadaan

${\displaystyle {\frac {a-b}{a+b}}={\frac {2\sin {\tfrac {1}{2}}\left(\alpha -\beta \right)\cos {\tfrac {1}{2}}\left(\alpha +\beta \right)}{2\sin {\tfrac {1}{2}}\left(\alpha +\beta \right)\cos {\tfrac {1}{2}}\left(\alpha -\beta \right)}}={\frac {\tan[{\frac {1}{2}}(\alpha -\beta )]}{\tan[{\frac {1}{2}}(\alpha +\beta )]}}.}$

• Sinilause
• Kosinilause
• Kotangenttilause