正弦定理

做一个边长为${\displaystyle a}$，${\displaystyle b}$，${\displaystyle c}$的三角形，对应角分别是${\displaystyle A}$，${\displaystyle B}$，${\displaystyle C}$。从角${\displaystyle C}$向${\displaystyle c}$边做垂线，得到一个长度为h的垂线和两个直角三角形。

显然：

${\displaystyle \sin A={\frac {h}{b}}}$

且

${\displaystyle \;\sin B={\frac {h}{a}}}$

故：

${\displaystyle h=b\,\sin A=a\,\sin B}$

且

${\displaystyle {\frac {\sin A}{a}}={\frac {\sin B}{b}}}$

同理可證：

${\displaystyle {\frac {\sin B}{b}}={\frac {\sin C}{c}}}$

作${\displaystyle \triangle ABC}$的外接圆，设半径为${\displaystyle R}$，${\displaystyle BC=a}$

由于${\displaystyle \angle A}$与${\displaystyle \angle D}$所对的弧都为${\displaystyle BC}$，根据圆周角定理可瞭解到

${\displaystyle \angle {\rm {A=\angle D}}}$

由于${\displaystyle BD}$为外接圆直径，

${\displaystyle {\rm {BD}}=2R,\ \angle {\rm {BCD}}={\pi \over 2}rad}$

所以

${\displaystyle \sin \angle D={a \over 2R}}$

${\displaystyle \sin \angle A={a \over 2R}}$

${\displaystyle {a \over \sin \angle A}=2R}$

因为${\displaystyle BC=a=2R}$，可以得到

${\displaystyle \sin \angle A=\sin {\pi \over 2}=1}$

所以可以证明

${\displaystyle {a \over \sin \angle A}=2R}$

线段${\displaystyle BD}$是圆的直径 根据圆内接四边形对角互补的性质

${\displaystyle \angle {\rm {D={\pi }-\angle BAC}}}$

所以

${\displaystyle \qquad \sin \angle BAC=\sin \angle D}$

因为${\displaystyle BD}$为外接圆的直径${\displaystyle BD=2R}$。根据正弦定义

${\displaystyle {\sin \angle BAC}={\sin \angle D}={a \over 2R}}$

变形可得

${\displaystyle {a \over \sin \angle BAC}=2R}$

根据以上的证明方法可以证明得到得到三角形的一条边与其对角的正弦值的比等于外接圆的直径，即

${\displaystyle {\frac {a}{\sin \angle A}}={\frac {b}{\sin \angle B}}={\frac {c}{\sin \angle C}}=2R}$

若三面角的三个面角分别为${\displaystyle \alpha }$、${\displaystyle \beta }$、${\displaystyle \gamma }$，它们所对的二面角分别为${\displaystyle A}$、${\displaystyle B}$、${\displaystyle C}$，则

${\displaystyle {\frac {\sin \alpha }{\sin A}}={\frac {\sin \beta }{\sin B}}={\frac {\sin \gamma }{\sin C}}}$[1]

${\displaystyle {\frac {OA}{\sin \angle OBA}}={\frac {OB}{\sin \angle OAB}},{\frac {OB}{\sin \angle OCB}}={\frac {OC}{\sin \angle OBC}},{\frac {OC}{\sin \angle ODC}}={\frac {OD}{\sin \angle OCD}},{\frac {OD}{\sin \angle OED}}={\frac {OE}{\sin \angle ODE}},{\frac {OE}{\sin \angle OAE}}={\frac {OA}{\sin \angle OEA}}}$

${\displaystyle {\frac {\sin \angle OAB\sin \angle OBC\sin \angle OCD\sin \angle ODE\sin \angle OEA}{\sin \angle OBA\sin \angle OCB\sin \angle ODC\sin \angle OED\sin \angle OAE}}={\frac {OB\cdot OC\cdot OD\cdot OE\cdot OA}{OA\cdot OB\cdot OC\cdot OD\cdot OE}}=1}$

${\displaystyle \sin \angle OAB\sin \angle OBC\sin \angle OCD\sin \angle ODE\sin \angle OEA=\sin \angle OBA\sin \angle OCB\sin \angle ODC\sin \angle OED\sin \angle OAE}$