91影视

We have been given that $伪+尾+纬=\mathrm{蟺}$.

We have to show that $\sin \left(2伪\right)+\sin \left(2尾\right)+\sin \left(2纬\right)=4\sin 伪\sin 尾\sin 纬$.

$\begin{array}{r}\sin 鈦�\left(2伪\right)+\sin 鈦�\left(2尾\right)+\sin 鈦�\left(2纬\right)=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬\\ 2\sin 鈦�\left(\frac{2伪+2尾}{2}\right)\cos 鈦�\left(\frac{2伪鈭�2尾}{2}\right)+\sin 鈦�\left(2纬\right)=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬\\ 2\sin 鈦�\left(\frac{2(伪+尾)}{2}\right)\cos 鈦�\left(\frac{2(伪鈭�尾)}{2}\right)+\sin 鈦�\left(2纬\right)=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬\\ 2\sin 鈦�\left(\frac{(谈伪+尾)}{2}\right)\cos 鈦�\left(\frac{(谈伪鈭�尾)}{\text{猝�}2}\right)+\sin 鈦�\left(2纬\right)=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬\\ 2\sin 鈦�(伪+尾)\cos 鈦�(伪鈭�尾)+\sin 鈦�\left(2纬\right)=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬\end{array}$

$$\begin{gathered} \begin{array}{r}2\sin 鈦�(伪+尾)\cos 鈦�(伪鈭�尾)+\sin 鈦�\left(2纬\right)=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬\\ 2\sin 鈦�(蟺鈭�纬)\cos 鈦�(伪鈭�尾)+\sin 鈦�\left(2纬\right)=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬\end{array} \\ 2\sin 鈦�(蟺鈭�纬)\cos 鈦�(伪鈭�尾)+\sin 鈦�\left(2纬\right)=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬 \end{gathered}$$

$$\begin{gathered} 2\sin 鈦�纬\cos 鈦�(伪鈭�尾)+\sin 鈦�\left(2纬\right)=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬 \\ 2\sin 鈦�纬\cos 鈦�(伪鈭�尾)+2\sin 纬\cos 纬=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬 \\ 2\sin 鈦�纬[\cos 鈦�(伪鈭�尾)+cos纬=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬 \\ \begin{array}{r}2\sin 鈦�纬[\cos 鈦�(伪鈭�尾)+\cos 鈦�纬]=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬\\ 2\sin 鈦�纬[\cos 鈦�(伪鈭�尾)+\cos 鈦�(蟺鈭�伪鈭�尾\left)\right]=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬\end{array} \end{gathered}$$

$\begin{array}{r}2\sin 鈦�纬[\cos 鈦�(伪鈭�尾)+\cos 鈦�(蟺鈭�伪鈭�尾\left)\right]=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬\\ 2\sin 鈦�纬[\cos 鈦�(伪鈭�尾)+\cos 鈦�(蟺鈭�(伪+尾)\left)\right]=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬\\ 2\sin 鈦�纬[\cos 鈦�(伪鈭�尾)鈭�\cos 鈦�(伪+尾\left)\right]=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬\end{array}$

$2\sin 鈦�纬[\cos 鈦�(伪鈭�尾)鈭�\cos 鈦�(伪+尾\left)\right]=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬$

$$\begin{gathered} 2\sin 鈦�纬[\cos 鈦�(伪鈭�尾)鈭�\cos 鈦�(伪+尾\left)\right]=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬 \\ 2\sin 鈦�纬[-2\sin \left(\frac{伪-尾+伪+尾}{2}\right)\sin \left(\frac{伪-尾-伪-尾}{2}\right)]=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬 \\ 2\sin 鈦�纬[-2\sin \left(\frac{2伪}{2}\right)\sin \left(\frac{-2尾}{2}\right)]=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬 \\ 2\sin 鈦�纬[-2\sin \left(伪\right)\sin \left(-尾\right)]=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬 \end{gathered}$$

Use $\sin (-A)=-\sin A$

$$\begin{gathered} 2\sin 鈦�纬[鈭�2\sin 鈦�伪\sin 鈦�(鈭�尾\left)\right]=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬 \\ 2\sin 鈦�纬[2\sin 鈦�伪\sin 鈦�尾]=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬 \\ 4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬=4\sin 鈦�伪\sin 鈦�尾\sin 鈦�纬 \end{gathered}$$