Problem 929

## Solution.

$\left(\sin 20^{\circ} \sin 40^{\circ}\right)\left(\sin 60^{\circ} \sin 80^{\circ}\right)=$

$$=\left[\sin x \sin y=\frac{1}{2}(\cos (x-y)-\cos (x+y))\right]=$$

$$\begin{aligned} & =\frac{1}{2}\left(\cos 20^{\circ}-\cos 60^{\circ}\right) \cdot \frac{1}{2}\left(\cos 20^{\circ}-\cos 140^{\circ}\right)= \\ & =\frac{1}{4}\left(\cos 20^{\circ}-\frac{1}{2}\right)\left(\cos 20^{\circ}-\cos \left(180^{\circ}-40^{\circ}\right)\right)= \\ & =\frac{1}{8}\left(2 \cos 20^{\circ}-1\right)\left(\cos 20^{\circ}+\cos 40^{\circ}\right)= \\ & =\frac{1}{8}\left(2 \cos ^{2} 20^{\circ}+2 \cos 20^{\circ} \cos 40^{\circ}-\cos 20^{\circ}-\cos 40^{\circ}\right)= \\ & =\left[\cos x \cos y=\frac{1}{2}(\cos (x-y)+\cos (x+y))\right]= \end{aligned}$$

$$\begin{aligned} & =\frac{1}{8}\left(2 \cos ^{2} 20^{\circ}+\cos 20^{\circ}+\cos 60^{\circ}-\cos 20^{\circ}-\cos 40^{\circ}\right)= \\ & =\frac{1}{8}\left(2 \cos ^{2} 20^{\circ}+\frac{1}{2}-\cos 2\left(20^{\circ}\right)\right)=\frac{1}{8}\left(2 \cos ^{2} 20^{\circ}+\frac{1}{2}-2 \cos ^{2} 20^{\circ}+1\right)=\frac{3}{16} . \end{aligned}$$

The equality holds.