820. Представив 75° как сумму 30° + 45°, вычислите: а) sin 75°; б) cos 75°.

1. \(\sin75^\circ=\sin(30^\circ+45^\circ)\).

   \(\sin75^\circ=\sin30^\circ\cos45^\circ+\cos30^\circ\sin45^\circ\)

   \(=\frac{1}{2}\cdot\frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}=\frac{\sqrt{2}+\sqrt{6}}{4}\).

2. \(\cos75^\circ=\cos(30^\circ+45^\circ)\).

   \(\cos75^\circ=\cos30^\circ\cos45^\circ-\sin30^\circ\sin45^\circ\)

   \(=\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}-\frac{1}{2}\cdot\frac{\sqrt{2}}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\).

Ответ: а) \(\frac{\sqrt{6}+\sqrt{2}}{4}\); б) \(\frac{\sqrt{6}-\sqrt{2}}{4}\).