Question

Как эта решить?
2sin22°30`×cos7°30`

Answer 1

$2\sin 22^{\circ}30'\cdot \cos 7^{\circ}30'=\sin(22,5^{\circ}+7,5^{\circ})+\sin(22,5^{\circ}-7,5^{\circ})=$

$=\sin 30^{\circ}+\sin15^{\circ}=\frac{1}{2}+\sin(45^{\circ}-30^{\circ})=\frac{1}{2}+\sin 45^{\circ}\cdot \cos 30^{\circ}-\cos 45^{\circ}\cdot \sin 30^{\circ}=$

$\frac{1}{2}+\frac{\sqrt{2}}{2}\cdot \frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\cdot \frac{1}{2}=\frac{2+\sqrt{6}-\sqrt{2}}{4}$