Ответ:

решение смотри на фотографии

Verified answer

[tex]\sin40^{\circ}+\sin100^{\circ}+\sin220^{\circ}+\sin160^{\circ}=\underbrace{(\sin40^{\circ}+\sin220^{\circ})}_{\sin\alpha+\sin\beta =2\sin\frac{\alpha +\beta }{2} \cdot\cos\frac{\alpha -\beta }{2} }+\underbrace{(\sin100^{\circ}+\sin160^{\circ})}_{\sin\alpha+\sin\beta =2\sin\frac{\alpha +\beta }{2} \cdot\cos\frac{\alpha -\beta }{2}}= \\ \\2\sin\frac{40^{\circ}+220^{\circ}}{2} \cdot\cos\frac{40^{\circ}-220^{\circ}}{2} +2\sin\frac{100^{\circ}+160^{\circ}}{2}\cdot\cos\frac{100^{\circ}-160^{\circ}}{2} = \\ = 2\sin130^{\circ}\cdot\cos90^{\circ}+2\sin130^{\circ}\cdot\cos30^{\circ}= \\ = 2\sin130^{\circ}\cdot0+\not2\sin130^{\circ}\cdot\frac{\sqrt{3} }{\not2} =0+\sin130^{\circ}\cdot\sqrt{3} =\bf{\sqrt{3}\cdot \sin130^{\circ}}[/tex]