Задание 16. Найдите значение выражения: 1) (7cos80°)/(sin10°) - 3

**Ответ:** 1) 4 2) 59 3) 28 4) 15 5) -6 6) 9 7) -7 8) 14 **Решение:** Для решения этих примеров используются формулы приведения: $\sin(90^\circ - \alpha) = \cos \alpha$ и $\cos(90^\circ - \alpha) = \sin \alpha$, а также периодичность функций $\sin(\alpha + 360^\circ) = \sin \alpha$ и $\cos(\alpha + 360^\circ) = \cos \alpha$. 1) $\frac{7\cos 80^\circ}{\sin 10^\circ} - 3 = \frac{7\cos 80^\circ}{\cos(90^\circ - 10^\circ)} - 3 = \frac{7\cos 80^\circ}{\cos 80^\circ} - 3 = 7 - 3 = 4$ 2) $\frac{51\cos 4^\circ}{\sin 86^\circ} + 8 = \frac{51\cos 4^\circ}{\cos(90^\circ - 86^\circ)} + 8 = \frac{51\cos 4^\circ}{\cos 4^\circ} + 8 = 51 + 8 = 59$ 3) $\frac{35\cos 11^\circ}{\sin 79^\circ} - 7 = \frac{35\cos 11^\circ}{\cos(90^\circ - 79^\circ)} - 7 = \frac{35\cos 11^\circ}{\cos 11^\circ} - 7 = 35 - 7 = 28$ 4) $\frac{2\cos 53^\circ}{\sin 37^\circ} + 13 = \frac{2\cos 53^\circ}{\cos(90^\circ - 37^\circ)} + 13 = \frac{2\cos 53^\circ}{\cos 53^\circ} + 13 = 2 + 13 = 15$ 5) $\frac{-6\sin 374^\circ}{\sin 14^\circ} = \frac{-6\sin(360^\circ + 14^\circ)}{\sin 14^\circ} = \frac{-6\sin 14^\circ}{\sin 14^\circ} = -6$ 6) $\frac{9\sin 32^\circ}{\sin 392^\circ} = \frac{9\sin 32^\circ}{\sin(360^\circ + 32^\circ)} = \frac{9\sin 32^\circ}{\sin 32^\circ} = 9$ 7) $\frac{-7\cos 387^\circ}{\cos 27^\circ} = \frac{-7\cos(360^\circ + 27^\circ)}{\cos 27^\circ} = \frac{-7\cos 27^\circ}{\cos 27^\circ} = -7$ 8) $\frac{14\cos 101^\circ}{\cos 461^\circ} = \frac{14\cos 101^\circ}{\cos(360^\circ + 101^\circ)} = \frac{14\cos 101^\circ}{\cos 101^\circ} = 14$