Задание 4. Найдите значение выражения: 1) 12sin60°cos150°...

1) **Ответ: −9** $12 \cdot \sin 60^{\circ} \cdot \cos 150^{\circ} = 12 \cdot \frac{\sqrt{3}}{2} \cdot \left(-\frac{\sqrt{3}}{2}\right) = 12 \cdot \left(-\frac{3}{4}\right) = -9$ 2) **Ответ: −24** $24\sqrt{2} \cos(-135^{\circ}) = 24\sqrt{2} \cos 135^{\circ} = 24\sqrt{2} \cdot \left(-\frac{\sqrt{2}}{2}\right) = -24 \cdot \frac{2}{2} = -24$ 3) **Ответ: −14,5** Используем формулу косинуса двойного угла: $\cos 2\alpha = \cos^2\alpha - \sin^2\alpha$. $\frac{29(\sin^2 81^{\circ} - \cos^2 81^{\circ})}{2\cos 162^{\circ}} = \frac{-29(\cos^2 81^{\circ} - \sin^2 81^{\circ})}{2\cos 162^{\circ}} = \frac{-29\cos 162^{\circ}}{2\cos 162^{\circ}} = -\frac{29}{2} = -14,5$ 4) **Ответ: 19** По формуле приведения $\sin 56^{\circ} = \cos(90^{\circ} - 56^{\circ}) = \cos 34^{\circ}$. $\frac{14\cos 34^{\circ}}{\cos 34^{\circ}} + 5 = 14 + 5 = 19$ 5) **Ответ: −13** $\sin 509^{\circ} = \sin(509^{\circ} - 360^{\circ}) = \sin 149^{\circ}$. $\frac{-13\sin 149^{\circ}}{\sin 149^{\circ}} = -13$ 6) **Ответ: −24** $\operatorname{tg} 75^{\circ} = \operatorname{ctg}(90^{\circ} - 75^{\circ}) = \operatorname{ctg} 15^{\circ}$. $-46 \cdot \operatorname{tg} 15^{\circ} \cdot \operatorname{ctg} 15^{\circ} + 22 = -46 \cdot 1 + 22 = -24$ 7) **Ответ: 3** $\cos^2 172^{\circ} = (-\cos 8^{\circ})^2 = \cos^2 8^{\circ} = \sin^2(90^{\circ} - 8^{\circ}) = \sin^2 82^{\circ}$. $\frac{18}{\cos^2 82^{\circ} + 5 + \sin^2 82^{\circ}} = \frac{18}{1 + 5} = \frac{18}{6} = 3$ 8) **Ответ: 3,5** $\cos^2 110^{\circ} = (-\cos 70^{\circ})^2 = \cos^2 70^{\circ} = \sin^2(90^{\circ} - 70^{\circ}) = \sin^2 20^{\circ}$. $\frac{28}{\cos^2 20^{\circ} + 7 + \sin^2 20^{\circ}} = \frac{28}{1 + 7} = \frac{28}{8} = 3,5$