Упростите выражение: 1) cos(α + π/6) - cos(α - π/6)...

189. Упростите выражение: 1) $\cos(\alpha + \frac{\pi}{6}) - \cos(\alpha - \frac{\pi}{6}) = -2\sin \alpha \sin \frac{\pi}{6} = -2\sin \alpha \cdot \frac{1}{2} = -\sin \alpha$ 2) $2\sin(\frac{\pi}{3} - \alpha) - \sqrt{3}\cos \alpha - \sin \alpha = 2(\sin \frac{\pi}{3}\cos \alpha - \cos \frac{\pi}{3}\sin \alpha) - \sqrt{3}\cos \alpha - \sin \alpha = 2(\frac{\sqrt{3}}{2}\cos \alpha - \frac{1}{2}\sin \alpha) - \sqrt{3}\cos \alpha - \sin \alpha = \sqrt{3}\cos \alpha - \sin \alpha - \sqrt{3}\cos \alpha - \sin \alpha = -2\sin \alpha$ 3) $\frac{\sin(45^\circ + \alpha) - \cos(45^\circ + \alpha)}{\sin(45^\circ + \alpha) + \cos(45^\circ + \alpha)} = \frac{\sin(45^\circ + \alpha) - \sin(45^\circ - \alpha)}{\sin(45^\circ + \alpha) + \sin(45^\circ - \alpha)} = \frac{2\sin \alpha \cos 45^\circ}{2\sin 45^\circ \cos \alpha} = \text{tg} \alpha$ 190. Упростите выражение: 1) $\cos 6\alpha \cos 4\alpha - \sin 6\alpha \sin 4\alpha = \cos(6\alpha + 4\alpha) = \cos 10\alpha$ 2) $\sin 14^\circ \cos 31^\circ + \cos 14^\circ \sin 31^\circ = \sin(14^\circ + 31^\circ) = \sin 45^\circ = \frac{\sqrt{2}}{2}$ 3) $\cos(24^\circ + \alpha)\cos(24^\circ - \alpha) + \sin(24^\circ + \alpha)\sin(24^\circ - \alpha) = \cos((24^\circ + \alpha) - (24^\circ - \alpha)) = \cos(2\alpha)$ 4) $\frac{\sin 21^\circ \cos 28^\circ + \cos 21^\circ \sin 28^\circ}{\cos 18^\circ \cos 31^\circ - \sin 18^\circ \sin 31^\circ} = \frac{\sin(21^\circ + 28^\circ)}{\cos(18^\circ + 31^\circ)} = \frac{\sin 49^\circ}{\cos 49^\circ} = \text{tg} 49^\circ$ 5) $\frac{\text{tg} 2^\circ - \text{tg} 47^\circ}{1 + \text{tg} 2^\circ \text{tg} 47^\circ} = \text{tg}(2^\circ - 47^\circ) = \text{tg}(-45^\circ) = -1$ 6) $\frac{\text{tg}(\frac{\pi}{6} + \alpha) + \text{tg}(\frac{\pi}{6} - \alpha)}{1 - \text{tg}(\frac{\pi}{6} + \alpha)\text{tg}(\frac{\pi}{6} - \alpha)} = \text{tg}((\frac{\pi}{6} + \alpha) + (\frac{\pi}{6} - \alpha)) = \text{tg}\frac{\pi}{3} = \sqrt{3}$ 191. Докажите тождество: 1) $\text{ctg} \alpha - \text{tg} \beta = \frac{\cos \alpha}{\sin \alpha} - \frac{\sin \beta}{\cos \beta} = \frac{\cos \alpha \cos \beta - \sin \alpha \sin \beta}{\sin \alpha \cos \beta} = \frac{\cos(\alpha + \beta)}{\sin \alpha \cos \beta}$. Тождество доказано. 2) $\frac{\sin(\alpha + \beta) - 2\cos \alpha \sin \beta}{2\cos \alpha \cos \beta - \cos(\alpha + \beta)} = \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta - 2\cos \alpha \sin \beta}{2\cos \alpha \cos \beta - (\cos \alpha \cos \beta - \sin \alpha \sin \beta)} = \frac{\sin \alpha \cos \beta - \cos \alpha \sin \beta}{\cos \alpha \cos \beta + \sin \alpha \sin \beta} = \frac{\sin(\alpha - \beta)}{\cos(\alpha - \beta)} = \text{tg}(\alpha - \beta)$. Тождество доказано. 3) $\sin 2\alpha + \cos 2\alpha \text{ctg} \alpha = 2\sin \alpha \cos \alpha + \frac{\cos 2\alpha \cos \alpha}{\sin \alpha} = \frac{2\sin^2 \alpha \cos \alpha + \cos 2\alpha \cos \alpha}{\sin \alpha} = \frac{\cos \alpha (2\sin^2 \alpha + \cos 2\alpha)}{\sin \alpha} = \frac{\cos \alpha (2\sin^2 \alpha + 1 - 2\sin^2 \alpha)}{\sin \alpha} = \frac{\cos \alpha}{\sin \alpha} = \text{ctg} \alpha$. Тождество доказано. 192. Преобразуйте в произведение: 1) $\text{tg} 63^\circ - \text{tg} 18^\circ = \frac{\sin(63^\circ - 18^\circ)}{\cos 63^\circ \cos 18^\circ} = \frac{\sin 45^\circ}{\cos 63^\circ \cos 18^\circ} = \frac{\sqrt{2}}{2\cos 63^\circ \cos 18^\circ}$ 2) $\text{tg} 14\varphi + \text{ctg} 2\varphi = \text{tg} 14\varphi + \text{tg}(90^\circ - 2\varphi) = \frac{\sin(14\varphi + 90^\circ - 2\varphi)}{\cos 14\varphi \cos(90^\circ - 2\varphi)} = \frac{\sin(12\varphi + 90^\circ)}{\cos 14\varphi \sin 2\varphi} = \frac{\cos 12\varphi}{\cos 14\varphi \sin 2\varphi}$ 3) $\text{tg}(\frac{\pi}{6} - 2\alpha) + \text{tg}(\frac{\pi}{3} + 2\alpha) = \frac{\sin((\frac{\pi}{6} - 2\alpha) + (\frac{\pi}{3} + 2\alpha))}{\cos(\frac{\pi}{6} - 2\alpha) \cos(\frac{\pi}{3} + 2\alpha)} = \frac{\sin \frac{\pi}{2}}{\cos(\frac{\pi}{6} - 2\alpha) \cos(\frac{\pi}{3} + 2\alpha)} = \frac{1}{\cos(\frac{\pi}{6} - 2\alpha) \cos(\frac{\pi}{3} + 2\alpha)}$ 4) $\frac{\sqrt{3}}{3} + \text{tg} \alpha = \text{tg} 30^\circ + \text{tg} \alpha = \frac{\sin(30^\circ + \alpha)}{\cos 30^\circ \cos \alpha} = \frac{2\sin(30^\circ + \alpha)}{\sqrt{3}\cos \alpha}$ 193. Найдите $\text{ctg} 75^\circ$: $\text{ctg} 75^\circ = \text{tg} 15^\circ = \text{tg}(45^\circ - 30^\circ) = \frac{\text{tg} 45^\circ - \text{tg} 30^\circ}{1 + \text{tg} 45^\circ \text{tg} 30^\circ} = \frac{1 - \frac{\sqrt{3}}{3}}{1 + \frac{\sqrt{3}}{3}} = \frac{3 - \sqrt{3}}{3 + \sqrt{3}} = \frac{(3 - \sqrt{3})^2}{9 - 3} = \frac{9 + 3 - 6\sqrt{3}}{6} = \frac{12 - 6\sqrt{3}}{6} = 2 - \sqrt{3}$ 194. Дано: $\cos \alpha = -\frac{5}{13}, 90^\circ < \alpha < 180^\circ$. Найдите $\cos(\alpha + 45^\circ)$: Так как $\alpha$ во II четверти, $\sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - (\frac{25}{169})} = \frac{12}{13}$. $\cos(\alpha + 45^\circ) = \cos \alpha \cos 45^\circ - \sin \alpha \sin 45^\circ = -\frac{5}{13} \cdot \frac{\sqrt{2}}{2} - \frac{12}{13} \cdot \frac{\sqrt{2}}{2} = -\frac{17\sqrt{2}}{26}$ 195. Дано: $\cos \alpha = 0,8, \cos \beta = -\frac{12}{13}, 270^\circ < \alpha < 360^\circ, 180^\circ < \beta < 270^\circ$. Найдите $\sin(\alpha - \beta)$: $\sin \alpha = -\sqrt{1 - 0,8^2} = -0,6$ (IV четверть). $\sin \beta = -\sqrt{1 - (-\frac{12}{13})^2} = -\frac{5}{13}$ (III четверть). $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta = (-0,6) \cdot (-\frac{12}{13}) - 0,8 \cdot (-\frac{5}{13}) = \frac{7,2}{13} + \frac{4}{13} = \frac{11,2}{13} = \frac{112}{130} = \frac{56}{65}$ 196. Найдите наименьшее значение выражения: 1) $\cos \alpha - \sin \alpha = \sqrt{2}(\frac{1}{\sqrt{2}}\cos \alpha - \frac{1}{\sqrt{2}}\sin \alpha) = \sqrt{2}\cos(\alpha + 45^\circ)$. Наименьшее значение: $-\sqrt{2}$. 2) $8\cos \alpha - 15\sin \alpha = 17(\frac{8}{17}\cos \alpha - \frac{15}{17}\sin \alpha) = 17\cos(\alpha + \varphi)$, где $\cos \varphi = \frac{8}{17}$. Наименьшее значение: $-17$.