Выполните задания 189-195 по теме тригонометрических преобразований.

189. Упростите выражение: 1) $\sin\left(\alpha - \frac{\pi}{3}\right) - \sin\left(\alpha + \frac{\pi}{3}\right) = \sin\alpha\cos\frac{\pi}{3} - \cos\alpha\sin\frac{\pi}{3} - \left(\sin\alpha\cos\frac{\pi}{3} + \cos\alpha\sin\frac{\pi}{3}\right) = -2\cos\alpha\sin\frac{\pi}{3} = -2\cos\alpha \cdot \frac{\sqrt{3}}{2} = -\sqrt{3}\cos\alpha$ 2) $2\cos\left(\frac{\pi}{3} + \alpha\right) + \sqrt{3}\sin\alpha + \cos\alpha = 2\left(\cos\frac{\pi}{3}\cos\alpha - \sin\frac{\pi}{3}\sin\alpha\right) + \sqrt{3}\sin\alpha + \cos\alpha = 2\left(\frac{1}{2}\cos\alpha - \frac{\sqrt{3}}{2}\sin\alpha\right) + \sqrt{3}\sin\alpha + \cos\alpha = \cos\alpha - \sqrt{3}\sin\alpha + \sqrt{3}\sin\alpha + \cos\alpha = 2\cos\alpha$ 3) $\frac{\sin(30^\circ + \alpha) - \cos(60^\circ + \alpha)}{\sin(30^\circ + \alpha) + \cos(60^\circ + \alpha)} = \frac{\sin(30^\circ + \alpha) - \sin(30^\circ - \alpha)}{\sin(30^\circ + \alpha) + \sin(30^\circ - \alpha)} = \frac{2\cos30^\circ\sin\alpha}{2\sin30^\circ\cos\alpha} = \text{ctg}30^\circ\text{tg}\alpha = \sqrt{3}\text{tg}\alpha$ 190. Упростите выражение: 1) $\cos2\beta\cos5\beta + \sin2\beta\sin5\beta = \cos(5\beta - 2\beta) = \cos3\beta$ 2) $\sin53^\circ\cos7^\circ + \cos53^\circ\sin7^\circ = \sin(53^\circ + 7^\circ) = \sin60^\circ = \frac{\sqrt{3}}{2}$ 3) $\cos(4^\circ + \alpha)\sin(\alpha - 41^\circ) + \cos(\alpha - 41^\circ)\sin(4^\circ + \alpha) = \sin((4^\circ + \alpha) + (\alpha - 41^\circ)) = \sin(2\alpha - 37^\circ)$ 4) $\frac{\cos63^\circ\cos22^\circ + \sin63^\circ\sin22^\circ}{\sin16^\circ\cos25^\circ + \cos16^\circ\sin25^\circ} = \frac{\cos(63^\circ - 22^\circ)}{\sin(16^\circ + 25^\circ)} = \frac{\cos41^\circ}{\sin41^\circ} = \text{ctg}41^\circ$ 5) $\frac{\text{tg}47^\circ - \text{tg}17^\circ}{1 + \text{tg}47^\circ\text{tg}17^\circ} = \text{tg}(47^\circ - 17^\circ) = \text{tg}30^\circ = \frac{\sqrt{3}}{3}$ 6) $\frac{\text{tg}\left(\frac{\pi}{8} + \alpha\right) + \text{tg}\left(\frac{\pi}{8} - \alpha\right)}{1 - \text{tg}\left(\frac{\pi}{8} + \alpha\right)\text{tg}\left(\frac{\pi}{8} - \alpha\right)} = \text{tg}\left(\left(\frac{\pi}{8} + \alpha\right) + \left(\frac{\pi}{8} - \alpha\right)\right) = \text{tg}\frac{\pi}{4} = 1$ 191. Докажите тождество: 1) $\text{tg}\alpha + \text{tg}\beta = \frac{\sin\alpha}{\cos\alpha} + \frac{\sin\beta}{\cos\beta} = \frac{\sin\alpha\cos\beta + \cos\alpha\sin\beta}{\cos\alpha\cos\beta} = \frac{\sin(\alpha + \beta)}{\cos\alpha\cos\beta}$. Тождество доказано. 2) $\frac{\cos(\alpha + \beta) + 2\sin\alpha\sin\beta}{2\sin\alpha\cos\beta - \sin(\alpha + \beta)} = \frac{\cos\alpha\cos\beta - \sin\alpha\sin\beta + 2\sin\alpha\sin\beta}{2\sin\alpha\cos\beta - (\sin\alpha\cos\beta + \cos\alpha\sin\beta)} = \frac{\cos\alpha\cos\beta + \sin\alpha\sin\beta}{\sin\alpha\cos\beta - \cos\alpha\sin\beta} = \frac{\cos(\alpha - \beta)}{\sin(\alpha - \beta)} = \text{ctg}(\alpha - \beta)$. Тождество доказано. 3) $\sin6\alpha\text{ctg}3\alpha - \cos6\alpha = 2\sin3\alpha\cos3\alpha \cdot \frac{\cos3\alpha}{\sin3\alpha} - \cos6\alpha = 2\cos^2 3\alpha - \cos6\alpha = (1 + \cos6\alpha) - \cos6\alpha = 1$. Тождество доказано. 192. Преобразуйте в произведение: 1) $\text{tg}14^\circ + \text{tg}16^\circ = \frac{\sin(14^\circ + 16^\circ)}{\cos14^\circ\cos16^\circ} = \frac{\sin30^\circ}{\cos14^\circ\cos16^\circ} = \frac{1}{2\cos14^\circ\cos16^\circ}$ 2) $\text{ctg}7\alpha - \text{tg}3\alpha = \frac{\cos7\alpha}{\sin7\alpha} - \frac{\sin3\alpha}{\cos3\alpha} = \frac{\cos7\alpha\cos3\alpha - \sin7\alpha\sin3\alpha}{\sin7\alpha\cos3\alpha} = \frac{\cos(7\alpha + 3\alpha)}{\sin7\alpha\cos3\alpha} = \frac{\cos10\alpha}{\sin7\alpha\cos3\alpha}$ 3) $\text{tg}\left(\frac{\pi}{4} + \frac{\alpha}{2}\right) - \text{tg}\left(\frac{\pi}{4} - \frac{\alpha}{2}\right) = \frac{\sin\left(\left(\frac{\pi}{4} + \frac{\alpha}{2}\right) - \left(\frac{\pi}{4} - \frac{\alpha}{2}\right)\right)}{\cos\left(\frac{\pi}{4} + \frac{\alpha}{2}\right)\cos\left(\frac{\pi}{4} - \frac{\alpha}{2}\right)} = \frac{\sin\alpha}{\frac{1}{2}(\cos\alpha + \cos\frac{\pi}{2})} = \frac{2\sin\alpha}{\cos\alpha} = 2\text{tg}\alpha$ 4) $\sqrt{3} - \text{tg}\alpha = \text{tg}60^\circ - \text{tg}\alpha = \frac{\sin(60^\circ - \alpha)}{\cos60^\circ\cos\alpha} = \frac{\sin(60^\circ - \alpha)}{\frac{1}{2}\cos\alpha} = \frac{2\sin(60^\circ - \alpha)}{\cos\alpha}$ 193. Найдите $\text{tg}105^\circ$: $\text{tg}105^\circ = \text{tg}(60^\circ + 45^\circ) = \frac{\text{tg}60^\circ + \text{tg}45^\circ}{1 - \text{tg}60^\circ\text{tg}45^\circ} = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} = \frac{(\sqrt{3} + 1)(1 + \sqrt{3})}{1 - 3} = \frac{3 + 2\sqrt{3} + 1}{-2} = \frac{4 + 2\sqrt{3}}{-2} = -2 - \sqrt{3}$ 194. Дано: $\sin\alpha = \frac{4}{5}$, $90^\circ < \alpha < 180^\circ$. Найдите $\sin(30^\circ + \alpha)$: Так как $\alpha$ во II четверти, $\cos\alpha = -\sqrt{1 - \sin^2\alpha} = -\sqrt{1 - \frac{16}{25}} = -\frac{3}{5}$. $\sin(30^\circ + \alpha) = \sin30^\circ\cos\alpha + \cos30^\circ\sin\alpha = \frac{1}{2} \cdot \left(-\frac{3}{5}\right) + \frac{\sqrt{3}}{2} \cdot \frac{4}{5} = \frac{4\sqrt{3} - 3}{10} = 0,4\sqrt{3} - 0,3$ 195. Дано: $\sin\alpha = 0,8$, $\cos\beta = -\frac{5}{13}$, $0^\circ < \alpha < 90^\circ$, $180^\circ < \beta < 270^\circ$. Найдите $\cos(\alpha + \beta)$: $\alpha$ в I четверти: $\cos\alpha = \sqrt{1 - 0,8^2} = 0,6$. $\beta$ в III четверти: $\sin\beta = -\sqrt{1 - \left(-\frac{5}{13}\right)^2} = -\sqrt{1 - \frac{25}{169}} = -\frac{12}{13}$. $\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta = 0,6 \cdot \left(-\frac{5}{13}\right) - 0,8 \cdot \left(-\frac{12}{13}\right) = -\frac{3}{13} + \frac{9,6}{13} = \frac{6,6}{13} = \frac{66}{130} = \frac{33}{65}$