Упростите выражение 189, 190; Докажите тождество 191; Преобразуйте в произведение 192; Найдите tg 105; Найдите sin(30+a) и cos(a+b).

Для решения этих заданий используются формулы сложения, приведения и тригонометрические тождества. **189. Упростите выражение:** 1) $\sin(\alpha - \frac{\pi}{3}) - \sin(\alpha + \frac{\pi}{3}) = \sin \alpha \cos \frac{\pi}{3} - \cos \alpha \sin \frac{\pi}{3} - (\sin \alpha \cos \frac{\pi}{3} + \cos \alpha \sin \frac{\pi}{3}) = -2 \cos \alpha \sin \frac{\pi}{3} = -2 \cos \alpha \cdot \frac{\sqrt{3}}{2} = -\sqrt{3} \cos \alpha$ 2) $2 \cos(\frac{\pi}{3} + \alpha) + \sqrt{3} \sin \alpha + \cos \alpha = 2(\cos \frac{\pi}{3} \cos \alpha - \sin \frac{\pi}{3} \sin \alpha) + \sqrt{3} \sin \alpha + \cos \alpha = 2(\frac{1}{2} \cos \alpha - \frac{\sqrt{3}}{2} \sin \alpha) + \sqrt{3} \sin \alpha + \cos \alpha = \cos \alpha - \sqrt{3} \sin \alpha + \sqrt{3} \sin \alpha + \cos \alpha = 2 \cos \alpha$ 3) $\frac{\sin(30^\circ + \alpha) - \cos(60^\circ + \alpha)}{\sin(30^\circ + \alpha) + \cos(60^\circ + \alpha)} = \frac{\sin(30^\circ + \alpha) - \sin(30^\circ - \alpha)}{\sin(30^\circ + \alpha) + \sin(30^\circ - \alpha)} = \frac{2 \sin \alpha \cos 30^\circ}{2 \sin 30^\circ \cos \alpha} = \text{tg} \alpha \cdot \text{ctg} 30^\circ = \sqrt{3} \text{tg} \alpha$ **190. Упростите выражение:** 1) $\cos 2\beta \cos 5\beta + \sin 2\beta \sin 5\beta = \cos(5\beta - 2\beta) = \cos 3\beta$ 2) $\sin 53^\circ \cos 7^\circ + \cos 53^\circ \sin 7^\circ = \sin(53^\circ + 7^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$ 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) $\frac{\cos 63^\circ \cos 22^\circ + \sin 63^\circ \sin 22^\circ}{\sin 16^\circ \cos 25^\circ + \cos 16^\circ \sin 25^\circ} = \frac{\cos(63^\circ - 22^\circ)}{\sin(16^\circ + 25^\circ)} = \frac{\cos 41^\circ}{\sin 41^\circ} = \text{ctg} 41^\circ$ 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) $\frac{\text{tg}(\frac{\pi}{8} + \alpha) + \text{tg}(\frac{\pi}{8} - \alpha)}{1 - \text{tg}(\frac{\pi}{8} + \alpha) \text{tg}(\frac{\pi}{8} - \alpha)} = \text{tg}((\frac{\pi}{8} + \alpha) + (\frac{\pi}{8} - \alpha)) = \text{tg}(\frac{2\pi}{8}) = \text{tg} \frac{\pi}{4} = 1$ **191. Докажите тождество:** 1) $\text{tg} \alpha + \text{tg} \beta = \frac{\sin \alpha}{\cos \alpha} + \frac{\sin \beta}{\cos \beta} = \frac{\sin \alpha \cos \beta + \cos \alpha \sin \beta}{\cos \alpha \cos \beta} = \frac{\sin(\alpha + \beta)}{\cos \alpha \cos \beta}$. *Ч.т.д.* 2) $\frac{\cos(\alpha + \beta) + 2 \sin \alpha \sin \beta}{2 \sin \alpha \cos \beta - \sin(\alpha + \beta)} = \frac{\cos \alpha \cos \beta - \sin \alpha \sin \beta + 2 \sin \alpha \sin \beta}{2 \sin \alpha \cos \beta - (\sin \alpha \cos \beta + \cos \alpha \sin \beta)} = \frac{\cos \alpha \cos \beta + \sin \alpha \sin \beta}{\sin \alpha \cos \beta - \cos \alpha \sin \beta} = \frac{\cos(\alpha - \beta)}{\sin(\alpha - \beta)} = \text{ctg}(\alpha - \beta)$. *Ч.т.д.* 3) $\sin 6\alpha \text{ctg} 3\alpha - \cos 6\alpha = 2 \sin 3\alpha \cos 3\alpha \cdot \frac{\cos 3\alpha}{\sin 3\alpha} - \cos 6\alpha = 2 \cos^2 3\alpha - \cos 6\alpha = (1 + \cos 6\alpha) - \cos 6\alpha = 1$. *Ч.т.д.* **192. Преобразуйте в произведение:** 1) $\text{tg} 14^\circ + \text{tg} 16^\circ = \frac{\sin(14^\circ + 16^\circ)}{\cos 14^\circ \cos 16^\circ} = \frac{\sin 30^\circ}{\cos 14^\circ \cos 16^\circ} = \frac{1}{2 \cos 14^\circ \cos 16^\circ}$ 2) $\text{ctg} 7\alpha - \text{tg} 3\alpha = \frac{\cos 7\alpha}{\sin 7\alpha} - \frac{\sin 3\alpha}{\cos 3\alpha} = \frac{\cos 7\alpha \cos 3\alpha - \sin 7\alpha \sin 3\alpha}{\sin 7\alpha \cos 3\alpha} = \frac{\cos(7\alpha + 3\alpha)}{\sin 7\alpha \cos 3\alpha} = \frac{\cos 10\alpha}{\sin 7\alpha \cos 3\alpha}$ 3) $\text{tg}(\frac{\pi}{4} + \frac{\alpha}{2}) - \text{tg}(\frac{\pi}{4} - \frac{\alpha}{2}) = \frac{\sin((\frac{\pi}{4} + \frac{\alpha}{2}) - (\frac{\pi}{4} - \frac{\alpha}{2}))}{\cos(\frac{\pi}{4} + \frac{\alpha}{2}) \cos(\frac{\pi}{4} - \frac{\alpha}{2})} = \frac{\sin \alpha}{\cos(\frac{\pi}{4} + \frac{\alpha}{2}) \cos(\frac{\pi}{4} - \frac{\alpha}{2})} = \frac{\sin \alpha}{\frac{1}{2}(\cos \frac{\pi}{2} + \cos \alpha)} = \frac{2 \sin \alpha}{\cos \alpha} = 2 \text{tg} \alpha$ 4) $\sqrt{3} - \text{tg} \alpha = \text{tg} 60^\circ - \text{tg} \alpha = \frac{\sin(60^\circ - \alpha)}{\cos 60^\circ \cos \alpha} = \frac{\sin(60^\circ - \alpha)}{0.5 \cos \alpha} = \frac{2 \sin(60^\circ - \alpha)}{\cos \alpha}$ **193. Найдите $\text{tg} 105^\circ$:** $\text{tg} 105^\circ = \text{tg}(60^\circ + 45^\circ) = \frac{\text{tg} 60^\circ + \text{tg} 45^\circ}{1 - \text{tg} 60^\circ \text{tg} 45^\circ} = \frac{\sqrt{3} + 1}{1 - \sqrt{3} \cdot 1} = \frac{(\sqrt{3} + 1)^2}{(1 - \sqrt{3})(1 + \sqrt{3})} = \frac{3 + 2\sqrt{3} + 1}{1 - 3} = \frac{4 + 2\sqrt{3}}{-2} = -2 - \sqrt{3}$ **194. Дано: $\sin \alpha = \frac{4}{5}, 90^\circ < \alpha < 180^\circ$. Найдите $\sin(30^\circ + \alpha)$:** Так как $\alpha$ во II четверти, $\cos \alpha < 0$: $\cos \alpha = -\sqrt{1 - (4/5)^2} = -\frac{3}{5}$. $\sin(30^\circ + \alpha) = \sin 30^\circ \cos \alpha + \cos 30^\circ \sin \alpha = \frac{1}{2} \cdot (-\frac{3}{5}) + \frac{\sqrt{3}}{2} \cdot \frac{4}{5} = \frac{-3 + 4\sqrt{3}}{10} = 0,4\sqrt{3} - 0,3$ **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, $\cos \alpha > 0$: $\cos \alpha = \sqrt{1 - 0,8^2} = 0,6$. $\beta$ в III четверти, $\sin \beta < 0$: $\sin \beta = -\sqrt{1 - (-5/13)^2} = -\frac{12}{13}$. $\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta = 0,6 \cdot (-\frac{5}{13}) - 0,8 \cdot (-\frac{12}{13}) = -\frac{3}{13} + \frac{9,6}{13} = \frac{6,6}{13} = \frac{33}{65}$