Упростите выражение sin(α - π/3) - sin(α + π/3)

189. Упростите выражение: 1) $\sin\left(\alpha - \frac{\pi}{3}\right) - \sin\left(\alpha + \frac{\pi}{3}\right) = 2 \cos\alpha \sin\left(-\frac{\pi}{3}\right) = 2 \cos\alpha \cdot \left(-\frac{\sqrt{3}}{2}\right) = -\sqrt{3} \cos\alpha$ 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) $\frac{\sin(30^\circ + \alpha) - \cos(60^\circ + \alpha)}{\sin(30^\circ + \alpha) + \cos(60^\circ + \alpha)} = \frac{\sin(30^\circ + \alpha) - \sin(30^\circ - \alpha)}{\sin(30^\circ + \alpha) + \sin(30^\circ - \alpha)} = \frac{2\cos 30^\circ \sin\alpha}{2\sin 30^\circ \cos\alpha} = \text{ctg } 30^\circ \text{tg } \alpha = \sqrt{3} \text{tg } \alpha$ 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) $\cos(4^\circ + \alpha)\sin(\alpha - 41^\circ) - \cos(\alpha - 41^\circ)\sin(4^\circ + \alpha) = \sin((\alpha - 41^\circ) - (4^\circ + \alpha)) = \sin(-45^\circ) = -\frac{\sqrt{2}}{2}$ 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) $\frac{\text{tg}\left(\frac{\pi}{8} + \alpha\right) + \text{tg}\left(\frac{\pi}{8} - \alpha\right)}{1 - \text{tg}\left(\frac{\pi}{8} + \alpha\right) \text{tg}\left(\frac{\pi}{8} - \alpha\right)} = \text{tg}\left(\left(\frac{\pi}{8} + \alpha\right) + \left(\frac{\pi}{8} - \alpha\right)\right) = \text{tg}\frac{\pi}{4} = 1$ 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) $\sin 6\alpha \text{ ctg } 3\alpha - \cos 6\alpha = 2\sin 3\alpha \cos 3\alpha \cdot \frac{\cos 3\alpha}{\sin 3\alpha} - \cos 6\alpha = 2\cos^2 3\alpha - (2\cos^2 3\alpha - 1) = 1$. Тождество доказано. 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) $\text{ctg } 7\alpha - \text{tg } 3\alpha = \frac{\cos 7\alpha}{\sin 7\alpha} - \frac{\sin 3\alpha}{\cos 3\alpha} = \frac{\cos 7\alpha \cos 3\alpha - \sin 7\alpha \sin 3\alpha}{\sin 7\alpha \cos 3\alpha} = \frac{\cos 10\alpha}{\sin 7\alpha \cos 3\alpha}$ 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) $\sqrt{3} - \text{tg } \alpha = \text{tg } 60^\circ - \text{tg } \alpha = \frac{\sin(60^\circ - \alpha)}{\cos 60^\circ \cos \alpha} = \frac{\sin(60^\circ - \alpha)}{\frac{1}{2}\cos \alpha} = \frac{2\sin(60^\circ - \alpha)}{\cos \alpha}$ 193. Найдите $\text{tg } 105^\circ$: $\text{tg } 105^\circ = \text{tg}(60^\circ + 45^\circ) = \frac{\text{tg } 60^\circ + \text{tg } 45^\circ}{1 - \text{tg } 60^\circ \text{tg } 45^\circ} = \frac{\sqrt{3} + 1}{1 - \sqrt{3}} = \frac{(\sqrt{3} + 1)^2}{1 - 3} = \frac{3 + 2\sqrt{3} + 1}{-2} = \frac{4 + 2\sqrt{3}}{-2} = -2 - \sqrt{3}$ 194. Дано: $\sin\alpha = \frac{4}{5}$, $90^\circ < \alpha < 180^\circ$. Найдите $\sin(30^\circ + \alpha)$: Т.к. $\alpha$ во II четверти, $\cos\alpha = -\sqrt{1 - \sin^2 \alpha} = -\sqrt{1 - (4/5)^2} = -\frac{3}{5}$. $\sin(30^\circ + \alpha) = \sin 30^\circ \cos \alpha + \cos 30^\circ \sin \alpha = \frac{1}{2} \cdot \left(-\frac{3}{5}\right) + \frac{\sqrt{3}}{2} \cdot \frac{4}{5} = \frac{4\sqrt{3} - 3}{10} = 0,4\sqrt{3} - 0,3$ 195. Дано: $\sin\alpha = 0,8$, $\cos\beta = -\frac{5}{13}$, $0^\circ < \alpha < 90^\circ$, $180^\circ < \beta < 270^\circ$. Найдите $\cos(\alpha + \beta)$: $\cos\alpha = \sqrt{1 - 0,8^2} = 0,6$; $\sin\beta = -\sqrt{1 - (-5/13)^2} = -\frac{12}{13}$. $\cos(\alpha + \beta) = \cos\alpha \cos\beta - \sin\alpha \sin\beta = 0,6 \cdot \left(-\frac{5}{13}\right) - 0,8 \cdot \left(-\frac{12}{13}\right) = -\frac{3}{13} + \frac{9,6}{13} = \frac{6,6}{13} = \frac{33}{65}$