Вычислите 10sin(7π/6) - 4cos(-5π/3)

Привет! Давай решим эти тригонометрические задания. 1. $10 \sin(\frac{7\pi}{6}) - 4 \cos(-\frac{5\pi}{3}) = 10 \cdot (-0,5) - 4 \cdot (0,5) = -5 - 2 = -7$. 2. $\cos \frac{7\pi}{3} - \sin \frac{17\pi}{6} = \cos(2\pi + \frac{\pi}{3}) - \sin(2\pi + \frac{5\pi}{6}) = \cos \frac{\pi}{3} - \sin \frac{5\pi}{6} = 0,5 - 0,5 = 0$. 3. $\sqrt{2} (\sin(-\frac{15\pi}{4}) + \cos(-\frac{17\pi}{4})) = \sqrt{2} (\sin(\frac{\pi}{4}) + \cos(-\frac{\pi}{4})) = \sqrt{2} (\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}) = \sqrt{2} \cdot \sqrt{2} = 2$. 4. $\frac{\sqrt{12}}{5} (\sin \frac{19\pi}{3} + \cos \frac{23\pi}{6}) = \frac{2\sqrt{3}}{5} (\sin(6\pi + \frac{\pi}{3}) + \cos(4\pi - \frac{\pi}{6})) = \frac{2\sqrt{3}}{5} (\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2}) = \frac{2\sqrt{3}}{5} \cdot \sqrt{3} = \frac{6}{5} = 1,2$. 5. $\frac{\sqrt{12}}{5} (\text{tg} \frac{59\pi}{6} - \text{ctg} \frac{40\pi}{3}) = \frac{2\sqrt{3}}{5} (\text{tg}(10\pi - \frac{\pi}{6}) - \text{ctg}(13\pi + \frac{\pi}{3})) = \frac{2\sqrt{3}}{5} (-\frac{\sqrt{3}}{3} - \frac{\sqrt{3}}{3}) = \frac{2\sqrt{3}}{5} \cdot (-\frac{2\sqrt{3}}{3}) = -\frac{4 \cdot 3}{15} = -\frac{12}{15} = -0,8$. 6. $\sqrt{108} (\text{tg} 390^\circ - 2\text{ctg} 480^\circ) = 6\sqrt{3} (\text{tg}(360^\circ + 30^\circ) - 2\text{ctg}(360^\circ + 120^\circ)) = 6\sqrt{3} (\frac{\sqrt{3}}{3} - 2 \cdot (-\frac{\sqrt{3}}{3})) = 6\sqrt{3} (\frac{\sqrt{3}}{3} + \frac{2\sqrt{3}}{3}) = 6\sqrt{3} \cdot \sqrt{3} = 18$. 7. $\text{tg} 45^\circ \cdot \text{ctg} 135^\circ = 1 \cdot (-1) = -1$. 8. $(\text{tg} 720^\circ + \text{ctg} 405^\circ) \cdot \sin 330^\circ = (0 + 1) \cdot (-0,5) = -0,5$.