Вычислите: а) sin(-π/4) + cos(π/3) + cos(-π/6)

Давай вычислим значения выражений, используя таблицу значений тригонометрических функций. **6.6.** а) $\sin\left(-\frac{\pi}{4}\right) + \cos\frac{\pi}{3} + \cos\left(-\frac{\pi}{6}\right) = -\sin\frac{\pi}{4} + \cos\frac{\pi}{3} + \cos\frac{\pi}{6} = -\frac{\sqrt{2}}{2} + \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3} - \sqrt{2}}{2}$ б) $\cos\frac{\pi}{6} \cdot \cos\frac{\pi}{4} \cdot \cos\frac{\pi}{3} \cdot \cos\frac{\pi}{2} = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} \cdot \frac{1}{2} \cdot 0 = 0$ в) $\sin\left(-\frac{\pi}{2}\right) - \cos(-\pi) + \sin\left(-\frac{3\pi}{2}\right) = -\sin\frac{\pi}{2} - \cos\pi - \sin\frac{3\pi}{2} = -1 - (-1) - (-1) = -1 + 1 + 1 = 1$ г) $\sin\frac{\pi}{6} \cdot \sin\frac{\pi}{4} \cdot \sin\frac{\pi}{3} \cdot \sin\frac{\pi}{2} = \frac{1}{2} \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} \cdot 1 = \frac{\sqrt{6}}{8}$ **6.7.** а) $\sin\left(-\frac{3\pi}{4}\right) + \cos\left(-\frac{\pi}{4}\right) + \sin\frac{\pi}{4} \cdot \cos\frac{\pi}{2} + \cos 0 \cdot \sin\frac{\pi}{2} = -\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \cdot 0 + 1 \cdot 1 = 0 + 0 + 1 = 1$ б) $\cos\frac{5\pi}{3} + \cos\frac{4\pi}{3} + \sin\frac{3\pi}{2} \cdot \sin\frac{5\pi}{8} \cdot \cos\frac{3\pi}{2} = \cos\left(2\pi - \frac{\pi}{3}\right) + \cos\left(\pi + \frac{\pi}{3}\right) + (-1) \cdot \sin\frac{5\pi}{8} \cdot 0 = \cos\frac{\pi}{3} - \cos\frac{\pi}{3} + 0 = 0$ **6.8.** а) $\text{tg}\frac{\pi}{4} + \text{ctg}\frac{5\pi}{4} = 1 + \text{ctg}\left(\pi + \frac{\pi}{4}\right) = 1 + \text{ctg}\frac{\pi}{4} = 1 + 1 = 2$ б) $\text{ctg}\frac{\pi}{3} \cdot \text{tg}\frac{\pi}{6} = \frac{\sqrt{3}}{3} \cdot \frac{\sqrt{3}}{3} = \frac{3}{9} = \frac{1}{3}$ в) $\text{tg}\frac{\pi}{6} - \text{ctg}\frac{\pi}{6} = \frac{\sqrt{3}}{3} - \sqrt{3} = \frac{\sqrt{3} - 3\sqrt{3}}{3} = -\frac{2\sqrt{3}}{3}$ г) $\text{tg}\frac{9\pi}{4} + \text{ctg}\frac{\pi}{4} = \text{tg}\left(2\pi + \frac{\pi}{4}\right) + 1 = \text{tg}\frac{\pi}{4} + 1 = 1 + 1 = 2$ **6.9.** а) $\text{tg}\frac{\pi}{4} \cdot \sin\frac{\pi}{3} \cdot \text{ctg}\frac{\pi}{6} = 1 \cdot \frac{\sqrt{3}}{2} \cdot \sqrt{3} = \frac{3}{2} = 1,5$ б) $2\sin\pi + 3\cos\pi + \text{ctg}\frac{\pi}{2} = 2 \cdot 0 + 3 \cdot (-1) + 0 = -3$ в) $2\sin\frac{\pi}{3} \cdot \cos\frac{\pi}{6} - \frac{1}{2}\text{tg}^2\frac{\pi}{3} = 2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{1}{2} \cdot (\sqrt{3})^2 = \frac{3}{2} - \frac{3}{2} = 0$ г) $2\text{tg}0 + 8\cos\frac{3\pi}{2} - 6\sin^2\frac{\pi}{3} = 2 \cdot 0 + 8 \cdot 0 - 6 \cdot \left(\frac{\sqrt{3}}{2}\right)^2 = 0 + 0 - 6 \cdot \frac{3}{4} = -\frac{18}{4} = -4,5$