Вычислите sin t и cos t, если:

Для решения задач на нахождение синуса и косинуса воспользуемся тригонометрической окружностью и периодичностью функций (период $2\pi$). **13.1** а) $t = 0$: $\sin 0 = 0$, $\cos 0 = 1$ б) $t = \frac{\pi}{2}$: $\sin \frac{\pi}{2} = 1$, $\cos \frac{\pi}{2} = 0$ в) $t = \frac{3\pi}{2}$: $\sin \frac{3\pi}{2} = -1$, $\cos \frac{3\pi}{2} = 0$ г) $t = \pi$: $\sin \pi = 0$, $\cos \pi = -1$ **13.2** а) $t = \frac{5\pi}{6}$: $\sin \frac{5\pi}{6} = \frac{1}{2}$, $\cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$ б) $t = \frac{5\pi}{4}$: $\sin \frac{5\pi}{4} = -\frac{\sqrt{2}}{2}$, $\cos \frac{5\pi}{4} = -\frac{\sqrt{2}}{2}$ в) $t = \frac{7\pi}{6}$: $\sin \frac{7\pi}{6} = -\frac{1}{2}$, $\cos \frac{7\pi}{6} = -\frac{\sqrt{3}}{2}$ г) $t = \frac{9\pi}{4} = 2\pi + \frac{\pi}{4}$: $\sin \frac{9\pi}{4} = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$, $\cos \frac{9\pi}{4} = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$ **13.3** а) $t = \frac{13\pi}{6} = 2\pi + \frac{\pi}{6}$: $\sin \frac{13\pi}{6} = \sin \frac{\pi}{6} = \frac{1}{2}$, $\cos \frac{13\pi}{6} = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$ б) $t = -\frac{8\pi}{3} = -2\pi - \frac{2\pi}{3} = -\frac{2\pi}{3} - 2\pi$: $\sin(-\frac{8\pi}{3}) = \sin(-\frac{2\pi}{3}) = -\frac{\sqrt{3}}{2}$, $\cos(-\frac{8\pi}{3}) = \cos(-\frac{2\pi}{3}) = -\frac{1}{2}$ в) $t = \frac{23\pi}{6} = 4\pi - \frac{\pi}{6}$: $\sin \frac{23\pi}{6} = \sin(-\frac{\pi}{6}) = -\frac{1}{2}$, $\cos \frac{23\pi}{6} = \cos(-\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ г) $t = -\frac{11\pi}{3} = -4\pi + \frac{\pi}{3}$: $\sin(-\frac{11\pi}{3}) = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$, $\cos(-\frac{11\pi}{3}) = \cos(\frac{\pi}{3}) = \frac{1}{2}$