Упр.6.3 ГДЗ Мордкович 10-11 класс (Алгебра)

а) t = 5пи/6;
6) t = 5пи/4;
в) t = 7пи/6;
г) t = 7пи/4.

1. $$t=\frac{5\pi}{6}$$

   Тогда

   $$\sin t=\sin\frac{5\pi}{6}=\frac12,$$

   $$\cos t=\cos\frac{5\pi}{6}=-\frac{\sqrt3}{2},$$

   $$\tg t=\tg\frac{5\pi}{6}=-\frac{1}{\sqrt3}.$$

2. $$t=\frac{5\pi}{4}$$

   Тогда

   $$\sin t=\sin\frac{5\pi}{4}=-\frac{\sqrt2}{2},$$

   $$\cos t=\cos\frac{5\pi}{4}=-\frac{\sqrt2}{2},$$

   $$\tg t=\tg\frac{5\pi}{4}=1.$$

3. $$t=\frac{7\pi}{6}$$

   Тогда

   $$\sin t=\sin\frac{7\pi}{6}=-\frac12,$$

   $$\cos t=\cos\frac{7\pi}{6}=-\frac{\sqrt3}{2},$$

   $$\tg t=\tg\frac{7\pi}{6}=\frac{1}{\sqrt3}.$$

4. $$t=\frac{7\pi}{4}$$

   Тогда

   $$\sin t=\sin\frac{7\pi}{4}=-\frac{\sqrt2}{2},$$

   $$\cos t=\cos\frac{7\pi}{4}=\frac{\sqrt2}{2},$$

   $$\tg t=\tg\frac{7\pi}{4}=-1.$$

Ответ

а) $$\frac12,\ -\frac{\sqrt3}{2},\ -\frac{1}{\sqrt3}$$; б) $$-\frac{\sqrt2}{2},\ -\frac{\sqrt2}{2},\ 1$$; в) $$-\frac12,\ -\frac{\sqrt3}{2},\ \frac{1}{\sqrt3}$$; г) $$-\frac{\sqrt2}{2},\ \frac{\sqrt2}{2},\ -1$$.