Упр.6.5 ГДЗ Мордкович 10-11 класс (Алгебра)

а) t = 13пи/6;
6) t = — 8пи/3;
в) t = 23пи/6;
г) t = — 11пи/4.

1. $$t=\frac{13\pi}{6}$$

   $$\sin t=\sin\left(\frac{13\pi}{6}\right)=\sin\left(\frac{13\pi}{6}-2\pi\right)=\sin\frac{\pi}{6}=\frac12$$
   
   $$\cos t=\cos\left(\frac{13\pi}{6}\right)=\cos\left(\frac{13\pi}{6}-2\pi\right)=\cos\frac{\pi}{6}=\frac{\sqrt3}{2}$$
   
   $$\tg t=\tg\left(\frac{13\pi}{6}\right)=\tg\frac{\pi}{6}=\frac{1}{\sqrt3}$$

2. $$t=-\frac{8\pi}{3}$$

   $$\sin t=\sin\left(-\frac{8\pi}{3}\right)=\sin\left(4\pi-\frac{8\pi}{3}\right)=\sin\frac{4\pi}{3}=-\frac{\sqrt3}{2}$$
   
   $$\cos t=\cos\left(-\frac{8\pi}{3}\right)=\cos\left(4\pi-\frac{8\pi}{3}\right)=\cos\frac{4\pi}{3}=-\frac12$$
   
   $$\tg t=\tg\left(-\frac{8\pi}{3}\right)=\sqrt3$$

3. $$t=\frac{23\pi}{6}$$

   $$\sin t=\sin\left(\frac{23\pi}{6}\right)=\sin\left(\frac{23\pi}{6}-2\pi\right)=\sin\frac{11\pi}{6}=-\frac12$$
   
   $$\cos t=\cos\left(\frac{23\pi}{6}\right)=\cos\left(\frac{23\pi}{6}-2\pi\right)=\cos\frac{11\pi}{6}=\frac{\sqrt3}{2}$$
   
   $$\tg t=\tg\left(\frac{23\pi}{6}\right)=-\frac{1}{\sqrt3}$$

4. $$t=-\frac{11\pi}{4}$$

   $$\sin t=\sin\left(-\frac{11\pi}{4}\right)=\sin\left(4\pi-\frac{11\pi}{4}\right)=\sin\frac{5\pi}{4}=-\frac{\sqrt2}{2}$$
   
   $$\cos t=\cos\left(-\frac{11\pi}{4}\right)=\cos\left(4\pi-\frac{11\pi}{4}\right)=\cos\frac{5\pi}{4}=-\frac{\sqrt2}{2}$$
   
   $$\tg t=\tg\left(-\frac{11\pi}{4}\right)=1$$

Ответ

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