Question

помогите срочно!!! 685-ое задание!!! даю 5 баллов!​

Answer 1

Решение:

а) $cos(\frac{2\pi}{3}-\alpha)cos(\frac{\pi}{3}+\alpha)-sin(\frac{2\pi}{3}-\alpha)sin(\frac{\pi}{3}+\alpha)=cos(\frac{2\pi}{3}-\alpha+\frac{\pi}{3}+\alpha) = cos\frac{3\pi}{3} = cos\pi=-1$

б) $sin(\frac{7\pi}{6}+\alpha)cos(\frac{\pi}{6}-\alpha)+cos(\frac{7\pi}{6}+\alpha)sin(\frac{\pi}{6}-\alpha)=sin(\frac{7\pi}{6}+\alpha+\frac{\pi}{6}-\alpha)=sin\frac{8\pi}{6}=sin\frac{4\pi}{3}=-\frac{\sqrt{3}}{2}$

в) $\frac{cos\frac{2\pi}{5}cos\frac{\pi}{15}+sin\frac{2\pi}{5}sin\frac{\pi}{15}}{sin\frac{7\pi}{30}cos\frac{4\pi}{15}+cos\frac{7\pi}{30}sin\frac{4\pi}{15}}=\frac{cos(\frac{2\pi}{5}-\frac{\pi}{15})}{sin(\frac{7\pi}{30}+\frac{4\pi}{15})}=\frac{cos(\frac{6\pi}{15}-\frac{\pi}{15})}{sin(\frac{7\pi}{30}+\frac{8\pi}{30})}=\frac{cos\frac{5\pi}{15}}{sin\frac{15\pi}{30}}=\frac{cos\frac{\pi}{3}}{sin\frac{\pi}{2}}=\frac{\frac{1}{2}}{1}=\frac{1}{2}$

г) $\frac{tg(\frac{\pi}{8}+\alpha)+tg(\frac{\pi}{8}-\alpha)}{1-tg(\frac{\pi}{8}+\alpha)tg(\frac{\pi}{8}-\alpha)}=tg(\frac{\pi}{8}+\alpha+\frac{\pi}{\alpha}-\alpha)=tg\frac{2\pi}{8}=tg\frac{\pi}{4}=1$