Question

помогите пожалуйста ))​

Answer 1

Ответ:

$\displaystyle 1)\ \ cos\Big(\frac{2\pi }{3}-a\Big)\cdot cos\Big(\frac{\pi }{3}+a\Big)-sin\Big(\frac{2\pi }{3}-a\Big)\cdot sin\Big(\frac{\pi }{3}+a\Big)=\\\\\star \ \ cosx\cdot cosy-sinx\cdot siny=cos(x+y)\ \ \star \\\\=cos\Big(\frac{2\pi }{3}-a+\frac{\pi }{3}+a\Big)=cos\pi =-1$

$\displaystyle 2)\ \ sin\Big(\frac{7\pi }{6}+a\Big)\cdot cos\Big(\frac{\pi }{6}-a\Big)+cos\Big(\frac{7\pi }{6}+a\Big)\cdot sin\Big(\frac{\pi }{6}-a\Big)=\\\\\star \ \ sinx\cdot cosy+cosx\cdot siny=sin(x+y)\ \ \star \\\\=sin\Big(\frac{7\pi }{6}+a+\frac{\pi }{6}-a\Big)=sin\frac{8\pi }{6}=sin\frac{4\pi }{3}=sin\Big(\pi +\frac{\pi }{3}\Big)=\\\\\star \ \ \ sin(\pi +a)=-sina\ \ \star \\\\=-sin\frac{\pi }{3}=-\frac{\sqrt3}{2}$