Question

765 а,б,в,г
Можно пожалуйста с объяснениями

Answer 1

$1)Cos(\frac{\pi }{6}+\alpha)*Cos(\frac{\pi }{6}-\alpha)=\frac{1}{2}[Cos(\frac{\pi }{6}+\alpha-\frac{\pi }{6}+\alpha)+Cos(\frac{\pi }{6}+\alpha+\frac{\pi }{6}-\alpha)]=\frac{1}{2}(Cos2\alpha+Cos\frac{\pi }{3})=\frac{1}{2}(Cos2\alpha+\frac{1}{2})=\frac{1}{2}Cos2\alpha+\frac{1}{4}$

$2)Cos(\frac{\pi }{8}+2\alpha )*Cos(\frac{\pi }{8}-2\alpha)=\frac{1}{2}[Cos(\frac{\pi }{8}+2\alpha-\frac{\pi }{8}+2\alpha)+Cos(\frac{\pi }{8}+2\alpha+\frac{\pi }{8}-2\alpha)]=\frac{1}{2}(Cos4\alpha+Cos\frac{\pi }{4})=\frac{1}{2}(Cos4\alpha+\frac{\sqrt{2}}{2})=\frac{1}{2}Cos4\alpha+\frac{\sqrt{2} }{4}$

$3)Sin(\frac{\pi }{3}-3\alpha)*Cos(\frac{\pi }{3}+3\alpha)=\frac{1}{2}[Sin(\frac{\pi }{3}-3\alpha-\frac{\pi }{3}-3\alpha)+Sin(\frac{\pi }{3}-3\alpha+\frac{\pi }{3}+3\alpha)]=\frac{1}{2}(Sin(-6\alpha)+Sin\frac{2\pi }{3})=\frac{1}{2}(\frac{\sqrt{3} }{2}-Sin6\alpha)=\frac{\sqrt{3} }{4}-\frac{1}{2}Sin6\alpha$

$4)2Cos(\frac{\pi }{4}-\alpha)*Sin(\frac{\pi }{4}+\alpha)=2*\frac{1}{2}[Sin(\frac{\pi }{4}+\alpha-\frac{\pi }{4}+\alpha)+Sin(\frac{\pi }{4} +\alpha+\frac{\pi }{4}-\alpha)]=Sin2\alpha+Sin\frac{\pi }{2}=Sin2\alpha +1$