Question

Помогите пожалуйста с 12 и 7.

Answer 1

$7)\\\\\boxed {sin^2\alpha =\frac{1-cos2a}{2}\; \; \Rightarrow \; \; \; 1-cos2\alpha =2sin^2\alpha }\\\\\\\boxed {cos^2\alpha =\frac{1+cos2\alpha }{2}\; \; \Rightarrow \; \; \; 1+cos2\alpha =2cos^2\alpha }\; \; \; \; \boxed {sin\alpha =cos(\frac{\pi}{2}-\alpha )}\\\\\\0<a<\frac{\pi}{2}:\; \; \sqrt\frac{1-sina}{1+sina}}-\sqrt{\frac{1+sina}{1-sina}}=\sqrt{\frac{1-cos(\pi /2-a)}{1+cos(\pi /2-a)}}-\sqrt{\frac{1+cos(\pi /2-a)}{1+cos(\pi /2-a)}}=$

$=\sqrt{\frac{sin^2(\pi /4-a/2)}{cos^2(\pi /4-a/2)}}-\sqrt{\frac{cos^2(\pi /4-a/2)}{sin^2(\pi /4-a/2)}}=tg(\frac{\pi}{4}-\frac{a}{2})-ctg(\frac{\pi}{4}-\frac{a}{2})=\\\\\star \; \; tgx-ctgx=\frac{sinx}{cosx}-\frac{cosx}{sinx}=\frac{sin^2x-cos^2x}{sinx\cdot cosx}=\frac{-cos2x}{1/2\cdot sin2x}=-2\, ctg2x\; \; \star \\\\=-2\, ctg(\frac{\pi}{2}-a)=-2\, tga$

$12)\; \; (sin3x-1)(sin3x1)=sin3x-cos^23x\\\\sin^23x-1=sin3x-cos^23x\\\\-(1-sin^23x)=sin3x-cos^23x\\\\-cos^23x=sin3x-cos^23x\\\\sin3x=0\\\\3x=\pi n,\; n\in Z\\\\x=\frac{\pi n}{3}\; ,\; n\in Z$