Question

Помогите пожалуйста с 12 и 7.

Answer 1

$7)\\boxed {sin^2alpha =frac{1-cos2a}{2}; ; Rightarrow ; ; ; 1-cos2alpha =2sin^2alpha }\\\boxed {cos^2alpha =frac{1+cos2alpha }{2}; ; Rightarrow ; ; ; 1+cos2alpha =2cos^2alpha }; ; ; ; boxed {sinalpha =cos(frac{pi}{2}-alpha )}\\\0<a<frac{pi}{2}:; ; sqrtfrac{1-sina}{1+sina}}-sqrt{frac{1+sina}{1-sina}}=sqrt{frac{1-cos(pi /2-a)}{1+cos(pi /2-a)}}-sqrt{frac{1+cos(pi /2-a)}{1+cos(pi /2-a)}}=$

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

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