Question

sin^2x+cos(п/3-x)cos(п/3+x)=1/4доказать тождество

Answer 1

$\sin^2x+\cos(\frac{\pi}{3}+x)\cos(\frac{\pi}{3}-x)=\frac{1}{4}$

$\sin^2x+(\cos({\frac{\pi}{3}})\cos(x)+\sin({\frac{\pi}{3}})\sin(x))(\cos({\frac{\pi}{3}})\cos(x)-\\-\sin({\frac{\pi}{3}})\sin(x))=\frac{1}{4}$

$\sin^2x+(\frac{1}{2}\cos(x)+\frac{\sqrt3}{2}\sin(x))(\frac{1}{2}\cos(x)-\frac{\sqrt3}{2}\sin(x))=\frac{1}{4}\\ \sin^2x+\frac{1}{4}\cos^2x-\frac{3}{4}\sin^2x=\frac{1}{4}$

$\frac{1}{4}(\cos^2x+\sin^2x)=\frac{1}{4}\\ \frac{1}{4}=\frac{1}{4}$

 

Answer 2

cos(п/3-x)cos(п/3+x)=1/2(cos(2п/3)+cos(2х))=-1/4+1/2-sin^2x

sin^2x-1/4+1/2-sin^2x=1/4

sin^2x-sin^2x=1/4+1/4-1/2

0=0

Тождество доказано.