Question

sin^2x+cos(п/3-x)cos(п/3+x)=1/4

доказать тождество

Answer 1

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

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

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

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

 

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.