Sin^2 x+cos(60°+x)cos(60°-x)=1/4

Question

Sin^2 x+cos(60°+x)cos(60°-x)=1/4

Answer 1

Sin^2 x+ 1/2*(cos2x+cos120)= sin^2x+ 1/2 cos2x- 1/4= sin^2 x+ 1/2*(1-2sin^2x)-1/4= sin^2 x +1/2 - sin^2 x - 1/4= 1/2-1/4= 1/4.
Ответ : 1/4

Answer 2

Cosa*cosb=1/2cos(a-b)+1/2cos(a+b)
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sin²x+cos(60+x)cos(60-x)=1/4
sin²x+1/2*cos2x+1/2cos120=1/4
sin²x+1/2*(1-2sin²x)+1/2*cos(180-60)=1/4
sin²x+1/2-sin²x+1/2*(-1/2)=1/4
1/2-1/4=1/4
1/4=1/4