Question

"sin2xcos2x=1/2sin4x" "cos^2x/2-sin^2x/2=cosx" Пожалуйста сделайте поясней

Answer 1

Sin2α=2sinα * cosα
sin2x *cos2x=(2sin2x *cos2x)/2=(sin(2*2x))/2=(1/2)*sin4x
(1/4)*sin4x=(1/4)*sin4x

cos2α=(cosα)²-(sinα)²
(cos(x/2))²-(sin (x/2))²=cos(2*(x/2))=cosx
cosx=cosx