(sin x*sin 3x)sin^2 x+(cos x*cos 3x)cos^2 x 如何变到1/2[(cos 2x-cos4x)sin^2x+(cos2x+cos4x)cos^2x]
问题描述:
(sin x*sin 3x)sin^2 x+(cos x*cos 3x)cos^2 x 如何变到1/2[(cos 2x-cos4x)sin^2x+(cos2x+cos4x)cos^2x]
答
用积化和差公式:sinasinb=(1/2)[cos(a-b)-cos(a+b)]
cosacosb=(1/2)[cos(a-b)+cos(a+b)]
答
sinx*sin3x=(1/2)[cos(x-3x)-cos(x+3x)]=(1/2)[cos2x-cos4x]
cosx*cos3x=(1/2)[cos(x-3x)+cos(x+3x)]=(1/2)[cos2x+cos4x]