cos4x×cos2x-cos눀3x等于多少?

2025-06-26 12:16:22
推荐回答(1个)
回答1:

原式=1/2*(cos6x+cos2x)-1/2(1+cos6x)
=(cos2x-1)/2
(cos2x-1)/2
=[(2cos^x-1)-1]/2
=(2cos^2x-2)/2
=cos^2x-1
=-(1-cos^2x)
=-sin^2x,