Use the power-reducing identities to write sin^2xcos^2xsin2xcos2x in terms of the first power of cosine?

1 Answer
Mar 13, 2018

sin^2xcos^2x=(1-cos(4x))/8sin2xcos2x=1cos(4x)8

Explanation:

sin^2x=(1-cos(2x))/2sin2x=1cos(2x)2

cos^2x=(1+cos(2x))/2cos2x=1+cos(2x)2

sin^2xcos^2x=((1+cos(2x))(1-cos(2x)))/4=(1-cos^2(2x))/4sin2xcos2x=(1+cos(2x))(1cos(2x))4=1cos2(2x)4

cos^2(2x)=(1+cos(4x))/2cos2(2x)=1+cos(4x)2

(1-(1+cos(4x))/2)/4=(2-(1+cos(4x)))/8=(1-cos(4x))/811+cos(4x)24=2(1+cos(4x))8=1cos(4x)8