Prove that cos^32x+3cos2x=4(cos^6-sin^6) Share with your friends Share 0 Muskaan Talwar answered this Dear Student, To prove: cos32x + 3 cos 2x=4(cos6x-sin6x)Now LHS=cos32x + 3 cos 2x = (cos2x-sin2x)3 + 3(cos2x-sin2x)=(cos2x)3-(sin2x)3 - 3 (cos2x)2 sin2x + 3 cos2x (sin2x)2 +3(cos2x-sin2x)=cos6x-sin6x -3 cos4x (1- cos2x) + 3(1-sin2x) sin4x + 3 cos2x-3 sin2x=cos6x-sin6x -3 cos4x +3 cos6x + 3sin4x-3 sin6x + 3 cos2x-3 sin2x=4 cos6x-4 sin6x +3 cos2x-3 cos4x -3 sin2x+ 3sin4x=4 cos6x-4 sin6x +3 cos2x(1- cos2x) -3 sin2x(1- sin2x)=4 cos6x-4 sin6x +3 cos2x sin2x-3 sin2x cos2x=4 cos6x-4 sin6x =4(cos6x-sin6x) = RHSHence provedIdentities used:1) cos 2x= cos2x-sin2x2) (a-b)3=a3-b3-3a2b+3ab23) sin2x + cos2x=1 Hope this information will clear your doubts about topic. If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible. Regards 1 View Full Answer
