Question

prove that cos2x cos(x/2)-cos3x cos(9x/2)=sin5x sin(5x/2)

Answer 1

cos2xcos(x/2)-cos3xcos(9x/2)
=(1/2)[2cos2xcos(x/2)-2cos3xcos(9x/2)]
=(1/2)[cos(2x+x/2)+cos(2x-x/2)-cos(3x+9x/2)-cos(3x-9x/2)]
[∵, 2cosAcosB=cos(A+B)+cos(A-B)]
=(1/2)[cos(5x/2)+cos(3x/2)-cos(15x/2)-cos(-3x/2)]
=(1/2)[cos(5x/2)+cos(3x/2)-cos(15x/2)-cos(3x/2)] [∵, cos(-A)=cosA]
=(1/2)[cos(5x/2)-cos(15x/2)]
=(1/2)[2sin{(5x/2+15x/2)/2}sin{(15x/2-5x/2)/2}] 
[∵, cosC-cosD=2sin(C+D)/2sin(D-C)/2]
=(1/2)×2[sin{(20x/2)/2}sin{(10x/2)/2}]
=sin(10x/2)sin(5x/2)
=sin5xsin(5x/2) (Proved)

Answer 2

Answer:

Check answer in given attachment