Question

Cos2x cosx/2-cos3x cos9x/2=sin5x sin5x/2

Answer 1

cos2xcosx/2-cos3xcos9x/2
=1/2(2cos2xcosx/2-2cos3xcos9x/2)
=1/2[{cos(2x-x/2)+cos(2x+x/2)}-{cos(3x-9x/2)+cos(3x+9x/2)}]
=1/2[(cos3x/2+cos5x/2)-(cos3x/2+cos15x/2)]
=1/2(cos3x/2+cos5x/2-cos3x/2-cos15x/2)
=1/2(cos5x/2-cos15x/2)
=1/2[{2sin(5x/2+15x/2)/2}sin{(15x/2-5x/2)/2}]
=1/2[2{sin(20x/2)/2}{sin(10x/2)/2}]
=1/2×2sin(10x/2)sin(5x/2)
=sin5xsin5x/2 (Proved)

Answer 2

Answer:

hope this helps you