Question

solve this ✔️

with explanation........​

Answer 1

Step-by-step explanation:

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[(cos5x/2+cos3x/2)-{cos15x/2 +cos(-3x/2)}]

=1/2[cos5x/2+cos3x/2-cos15x/2-cos3x/2

=1/2[cos5x/2-cos15x/2]

=1/2[-2sin{(5x/2+15x/2)/2}sin{(5x/2-15x/2)/2}]

=1/2[-2sin(20x/4)sin(-10x/4)]

=1/2[2sin5xsin(5x/2)]

=sin5xsin5x

2