Question

Question #b594a

Answer 1

`1/128(cos 8x -8cos 6x+28 cos 4x-56 cos 2x+35)`

Explanation:

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

=`1/16 (1-2cos 2x +cos^2 2x)^2`

=`1/16(1-2cos 2x +((1+cos 4x)/2)^2`

=`1/64(3-4cos 2x+cos 4x)^2`

=`1/64 (9+16cos^2 2x +cos^2 4x -24 cos 2x +6 cos 4x -8cos2x cos 4x)`

=`1/64(9+ 16 (1+cos4x)/2 +(1+cos 8x)/2 -24cos 2x +6cos 4x-4(cos 2x +cos 6x))`

=`1/128(18+16 +16cos 4x +1+cos 8x -48 cos 2x +12 cos 4x-8cos 2x -8 cos 6x)`

`1/128(cos 8x -8cos 6x+28 cos 4x-56 cos 2x+35)`