prove sin theta-2sin cube theta/2cos cube theta-cos theta=tan theta

let theta =x

so __sin x - 2sin ^{3}x __

2cos^{3}x - cosx = tanx

= __sinx(1 - 2sin ^{2}x) __

cosx(2cos^{2}x - 1)

=__sinx[1 -2(1 -cos ^{2}x)]__

cosx(2cos^{2}x - 1)

=__sinx[1 - 2 + 2cos ^{2}x]__

cosx(2cos^{2}x - 1)

=__sinx____[2cos ^{2}x - 1]__

cosx~~(2cos~~)^{2}x - 1

= __sinx__

cosx

= tanx = rhs