Show that this: 6 sin^{2}(x) * cosx - 3cosx

Is equal to: -3 cos (x) cos (2x)

Guest Aug 5, 2017

#1**+2 **

Show that 6 sin^{2}(x) * cosx - 3cosx = -3 cos (x) cos (2x)

Now cos(2x) = 1 - 2sin^{2}(x) so the RHS of the above becomes:

-3 cos (x) cos (2x) → -3 cos (x) (1 - 2sin^{2}(x)) → -3 cos (x) + 6cos(x)*sin^{2}(x) which equals the LHS of the original.

Alan Aug 5, 2017

#3**+1 **

6 sin^2(x) * cosx - 3cosx = -3 cos (x) cos (2x)

We can work on both sides and show that they are equal.....

First, factor out cos x on both sides and write cos 2x as 1 - 2sin^2 x

cos x [ 6sin^2 x - 3 ] = cos x [ -3 (1 - 2sin^2 x) ]

Simplify

cos x [ 6sin^2 x - 3 ] = cos x [ 6sin^2 x - 3 ]

CPhill Aug 5, 2017