Show that this: 6 sin2(x) * cosx - 3cosx
Is equal to: -3 cos (x) cos (2x)
Show that 6 sin2(x) * cosx - 3cosx = -3 cos (x) cos (2x)
Now cos(2x) = 1 - 2sin2(x) so the RHS of the above becomes:
-3 cos (x) cos (2x) → -3 cos (x) (1 - 2sin2(x)) → -3 cos (x) + 6cos(x)*sin2(x) which equals the LHS of the original.
I looked up the relationship between cos and sin, where I found this: https://www.afit.edu/KNEEBOARD/images/halfangle2.gif
Which, I believe, is exactly the rule you used. At what age are you expected to know this, because I've never heard of such a thing.
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) ]
cos x [ 6sin^2 x - 3 ] = cos x [ 6sin^2 x - 3 ]