+0

# How to simplify? Sin and cos

0
117
4

Show that this: 6 sin2(x) * cosx - 3cosx

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

Guest Aug 5, 2017
Sort:

#1
+26240
+2

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.

Alan  Aug 5, 2017
#2
0

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.

Guest Aug 5, 2017
#4
+26240
+1

If you are being asked to tackle the question you specified then you should already have come across the half-angle trigonometric formulae.  Such questions don't make sense otherwise!

Alan  Aug 5, 2017
#3
+76929
+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
edited by CPhill  Aug 5, 2017

### 18 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details