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

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

Guest Aug 5, 2017
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4+0 Answers

 #1
avatar+25995 
+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
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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
avatar+25995 
+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
avatar+75333 
+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 ]

 

 

cool cool cool

CPhill  Aug 5, 2017
edited by CPhill  Aug 5, 2017

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