SOLUTION: Good evening, I'm stuck on this question, sin(x)cos(x)cos(2x)=(sin(4x))/4 I've been able to prove the left side from the right side, but I cant prove the right side from the l

Question 635383: Good evening,
I'm stuck on this question,
sin(x)cos(x)cos(2x)=(sin(4x))/4
I've been able to prove the left side from the right side, but I cant prove the right side from the left.
This is as far as I could get,
sin(x)cos(x)(1-2sin^2(x)
=> sin(x)cos(x)-sin^3(x)cos(x)
=> cos(x)(sin(x)-sin^3(x))
Am I on the right track?
Thankyou
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
As far a being on the right track, the short answer is: No. In fact you're farther away from the end that when you started. Yu want to end up with a argument of 4x. You've changed the argument's to all x's. At the start you had a 2x, at least, which is close to 4x than x.

If you were able to prove the left side from the right, then just do everything backwards! (Most operations in Math can be done forwards and backwards.) Let's look at what you probably have done:

Now let's look at doing that backwards:

"Uncancel" a 4. (IOW, multiply the numerator and denominator by 4:

"Unmultiply" (IOW factor) the 4 in the numerator:

Replace 2sin(x)cos(x) with sin(2x) (instead of the other way around):

Replace 2sin(2x)cos(2x) with sin(4x) (instead of the other way around):

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