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I have a problem that I cannot seem to solve. I need to prove the following identity,

$$\sin(11x) = 2\sin(8x)\cos(3x)-\sin(5x)$$

I do not understand how to go about this problem, as clearly expanding each term using compound and double angle formulas would be inefficient and take up a lot of space. I feel like there must be a shortcut to take. I have a test tomorrow and this is one of the practice questions. Therefore, I would immensely appreciate any help right now!

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    $\begingroup$ $11=3+8$ and $8=3+5$. Try compound angle formulas for this, and maybe you'll get it $\endgroup$ – Yuriy S Apr 14 '16 at 0:35
  • $\begingroup$ ^ Those formulas are also known as Angle-Sum Formulas. $\endgroup$ – Justin Benfield Apr 14 '16 at 2:41
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The main formula (you either know it or you don't - meaning either it has been taught in class or your book, or if not then it's an unfair problem) is this:

$\sin A + \sin B = 2 \sin \frac {A+B} 2 \cos \frac {A-B} 2$.

Have you seen this before? And if so, can you use it now?

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    $\begingroup$ But it gives a very strong hint of how to do the problem...So to the OP: Bring over that sin5x term to the left and this problem is easy... $\endgroup$ – imranfat Apr 14 '16 at 1:02
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By Werner Formula, $$2\sin8x\cos3x=\sin(8x+3x)+\sin(8x-3x)$$

Now rearrange

OR

by Prosthaphaeresis Formula, $$\sin11x+\sin5x=2\sin\dfrac{11x+5x}2\cos\dfrac{11x-5x}2=?$$

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