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I was going through some of my notes and found they said we can express $\cos(\frac{x}{2})\sin(x)$ as a linear combination of even multiples of $\frac{x}{2}$ in sin and odd ones in cos. However, I can't see how this is possible, and can only express it as:

$\cos(\frac{x}{2})\sin(x) = \sin(\frac{3x}{2})+\sin(\frac{x}{2})$

So is there another way to express the original expression, so that we would be summing up terms like $\sin(x), \sin(2x), \sin(3x)...$ and $\cos(\frac{x}{2}), \cos(\frac{3x}{2}), \cos(\frac{5x}{2})...$?

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Perhaps this relates to the Dirichlet kernel? –  A Walker Mar 22 '13 at 7:09
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That should be $2\cos(\frac{x}{2})\sin(x) = \sin(\frac{3x}{2})+\sin(\frac{x}{2})$. –  Inceptio Mar 22 '13 at 7:34
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2 Answers 2

Hint:

$$\sin C+\sin D= 2 \sin \left(\frac{(C+D)}{2}\right)\cdot\cos\left(\frac{C-D}{2}\right)$$

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Yes, that is what I used, but then C and D won't be integer multiples of x. –  Ryker Mar 22 '13 at 14:54
    
@Ryker, And there is an error in your calculation. It should $2\cos(\frac{x}{2})\sin(x) = \sin(\frac{3x}{2})+\sin(\frac{x}{2})$ –  Inceptio Mar 22 '13 at 15:45
    
Yeah, I realized that, but I guess "up to a factor" would be sufficient for me anyway :) –  Ryker Mar 22 '13 at 19:11
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The left hand side is an odd function. Hence, no linear combination involving $\cos a x$ may be equal, whatever the value of $a \neq 0$.

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