Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\cos(\frac{x}{2}), \cos(\frac{3x}{2}), \cos(\frac{5x}{2})...$?

share|cite|improve this question
Perhaps this relates to the Dirichlet kernel? – A Walker Mar 22 '13 at 7:09
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

The left hand side is an odd function. Hence, no linear combination involving $\cos a x$ may be equal, whatever the value of $a \neq 0$.

share|cite|improve this answer


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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.