Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

How do you work out the sum of the series: $\cos{x}+\cos{2x}+\cdots+\cos{(n-1)x}$ by multiplying through by $2\sin(x/2)$? I am supposed to find the sum using only this method and I'm not completely sure what the sum at the end would look like. Can anyone help? Thanks.

share|cite|improve this question
@Adam: Why not post that as an answer? I'd up-vote it. – Clive Newstead Jan 8 '13 at 12:15
Adam gave you the correct answer. Another way is to use that $$e^{ix}+e^{2ix}+\cdots+e^{(n-1)ix}=e^{ix}\dfrac{e^{(n-1)ix}-1}{e^{ix}-1}$$ and Euler's identity. – P.. Jan 8 '13 at 12:20
@Pambos: You didn't need to delete your answer. It's OK to have multiple answers showing different ways of solving a problem. – Rahul Jan 9 '13 at 18:33

Hint: Use identity $$2\cos(\theta)\sin(\phi) = \sin(\theta + \phi)-\sin(\theta-\phi)$$

share|cite|improve this answer
Style: the \ is missing from the second sine! I tried to put that in but it wouldn't let me! :[ – NeverBeenHere Jan 8 '13 at 12:49
@Artist Thanks, corrected :) – Adam Jan 8 '13 at 12:56

Another way is to use that $$e^{ix}+e^{2ix}+\cdots+e^{(n-1)ix}=e^{ix}\dfrac{e^{(n-1)ix}-1}{e^{ix}-1}$$ and Euler's identity.

share|cite|improve this answer

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.