So I'm working on proving Lagrange's trig identity

$$\sum_{k=0}^n\cos k\theta=\frac{1}{2}+\frac{\sin(2n+1)\frac{\theta}{2}}{2\sin\frac{\theta}{2}}$$

Skipping some steps I arrive here:




Both my book and other online sources all simply skip the trig manipulation required to finish proving the identity as if it's trivial. I honestly see no way to manipulate any of the above equations to the desired result and am wondering what I'm missing.

  • $\begingroup$ Do you want to prove the result from the given expressions or a simple alternative proof would also suffice? $\endgroup$ – Math Lover Sep 6 '17 at 1:54
  • $\begingroup$ I'm aware there are proofs involving telescoping sums etc. but I'm asked to do it this way $\endgroup$ – Lanier Freeman Sep 6 '17 at 1:54
  • $\begingroup$ I'm not sure about what you mean by "this" way. You may find the following expressions useful: $2\cos(A)\cos(B) = \cos(A+B) + \cos(A-B)$, $2\sin(A)\sin(B) = \cos(A-B) - \cos(A+B)$, and $2\sin(A)\cos(B) = \sin(A+B) + \sin(A-B)$. $\endgroup$ – Math Lover Sep 6 '17 at 2:06
  • $\begingroup$ By "this way" I meant $\Re\sum_{k=0}^ne^{ik\theta}$ $\endgroup$ – Lanier Freeman Sep 6 '17 at 2:18

Note that $$\sum_{k=0}^{n}e^{i k \theta} = \frac{e^{i (n+1)\theta}-1}{e^{i \theta}-1} = \frac{e^{i(n+1)\theta/2}}{e^{i \theta/2}}\cdot\frac{e^{i (n+1)\theta/2}-e^{-i (n+1)\theta/2}}{e^{i \theta/2}-e^{-i\theta/2}} = e^{i n\theta/2}\frac{\sin\left(\frac{(n+1)\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}.$$ Consequently, $$\Re\left(\sum_{k=0}^{n}e^{i k \theta}\right) = \sum_{k=0}^{n}\cos(k \theta) = \frac{\cos(n\theta/2)\sin((n+1)\theta/2)}{\sin(\theta/2)}.$$ The result immediately follows by noting that $$\sin(A)\cos(B) = \frac{\sin(A+B)+\sin(A-B)}{2}.$$

  • $\begingroup$ Although I see how you've manipulated the first line, where does the $\cos$ term in line two come from? Is that just $\Re(e^{in\theta/2})$? $\endgroup$ – Lanier Freeman Sep 6 '17 at 3:37
  • $\begingroup$ Damn I see it now, this is really nice. Thanks! $\endgroup$ – Lanier Freeman Sep 6 '17 at 3:38

Or try induction.

$\sum_{k=0}^n\cos k\theta=\frac{1}{2}+\frac{\sin(2n+1)\frac{\theta}{2}}{2\sin\frac{\theta}{2}} $ is true for $n=0$.

If this is true for $n$, then, putting $n+1$ for $n$, you want to prove $\sum_{k=0}^{n+1}\cos k\theta=\frac{1}{2}+\frac{\sin(2n+3)\frac{\theta}{2}}{2\sin\frac{\theta}{2}} $.

But, by the induction hypothesis, $\sum_{k=0}^{n+1}\cos k\theta =\sum_{k=0}^{n}\cos k\theta+\cos((n+1)\theta) =\frac{1}{2}+\frac{\sin(2n+1)\frac{\theta}{2}}{2\sin\frac{\theta}{2}}+\cos((n+1)\theta) $ so you want to prove $\frac{1}{2}+\frac{\sin(2n+1)\frac{\theta}{2}}{2\sin\frac{\theta}{2}}+\cos((n+1)\theta) =\frac{1}{2}+\frac{\sin(2n+3)\frac{\theta}{2}}{2\sin\frac{\theta}{2}} $ or $\sin\frac{(2n+1)\theta}{2}+2\sin\frac{\theta}{2}\cos((n+1)\theta) =\sin\frac{(2n+3)\theta}{2} $ or $2\sin\frac{\theta}{2}\cos((n+1)\theta) =\sin\frac{(2n+3)\theta}{2}-\sin\frac{(2n+1)\theta}{2} $.

Using the formula for $\sin(a)-\sin(b)$ will finish the proof.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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