State the sum of the series $z+z^2+z^3+\cdots+z^n$, for $z\neq1$.

By letting $z=\cos\theta+i\sin\theta$, show that

$$\cos \theta+\cos 2\theta+\cos 3\theta+\cdots+\cos n\theta=\frac{\sin\frac12n\theta}{\sin\frac12\theta}\cos\frac12(n+1)\theta$$

Where $\sin\frac12\theta\neq0$.

I know the first part, The second part Im kind of stuck in showing that

My Attempt:


I realized that this is a Geometric Progression, so its in the form:

$a+ar+ar^2+....+ar^n$ , where $a=(\cos\theta+i\sin\theta)$ and $r=(\cos\theta+i\sin\theta)$

So I will apply the formula for the Sum of a G.P to my problem.

$$\begin{align}\Re \frac{(\cos\theta+i\sin\theta)(1-(\cos\theta+i\sin\theta)^n)}{1-(\cos\theta+i\sin\theta)}\end{align}$$

I applied the De Movire Theorem and simplified as follows:

$$\begin{align}\Re \frac{\cos\theta+i\sin\theta-\cos(n+1)\theta+i\sin(n+1)\theta}{1-(\cos\theta+i\sin\theta)}\end{align}$$

$$\begin{align}\Re \frac{\cos\theta+i\sin\theta-\cos(n+1)\theta+i\sin(n+1)\theta}{1-(\cos\theta+i\sin\theta)}\frac{(1+(cos\theta+isin\theta))}{(1+(cos\theta+isin\theta))}\end{align}$$

$$\begin{align}\Re \frac{\cos\theta+i\sin\theta-\cos(n+1)\theta+i\sin(n+1)\theta+cos^2\theta+2icos\theta sin\theta-sin^2\theta-cos\theta cos(n+1)\theta- isin \theta cos(n+1)\theta+i cos\theta sin(n+1)\theta-sin\theta sin(n+1)\theta}{1-cos^2\theta-2i cos\theta sin\theta+sin^2\theta}\end{align}$$

$$\begin{align}\Re \frac{\cos\theta+i\sin\theta-\cos(n+1)\theta+i\sin(n+1)\theta+cos (2\theta) +i sin (2\theta)- cos(n+1)\theta(cos\theta+isin \theta)+(i cos\theta -sin \theta) sin(n+1)\theta}{1-cos^2\theta-2i cos\theta sin\theta+sin^2\theta}\end{align}$$

$$\begin{align}\Re \frac{\cos\theta+i\sin\theta+cos (2\theta) +i sin (2\theta)- cos(n+1)\theta(1+cos\theta+isin \theta)+(1+i cos\theta -sin \theta) sin(n+1)\theta}{2sin^2\theta-2i cos\theta sin\theta}\end{align}$$

$$\begin{align}\Re \frac{[\cos\theta+i\sin\theta+cos (2\theta) +i sin (2\theta)- cos(n+1)\theta(1+cos\theta+isin \theta)+(1+i cos\theta -sin \theta) sin(n+1)\theta]}{2sin\theta(sin\theta-i cos\theta)} \frac{(sin\theta+i cos\theta)}{(sin\theta+i cos\theta)}\end{align}$$

$$\begin{align}\frac{[-cos(n+1)\theta(1+cos\theta+isin \theta)+(1+i cos\theta -sin \theta) sin(n+1)\theta]}{2sin\theta} \frac{(sin\theta+i cos\theta)}{1}\end{align}$$

$$\begin{align}\frac{[-cos(n+1)\theta(sin\theta+cos\theta sin\theta - cos\theta sin \theta)+(sin\theta-cos^2\theta -sin^2 \theta) sin(n+1)\theta]}{2sin\theta} \end{align}$$

$$\begin{align}\frac{[-cos(n+1)sin\theta+(sin\theta-1) sin(n+1)\theta]}{2sin\theta} \end{align}$$

I have no idea right now where I am taking this, I just dont know what the next step I should take. Please dont send me the solution (at least yet). Can anyone give me a hint (a little boost to my little mind) as to what I should do next, (make sure its a small hint, Don't give me the full next step) Just the help in order for me to construct the next step and carry on.

  • $\begingroup$ Yes, I know and even I thought this at the start of doing the problem and it would take less time to even type in latex. But for some reason (I donno) I always prefer going with $cos\theta+isin\theta$. So yes right now wondering how to continue in this form. $\endgroup$
    – M.S.E
    Sep 16, 2014 at 8:26
  • $\begingroup$ You havent taken the real part correctly after applying the geometric series formula $\endgroup$
    – Jasser
    Sep 16, 2014 at 8:27
  • $\begingroup$ @user291957 , My bad :D Got it , it was wrong of me to remove the $isin\theta$ from the denominator. $\endgroup$
    – M.S.E
    Sep 16, 2014 at 9:03
  • $\begingroup$ Two proofs at the end of my answer here (one with complex numbers, one with trigonometry only) $\endgroup$ Sep 16, 2014 at 11:02
  • $\begingroup$ LaTeX note: Write \sin and \cos instead of sin and cos in your code. (This renders them as $\sin,\cos$ rather than $sin,cos$.) $\endgroup$ Nov 18, 2014 at 11:13

3 Answers 3


Here are the main steps.

You may write $$ \begin{align} \sum_{k=1}^{n} \cos (k\theta)&=\Re \sum_{k=1}^{n} e^{ik\theta}\\\\ &=\Re\left( e^{i\theta}\frac{e^{in\theta}-1}{e^{i\theta}-1}\right)\\\\ &=\Re\left( e^{i\theta}\frac{e^{in\theta/2}\left(e^{in\theta/2}-e^{-in\theta/2}\right)}{e^{i\theta/2}\left(e^{i\theta/2}-e^{-i\theta/2}\right)}\right)\\\\ &=\Re\left( e^{i\theta}\frac{e^{in\theta/2}\left(2i\sin(n\theta/2)\right)}{e^{i\theta/2}\left(2i\sin(\theta/2)\right)}\right)\\\\ &=\Re\left( e^{i(n+1)\theta/2}\frac{\sin(n\theta/2)}{\sin(\theta/2)}\right)\\\\ &=\Re\left( \left(\cos ((n+1)\theta/2)+i\sin ((n+1)\theta/2)\right)\frac{\sin(n\theta/2)}{\sin(\theta/2)}\right)\\\\ &=\frac{\sin(n\theta/2)}{\sin(\theta/2)}\cos ((n+1)\theta/2). \end{align} $$

  • $\begingroup$ This is perfect! (Upvoted) and thank you for sharing this. However, I'm looking to solve it without using $e^{in\theta}$ , and rather a continuation of my work (The way I have gone through the question, I want to continue it- Not a whole different way). $\endgroup$
    – M.S.E
    Sep 16, 2014 at 9:37
  • $\begingroup$ @Tharindu Writing $e^{i\alpha}$ or $\cos\alpha+i\sin\alpha$ is exactly the same thing, but the former is handier, because it suggests operations that are harder to see with the latter form. In particular, whenever one has $e^{i\alpha}-1$, writing this as $e^{i\beta}(e^{i\beta}-e^{-i\beta})$, where $2\beta=\alpha$, ofter reveals a very useful trick (it's similar to writing down $e^{i\alpha}-1$ in trigonometric form, if you think to it). $\endgroup$
    – egreg
    Sep 16, 2014 at 11:58
  • $\begingroup$ @egreg, I understand it is the same thing, but I mentioned in the question that please help me continue from what Im doing. This was my question, I didnt ask for a full solution (rather to help me tell where I went wrong and a Hint). $\endgroup$
    – M.S.E
    Sep 16, 2014 at 12:01
  • $\begingroup$ @egreg I like to do and know things in the hard (a more difficult method of handling) way too. $\endgroup$
    – M.S.E
    Sep 16, 2014 at 12:04
  • $\begingroup$ @Tharindu Thank you for your touching gesture. Hoping you will come back. $\endgroup$ Nov 24, 2014 at 16:56

Using the appropriate prosthaphaeresis formula,

$$\begin{align} \sin{\frac{\theta}{2}}\cos{k\theta} &=\frac12\left[\sin{\left(\frac{\theta}{2}+k\theta\right)}+\sin{\left(\frac{\theta}{2}-k\theta\right)}\right]\\ &=\frac12\left[\sin{\left(\left(k+\frac{1}{2}\right)\theta\right)}-\sin{\left(\left(k-\frac{1}{2}\right)\theta\right)}\right]. \end{align}$$

In this form, summation of $k$ now transparently telescopes:

$$\begin{align} \sum_{k=1}^{n}\sin{\frac{\theta}{2}}\cos{k\theta} &=\sum_{k=1}^{n}\frac12\left[\sin{\left(\left(k+\frac{1}{2}\right)\theta\right)}-\sin{\left(\left(k-\frac{1}{2}\right)\theta\right)}\right]\\ &=\frac12\left[\sin{\left(\left(n+\frac{1}{2}\right)\theta\right)}-\sin{\left(\left(1-\frac{1}{2}\right)\theta\right)}\right]\\ &=\frac12\left[\sin{\left(\left(n+\frac{1}{2}\right)\theta\right)}-\sin{\left(\frac{\theta}{2}\right)}\right]\\ &=\sin{\left(\frac{n\theta}{2}\right)\cos{\left(\frac{(n+1)\theta}{2}\right)}}. \end{align}$$

Now divide through by $\sin\frac{\theta}{2}$.

  • $\begingroup$ Im so sorry, I dont understand this, did you see the part of the question where it says to use $cos\theta+isin\theta$. $\endgroup$
    – M.S.E
    Sep 16, 2014 at 9:35
  • $\begingroup$ Im sorry, why did you take the summation of $sin\frac{\theta}{2} cos (k\theta)$ ??? $\endgroup$
    – M.S.E
    Sep 16, 2014 at 9:42
  • $\begingroup$ @Tharindu Because it's easier to calculate. Note that $\sum\limits_k\sin\frac\theta2\cos k\theta=\sin\frac\theta2\sum\limits_k\cos k\theta$, so he can just divide both sides by $\sin\frac\theta2$ and be done. (This is a different method than what your teacher wanted you to do, but it works.) $\endgroup$ Nov 18, 2014 at 11:11
  • $\begingroup$ @columbus8myhw Please take a minute to visit my User page, it was an accident to not mention it on placing the bounty. Actually, I am leaving this site, so I am giving away my reputation to the ones who have helped me so much as a way of appreciation. $\endgroup$
    – M.S.E
    Nov 18, 2014 at 11:13

Hint : You have not taken the real part correctly as you have missed the $isin\theta$ in the denominator

  • $\begingroup$ Yes, thank you. $\endgroup$
    – M.S.E
    Sep 16, 2014 at 9:23
  • $\begingroup$ Fixed my mistake, now the problem became even harder :p, take a look at my edit. $\endgroup$
    – M.S.E
    Sep 16, 2014 at 9:31
  • $\begingroup$ try by taking the conjugate of $1+cos\theta+isin\theta$ and why dont you take it as $e^{i\theta}=cos\theta+isin\theta$ which smplifies the result as you can see. $\endgroup$
    – Jasser
    Sep 16, 2014 at 9:48
  • $\begingroup$ I'm doing it :) I still end up with a term containing $i$. I want to try without taking $e^{i\theta}$ ;) $\endgroup$
    – M.S.E
    Sep 16, 2014 at 10:00

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