How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? For example here is the sum of $\cos$ series:

$$\sum_{k=0}^{n-1}\cos (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \cos \biggl( \frac{ 2 a + (n-1)\cdot d}{2}\biggr)$$

There is a slight difference in case of $\sin$, which is: $$\sum_{k=0}^{n-1}\sin (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \sin\biggl( \frac{2 a + (n-1)\cdot d}{2}\biggr)$$

How do we prove the above two identities?

  • 2
    $\begingroup$ Hint: reverse the series and sum it up term by term with the original series. So $\cos(a)+\cos(a+(n-1)\cdot d)$, etc... And use the Simpson formula for sums of cosines (and sines for the other identity). $\endgroup$ Commented Jan 18, 2011 at 10:03
  • $\begingroup$ Alternative hint: make an induction proof. $\endgroup$ Commented Jan 18, 2011 at 10:04
  • 7
    $\begingroup$ Simpson's formula?! Do you mean this: mathworld.wolfram.com/ProsthaphaeresisFormulas.html $\endgroup$
    – Quixotic
    Commented Jan 18, 2011 at 10:04
  • 1
    $\begingroup$ Yes,that's the formulas I meant. $\endgroup$ Commented Jan 18, 2011 at 10:18
  • 1
    $\begingroup$ Here is my proof. math.stackexchange.com/a/3787528/577710 $\endgroup$ Commented Aug 11, 2020 at 18:42

8 Answers 8


Let $$ S = \sin{(a)} + \sin{(a+d)} + \cdots + \sin{(a+nd)}$$ Now multiply both sides by $\sin\frac{d}{2}$. Then you have $$S \times \sin\Bigl(\frac{d}{2}\Bigr) = \sin{(a)}\sin\Bigl(\frac{d}{2}\Bigr) + \sin{(a+d)}\cdot\sin\Bigl(\frac{d}{2}\Bigr) + \cdots + \sin{(a+nd)}\cdot\sin\Bigl(\frac{d}{2}\Bigr)$$

Now, note that $$\sin(a)\sin\Bigl(\frac{d}{2}\Bigr) = \frac{1}{2} \cdot \biggl[ \cos\Bigl(a-\frac{d}{2}\Bigr) - \cos\Bigl(a+\frac{d}{2}\Bigr)\biggr]$$ and $$\sin(a+d) \cdot \sin\Bigl(\frac{d}{2}\Bigr) = \frac{1}{2} \cdot \biggl[ \cos\Bigl(a + d -\frac{d}{2}\Bigr) - \cos\Bigl(a+d+\frac{d}{2}\Bigr) \biggr]$$

Then by doing the same thing you will have some terms cancelled out. You can easily see which terms are going to get Cancelled. Proceed and you should be able to get the formula.

I tried this by seeing this post. This has been worked for the case when $d=a$. Just take a look here:

  • 3
    $\begingroup$ It appears that the mathforum.org link is broken. $\endgroup$ Commented Mar 4, 2021 at 4:32
  • $\begingroup$ The Wayback Machine doesn't appear to be accepting that link. It doesn't load. $\endgroup$
    – GSmith
    Commented Jun 4 at 7:35

Here's a trigonograph for $a = 0$ and $d = 2\theta$:

enter image description here

  • 8
    $\begingroup$ Beautiful! One doesn't see that very often. $\endgroup$
    – Andreas
    Commented Jan 4, 2019 at 17:03
  • 1
    $\begingroup$ @onepound: The big right triangle (with "trigonography.com" along its hypotenuse) has a hypotenuse length of $\sin n\theta/\sin\theta$. The triangle's acute angle on the left is an inscribed angle in the circular arc, so its measure is half the corresponding central angle, $2(n-1)\theta$. Thus, the horizontal and vertical legs of that right triangle are, respectively, $\text{(hypotenuse)}\cdot \cos(n-1)\theta$ and $\text{(hypotenuse)}\cdot \sin(n-1)\theta$. $\endgroup$
    – Blue
    Commented Feb 5, 2020 at 15:01
  • 1
    $\begingroup$ I am honestly embarrased of myself for not paying a closer look to the diagram itself. Still I want to thank you so much for such a beautiful geometric representation and the fact that you took your time out to respond so quickly to my comment on such an old post. Thanks a lot!! $\endgroup$ Commented Jul 25, 2023 at 19:18
  • 1
    $\begingroup$ @GSmith: "Did you create that image in the trigonography site?" Yep. I'm the Trigonographer. ... "Nice work!!!" Thanks!!! :) :) :) $\endgroup$
    – Blue
    Commented Jun 4 at 10:02
  • 1
    $\begingroup$ @Blue I figured that out from this pdf Almost Everything you Need to Know About Trig. Anyways, great answer! $\endgroup$
    – GSmith
    Commented Jun 4 at 10:17

Writing $\cos x = \frac12 (e^{ix} + e^{-ix})$ will reduce the problem to computing two geometric sums.

  • $\begingroup$ and the $\sin$ one ? $\endgroup$
    – Quixotic
    Commented Jan 18, 2011 at 10:25
  • 4
    $\begingroup$ The same trick, but with $\sin x=\frac{1}{2i} (e^{ix}-e^{-ix})$ instead. $\endgroup$ Commented Jan 18, 2011 at 11:14
  • 80
    $\begingroup$ Or perhaps more simply, just sum up $e^{ix}$ and extract the real and imaginary parts... $\endgroup$
    – Aryabhata
    Commented Jan 18, 2011 at 23:02

From Euler's Identity we know that $\cos (a+kd) = \text{Re}\{e^{i(a+kd)}\}$ and $\sin (a+kd) = \text{Im}\{e^{i(a+kd)}\}$.$\,$ Thus,

$$\begin{align} \sum_{k=0}^{n-1} \cos (a+kd) &= \sum_{k=0}^{n-1} \text{Re}\{e^{i(a+kd)}\}\\\\ &=\text{Re}\left(\sum_{k=0}^{n-1} e^{i(a+kd)}\right)\\\\ &=\text{Re}\left(e^{ia} \sum_{k=0}^{n-1} (e^{id})^{k} \right)\\\\ &=\text{Re} \left( e^{ia} \frac{1-e^{idn}}{1-e^{id}}\right) \\\\ &=\text{Re} \left( e^{ia} \frac{e^{idn/2}(e^{-idn/2}-e^{idn/2})}{e^{id/2}(e^{-id/2}-e^{id/2})}\right) \\\\ &=\frac{\cos(a+(n-1)d/2)\sin(nd/2)}{\sin(d/2)} \end{align}$$

as was to be shown. Likewise for the sine function identity, follow the same procedure and take the imaginary part of the sum rather than the real part.


This is similar to the currently accepted answer, but more straightforward. You can use the trig identity \begin{equation*} \sin(\alpha + \beta) - \sin(\alpha - \beta) = 2\sin \beta \cos \alpha. \end{equation*}

Let $a_n = a + 2dk$ be an arithmetic sequence of difference $2d$, and set $b_n = a_n - d = a + d(2k - 1)$. Note that $\{b_n\}$ is also an arithmetic sequence of difference $2d$, hence $a_n + d = b_n + 2d = b_{n + 1}$. Therefore

\begin{equation*} 2 \sin d \cos a_n = \sin(a_n + d) - \sin(a_n - d) = \sin b_{n + 1} - \sin b_n. \end{equation*}

Summing both sides from $0$ to $n$ yields

\begin{align*} 2 \sin d \sum_{k = 0}^n \cos a_k &= \sin b_{n + 1} - \sin b_0 \\ &= \sin(a + d(2n + 1)) - \sin(a - d). \end{align*}

From our original trig identity, \begin{equation*} 2\sin((n + 1)d) \cos(a + nd) = \sin(a + d(2n + 1)) - \sin(a - d). \end{equation*} Thus, if $\sin d \neq 0$, we can rewrite our result as \begin{equation*} \sum_{k = 0}^n \cos (a + 2dk) = \frac{\sin((n + 1)d) \cos(a + nd)}{\sin d}. \end{equation*} This is OP's formula with $2d$ and $n$ instead of $d$ and $n - 1$. A similar process will yield the formula for $\sum_{k = 0}^n \sin(a + 2dk)$.


Small comment:

If we have one of the identities then we can derive the other!


$$\sum_{k=0}^{n-1}\cos (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \cos \biggl( \frac{ 2 a + (n-1)\cdot d}{2}\biggr)$$

Take the derivative on both sides with $a$ while keeping everything else constant:

$$\sum_{k=0}^{n-1}\sin (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \sin \biggl( \frac{ 2 a + (n-1)\cdot d}{2}\biggr)$$


There's also a nice proof of the cosine formula by Michael Knapp in Mathematics Magazine (vol 82, number 5, December 2009, p. 371-372). Download a free version here: https://www.maa.org/sites/default/files/Knapp200941575.pdf

Knapp mentions in his article that a proof for the sine formula was given by Samuel Greitzer in the obscure student-oriented math journal Arbelos from 1986.


I did this by using $\Large{\frac{e^{i\theta}+e^{-i\theta}}{2}}$ formula to their corresponding cos terms




using geometric progression formula



=$\Large{(\frac{e^{i\alpha(n+1)}-1}{e^{i\alpha}-1})(\frac{e^{i(2\theta+n\alpha)}+1}{2e^{i(\theta+n\alpha)}})}$ ........(1)

Now $\Large{(\frac{e^{i\alpha(n+1)}-1}{e^{i\alpha}-1})*\frac{(e^{-i\alpha}-1)}{(e^{-i\alpha}-1)}}$

=$\Large{\frac{e^{i\alpha n}-e^{-i\alpha}+1-e^{-i\alpha(n+1)}}{2-2cos(\alpha)}}$

and $\Large{(\frac{e^{i(2\theta+n\alpha)}+1}{2e^{i(\theta+n\alpha)}})*\frac{(2e^{-i(\theta+n\alpha)})}{(2e^{-i(\theta+n\alpha)})}}$

=$\Large{\frac{e^{i\theta}+e^{-i(\theta+\alpha n)}}{2}}$

So from (1)

$\Large{(\frac{e^{i\alpha n}-e^{-i\alpha}+1-e^{-i\alpha(n+1)}}{2-2cos(\alpha)})}*\Large{(\frac{e^{i\theta}+e^{-i(\theta+\alpha n)}}{2})}$

=$\Large{\frac{e^{i(\alpha n + \theta)}-e^{-i(\alpha n + \theta)}-e^{i(\theta-\alpha)}-e^{-i(\theta-\alpha)}+e^{i\theta}+e^{-i\theta}-e^{i(\theta+\alpha(n+1))}-e^{-i(\theta+\alpha(n+1))}}{4(1-cos(\alpha))}}$

=$\Large{\frac{cos(\alpha n+\theta)-cos(\theta-\alpha)+cos(\theta)-cos(\theta+\alpha(n+1))}{2(1-cos(\alpha))}}$ ..........(2)




$cos(\alpha n+\theta)-cos(\theta+\alpha(n+1))=2sin(\frac{2\alpha n +\alpha+2\theta}{2})(sin(\frac{\alpha}{2}))$

also $(1-cos(\frac{\alpha}{2})=2sin^{2}(\frac{\alpha}{2}))$

from (2)

$\Large{\frac{sin(\frac{\alpha}{2}-\theta)+sin(\theta+\alpha n+\frac{\alpha}{2})}{2sin(\frac{\alpha}{2})}}$

=$\Large{\frac{sin(\frac{\alpha(n+1)}{2})cos(\theta+\frac{\alpha n}{2})}{sin(\frac{\alpha}{2})}}$

The sine series can also be done in this way


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .