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?

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    $\begingroup$ 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). $\endgroup$ – Raskolnikov Jan 18 '11 at 10:03
  • $\begingroup$ Alternative hint: make an induction proof. $\endgroup$ – Raskolnikov Jan 18 '11 at 10:04
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    $\begingroup$ Simpson's formula?! Do you mean this: mathworld.wolfram.com/ProsthaphaeresisFormulas.html $\endgroup$ – Quixotic Jan 18 '11 at 10:04
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    $\begingroup$ Yes,that's the formulas I meant. $\endgroup$ – Raskolnikov Jan 18 '11 at 10:18
  • $\begingroup$ @Raskolnikov:I don't know if I understand you,say I have $ \cos a + \cos (a+d) + \cos (a+2d) $ using your idea,I am getting $2\cos (a+d) cos d + cos(a+d) $ then $ \cos (a+d) \times ( 2 \cos (d) +1) $ what next? $\endgroup$ – Quixotic Jan 18 '11 at 10:23

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:

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    $\begingroup$ It appears that the mathforum.org link is broken. $\endgroup$ – Gregory Dresden Mar 4 at 4:32

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

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

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

enter image description here

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    $\begingroup$ Beautiful! One doesn't see that very often. $\endgroup$ – Andreas Jan 4 '19 at 17:03
  • $\begingroup$ I've understood where the $\frac{1}{2}\frac{1}{\sin\theta}\sin n\theta$ and $\cos2(n-1)\theta$come from but I do not see visually from the diagram why the product of the two gives the answer - is the product area? Please could you explain why the product of the two on the diagram gives the answer. $\endgroup$ – onepound Feb 5 '20 at 14:45
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    $\begingroup$ @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 Feb 5 '20 at 15:01
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    $\begingroup$ thanks for the link and the explaination - very useful diagram. $\endgroup$ – onepound Feb 5 '20 at 15:11

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 one 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


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