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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    $\begingroup$ You probably meant: $\sum_{k=0}^{n-1}\cos (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \cdot \cos( \frac{ a + (n-1)\cdot d}{2})$ $\endgroup$ – Raskolnikov Jan 18 '11 at 9:57
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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
  • $\begingroup$ Yes,that's the formulas I meant. $\endgroup$ – Raskolnikov Jan 18 '11 at 10:18

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 $d=1$. Just take a look here:

  • $\begingroup$ You mean "for the case when $d=a$". $\endgroup$ – user236182 Feb 21 '16 at 9:22

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 at 17:03

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 $\sum_{k = 0}^n \sin(a + 2dk)$.


protected by user99914 Jan 28 '18 at 4:19

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