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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?

  • 0
    I found this video: https://www.youtube.com/watch?v=7LBQTpiK-Xg helpful when trying to get an intuition for how this would work. Dr. Peyam takes on the case for when $a=1$ and $d=1$.2018-04-22

5 Answers 5

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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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    You mean "for the case when $d=a$".2016-02-21
71

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

  • 38
    Or perhaps more simply, just sum up $e^{ix}$ and extract the real and imaginary parts...2011-01-18
59

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

enter image description here

40

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.

7

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