How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?

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?

• 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). Jan 18, 2011 at 10:03
• Alternative hint: make an induction proof. Jan 18, 2011 at 10:04
• Simpson's formula?! Do you mean this: mathworld.wolfram.com/ProsthaphaeresisFormulas.html Jan 18, 2011 at 10:04
• Yes,that's the formulas I meant. Jan 18, 2011 at 10:18
• Here is my proof. math.stackexchange.com/a/3787528/577710 Aug 11, 2020 at 18:42

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:

• It appears that the mathforum.org link is broken. Mar 4, 2021 at 4:32

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

• Beautiful! One doesn't see that very often. Jan 4, 2019 at 17:03
• 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. Feb 5, 2020 at 14:45
• @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 $\text{(hypotenuse)}\cdot \sin(n-1)\theta$.
– Blue
Feb 5, 2020 at 15:01
• I know I am asking this very late but where does the factor of "1/2" come from in the length of the hypoteneuse of the triangle with angle "θ". Shouldn't it just be "1/sinθ" (as sin θ = P/H, so H = P/sin θ and P is just cos 0 which is 1)? Also, I don't think it is related to the area in anyway because we are unware of the base of that triangle. Jul 25, 2023 at 18:32
• 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!! Jul 25, 2023 at 19:18

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

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

Consider:

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

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

L.H.S

=$$\Large{cos{(\theta)}+cos{(\theta+\alpha)}+......+cos(\theta+n\alpha)}$$

=$$\Large{\frac{e^{i\theta}}{2}(1+e^{i\alpha}+e^{i2\alpha}+......+e^{in\alpha})+\frac{e^{-i\theta}}{2}(1+e^{-i\alpha}+e^{-i2\alpha}+......+e^{-in\alpha})}$$

using geometric progression formula

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

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

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

Now

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

and

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

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