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

How is it possible to show for integer $m$:

$$\frac{1}{M}\sum_{k=1}^{M}\sin(m\cdot y_{k})=0$$

Thank you very much

Interval $[-\pi,\pi]$ split into $M$ equal intervals, with the mid point of interval is $y_{k}$

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marked as duplicate by Gerry Myerson, Jyrki Lahtonen, Jennifer Dylan, William, wentaway Sep 7 '12 at 12:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

what is m, N? did you mean "for integer m"? if so, what about m=1? – tomasz Jun 17 '12 at 20:28
@Gerry: while abstractly I agree it is a duplicate, in this particular case the question is about summing an odd function at points symmetric about zero. – Willie Wong Jun 20 '12 at 8:01
up vote 4 down vote accepted

The sine function is odd function, meaning that in general $\sin(-t)=-\sin(t)$.

The interval $[-\pi,\pi]$ is symmetrical about $0$. So if $x$ is one of the $x_k$, then $-x$ also is one of the $x_k$, and the values of $\sin(mx)$ and $\sin(-mx)$ add up to $0$. (If $N$ is odd, there isn't the perfect twinning, but the untwinned point is $0$, and $\sin(0)=0$.)

Exactly the same argument works for any odd function and any interval symmetric about the origin.

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but how to proove this? – nanme Jun 17 '12 at 20:37
How to prove that $\sin x$ is an odd function? – André Nicolas Jun 17 '12 at 20:40
thank you very much but problem I try to apply concept to new problem but cosine function not symmmetric and so don't know how to show the statements truth.$$ \frac{1}{M}\sum_{j=1}^{M}\cos(mx_{j})=\begin{cases} 1, & \ m \equiv 0\pmod{M}\\ 0, & \text{else} \end{cases}$$ – nanme Jun 17 '12 at 20:52
@nanme: Different functions may require different techniques. My answer used a technique that works for all odd functions. The cosine problem above is not yet clear enough for me to give an answer. – André Nicolas Jun 17 '12 at 20:59
I understand. it follows from the same description as the original sin question. I try to find technique that is general and so can be used in many cases but am not sure if there is one that does exist. thank you again very much – nanme Jun 17 '12 at 21:09

When you sum over the $y_k$'s backwards, you get the both the same value and its negative:

$$\frac{1}{N}\sum_{k=1}^N \sin(m y_k) = \frac{1}{N} \sum_{k=1}^N \sin(m y_{N+1-k}) = \frac{1}{N} \sum_{k=1}^N \sin(-m y_k) = - \frac{1}{N} \sum_{k=1}^N \sin(m y_k)$$

so the sum is zero.

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hi and thank you for response. however i don't seem to follow your logic thank you again – nanme Jun 17 '12 at 20:32
The $y_k$ are symmetric about 0 and $\sin(-x) = -\sin(x)$ for any $x$. – Cocopuffs Jun 17 '12 at 20:42

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