# Proof explanation for a sum of sines [duplicate]

Looking through proofs and there's one part I'm confused about, that if solved will get me straight to the answer.

I need to show that $$\sum_{m=0}^{N-1} \sin((m+\frac{1}{2})x) = \frac{\sin(Nx/2)^2}{\sin(x/2)}$$

This is quite a simple solution based on telescoping – see e.g. the Wikpiedia article.

Let's multiply both sides by the denominator first: $$\sin(x/2) \sum\limits_{m=0}^{N-1} \sin((m+\tfrac{1}{2})x) = \sin(Nx/2)^2$$

Then we include the leading term into the sum and utilize a well-known trigonometric identity $$\sin(a)\sin(b)=\tfrac 12(\cos(a-b)-\cos(a+b))$$ to transform products into sums:

$$LHS = \sum\limits_{m=0}^{N-1} \left(\sin(x/2) \sin((m+\tfrac{1}{2})x)\right) =$$ $$\frac 12\,\sum\limits_{m=0}^{N-1} \left(\cos(x/2 - (m+\tfrac{1}{2})x) - \cos(x/2 + (m+\tfrac{1}{2})x)\right) =$$ $$\frac 12\,\sum\limits_{m=0}^{N-1} \left(\cos(-mx) - \cos((m+1)x)\right) =$$ $$\frac 12\,\sum\limits_{m=0}^{N-1} \left(\cos(mx) - \cos((m+1)x)\right)$$

The second term of each sum cancels the first term of the next sum, and the whole sum telescopes so that only the first and last terms remain:

$$= \frac 12\, \left(\cos(0x) - \cos(Nx)\right) =$$

which we transform back into a product by the same identity mentioned above:

$$= \sin(Nx/2)\sin(Nx/2) = \sin^2(Nx/2) = RHS.$$

Q.E.D.

• This is an interesting approach!
– mjw
Mar 4 at 21:50

One way:

Recall that$$\sin p = \frac{1}{2i}[e^{ip}-e^{-ip}]$$

Also, $$\sum_{m=0}^{N-1} a^m = \frac{1-a^N}{1-a}.$$ Thus $$\sum_{m=0}^{N-1} \sin [(m+\frac{1}{2})x]= \frac{1}{2i} \sum_{m=0}^{N-1}\left\{ \exp [(m+\frac{1}{2})ix]- \exp [-(m+\frac{1}{2})ix] \right\}.$$

Now you have two geometric series. Compute the sum each of them and combine $$\cdots.$$

• Easier way is to say that $\sin x = \operatorname{Im}e^{ix}$. Mar 4 at 21:09
• @vitamind, agreed. That is a cool way to do it!
– mjw
Mar 4 at 21:10