$\left(\frac{\sin(\frac{n\theta}{2})}{\sin(\frac{\theta}{2})}\right)^2=\left|\sum_{k=1}^{|n|}e^{ik\theta}\right|^2$ I'm having trouble proving $$\left(\frac{\sin(\frac{n\theta}{2})}{\sin(\frac{\theta}{2})}\right)^2=\left|\sum_{k=1}^{|n|}e^{ik\theta}\right|^2$$ where $n\in\mathbb{Z}$ and $\theta\in\mathbb{R}$. Can anyone suggest a hint?
 A: Using Euler's Formula , $e^{ix}= \cos x+i\sin x$
So, $e^{-ix}= \cos(-x)+i\sin (-x)=\cos x-i\sin x\implies 2i\sin x=e^{ix}-e^{-ix}$ 
$$\text{ If }n>0,\sum_{k=1}^{|n|}e^{ik\theta}= \sum_{k=1}^n e^{ik\theta}=e^{i\theta}\left( \frac{e^{in\theta}-1}{e^{i\theta}-1}\right)$$
$$=\frac{e^{in\frac\theta2}}{e^{i\frac\theta2}}\frac{(e^{in\frac\theta2}-e^{-in\frac\theta2})}{(e^{i\frac\theta2}-e^{-i\frac\theta2})}$$
$$=e^{i\frac{(n-1)\theta}2}\frac{2i\sin \frac{n\theta}2}{2i\sin \frac{\theta}2}$$
$$=\left(\cos \frac{(n-1)\theta}2+i\sin \frac{(n-1)\theta}2\right) \frac{\sin \frac{n\theta}2}{\sin \frac{\theta}2}$$
Taking modulus  $$\left|\sum_{k=1}^{|n|}e^{ik\theta}\right|$$
$$=\left|\left(\cos \frac{(n-1)\theta}2+i\sin \frac{(n-1)\theta}2\right) \frac{\sin \frac{n\theta}2}{\sin \frac{\theta}2}\right|$$
$$=\left| \cos \frac{(n-1)\theta}2+i\sin \frac{(n-1)\theta}2 \right|\left|\frac{\sin \frac{n\theta}2}{\sin \frac{\theta}2}\right|$$
$$=\left|\frac{\sin \frac{n\theta}2}{\sin \frac{\theta}2}\right|$$
Similarly, for $n<0$
A: Try this-
We know the Euler's Formula-$e^{ix} = \cos x + i\sin x \ $-http://en.wikipedia.org/wiki/Euler%27s_formula.
Link this to summation of sines in as arithmetic progression. Try Proving this!
From wiki:-
Sum of sines and cosines with arguments in arithmetic progression:if $\alpha\ne0$, then
\begin{align} & \sin{\varphi} + \sin{(\varphi + \alpha)} + \sin{(\varphi + 2\alpha)} + \cdots {} \\[8pt] & {} \qquad\qquad \cdots + \sin{(\varphi + n\alpha)} = \frac{\sin{\left(\frac{(n+1) \alpha}{2}\right)} \cdot \sin{(\varphi + \frac{n \alpha}{2})}}{\sin{\frac{\alpha}{2}}} \quad\hbox{and}\\[10pt] & \cos{\varphi} + \cos{(\varphi + \alpha)} + \cos{(\varphi + 2\alpha)} + \cdots {} \\[8pt] & {} \qquad\qquad \cdots + \cos{(\varphi + n\alpha)} = \frac{\sin{\left(\frac{(n+1) \alpha}{2}\right)} \cdot \cos{(\varphi + \frac{n \alpha}{2})}}{\sin{\frac{\alpha}{2}}}. \end{align} 
A: Using the sum of sines and cosines with arguments in arithmetic progression as given above: if $\theta\ne0$ and let $\varphi =0$, then we have,
\begin{align} &S =\sin{(\theta)} + \sin{(2\theta)} + \cdots + \sin{(n\theta)} = \frac{\sin{\left(\frac{(n+1) \theta}{2}\right)} \cdot \sin{(\frac{n \theta}{2})}}{\sin{\frac{\theta}{2}}} \quad\hbox{and}\\[10pt] &C =\cos{(\theta)} + \cos{(2\theta)} + \cdots+ \cos{(n\theta)} = \frac{\cos{\left(\frac{(n+1) \theta}{2}\right)} \cdot \sin{(\frac{n \theta}{2})}}{\sin{\frac{\theta}{2}}}. \end{align} 
From Euler's Formula and the definition of absolute value of a complex number, we can write $$\left|\sum_{k=1}^{|n|}e^{ik\theta}\right|^2 = C^2 + S^2$$ 
$$= \frac{\sin{\left(\frac{n \theta}{2}\right)}^2}{\sin{\frac{\theta}{2}}^2}\cdot\left(\cos{(\frac{(n+1) \theta}{2})}^2 + \sin{(\frac{(n+1) \theta}{2})}^2\right)$$
$$ = \left(\frac{\sin{\frac{n \theta}{2}}}{\sin{\frac{\theta}{2}}}\right)^2$$
A: In order to proof 
$$\sum_{k=1}^{n}{\sin \left(\varphi + k\alpha \right)}=\frac{\sin\left(\frac{\left(n+1\right)\alpha}{2}\right)\cdot\sin\left(\varphi+\frac{n\alpha}{2}\right)}{\sin \frac{\alpha}{2}}$$
observe
$$\sin \frac{\alpha}{2}\cdot \sin \left(\varphi + k\alpha\right)=\frac{1}{2}\left[\cos \left(\varphi + k\alpha - \frac{\alpha}{2}\right)-\cos \left(\varphi+k\alpha+\frac{\alpha}{2}\right)\right]$$
because $\sin u \cdot \sin v = \frac{1}{2}\left[\cos \left(u-v\right) - \cos \left(u + v\right)\right]$.
Then $\sin \frac{\alpha}{2}\cdot\sum_{k=1}^{n}{\sin \left(\varphi + k\alpha \right)}$ is a telescopic series:
$$\sin \frac{\alpha}{2}\cdot\sum_{k=1}^{n}{\sin \left(\varphi + k\alpha \right)} = \frac{1}{2}\left[\cos \left(\varphi + \frac{\alpha}{2}\right)-\cos \left(\varphi + n\alpha +\frac{\alpha}{2}\right)\right].$$
And use $\sin u \cdot \sin v = \frac{1}{2}\left[\cos \left(u-v\right) - \cos \left(u + v\right)\right]$ again in order to reduce the last expresion.
Also, for the identity
$$\sum_{k=1}^{n}{\cos \left(\varphi + k\alpha \right)}=\frac{\sin\left(\frac{\left(n+1\right)\alpha}{2}\right)\cdot\cos\left(\varphi+\frac{n\alpha}{2}\right)}{\sin \frac{\alpha}{2}}$$
we can use the identity $\cos u \sin v =\frac{1}{2}\left[\sin \left(u+v\right)-\sin \left(u-v\right)\right]$:
$$\cos \left(\varphi + k\alpha\right)\cdot \sin \frac{\alpha}{2}=\frac{1}{2}\left[\sin \left(\varphi + k\alpha + \frac{\alpha}{2}\right) - \sin \left(\varphi + k\alpha - \frac{\alpha}{2}\right)\right].$$
And so on.
