Relationship between e and sine I am reading a physics text about scattering but I am stuck with the maths.
I think the solution might be related to Euler's formula where:
$$\frac{e^{i\theta} - e^{-i\theta} }{2i} = \sin\theta$$
The text says that:
$$I(Q) = \bigg|\sum_{n=0}^{N-1}b e^{ikna\sin\theta}\bigg|^2$$
and
$$\sum_{n=0}^{N-1} u^{n} = \frac{1-u^N}{1-u}$$
where $u = e^{ika\sin\theta}$  and they say that this means:
$$I(Q) = b^2\frac{\sin^2(\tfrac12kNa\sin\theta)}{\sin^2(\tfrac12ka\sin\theta)}$$
Can someone show this?
Thanks
 A: We shall use that for all $s\in\mathbb R$,
$$|1-e^{is}|=|e^{is}-1|=|e^{is/2}(e^{is/2}-e^{-is/2})|=|2ie^{is/2}\sin(s/2)|=|2\sin(s/2)|.$$
Let $t=ka\sin\theta$.
$$\begin{align}I(Q)&=b^2\left|\sum_{n=0}^{N-1}(e^{it})^n\right|^2\\&=b^2\left|\frac{1-e^{iNt}}{1-e^{it}}\right|^2
\\&=b^2\left|\frac{2\sin(Nt/2)}{2\sin(t/2)}\right|^2
\\&=b^2\frac{\sin^2(Nt/2)}{\sin^2(t/2)}.
\end{align}$$
A: Let $x=kna \sin(\theta)$.  Then we can write
$$\begin{align}
\sum_{n=0}^{N-1}be^{ikna \sin(\theta)}&=b\sum_{n=0}^{N-1}be^{ix}\\\\
&=b\frac{1-e^{iNx}}{1-e^{ix}}\\\\
&=b\frac{e^{-iNx/2}(e^{iNx/2}-e^{-iNx/2})}{e^{-ix/2}(e^{ix/2}-e^{-ix/2})}\\\\
&=be^{-ix(N-1)/2}\frac{2i\sin(Nx/2)}{2i\sin(x/2)}\\\\
&=be^{-ix(N-1)/2}\frac{\sin(Nx/2)}{\sin(x/2)}
\end{align}$$
Now, taking the magnitude and squaring and ssuming $b$ and $kna\sin(\theta)$ are real yields
$$\begin{align}
\left|\sum_{n=0}^{N-1}be^{ikna \sin(\theta)}\right|^2&=\left|b\sum_{n=0}^{N-1}be^{ix}\right|^2\\\\
&=|b|^2\,\underbrace{\left|e^{-ix(N-1)/2}\right|^2}_{=1} \left|\frac{\sin(Nx/2)}{\sin(x/2)}\right|^2\\\\
&=b^2 \frac{\sin^2(Nx/2)}{\sin^2(x/2)}\\\\
&=b^2 \frac{\sin^2(Nkna\sin(\theta)/2)}{\sin^2(kna\sin(\theta)/2)}
\end{align}$$
as was to be shown!
A: $$
I(Q) = \left|\sum_{n=0}^{N-1}b e^{ikna\sin\theta}\right|^2
$$
Assume $a$, $b$, and $\theta$ are real.  The sum here is a geometric series (take $u=e^{i k a \sin \theta}$ so that $u^n = e^{ikna\sin\theta}$).  So
\begin{align}
I(Q) &= \left| b \frac{1-e^{i N k a \sin \theta}}{1-e^{i k a \sin \theta}}\right|^2
\\ &=
\left|b\frac{e^{i\frac12 N k a\sin\theta}}{e^{i\frac12 k a\sin\theta}}
\;\frac{e^{-i \frac12 N k a \sin \theta}-e^{i \frac12 N k a \sin \theta}}{e^{-i \frac12 k a \sin \theta}-e^{i \frac12 k a \sin \theta}}\right|^2
\\ &=
|b|^2\cdot 1^2\cdot \left(\frac{\sin (\frac12 N k a \sin \theta)}{\sin( \frac12 k a \sin \theta)}\right)^2\cdot\left(\frac{2i}{2i}\right)^2
\\ &=
b^2\frac{\sin^2(\tfrac12kNa\sin\theta)}{\sin^2(\tfrac12ka\sin\theta)}
\end{align}
A: So first, we have the identities
$$|x+iy|^2 = x^2 + y^2 \text{ for } x,y\in\mathbb{R}$$
$$\left|\frac{x}{y}\right| = \frac{|x|}{|y|}$$
$$1-\cos(x) = 2\sin^2(x/2)$$
$$\cos^2(x) + \sin^2(x) = 1$$
$$\sum_{j=0}^{N-1} r^j = \frac{1-r^N}{1-r}$$
Also for any $x$,
$$
\begin{align*}
\left|1-e^{ix}\right| &= |1-\cos(x)-i\sin(x)|^2 \\
&= (1-\cos(x))^2 + \sin^2(x) \\
&= 1 - 2\cos(x) + \cos^2(x) + \sin^2(x) \\
&= 2 - 2\cos(x) \\
&= 4\sin^2(x/2)
\end{align*}
$$
Let $\phi = ka\sin(\theta)$
Then
$$
\begin{align*}
I &= b^2\left|\sum_{n=0}^{N-1} e^{i\phi n}\right|^2 \\
&= b^2\left|\frac{1-e^{i\phi N}}{1-e^{i\phi}}\right| \\
&= b^2 \frac{|1-e^{i\phi N}|}{|1-e^{i\phi}|} \\
&= b^2 \frac{\sin^2(N\phi/2)}{\sin^2(\phi/2)} \\
&= b^2 \frac{\sin^2(\frac{1}{2}Nka\sin(\theta))}{\sin^2(\frac{1}{2}ka\sin(\theta))}
\end{align*}
$$
