What trigonometric identities produce $\sum_{n=0}^4e^{-j\omega n}=e^{-j2\omega}\frac{\sin(5\omega/2)}{\sin(\omega/2)}$?

Question

Reading through my digital signal processing text, I came across this statement: $$X(e^{j\omega})=\sum_{n=0}^4e^{-j\omega n}=e^{-j2\omega}\frac{\sin(5\omega/2)}{\sin(\omega/2)}$$ I was able to reduce it to the following using Euler's formula: $$e^{-2j\omega}(1+2\cos\omega+2\cos2\omega)$$ I'm thinking something along the lines of a double-angle identity, but I'm hoping there's an easier way that I'm not realizing.

Answer 1

Summing the geometric series gives $$\sum_{n=0}^4e^{-nj\omega}=\frac{1-e^{-5j\omega}}{1-e^{-j\omega}}$$ Now we have $$1-e^{jx}=2\sin(x/2)e^{j(x-\pi)/2}{}^*$$ and thus $$\frac{1-e^{-5j\omega}}{1-e^{-j\omega}}=\frac{2\sin(-5\omega/2)e^{j(-5\omega-\pi)/2}}{2\sin(-\omega/2)e^{j(-\omega-\pi)/2}}=e^{-2j\omega}\frac{\sin5\omega/2}{\sin\omega/2}$$

*A geometric derivation of this identity is shown above. The magnitude of $1-e^{-jx}$ can be calculates using the cosine rule as $\sqrt{1^2+1^2-2(1)(1)\cos x}$, which simplifies to $\sqrt{2-2\cos x}=2\sin\frac x2$; the argument is an easy application of angles in a triangle and is $\frac{x-\pi}2$.