<s>[INST] Prove that $ \frac{\sin(2m+1) \theta}{\sin \theta} = 1 + 2 \cos2\theta + 2\cos4\theta+\dots+2 \cos2m\theta$ I am struggling with a part of my textbook can anyone help me? It states:
$$ \frac{\sin(2m+1) \theta}{\sin \theta} = 1 + 2 \cos2\theta + 2\cos4\theta+\dots+ 2\cos2m\theta$$
Why is this obvious? Do I have to perform a Taylor series?
Any help appreciated
 [/INST] You can use the standard formula:
$$1+\cos x+\cos 2x+\dots +\cos mx=\frac{\sin\dfrac{(m+1)x}2}{\sin\dfrac{x}2}\,\cos\dfrac{mx}2.$$
Here, with $x=2\theta$, one obtains
\begin{align} 
1 +{} & 2 \cos2\theta + \dots+2 \cos2m\theta =2(1 + \cos2\theta + \dots+\cos2m\theta)-1 \cr
={} &\frac{2\sin(m+1)\theta}{\sin\theta}\,\cos m\theta -1 = \frac{\sin\bigl((m+1)+m\bigr)\theta+\sin\bigl((m+1)-m\bigr)\theta}{\sin\theta}-1 \cr
={}&\frac{\sin(2m+1)\theta+\sin\theta}{\sin\theta}-1 =\frac{\sin(2m+1)\theta}{\sin\theta}.
\end{align}
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