Prove that $ \frac{\sin(2m+1) \theta}{\sin \theta} = 1 + 2 \cos2\theta + 2\cos4\theta+\dots+2 \cos2m\theta$

Question

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

Answer 1

Because $$\sin\theta\left(1 + 2 \cos2\theta + 2\cos4\theta+ ...+2 \cos2m\theta\right)=$$ $$=\sin\theta+\sin3\theta-\sin\theta+\sin5\theta-\sin3\theta+...+\sin(2m+1)\theta-\sin(2m-1)\theta=$$ $$=\sin(2m+1)\theta.$$

Answer 2

Hint...Consider that $$2\cos r\theta\sin \theta=\sin(r\theta+\theta)-\sin(r\theta-\theta)$$

So if you sum both sides from $r=1$ to $m$ you have a telescoping series which gives you the result.

Answer 3

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}