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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

Michael Rozenberg
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mathpieuler
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3 Answers3

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

Michael Rozenberg
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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.

David Quinn
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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}

Bernard
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  • Thanks, combining the solution of Michael and the hint of David, I do understand it now! – mathpieuler Feb 26 '18 at 23:47
  • The formula I mention is easy to prove (if you don't know it), considering the geometric series with ratio $\mathrm e^{ix}$,summing the first $m$ terms, then taking the real part of the result. – Bernard Feb 27 '18 at 00:38
  • It took me a few hours to understand your solution, But I think I have got it. I didn't know the RHS in your first equation. – mathpieuler Feb 27 '18 at 15:21
  • It was that hard, really? Note there's a similar formula for the sum of sines in arithmetic progression, where the only difference is the second factor ($\sin\frac{mx}2$ in the place of $\cos\frac{mx}2$). – Bernard Feb 27 '18 at 15:59
  • This proof is unacceptable because the initial formula you quote is relying on the general property which needs to be proved. – Tom Apr 28 '18 at 02:12
  • @Tomp I rather think the formula to be proved is a specialisation of the general formula I quote (which is a high school formula anyway). – Bernard May 17 '18 at 18:34