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Use the identity $$\cos(A-B) -\cos(A+B) = 2\sin(A)\sin(B)$$ to prove that:


This seemed almost too easy, so I am wondering if I am missing something?

I just let $A = j\theta$ and $B = \frac{1}{2}\theta$, and substituted it directly into the identity. Since the question has asked me to use the identity, I am assuming that we can use it directly.

Have I missed anything?

The next part of the question asks:

Deduce that

$$\sum^n_{j=1}\sin(j\theta) = \frac{\cos(\frac{1}{2}\theta)-\cos((n+\frac{1}{2})\theta)}{2\sin(\frac{1}{2}\theta)}, \text{ if $\theta$ is not a multiple of $2\pi$}$$

Writing this out in general terms I see that the middle terms all cancel out (i have omitted the denominator here):

$$\cos((j-\frac{1}{2})\theta)-\cos((j+\frac{1}{2})\theta) + \cos(((j+1)-\frac{1}{2})\theta)-\cos(((j+1)+\frac{1}{2})\theta) + \cos(((j+2)-\frac{1}{2})\theta)-\cos(((j+2)+\frac{1}{2})\theta)+\dotsb+ \cos((n-\frac{1}{2})\theta)-\cos((n+\frac{1}{2})\theta)$$

I notice that if $\theta$ is a multiple of $2\pi$, then I would lose the 1st term, thus $\theta$ cannot be a multiple of $2\pi$.

Can anyone give me some direction to make this more mathematically rigorous? I feel like I am just a few steps away from answering this correctly.

Thanks in advance!

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The first part is indeed just substituting in the given identity. The problem with $\theta$ a multiple of $2\pi$ is that to finish you would need to divide by $0$. The idea of the proof is right, it is just a telescoping sum. – André Nicolas Aug 10 '12 at 6:15
Thanks for your comment! – JackReacher Aug 11 '12 at 10:47
up vote 3 down vote accepted

Your proof of the second identity is correct. If $\sin \left( \frac{1}{2}\theta \right) \neq 0$ (what does this mean in terms of $\theta$?) this identity is equivalent to $$\begin{equation*} \sin \left( j\theta \right) =\frac{\cos \left( \left( j-\frac{1}{2}\right) \theta \right) -\cos \left( \left( j+\frac{1}{2}\right) \theta \right) }{ 2\sin \left( \frac{1}{2}\theta \right) }. \end{equation*}$$

The first terms of the numerators of the sum, which you didn't wrote in your proof, are $$\cos \left( \frac{1}{2} \theta \right) -\cos \left( \frac{3}{2} \theta \right) +\cos \left( \frac{3}{2} \theta \right) -\cos \left( \frac{5}{2} \theta\right) +\cdots $$

Evaluation of the sum, which is a telescoping sum, as noticed by André Nicolas \begin{eqnarray*} \sum_{j=1}^{n}\sin \left( j\theta \right) &=&\sum_{j=1}^{n}\frac{\cos \left( \left( j-\frac{1}{2}\right) \theta \right) -\cos \left( \left( j+ \frac{1}{2}\right) \theta \right) }{2\sin \left( \frac{1}{2}\theta \right) } \\ &=&\frac{1}{2\sin \left( \frac{1}{2}\theta \right) }\sum_{j=1}^{n}\cos \left( \left( j-\frac{1}{2}\right) \theta \right) -\cos \left( \left( j+ \frac{1}{2}\right) \theta \right) \\ &=&\frac{1}{2\sin \left( \frac{1}{2}\theta \right) } \sum_{j=1}^{n}a_{j}-a_{j+1},\qquad a_{j}=\cos \left( \left( j-\frac{1}{2} \right) \theta \right) \\ &=&\frac{1}{2\sin \left( \frac{1}{2}\theta \right) }\left( a_{1}-a_{n+1}\right) ,\qquad \text{See below} \\ &=&\cdots \end{eqnarray*}

For the telescoping sum $\sum_{j=1}^{n}a_{j}-a_{j+1}$ we have \begin{eqnarray*} \sum_{j=1}^{n}a_{j}-a_{j+1} &=&\left( a_{1}-a_{2}\right) +\left( a_{2}-a_{3}\right) +\cdots +\left( a_{n-1}-a_{n}\right) +\left( a_{n}-a_{n+1}\right) \\ &=&a_{1}-a_{n+1}. \end{eqnarray*}

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