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I am having a problem with this question.

I need to prove by induction that: $$\sum_{k=1}^n \sin(kx)=\frac{\sin(\frac{n+1}{2}x)\sin(\frac{n}{2}x)}{\sin(\frac{x}{2})}$$

The relation is obvious for n=1

Now I suppose that the relation is true for a natural number n and I want to show that $$\sum_{k=1}^{n+1} \sin(kx)=\frac{\sin(\frac{n+2}{2}x)\sin(\frac{n+1}{2}x)}{\sin(\frac{x}{2})}$$

We have $$\sum_{k=1}^{n+1} \sin(kx)=\sin[(n+1)x]+\sum_{k=1}^{n} \sin(kx) =\sin[(n+1)x]+\frac{\sin(\frac{n+1}{2}x)\sin(\frac{n}{2}x)}{\sin(\frac{x}{2})}= \frac{\sin[(n+1)x]\sin(\frac{n+1}{2}x)\sin(\frac{n}{2}x)}{\sin(\frac{x}{2})}$$

I am unable to simplify the expression using the trigonometric identities. I keep turning around in circles.

Can somebody help me please.

Thank you in advance

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Why dont you simpliy add $\sin\big((n+1) x\big)$ to both sides of the induction hypotesis? –  Pragabhava Oct 29 '12 at 19:25
1  
Why do you think $\sum_{k=1}^{n+1} a_k = a_{n+1}\sum_{k=1}^n a_k$? Shouldn't it be $a_{n+1} + \sum_{k=1}^n a_k$? –  Thomas Andrews Oct 29 '12 at 19:31
    
The last equality is incorrect. It should read $$\frac{\sin[(n+1)x]\sin[\frac{x}{2}] + \sin[\frac{n+1}{2} x]\sin[\frac{n}{2} x]}{\sin[\frac{x}{2}]}$$ –  Pragabhava Oct 29 '12 at 20:40

2 Answers 2

up vote 2 down vote accepted

Sign $"+"$ is missing: $$\sum\limits_{k=1}^{n+1} \sin(kx)=\sin[(n+1)x]\color{red}{+}\sum\limits_{k=1}^{n} \sin(kx) =\sin[(n+1)x]\color{red}{+}\dfrac{\sin(\frac{n+2}{2}x)\sin(\dfrac{n+1}{2}x)}{\sin(\dfrac{x}{2})}=\ldots$$

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You can actually use telescopy, which is just induction in disguise.

We have that

$$\cos b - \cos a = 2\sin \frac{{a + b}}{2}\sin \frac{{a - b}}{2}$$ Now let $$b=\left(k+\frac 1 2 \right)x$$ $$b=\left(k-\frac 1 2 \right)x$$

Then

$$\cos \left( {k + \frac{1}{2}} \right)x - \cos \left( {k - \frac{1}{2}} \right)x = 2\sin kx\sin \frac{x}{2}$$

Now sum through $k=1,\dots,n$, to get

$$\sum\limits_{k = 1}^n {\cos \left( {k + \frac{1}{2}} \right)x - \cos \left( {k - \frac{1}{2}} \right)x} = 2\sin \frac{x}{2}\sum\limits_{k = 1}^n {\sin kx} $$

$$\cos \left( {n + \frac{1}{2}} \right)x - \cos \frac{x}{2} = 2\sin \frac{x}{2}\sum\limits_{k = 1}^n {\sin kx} $$

whence

$$\frac{{\cos \left( {n + \frac{1}{2}} \right)x - \cos \frac{x}{2}}}{{2\sin \frac{x}{2}}} = \sum\limits_{k = 1}^n {\sin kx} $$

But using our first formula once more, we have

$$\cos \left( {n + \frac{1}{2}} \right)x - \cos \frac{x}{2} = 2\sin \frac{{\left( {n + 1} \right)x}}{2}\sin \frac{{nx}}{2}$$ so finally

$$\frac{{\sin \frac{{\left( {n + 1} \right)x}}{2}\sin \frac{{nx}}{2}}}{{\sin \frac{x}{2}}} = \sum\limits_{k = 1}^n {\sin kx} $$

as desired.

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