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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

  • 0
    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}]}$2012-10-29

2 Answers 2

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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.