trigonometry and rows

**mahadeva** · January 21st 2006, 07:23 AM

I need to prove that:
sin x + sin 2x+ ...........sin nx = (sin nx/2 * sin (n+1)x/2) / sinx/2

it is specified that sin x/2 is not 0 ,and that n is >=1

I used various trigonometry theoremes but i just cant find the algoritamto solve my problem so please if anyone knows how to solve it please reply

**TD!** · January 21st 2006, 07:32 AM

Do you know the exponential sum formula?

**mahadeva** · January 21st 2006, 10:02 AM

Probably.But i dont really know to tranlaste exponential sum into my language.I dont speak English very well,sorry.It's forumula about what?

If it's something about integrals then I dont know,but I would like to know.

**TD!** · January 21st 2006, 10:22 AM

Well, we have that:

$\sum\limits_{n = 0}^{N - 1} {r^n } = \frac{{1 - r^N }} {{1 - r}}$

Applying that to e:

$\sum\limits_{n = 0}^{N - 1} {e^{inx} } = \frac{{1 - e^{iNx} }} {{1 - e^{ix} }}$

Here, i stands for the imaginary unit. We also have that:

$e^{ix} = \cos x + i\sin x$

So by taking the imaginary part of such a complex exponential, we obtain the sine.

$\sum\limits_{n = 0}^N {\sin x} = \Im \left( {\sum\limits_{n = 0}^N {e^{inx} } } \right) = \Im \left( {\frac{{e^{i\left( {N + 1} \right)x} - 1}} {{e^{ix} - 1}}} \right)$

Now we need a bit of algebraic manipulation to obtain our result

$\Im \left( {\frac{{e^{i\left( {N + 1} \right)x} - 1}} {{e^{ix} - 1}}} \right) = \Im \left( {\frac{{e^{i\left( {N + 1} \right)x/2} }} {{e^{ix/2} }}\frac{{e^{i\left( {N + 1} \right)x/2} - e^{ - i\left( {N + 1} \right)x/2} }} {{e^{ix/2} - e^{ - ix/2} }}} \right)$

If you don't see this, work out the last fractions (multiplying again) to see it's correct.
We can now take the imaginary part of the first fraction to get sines and simplify the second fraction.

$\frac{{\sin \left( {\frac{1} {2}\left( {N + 1} \right)x} \right)}} {{\sin \left( {\frac{1} {2}x} \right)}}\Im \left( {e^{iNx/2} } \right) = \frac{{\sin \left( {\frac{1} {2}\left( {N + 1} \right)x} \right)\sin \left( {\frac{1} {2}Nx} \right)}} {{\sin \left( {\frac{1} {2}x} \right)}}$

So we conclude

$\sum\limits_{n = 0}^N {\sin x} = \frac{{\sin \left( {\frac{1} {2}\left( {N + 1} \right)x} \right)\sin \left( {\frac{1} {2}Nx} \right)}} {{\sin \left( {\frac{1} {2}x} \right)}}$

**ThePerfectHacker** · January 21st 2006, 02:20 PM

Prove it with mathematical induction.
Use the identity $\sin x+\sin y=2\sin\frac{x+y}{2}\cos\frac{x-y}{2}$
First we show it works for $n=2$ .
Because,
$\sin x+\sin 2x=\frac{\sin x \sin (3x/2)}{\sin (x/2)}$
If and only if,
$2\sin (3x/2)\cos (x/2)=\frac{\sin x \sin (3x/2)}{\sin (x/2)}$
If and only if,
$2\cos (x/2)=\frac{\sin x}{\sin (x/2)}$
If and only if,
$2\sin (x/2)\cos (x/2)=\sin x$
True, by the double angle identity.

Now assume it is true for $k$
Thus,
$\sin x+\sin 2x +...+\sin kx=\frac{\sin (kx/2) \sin ((k+1)x/2)}{\sin (x/2)}$

BEWARE!, this is going to be messy.
Thus, we need to show that
$ \frac{\sin (kx/2)\sin ((k+1)x/2)}{\sin (x/2)}+\sin (k+1)x=$ $\frac{\sin ((k+1)x/2)\sin ((k+2)x/2)}{\sin (x/2)}$
If and only if,
$\sin (kx/2)\sin ((k+1)x/2)+\sin (x/2) \sin (k+1)x=$ $\sin ((k+1)x/2)\sin ((k+2)x/2)$
If and only if,
$\sin (kx/2)\sin ((k+1)x/2)+\sin (x/2) \sin (k+1)x-$ $\sin ((k+1)x/2)\sin ((k+2)x/2)=0$
If and only if,
$\sin \frac{(k+1)x}{2} \left( \sin\frac{kx}{2} -\sin\frac{(k+2)x}{2}\right)+$ $\sin (x/2) \sin (k+1)x=0$
If and only if,
$\sin \frac{(k+1)x}{2}\left( 2\cos \frac{(k+1)x}{2}\sin \frac{-x}{2}\right)+$ $\sin (x/2) \sin (k+1)x=0$
If and only if,
$-2\sin \frac{(k+1)x}{2}\cos \frac{(k+1)x}{2}\sin \frac{x}{2}+$ $\sin (x/2) \sin (k+1)x=0$
If and only if,
$-2\sin \frac{(k+1)x}{2}\cos \frac{(k+1)x}{2}\sin \frac{x}{2}+$ $2\sin \frac{(k+1)x}{2}\cos \frac{(k+1)x}{2}\sin \frac{x}{2}=0$
This is true.
Q.E.D.

Note in the proof I used the Sine Sum formula stated in the begining.
The Sine difference formula which is,
$\sin x-\sin y=2\cos \frac{x+y}{2}\sin \frac{x-y}{2}$
And the double angle,
$\sin x=2\sin (x/2)\cos (x/2)$

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