De Moivre's Theorem

**Modernized** · August 30th 2009, 03:58 AM

Q: Prove that,
(1 + sin 0 + i cos 0)^n
-------------------- = cos n (pi/2 - 0) + i sin n(pi/2 - 0)
(1 + sin 0 - i cos 0)^n

where, 0 = theta (Cant type it)

Thanks !

I've thought about multipying the conjugate (sin 0 + i cos 0) but it didnt work out nicely.. so I need some help with this! Thanks!

**ynj** · August 30th 2009, 05:05 AM

$\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=\frac{1+e^{(\frac{\pi}{2}-\theta)i}}{1+e^{(\theta-\frac{\pi}{2})i}}=e^{(\frac{\pi}{2}-\theta)i}=\cos (\frac{\pi}{2}-\theta)+i\sin (\frac{\pi}{2}-\theta)$

**Modernized** · August 30th 2009, 06:56 AM

Argh.. my textbook never went into the Eulers rule..
so I dont really know how you got that..
I did a search on Wiki
it says,
e^i(nx) = cos (nx) + i sin (nx)
But this never appears in my textbook! Argh~

Is there any other ways to do it?

Thanks anyway though!!!

**ynj** · August 30th 2009, 07:49 AM

Originally Posted by Modernized

Argh.. my textbook never went into the Eulers rule..
so I dont really know how you got that..
I did a search on Wiki
it says,
e^i(nx) = cos (nx) + i sin (nx)
But this never appears in my textbook! Argh~

Is there any other ways to do it?

Thanks anyway though!!!

$\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}=\frac{(1+\sin\theta+i\cos\theta)(\sin \theta+i\cos\theta)}{(1+\sin\theta-i\cos\theta)(\sin\theta+i\cos\theta)}$
$=\frac{(1+\sin\theta+i\cos\theta)(\sin\theta+i\cos \theta)}{\sin\theta+i\cos\theta+\sin\theta^2+\cos\ theta^2}=\frac{(1+\sin\theta+i\cos\theta)(\sin\the ta+i\cos\theta)}{\sin\theta+i\cos\theta+1}$
$=\sin\theta+i\cos\theta$
understand now?

**Modernized** · August 31st 2009, 02:52 AM

Yes. I got till that step.. because I said I tried to times the conjugate like what you've done.. and we got the same answer, but I just dont understand how to reduce it to the final answer that they want me to show.

**ynj** · August 31st 2009, 02:59 AM

Originally Posted by Modernized

Yes. I got till that step.. because I said I tried to times the conjugate like what you've done.. and we got the same answer, but I just dont understand how to reduce it to the final answer that they want me to show.

oh...there may be some mistake in the problem??
I think it shoud be:
$(\sin\theta+i\cos\theta)^n=\sin n\theta+i\cos n\theta=\cos (\frac{\pi}{2}-n\theta)+i\sin (\frac{\pi}{2}-n\theta)$

**pacman** · August 31st 2009, 06:45 AM

ynj, i think your observation is correct, the 1 should not be there.

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