de Moivre and Binomial Theorem

**PTL** · August 5th 2009, 02:59 PM

Use de Moivre's Theorem and the Binomial Theorem to find integers A,B,C,D,E,F such that
$\sin(5\theta)=A\sin^{5}\theta + B\sin^{3}\theta+ C\sin\theta$
$16\sin^{5}\theta=D\sin\theta+E\sin(3\theta)+F\sin( 5\theta)$ .

Any help much appreciated

**Bruno J.** · August 5th 2009, 03:14 PM

Do you know the statement of De Moivre's theorem? Do you know how to expand $(x+y)^5$ ? If so just do it and set $x=\cos\theta, \: y=i\sin\theta$ ...

**PTL** · August 6th 2009, 11:21 AM

So I get
$\cos^{5}\theta+5\cos^{4}\theta\sin\theta + 10\cos^{3}\theta\sin^{2}\theta + 10\cos^{2}\theta\sin^{3}\theta+ 5\cos\theta\sin^{4}\theta +\sin^{5}\theta$

$(\cos x + i \sin x)^{n}= \cos n\theta + i \sin n\theta$

I still don't see where I'm supposed to go with it...

Is it something to do with matching up the real parts on both sides?

**Bruno J.** · August 6th 2009, 11:36 AM

Originally Posted by PTL

So I get
$\cos^{5}\theta+5\cos^{4}\theta\sin\theta + 10\cos^{3}\theta\sin^{2}\theta + 10\cos^{2}\theta\sin^{3}\theta+ 5\cos\theta\sin^{4}\theta +\sin^{5}\theta$

$(\cos x + i \sin x)^{n}= \cos n\theta + i \sin n\theta$

I still don't see where I'm supposed to go with it...

Is it something to do with matching up the real parts on both sides?

You forgot $i$ my friend!

you should get

$(\cos x + i \sin x)^{5} =$
$\cos^{5}\theta+5i\cos^{4}\theta\sin\theta - 10\cos^{3}\theta\sin^{2}\theta - 10i\cos^{2}\theta\sin^{3}\theta+ 5\cos\theta\sin^{4}\theta +i\sin^{5}\theta$

(just alternate $1,i,-1,-i,...$ )

Then equate the imaginary parts because the real part is $\cos 5\theta$ which is not what you want.

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