1. ## De Moivre's theorem

Question: Use De Moivre's theorem to turn the functikons of the multiple angle into a polynomial in the sine or cosine of the angle theta as indicated.

(a) cos(6theta) in terms of cos(theta)

I have no idea at all where to begin on this, my lecture notes are unhelpful... any help would be most appreciated thanks!

2. Hello !

Originally Posted by DefilerAlpha
Question: Use De Moivre's theorem to turn the functikons of the multiple angle into a polynomial in the sine or cosine of the angle theta as indicated.

(a) cos(6theta) in terms of cos(theta)

I have no idea at all where to begin on this, my lecture notes are unhelpful... any help would be most appreciated thanks!
De Moivre's theorem states :

$(\cos x+i \sin x)^n=\cos nx+i \sin nx$

--> $\cos nx=\Re (\cos nx+i \sin nx)=\Re (\cos x+i \sin x)^n \qquad \Re \ : \ \text{Real part}$

$\implies \cos 6 \theta=\Re (\cos \theta+i \sin \theta)^6$

Now use Newton's Binomial formula to get it...

3. Originally Posted by DefilerAlpha
Question: Use De Moivre's theorem to turn the functikons of the multiple angle into a polynomial in the sine or cosine of the angle theta as indicated.

(a) cos(6theta) in terms of cos(theta)

I have no idea at all where to begin on this, my lecture notes are unhelpful... any help would be most appreciated thanks!
His theorem states:

$[\cos(\theta)+i\sin(\theta)]^n=\cos(n\theta)+i\sin(n\theta)$

Using n=6 we get

$[\cos(\theta)+i\sin(\theta)]^6=\cos(6\theta)+i\sin(6\theta)$

Expanding the left hand side with the binomial theorem (you could do it by hand, but I'm lazy) we get

$\cos^6(\theta)+i\cos^5(\theta)\sin(\theta)-\cos^4(\theta)\sin^2(\theta)-i\cos^3(\theta)\sin^3(\theta)+$
$\cos^2(\theta)\sin^4(\theta)+i\cos(\theta)\sin^5(\ theta)-\sin^6(\theta) =\cos(6\theta)+i\sin(6\theta)$

Now we can equate the real part of both sides of the equation to get

$\cos^{6}(\theta)-\cos^{4}(\theta)\sin^2(\theta)+\cos^2(\theta)\sin^ 4(\theta)-\sin^6(\theta)=\cos(6\theta)$

Yeah!!