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let $z = \cos(\theta) + i\sin(\theta)$.

a) Find $z^3$ using binomial expansion.

b) Use De Moivre's theorem to show that: \begin{equation*} \cos(3\theta) = 4\cos(3\theta)-3\cos(\theta)~\text{and}~\sin(3\theta) = 3\sin(\theta)-4\sin(3\theta). \end{equation*} I found the first part to be: \begin{equation*} \cos^3(\theta) +3\cos^2(\theta)(i\sin(\theta)) - 3\cos(\theta)\sin^2(\theta) -i\sin^3(\theta) \end{equation*} but I'm stuck on part b.

i tried doing it by equating the binomial expansion answer to the answer we get from De Moivre's theorem and then using double angle identities, but I got no where with that.

I've spent a really long time trying to do this! it would be really nice if i could just get some direction on what i should do next.

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For part a), use the Binomial expansion \begin{equation*} (x+y)^n=\binom{n}{0}x^ny^0_+\binom{n}{1}x^{n-1}y^1+\binom{n}{2}x^{n-2}y^2+...+\binom{n}{n-1}x^1y^{n-1}+\binom{n}{n}x^0y^n \end{equation*} with $n=3,~x=\cos(\theta)$ and $y=i\sin(\theta)$. This gives \begin{equation*} \cos^3(\theta)+3i\cos^2(\theta)\sin(\theta)-3\cos(\theta)\sin(\theta)-i\sin^3(\theta). \end{equation*} For part b), these identities do not hold. If you rearrange them you get \begin{equation*} 3\cos(\theta)=3\cos(3\theta)~\text{and}~5\sin(3\theta)=3\sin(\theta) \end{equation*} which do not hold. There are identities which are similar to the one you want: \begin{equation*} \sin(3\theta)=3\sin(\theta)-4\sin^3(\theta), \\ \cos(3\theta)=4\cos^3(\theta)-3\cos(\theta) \end{equation*} which you can prove by manipulating both sides (start with the addition formulas \begin{equation*} \sin(\theta+2\theta)=\sin(\theta)\cos(2\theta)+\sin(2\theta)\cos(\theta)~\text{and}~ \\ \cos(\theta+2\theta)=\cos(\theta)\cos(2\theta)-\sin(\theta)\sin(2\theta) \end{equation*} to expand the left side). Does this help? I'll go into anything in more detail if you like.

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  • $\begingroup$ omg! thank you so much! i think i just forgot to split theta in terms of 2theta and theta. but thanks a lot! $\endgroup$ – niharixxx Jun 7 '15 at 8:49
  • $\begingroup$ Sure. No problem $\endgroup$ – user230715 Jun 28 '15 at 18:05

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