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I have seen a few examples of finding $\cos 5\theta$ before on here but I was wondering how to just find it by equating the real parts?

So far I have: $$\cos5\theta=\Re(cos\theta+i\sin\theta)^5$$

But how do I know which is the real part and expand from this further? Any hints would be appreciated!

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  • $\begingroup$ Use the binomial theorem to expand the bracket. Using $i^2=-1$ you get an expression $a+bi$ where $a$ and $b$ are expressions in sin and cos. $a$ is the real part and $b$ is the imaginary part. $\endgroup$ May 22 '19 at 20:25
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So here we will use De Moivre's theorem here:

As you stated we will use the fact that $\cos(5\theta)+i\sin(5\theta)=(\cos(\theta)+i\sin(\theta))^5$, so get this just in terms of $\cos(5\theta)$ we can write $\cos(5\theta)=\Re(\cos(\theta)+i\sin(\theta))^5$

So expanding this we have:

$$\cos(5\theta)=\Re(\cos^5\theta+5i\cos^4\theta\sin\theta+10i^2\cos^3\theta \sin^2\theta+10i^3\cos^2 \theta \sin^3\theta +5i^4\cos\theta \sin^4\theta+i^5\sin^5\theta)$$

Since the real parts cannot have an odd power of $\sin\theta$ (since it will not give a real number) we get :

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

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

$$\cos(5\theta)=\cos^5\theta-10\cos^3\theta(1-\cos^2\theta)+5\cos\theta (1-\cos^2\theta)^2$$

$$\cos(5\theta)=\cos^5\theta-10\cos^3\theta+10\cos^5\theta+5\cos\theta-10\cos^3\theta+5\cos^5\theta$$

$$\cos(5\theta)=16\cos^5\theta-20\cos^3\theta+5\cos\theta$$

Hope this helps :)

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Now, use $$(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5.$$ Can you end it now?

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