Show that $z+\frac{1}{z} = 2\cos(\theta)$ and $z^n+\frac{1}{z^n} = 2\cos (n\theta)$. Hence express $\cos^6 (\theta)$

Question

Show that $z+ \frac{1}{z} = 2\cos (\theta)$ and $z^n+\frac{1}{z^n} = 2\cos (n\theta)$

Hence express $\cos^6(\theta)$

So I have nailed down the proof:

Setting the value of $z=\cos(\theta) + i\sin(\theta)$, $$z^n+\frac{1}{z^n}$$$$=z^n+z^{-n}$$ $$=(\cos\theta+i\sin\theta)^n+(\cos\theta+i\sin\theta)^{-n}$$ Using d-Moivre's theorem, $$=\cos(n\theta)+i\sin(n\theta)+\cos(-n\theta)+i\sin(-n\theta)$$ $$=\cos(n\theta)+i\sin(n\theta)+\cos(n\theta)-i\sin(n\theta)$$ $$=2\cos(n\theta)$$

I'm stuck from here since, I assume we somehow need to use the theorem. and equate $\cos^6$ to something based on $z^n+\frac{1}{z^n}$

Answer 1

I assume you want to express $\cos^6(x)$ in terms of a linear combination of terms from $\{\cos(kx)\}_{k=0}^6$, as this is a standard question. By the binomial theorem, $$2^6\cos^6(x)=(z+z^{-1})^6=(z^6+z^{-6})+6(z^4+z^{-4})+\binom{6}{2}(z^2+z^{-2})+\binom{6}{3}$$ Which reduces to $$2^5\cos^6(x)=\cos(6x)+6\cos(4x)+15\cos(2x)+10$$ The overarching point is that we can form a recursive relation between $\cos^nx$ and $\cos^{n-2}(x)$ to prove any power of cosine is expressible as a linear combination of $\{\cos(kx)\}_{k=0}^n$. The polynomial $T_n(x)$ for which $T_n(\cos(x))=\cos(nx)$ is known as the Chebyshev polynomial of the first kind.

Answer 2

You can also express it as such,

$2^6\cos^6(\theta)=(z+z^{-1})^6=(e^{i\theta}+e^{-i\theta})^6$

So, $\cos^6(\theta)=\frac{e^{-6i\theta}}{2^6}(1+e^{2i\theta})^6$