Prove $x^2+\frac{1}{x^2}=2\cos(2\theta)$ Prove $x^2+\frac{1}{x^2}=2\cos(2\theta)$ and $x^3+\frac{1}{x^3}=2\cos(3\theta)$
knowing that there exist a number $x$ given angle $\theta$ such that $x+\frac{1}{x}=2\cos(\theta)$
Doesn't really know how to start this problem, thought that I would some how need to use the double angle identities 

$$\left(x+\frac 1x\right)^2 = x^2+2+\frac 1{x^2} = 4\cos^2\theta \\\implies x^2+\frac{1}{x^2} = 2(2\cos^2\theta-1) = 2\cos2\theta$$
Similarly
$$\left(x+\frac 1x\right)^3 = x^3 + 3\left(x+\frac 1x\right) + \frac{1}{x^3} = x^3+\frac{1}{x^3} + 6\cos\theta = 8\cos^3\theta$$
$$ \implies x^3+\frac{1}{x^3} = 2(4\cos^3\theta-3\cos\theta) = 2\cos3\theta$$
