Show that $\forall n \in \mathbb{N} \left ( \left [(2+i)^n + (2-i)^n \right ]\in \mathbb{R} \right )$ Show that  $\forall n \in \mathbb{N} \left ( \left [(2+i)^n + (2-i)^n  \right ]\in \mathbb{R} \right )$
My Trig is really rusty and weak so I don't understand the given answer:
$(2+i)^n + (2-i)^n $
$= \left ( \sqrt{5} \right )^n \left (\cos n\theta + i \sin n\theta  \right ) + \left ( \sqrt{5} \right )^n \left (\cos (-n\theta) + i \sin (-n\theta)  \right ) $
$= \left ( \sqrt{5} \right )^n \left ( \cos n\theta + \cos (-n\theta) + i \sin n\theta + i \sin (-n\theta) \right ) $
$= \left ( \sqrt{5} \right )^n 2\cos n\theta$
Could someone please explain this?

Hint $\ $ Scaling the equation by $\sqrt{5}^{\:-n}$ and using Euler's $\: e^{{\it i}\:\!x} = \cos(x) + {\it i}\: \sin(x),\ $ it becomes
$$\smash[b]{\left(\frac{2+i}{\sqrt{5}}\right)^n + \left(\frac{2-i}{\sqrt{5}}\right)^n} =\: (e^{{\it i}\:\!\theta})^n + (e^{- {\it i}\:\!\theta})^n $$ 
But
$$\smash[t]{ \left|\frac{2+i}{\sqrt{5}}\right| = 1\ \Rightarrow\ \exists\:\theta\!:\ e^{{\it i}\:\!\theta} = \frac{2+i}{\sqrt{5}} \ \Rightarrow\ e^{-{\it i}\:\!\theta} = \frac{1}{e^{i\:\!\theta}} = \frac{\sqrt{5}}{2+i} = \frac{2-i}{\sqrt 5}}$$
Remark $\ $ This is an example of the method that I describe here, of transforming the equation into a simpler form that makes obvious the laws or identities needed to prove it. Indeed, in this form, the only nontrivial step in the proof becomes obvious, viz. for complex numbers on the unit circle, the inverse equals the conjugate: $\: \alpha \alpha' = 1\:\Rightarrow\: \alpha' = 1/\alpha.$
