Prove a relation for Chebyshev polynomials The Chebyshev polynomials of the first kind are: 
$$T_n(x)=\cos(n\theta)$$ where $x=\cos(\theta)$.
Prove relation:
$$T_n(x)=\frac{1}{2}[(x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n].$$
I tried but I don't know how to get this ($\sqrt{x^2-1}$) inside the relation.
If someone has any idea or proof, please tell!
 A: We have $T_{n+2}(x)=2xT_{n+1}(x)-T_{n}(x)$. Like linear numerical recurrences, the solution comes from the solution of the characteristic equation  $u^2= 2xu-1$, which is $u=x\pm\sqrt{x^2-1}$.
Then $u^{n+2}=2xu^{n+1}-u^n$ and so any linear combination of $(x\pm\sqrt{x^2-1})^n$ will also satisfy $u^2= 2xu-1$. It only remains to check that the initial conditions hold, that is,
$$
T_n(x)=\frac{1}{2}[(x+\sqrt{x^2-1})^n+(x-\sqrt{x^2-1})^n].
$$
for $n=0$ and $n=1$.
A: Observe that
 $$\begin{align}
\cos(n\theta) &= \mathfrak{Re} (e^{n\theta\ i})\\
&= \frac12(e^{n(\theta i)}+ e^{n (-\theta i)})
\end{align}$$ 
and note that $\sqrt{x^2 - 1} = i\sqrt{1-x^2} = i|\!\sin \theta|$; you don’t have to care about the sign of $\sin \theta$ since both possibilities appear in the final formula, so you’re done.

Explaining a previous answer, you could take the (linear) $\mathbb R^2$ rotation operator $R_\theta(v) = \begin{bmatrix}\cos\theta & -\sin\theta \\ \sin \theta & \cos \theta \end{bmatrix} v$. The characteristic polynomial of this matrix is $p(\lambda) = (\cos\theta - \lambda)^2 + \sin^2\theta$; so its the eigenvalues of $R_\theta$ are $\cos \theta \pm i|\!\sin\theta| = \cos\theta \pm i\sin\theta.$ Note that those are different iff $\sin \theta \neq 0$; i.e., $\theta \not \equiv 0,\pi\,\, \mathrm{mod}\,2\pi$. If they're different, we know for sure that $R_\theta$ is diagonalizable. So the eigenvector equation becomes
$$\begin{bmatrix} \mp i \sin \theta & -\sin \theta \\  \sin \theta & \mp i\sin \theta \end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix} = \mathrm O,$$
with associated eigenspace $E_\lambda = \mathrm{Span}((\mp i, 1))$, for $\lambda = \cos \theta \pm i \sin \theta$. So, defining $P = \begin{bmatrix} -i & i \\ 1 & 1 \end{bmatrix}$, we have $$R_\theta= P^{-1}\begin{bmatrix}\cos\theta + i\sin \theta & 0 \\ 0 & \cos \theta - i \sin \theta\end{bmatrix}P.$$
So we can take powers of this matrix easily (and geometrically, the $n$-th power is $R_{n\theta}$). Using DeMoivre's formula, the fact that $\operatorname{tr}$ is invariant under change of basis, we deduce the same fact: $2\cos(n\theta) = (\cos \theta + i \sin \theta)^n + (\cos \theta - i \sin \theta).$
And using the same substitutions, you'll get the same result.
A: Hint: The quadratic formula. You have $x\pm\sqrt{x^2-1}$.
Also, it also looks like the trace of the $n^{th}$ power of a 2 by 2 matrix with  $x+\sqrt{x^2-1}$ and $x-\sqrt{x^2-1}$ on the diagonal. 
