Chebyshev polynomials are defined as such:
$$T_n(x)=\cos(n\arccos(x))$$
I'm asked to show that $\deg(T_j(x))=j$ and that $T_0,T_1,T_2,\ldots,T_n$ are an orthogonal basis of $\Bbb R_n[x]$.
I think I can show that they are an orthogonal base. Right now I'm stuck on the degree thing. I read in wikipedia that $T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$ and hat indeed solves the problem, but I don't understand why that's true.
if $T_n(x)=\cos(n\arccos(x))$ then $\begin{aligned}T_{n+1}(x)&=\cos(n\arccos(x)+\arccos(x))\\&= \cos(n\arccos(x))\cos(\arccos(x))-\sin(n\arccos(x))\sin(\arccos(x))\\&= \cos(n\arccos(x))T_n(x)-\sin(n\arccos(x))\sin(\arccos(x))\end{aligned}$
I don't see why that is equal to $2xT_n(x)-T_{n-1}(x)$
Let $x \in [-1,1]$ and, for notational convenience, set $\theta = \arccos x$, so that $x = \cos \theta$ and $T_k(x) = \cos(k\theta)$. But then, $$ T_{n+1}(x) + T_{n-1}(x) = \cos((n+1)\theta) + \cos((n-1)\theta) = ? $$