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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 $T_0,T_1,T_2,...,T_n$ are an orthogonal basis of $\mathbb R_n[x]$.

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 $T_{n+1}(x)=cos(n*arccos(x)+arccos(x)) = cos(n*arccos(x))cos(arccos(x))-sin(n*arccos(x))sin(arccos(x)) = cos(n*arccos(x))*T_n(x)-sin(n*arccos(x))sin(arccos(x))$

I don't see why that is equal to $2xT_n(x)-T_{n-1}(x)$

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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, $$ T_{n+1}(x) + T_{n-1}(x) = \cos((n+1)\theta) + \cos((n-1)\theta) = ? $$

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  • $\begingroup$ Splitting the cosines is a nice trick. It may also be helpful to remind that: $T_0(x) = cos(0 \times \theta) = 1$ and $T_1(x) = cos(1\times\theta) = cos(arccos(x))=x$ $\endgroup$ – eigenjohnson Nov 8 '17 at 1:27

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