Chebyshev Polynomials

Plugging in values will only prove finitely many instances.

This is a sequence of trigonometric identities. Since it's a definition by recursion, you do the proof by mathematical induction. It's obviously true if $n=0$ or $1$. So suppose it's true in the first $n$ cases, so you know $$ T_n(\cos\theta) = \cos(n\theta). $$ and similarly for $n-1$.

You want to prove $$ T_{n+1}(\cos\theta) = \cos((n+1)\theta). $$

So write $$ \begin{align} T_{n+1}(\cos\theta) & = 2(\cos\theta) T_n(\cos\theta) - T_{n-1}(\cos\theta) \\[8pt] & = 2(\cos\theta)(\cos(n\theta)) - \cos((n-1)\theta) \\[8pt] & = 2(\cos\theta)(\cos(n\theta)) - \Big( \cos(n\theta)\cos\theta + \sin(n\theta)\sin\theta \Big) \\[8pt] & = (\cos\theta)(\cos(n\theta)) -\sin(n\theta)\sin\theta \\[8pt] & = \cos((n+1)\theta). \end{align} $$

(Of course, you have to remember some basic trigonometric identities to follow this.)

I had to answer this question even though its years late :P

$$ \cos\left((n+1)\theta\right) = \cos(n\theta + \theta) = \cos(n\theta)\cos(\theta) - \sin(n\theta) \sin(\theta) \\ \cos\left((n-1)\theta\right) = \cos(n\theta - \theta) = \cos(n\theta)\cos(\theta) + \sin(n\theta) \sin(\theta) $$

Now if you add the left and right sides of this equation you get:

$$ \cos((n+1)\theta) + \cos((n-1)\theta) = 2 \cos(n\theta) \cos(\theta) $$

If you let $x = \cos(\theta)$ and $T_n(x) = \cos(n\theta)$, you get:

$$ T_{n+1} + T_{n-1} = 2 x T_{n} $$

So we have:

$$ T_{n+1} = 2 x T_{n} -T_{n-1} $$

Which is equivalent to what you asked :)