Chebyshev Polynomials I am trying to prove a something regarding Chebyshev polynomials. Given the polynomials $T_n(x), n = 0, 1, \ldots$ which are recursively defined by
$$\begin{cases} T_0(x) = 1\\ T_1(x) = x \\T_n(x) = 2x T_{n−1}(x) − T_{n−2}(x), & \text{for } n \geq 2\end{cases}$$
I want to show that
For every $n$, $$T_n(x) = \cos(n \arccos(x))$$
Is there some type of proof I could use or is just plugging in values the only way?
 A: 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.)
A: 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 :)
