$\cos(nx)=Q_n(\cos(x))$ for polynomial $Q_n$ of degree $n$ Are there any proofs of this equality online? I'm just looking for something very simply that I can self-verify. My textbook uses the result without a proof, and I want to see what a proof would look like here.
 A: \begin{align}
e^{inx} &= (\cos x + i\sin x)^n 
\\&= \sum_{k=0}^n \binom nk\cos^{n-k}(x) i^k \sin^k (x) 
\\&= \sum_{k=0}^{\lfloor n/2\rfloor} \binom n{2k}\cos^{n-2k}(x) i^{2k} \sin^{2k} (x) + 
 \sum_{k=0}^{\lfloor (n-1)/2\rfloor} \binom n{2k+1}\cos^{n-2k-1}(x)
 i^{2k+1} \sin^{2k+1} (x) 
\end{align}
Take the real part of both sides and you get
$$
\cos nx =\sum_{k=0}^{\lfloor n/2\rfloor} \binom n{2k}\cos^{n-2k}(x)
 (-1)^k (1-\cos^2x)^k
$$
The last expression is clearly a polynomial of degree $n$ in $\cos (x)$.

Alternative:
as $$
\cos ((n+1)x) + \cos ((n-1)x) = 2\cos (x) \cos (nx)
$$
you can prove it directly using induction.
A: We have
\begin{align}
\cos((n+1)x) & = \cos(x) \cos(nx) - \sin(x) \sin(nx)\\
& = \cos(x) \cos(nx) + \dfrac{\cos((n+1)x) - \cos((n-1)x)}2
\end{align}
Hence,
$$2 \cos((n+1)x) = 2\cos(x) \cos(nx) + \cos((n+1)x) - \cos((n-1)x)$$
Therefore,
$$\cos((n+1)x) = 2\cos(x) \cos(nx) - \cos((n-1)x)$$
Now use induction to conclude what you want.
A: I am not sure which proof you are looking for.
Your polynomial $Q_n$ is called the Chebyshev polynomial of degree $N$. 
Probably the wikipedia page (http://en.wikipedia.org/wiki/Chebyshev_polynomials) will help you out in finding what you are looking for. 
