Link between a particular function and cos(nt) There was a particular question we were given that I can't work out. The question's fine but the difficulty is the extra bit. So the original question is:
Define the polynomials $f_n$ by
$$ f_0(x) = 1, $$
$$ f_1(x) = x, $$
$$ f_n(x) = 2x f_{n-1}(x) - f_{n-2}(x) \quad  (n = 2,3,\ldots). $$
Let $x$ be a real number. Prove that $f_n(-x) = (-1)^n f_n(x)$.
The question itself is quite simple using the minimal criminal idea.
But there's an extra bit which asked: "What is the connection between $f_n(x)$ and $\cos(nt)$?"
I can't figure it out! It's only optional so I don't have to do it but for completeness purposes I've tried and I can't figure it out. I've asked all my friends and they're stuck too! Any help appreciated!
 A: If you compute a few of the values of $f_n(x)$, i.e. for a few values of $n=2,3,4$, the familiarity of the formulae for $\cos(nx)$ in terms of $\cos(x)$ would lead you to the answer.
$f_2(x)=2xf_1(x)-f_0(x) = 2x^2-1,\ ( \text{Note } \cos(2\theta)=2\cos^2(\theta)-1) \\ f_3(x)=2xf_2(x)-f_1(x)=2x(2x^2-1)-x=4x^3-3x, \ (\cos(3\theta)=4\cos^3(\theta)-3\cos(\theta))$
So, taking $x=\cos(\theta), f_n(x)=\cos(n\theta)$, which you can prove by induction easily, over $n$.
Assuming $f_1(x)=\cos(\theta), f_2(x), \cdots,f_k(x)$ to be the representations of $f_t(x)=cos(t\theta)$ (base case for $k=2$ checked above), $$f_{k+1}(x) = 2xf_{k}(x) -f_{k-1}(x) = 2\cos(\theta)\cos(k\theta)-\cos((k-1)\theta)\\ =2\cos(\theta)\cos(k\theta)-\cos(k\theta - \theta)\\=2\cos(\theta)\cos(k\theta)-\cos(k\theta)\cos(\theta)-\sin(k\theta)\sin(\theta) \\ =\cos(k\theta)\cos(\theta)-\sin(k\theta)\sin(\theta) \\ = \cos(k\theta+\theta) = \cos((k+1)\theta)$$ proves the statement, by the strong version of mathematical induction.
The recursive definition of $f_n(x)$ you've been given is the way Chebyshev polynomials of the first kind are defined, and they are quite important in numerical methods, particularly among the class of orthogonal polynomials.
