I am stuck with understanding the meaning of the question, which states:
Show that $\cos(n\theta)=f_n(\cos\theta)$ for polynomials $f_n(x)$ satisfying $$f_{n+1}(x)=2xf_n(x)-f_{n-1}(x) \tag{1}$$ Find all roots of $f_2(x)+f_3(x)=0$, and write them in the form $\cos(\phi)$ for suitable $\phi$
How should I solve this question? Does the first part mean 'given the relation $(1)$, deduce $\cos(n\theta)=f_n(\cos\theta)$'? Or does it mean to show $\cos(n\theta)=f_n(\cos\theta)$ satisfy $(1)$?
On
The question is:
Given the relationship
$$f_{n+1}(x)=2xf_n(x)-f_{n-1}(x)$$
demonstrate that $\cos(n\theta)=f_n(\cos\theta)$. To show this, induction may be useful.
The second part can probably be done through the substitution $x = \cos \theta$ and the relationship from above.
On
Christian's answer nicely addresses how to solve the second part, but let me see if I can give some additional insight on why we should interpret the first part as he did. (A.G.'s comment alludes to the problem, too.)
If we are supposed to assume only that $(1)$ holds for all (appropriate) $n$, then it is not possible to verify that $\cos(n\theta)=f_n(\cos\theta)$ hold in general. In fact, given any sequence of functions $f_n$ satisfying $(1)$ for all (appropriate) $n$, we have for any function $g(x)$ that the sequence of functions $g(x)f_n(x)$ also satisfies $(1),$ but on the other hand, $\cos(n\theta)=g(\cos\theta)f_n(\cos\theta)$ need not hold in general for all such $g$, even if $\cos(n\theta)=f_n(\cos\theta)$ holds in general. Thus, we must conclude that the question is poorly written, that and we are instead supposed to assume (or demonstrate) the existence of a sequence of polynomials satisfying $\cos(n\theta)=f_n(\cos\theta)$ in general, then prove that such a sequence satisfies $(1).$
One has $$\cos(0x)=1,\quad \cos(1x)=\cos x,\qquad\cos(2x)=2\cos^2x-1\ .$$ This shows that for $0\leq n\leq 2$ the functions $x\mapsto \cos(nx)$ can be written "as polynomials in $\cos x\>$". Furthermore, it follows from the addition theorem for $\cos$ that $$\cos\bigl((n+1)x\bigr)+\cos\bigl((n-1)x\bigr)=2\cos(nx)\cos x\ ,$$ since the $\sin$ terms cancel. The last formula can be written as $$\cos\bigl((n+1)x\bigr)=2\cos x\cos(nx)-\cos\bigl((n-1)x\bigr)\qquad(n\geq1),\tag{1}$$ and serves now as a recursion formula for the successive $\cos(n\cdot)$. In particular it provides an induction proof that all functions $x\mapsto \cos(nx)$, $n\in{\mathbb N}_{\geq0}$, can be written "as polynomials in $\cos x\>$".
In order to describe this in a more abstract way one introduces an "indeterminate" $X$ (this is just a letter, and is not to be interpreted a priori as a real variable), and defines a sequence of polynomials $(f_n)_{n\geq0}$ in this indeterminate recursively as follows: $$f_0(X)=1, \quad f_1(X)=X,\qquad f_{n+1}(X):=2X f_n(X)-f_{n-1}(X)\ .\tag{2}$$ The polynomials $f_n$ produced in this way are called Chebyshev polynomials. It is easy to see that ${\rm deg}\,(f_n)=n$ for all $n$. Above all, a comparison of $(1)$ and $(2)$ immediately shows that $$\cos(nx)=f_n(\cos x)\qquad(x\in{\mathbb R})$$ for all $n\geq0$.
One obtains $f_2(X)=2X^2-1$ and $f_3(X)=2Xf_2(X)-f_1(X)=4X^3-3X$. We now are told to solve $$4X^3+2X^2-3X-1=0\tag{3}$$ for (complex) $X$. One immediately checks that $X=-1$ is a solution, so that one could deflate $(3)$ to a quadratic equation – I leave that to you. But we are given a hint: Our equation is "special". The substitution $$X:=\cos x\qquad (0\leq x\leq\pi)$$ transforms it into $$0=\cos(2x)+\cos(3x)=2\cos{5x\over2}\cos{x\over2}\ .$$ This is fulfilled if one of the following is true: $${5x\over2}={\pi\over2},\quad{5x\over2}={3\pi\over2},\quad{x\over2}={\pi\over2}\ ,$$ or $$x\in\left\{{\pi\over5}, {3\pi\over5}, \pi\right\}\ .$$ It follows that $$X=\cos x\in\left\{\cos{\pi\over5}, -\cos{2\pi\over5}, -1\right\}\ .$$ Since we have found three different values for $X$ we don't need to look further. (This "geometric" approach only brings solutions $X\in[{-1},1]$ to the fore.)