discrete mathematics - Proof by induction involving trigonometric identities

Show that the sequence of polynomials defined by the rules:

$$p_0(x) = 1,\hspace{.3cm} p_1(x) = x,\hspace{.3cm} p_{n+1}(x) = 2xp_n(x) − p_{n−1}(x)$$

is the same as the polynomials defined as follows:

(a) $x=\cos(t)$

(b) $p_n(x)$ is the polynomial in $x$ which relates $\cos(nt)$ to $\cos(t)$.

I assume that I want to prove these polynomial descriptions are equivalent by showing how $\cos(nt)$ is tied to $\cos((n-1)t)$ and $\cos((n+1)t)$ then I could use trigonometric identities to carry my proof and then convert that expression back to $x$ then use proof by induction to show these two polynomials are the same. Is this the right way to approach this? One place of my confusion is how to apply trigonometric identities to show the relationship.

Answer

It's exactly that: the addition formulæ yield

$$\cos (a-b)+\cos(a+b)=2\cos a\cos b,$$
whence $$\cos(n-1)x+\cos(n+1)x=\cos(nx-x)+\cos(nx+x)=2\cos x\cos nx.$$