Prove that $\cos (n \arccos (x))$ is a polynomial of $n$-th degree So, we got an assignment from the mathematical analysis class to prove that $\cos (n \arccos (x))$ is a polynomial of $n$th degree.
I tried to prove it with mathematical induction and so far what I got is:
basis :- $T(1) - \cos (1\cdot\arccos (x)) = \cos(\arccos(x)) = x$
Inductive step : $T(n+1)$ - assume it holds for $T(n)$, then it must hold for $T(n+1)$
$$\begin{align}\cos[(n+1) \arccos(x)] = &\cos[n \arccos(x) + \arccos(x)] &\\= 
&\cos[n \arccos(x)] \cos[\arccos(x)] - \sin[n \arccos(x)] \sin[\arccos(x)] 
&\\=& \cos[n\arccos(x)] x - \sin[n \arccos(x)] \sin[\arccos(x)]\end{align}$$
and this is where I got stuck because I don't know what to do with the sine part. like, I get that the $x\cos(n \arccos(x))$ holds because, by inductive step, it holds that $\cos(n\arccos(x))$ is the polynomial of nth degree, so $x\cos(n \arccos(x))$ is polynomial of $(n+1)$th degree. But I don't know if this is enough to prove this or if I need to do something with the sine too.
 A: \begin{eqnarray*}
T_{n+1}(x) = \cos( n \cos^{-1}(x) )  \cos(  \cos^{-1}(x) ) - \sin( n \cos^{-1}(x) )  \sin(  \cos^{-1}(x) ) \\
T_{n-1}(x) = \cos( n \cos^{-1}(x) )  \cos(  \cos^{-1}(x) ) + \sin( n \cos^{-1}(x) )  \sin(  \cos^{-1}(x) ) \\
\end{eqnarray*}
Add these and we have $T_{n+1}(x) =2x T_n(x) -T_{n-1}(x)$.
A: If $P_n(x) = \cos(n\arccos(x))$, note that 
$$\sin(n\arccos(x))=\frac1n\sqrt{1-x^2}P^{\prime}_n(x),$$
so $$\sin(n\arccos(x))\sin(\arccos(x))=\frac1n(1-x^2)P^{\prime}_n(x)P^{\prime}_1(x).$$
You know $P_1'(x) \equiv 1$, and your induction hypothesis implies that $P_n'(x)$ is a polynomial of degree $n-1$.

I noticed a gap in the above.  This would give you that each of the two terms added is a polynomial of degree $n+1$ by the induction hypothesis, but it wasn't explained why the leading coefficients don't cancel out leaving a polynomial of lower degree.  
But there's no problem once you look: If the leading coefficient of $P_n$ is $a$, then the leading coefficient of your cosine term is $a$, and the leading coefficient of your sine term is also $a$: $$-\frac1n(1-x^2)(nax^{n-1}+\cdots)=ax^{n+1}+\cdots.$$
A: The way you've used the symbol $T$ in the question could lead to confusion,
because your function $\cos (n \arccos (x))$ is what's called a
Chebyshev polynomial, written $T_n(x)$.
(That's the letter $T$ used as the name of a polynomial,
whereas you used it as the name of a proposition of logic.)
The trick here is you don't just use ordinary ("weak") induction to
prove that $T_n(x)=\cos (n \arccos (x))$ is a polynomial over $[-1,1]$
for all $n\geq 1;$
you prove that both $T_n(x)=\cos (n \arccos (x))$
and $T_{n-1}(x)=\cos ((n-1) \arccos (x))$ are polynomials for all $n \geq 1.$
This means that your base case needs to show more than just that
$T_1(x)$ is a polynomial.
The usual approach is to also look at $T_0(x).$
