# 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.

\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 $T_{n+1}(x) =2x T_n(x) -T_{n-1}(x)$.

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 $a$: $$-\frac1n(1-x^2)(nax^{n-1}+\cdots)=ax^{n+1}+\cdots.$$

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 $T_0(x).$

• oh thanks! this is my freshman year of college and we really havent learned a lot from analysis so far, so we still havent covered the chebyshev polynomial. thank you for this information and i will try to use it to better understand other answers! – i dont know much about algebra Oct 23 '17 at 21:28
• 2 base cases need if you're following the approach in Donald Splutterwit's answer. 1 base case suffices for ordinary induction which also works. I'm confused by "...you prove that both $T_n(x)$..."; do you mean "...your induction hypothesis is that both $T_n(x)$..."? – Jonas Meyer Oct 23 '17 at 21:33
• By "prove both" I mean you set out to prove the proposition $P(k)$ defined by "$T_k$ is a polynomial and $T_{k-1}$ is a polynomial." From then on everything is ordinary induction. ($P(1)$ is just one base case, but it says things about both $T_1$ and $T_0$.) But your answer is the first time I've seen your technique for dealing with the sines; it's very nice. – David K Oct 23 '17 at 21:49