So, I've got an assignment to prove that $ \displaystyle f(x)=\cos{(n \cdot \arccos{x})} $ is a polynomial for $ \displaystyle \forall n \in \mathbb{N} $. Also, we were suggested to use mathematical induction. So, I've tried:

Base step: $ \displaystyle n=1 \implies f(x)=\cos{(\arccos{x})}=x$
Assumption step: $ \displaystyle f(x)=\cos{(n \cdot \arccos{x})}, \forall n \in \mathbb{N} $
Induction step: $ \displaystyle f(x)=\cos{((n+1) \cdot \arccos{x})}=\cos{(n \arccos{x}+\arccos{x})}=\cos{(n \arccos{x})}\cos{( \arccos{x})}-\sin{(n \arccos{x})}\sin{( \arccos{x})}=f(x) \cdot x -\sin{(n \arccos{x})}\sin{( \arccos{x})}$

And I don't know what to do with sine.

Originally Posted by I like Serena

Hi karseme!

How about assuming that $\sin{(n \arccos{x})}\sin{( \arccos{x})}$ is a polynomial in $x$?

\begin{aligned}\sin{((n+1) \arccos{x})}\sin{(\arccos{x})}
&= \big(\sin{(n \arccos{x})}\cos{(\arccos{x})} + \cos{(n \arccos{x})}\sin{(\arccos{x})} \big)\sin{( \arccos{x})} \\
&= x \sin{(n \arccos{x})}\sin{( \arccos{x})} + \cos{(n \arccos{x})}\sin^2{(\arccos{x})} \\
&= x \sin{(n \arccos{x})}\sin{( \arccos{x})} + \cos{(n \arccos{x})}(1 - \cos^2{(\arccos{x})})
\end{aligned}

But, how can we assume that? It's like let's assume that every number is divisible by 3 for the sake of the convenicence. I don't see how can we assume that here. Maybe it is polynomial, I don't know. Anyway what can we achieve by assuming that, you're still left with $ x \sin{(n \arccos{x})}\sin{( \arccos{x})}$. And who says that $\sin{(n \arccos{x})}\sin{( \arccos{x})}$ is a polynomial. How to prove that.

$$\cos[(n+1)\arccos(x)]=\cos(n\arccos(x))\cos(\arccos(x))-\sin(n\arccos(x))\sin(\arccos(x))$$

$$=x\cos(n\arccos(x))+\frac{\cos[(n+1)\arccos(x)]-\cos[(n-1)\arccos(x)]}{2}$$

$$\cos[(n+1)\arccos(x)]=2x\cos(n\arccos(x))-\cos[(n-1)\arccos(x)]$$

I believe that's sufficient.