Step to prove that $\cos (n \arccos (x))$ is a polynomial of $n$-th degree

Question

I am confronted to the same problem stated in that question, namely to prove that cos(arccos()) is a polynomial of -th degree.

However to begin with I don't understand how

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

Answer 1

Use $x=n\arccos(x)$ and $y=\arccos(x)$. Then $\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$. This a simple trigonometric identity. You can find a possible proof at http://www.math.ubc.ca/~feldman/m100/trigId.pdf, but just search for it and you get many variants.