0
$\begingroup$

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

However to begin with I don't understand how

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.