Binomial theorem or De Moivre's theorem, or either?

Question

How to prove $x^2+1$ is a factor of $f(x)=(\cos(a)+x\sin(a))^n-\cos(na)-x\sin(na)$. I am not sure if I can expand $(\cos(a)+x\sin(a))^n$ by using Binomial theorem or De Moivre's theorem. I am stuck, please help.

Thank you!

Answer 1

Note that $x^2+1=(x+i)(x-i)$. We can see that $$f(i) = [\cos a + i\sin a]^n - \cos(na)-i\sin(na) = [\cos na + i\sin na]-[\cos na + i\sin na]=0$$ through de Moivre’s formula.

Similar is the case for proving that $f(-i)=0$.

Answer 2

If $x^2+1$ is a factor of $f(x)$ this implies that $f(i)=f(-i)=0$. Can you take it further (using De Moivre's theorem)?