cos nx from cos x in fractional form

Question

I am given $\cos x$ in fractional form. How to find $\cos nx$ in fractional form?

I tried using half angle formula repeatedly, but the extra $x$ introduced a sin x, which might not be rational.

I can't directly find out $x$, as I will lose precision when finding $\cos nx$.

Answer 1

You have to use Chebyshev's polynomials; $\cos nx$ is a polynomial in $\cos x$: $T_n(cos x)$. One shows the polynomial has leading term $2^{n-1}t^n$, and it can be calculated recursively with the second-order recurrence: $$T_{n+1}(t)=2t T_n(t)-T_{n-1}(t).$$ (deduced from the factorisation formula:$$\cos (n+1)x+\cos (n-1)x=2\cos(nx)\cos x.)$$

Answer 2

HINT

You can use Viete's formula

$$\cos(n\theta) = \sum_{k\text{ even}} (-1)^\frac{k}{2} {n \choose k}\cos^{n-k} \theta \sin^k \theta$$