Write $\cos(5x)$ as a function of $\cos(x)$, answer with a polynomial.

You have by De Moivre's formula that $$(\cos(x)+i\sin(x))^5=\cos(5x)+i\sin(5x)$$

but $$(\cos(x)+i\sin(x))^5=\cos^5(x)-10\sin^2(x)\cos^3(x)+5\sin^4(x)\cos(x)$$ $$+i(\sin^5(x)+5\sin(x)\cos^4(x)-10\sin^3(x)\cos^2(x)).$$

Now compare real parts and use the identity $\sin^{2}(x)=1-\cos^2(x).$

You get $$\cos(5x)=16\cos^5(x)-20\cos^3(x)+5\cos(x).$$

For a given integer $n$,

$$\cos (n+1)t+\cos(n-1)t=2\cos(t)\cos(nt)$$

Here I use the general formula $\cos p+\cos q=2\cos\frac{p+q}2\cos\frac{p-q}2$.

Hence, if $T_n(x)$ is a polynomial such that $T_n(\cos(t))=\cos(nt)$, then

$$T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$$

With trivially $T_0(x)=1$ and $T_1(x)=x$.

Then

$$T_2(x)=2x^2-1$$ $$T_3(x)=4x^3-3x$$ $$T_4(x)=8x^4-8x^2+1$$ $$T_5(x)=16x^5-20x^3+5x$$

And

$$\cos(5t)=16\cos^5(t)-20\cos^3(t)+5\cos(t)$$

See also https://en.wikipedia.org/wiki/Chebyshev_polynomials