$\cos (n \phi)$ in terms of $\cos (\phi)$

Question

Consider the following quantity:

$$\cos(n \phi), \ n \in \mathbb{N}, \ n > 1$$

I know it is possible to alternatively express it as a polynomial of degree $n$, with powers of $\cos (\phi)$ and $\sin (\phi)$.

But is it possible to express $\cos(n \phi)$ in terms of $\cos(\phi)$ only? In other words, is it possible to write

$$\cos(n \phi) = A \cos (\phi + \alpha)$$

for some real constants $A, \alpha$, or as a linear combination

$$\cos(n \phi) = A \cos (\phi + \alpha) + B \cos (\phi + \beta) + \ldots$$

?

My guess: no, because the base frequency of $\cos (\phi)$ is too low, therefore unable to represent faster variations with respect to $\phi$, as it happens in $\cos(n \phi)$.

Answer 1

No. The right-hand side asds up to one big $Z\cos(\phi+\omega)$.
$$A\cos(\phi+\alpha)+B\cos(\phi+\beta)+\ldots\\ =\cos\phi(A\cos\alpha+B\cos\beta+\ldots)-\sin\phi(A\sin\alpha+B\sin\beta+\ldots)\\ =P\cos\phi+Q\sin\phi=Z\cos(\phi+\omega)$$

Answer 2

No, it is not possible. Consider differentiating the equation with respect to $\phi$ four times. The LHS gets multiplied by $n^4$ but the RHS does not - a contradiction.