(dis)proving $Span((\cos nx)_{n\in\mathbb{N}})=Span((\cos^nx)_{n\in\mathbb{N}})$ in $\mathbb{R}^\mathbb{R}$ I am trying to show that $Span((\cos nx)_{n\in\mathbb{N}})=Span((\cos^nx)_{n\in\mathbb{N}})$ in $\mathbb{R}^\mathbb{R}$. ($0\in\mathbb{N}$)
I immediately thought of the Chebyshev polynomials :
$T_n(x)= \displaystyle\sum_{k=0}^{\lfloor n/2\rfloor} \binom{n}{2k} (x^2-1)^k x^{n-2k}.$
However, this would only tell me that some elements (the ones corresponding to the coefficients) are in common.
How do i show that the equality holds ? (or that it doesn't !)

Let $\mathcal V =\text{Span}\{\cos nx / n \in \mathbb N\}$  and  $\mathcal W =\text{Span}\{\cos^nx / n \in \mathbb N\}$
$(\cos x + i  \sin x)^n =\cos nx + i \sin nx $ gives : $\cos nx = \mathcal  Re(\cos x + i  \sin x)^n $
$$(1) \quad \cos nx=\mathcal Re(\sum_{k=0}^n C_n^k \cos^{n-k} x\, i^k \sin^kx) = \sum_{\ell=0}^{m} C_n^{2\ell} \cos^{n-2\ell}(-1)^{\ell}(1-\cos^2 x)^{\ell}$$ where $m=\rm Ent(n/2)$
Let us put for all $n \in \mathbb N:$ $$\mathcal V_n= \text {Span}\{\cos kx / 0 \leq k \leq n\}$$ and  $$\mathcal W_n = \text {Span}\{cos^k x / 0 \leq k \leq n \}$$
$(1)$ above shows that $$(2) \quad \mathcal V_n \subset \mathcal W_n$$
It's easy to show that $(\cos kx)_{0 \leq k \leq n}$ and $(\cos^kx)_{0 \leq k \leq n}$ are free families, then there are  respectivelly basis of $\mathcal V_n$ and  $\mathcal W_n$. By dimesion theory, since $(2)$ above and  $\dim \mathcal V_n = \dim \mathcal W_n= n+1$ , we have $$\forall n \in \mathbb N \quad \mathcal V_n = \mathcal W_n$$  then : $$\mathcal V = \mathcal W$$
