What would a change of basis matrix look like between the abstract vector spaces of $\cos^k(x)$ and $\cos(kx)$? My friend mentioned that creating this matrix is possible and useful for difficult integrals with high power cosine functions. However, I am having trouble actually creating it.
My thoughts are to use the formula: $$\cos(2x)=2\cos^2(x)-1$$
But that is the end of it. I imagine there is some simple pattern I am missing. If someone could highlight it that would be great!
In case my wording is confusing, I am imagining some abstract vector space W where the first entry is the coefficient on $\ \cos^0(x) $, the second entry is the coefficient on $\ \cos^1(x) $, the third entry is the coefficient on $\ \cos^2(x) $, etc. and another abstract vector space H where the first entry in any vector is the coefficient on $\ \cos(0x) $, the second entry is the coefficient on $\ \cos(1x) $, the third entry is the coefficient on $\ \cos(2x) $, etc. The matrix I am trying to find, but currently failing at, is the change of basis matrix from W$\ \rightarrow$ H.
This would make integrals with high cosine powers easier to integrate, since a linear u-sub is extremely simple.
 A: To find that matrix, you may use the Euler's formula to help you.
$$ e^{ix} = \cos(x) + i\sin(x)$$
Since $ e^{inx}=\cos(nx)+i\sin(nx)$
And we know that $e^{inx} = {e^{ix}}^n = {(\cos(x) + i\sin(x))}^n $
So $$\cos(nx)+i\sin(nx) = {(\cos(x) + i\sin(x))}^n $$
If we take $n=2$ , using the binomial theorem on RHS, and taking the Real Part on both sides. After comparing, you can find that $\cos(2x)=2\cos^2(x)-1$. 
For $n=3,4,\dots$ you can use the same technique to find out what $\cos(kx)$ is, in terms of linear combination of $1,\cos(x),\cos^2(x),\dots,\cos^k(x)$
So you can find out more columns of the change of coordinate matrix now.
And this video from Dr.Peyam may be useful. 1
A: The main fact is that $\cos(n\theta)=T_n(\cos(\theta))$ where $T_n$ is the $n$th degree Chebyshev polynomial of the first kind. There's no simple closed form for these polynomials but you can compute them by the recursion $T_0(x)=1,T_1(x)=x,T_{n+1}(x)=2xT_n(x)-T_{n-1}(x)$. 
Thus the coefficients of $T_n$ enter the coefficient matrix mapping $\cos(n\theta)$ into $\cos^n(\theta)$, which is the opposite of the transformation you would usually want to do in trig integration. Inverting this matrix lets you go the other way.
