I was wondering if a general identity of $\cos^{2m}(x)=\,$[in terms of] $\cos^{2m-1}$ exists?
One particular example where $m=1$ is:
$$\cos^2(x)=\frac12(\cos(2x)+1).\tag{1}$$
but can this be generalised to an even power of $\cos^{2m}(x)$ expressed in terms of the next odd power down $\cos^{2m-1}$?
So say for a random example maybe $\cos^{42}(x)$ equals something in terms of $\cos^{41}$ .
Let's try it for the cube.
$\cos(x)^3-(a\cos(bx)^2+c)=\frac 14\cos(3x)+\frac 34\cos(x)-\frac a2\cos(2bx)-\frac a2-c$
So to make the most things cancel, set $a=\frac 12,\ b=\frac 32,\ c=-\frac 14$
$$\cos(x)^3-\left(\frac 12\cos(\frac{3x}2)^2-\frac 14\right)=\frac 34\cos(x)$$
And we cannot get rid of the term in $\cos(x)$.
Eventually we can replace it by $\ \frac 32\cos(\frac x2)^2-\frac 34\ $ to have an expression in term of $\sum a_i\cos(b_ix)^2$
$$\cos(x)^3= \frac 12\cos\left(\frac{3x}2\right)^2+\frac 32\cos\left(\frac x2\right)^2-1$$
So this is going to be recursive, and you can find such a combinaison for any degree, I suppose you can try to work out a closed form from Chebychev polynomial, not sure if this is the right course of action though or even if such a lenghty expression is desirable. You wanted a short formula for $\cos^{m+1}$ in term of $\cos^m$ it ends up being not so short...