Suppose we have a function $f: \theta\rightarrow f(\theta)$ whose domain is $[0,180°]$. The course I follow, says that we can use the change of variable $x=cos(\theta)$ in the function $f$ (because $cos$ is bijective from $[0,180°]$ to $[-1,1]$).
I don't understand.. If we have, say, $$f(\theta) = \theta^2 +1$$
Then $f(x)=f(cos(\theta))= cos^2(\theta) +1$
Then if I evaluate $f$ on $0$, I have $f(0)=1$ according to the first expression and $f(0)=2$ according to the second expression.
I you have any idea how this variable change works, I'll gladly try my best to understand it
PS: I've tried to think about this problem in terms of matrices. We start from $AX$ and $cos$ is like a $B$ matrix that we put in-between $A$ and $X$ so the whole is tantamount to $AX'$ with $X'=BX$
Edit: Here's the context ->
I have the differential equation:
$$sin(θ)\frac{d}{d\theta}\left(sin(\theta)\frac{dP_l^m}{d\theta}\right) +\left(l(l+1)sin^2(\theta)-m^2\right)P_l^m =0 $$
And $P_l^m$ is a function of $\theta$. The variable change used is $x=cos(\theta)$
On the other hand, $\cos^2\theta+1=1$ when $\theta=90^\circ$. So, since $\cos(90^\circ)=0$, you have $f\bigl(\cos(90^\circ)\bigr)=f(0)=1$.
Saying that you can use the substitution $x=\cos\theta$ does not mean that $f$ and $f\circ\cos$ are the same function. But it implies that they have the same ranges. For a more complete answer, it would be useful to know why is it that you want to do this substitution.