# Bijective change of variable

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.
• Yes I also think $f$ is different from $f\circ cos$. You have plugged in $90°$ because $Arcos(0)=90°$ ? What surprises me from the paper I learn from is that there is no $Arcos$ in the final answer... Jun 10 at 16:30
• No. I plugged in $90^\circ$ because $\cos(90^\circ)=0$. Jun 10 at 16:33
• There is nothing peculiar about this change of variable. For instance, computing $\int_{-1}^1\sqrt{1-x^2}\,\mathrm dx$ can be done doing the substitution $x=\cos\theta$ and $\mathrm dx=-\sin\theta\,\mathrm d\theta$. Jun 10 at 19:45
• In the proof I know of the integral change of variable, we use $(f\circ g)'=g'f'\circ g$ Jun 11 at 19:01