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 $x=cos(\theta)$


1 Answer 1


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.

  • $\begingroup$ 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... $\endgroup$
    – niobium
    Jun 10 at 16:30
  • $\begingroup$ No. I plugged in $90^\circ$ because $\cos(90^\circ)=0$. $\endgroup$ Jun 10 at 16:33
  • $\begingroup$ I have to say I don't understand... I have learnt in school about change of variable in integral, but this one..... looks very different. Is there a name for this sort of change of variable ? (so I can learn from it in books, or internet?) $\endgroup$
    – niobium
    Jun 10 at 19:38
  • $\begingroup$ 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 $\mathrm dx=-\sin\theta\,\mathrm d\theta$. $\endgroup$ Jun 10 at 19:45
  • $\begingroup$ In the proof I know of the integral change of variable, we use $(f\circ g)'=g'f'\circ g$ $\endgroup$
    – niobium
    Jun 11 at 19:01

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