I am trying to answer the following question:
- What is the inverse function of $y=\cos^2(x)$, $x\in[0 , \pi/2]$?
- What is the inverse function of $y=\cos^2(x)$, $x\in[\pi/2 , \pi]$?
I answered the first by "solving for $x$" and obtained $x=\arccos(\sqrt{y})$, which seems to work just fine. As $0=\arccos(\sqrt{1})$ and $\pi/2=\arccos(\sqrt{0})$ but for the other ones, it didn't work very well. I tried to write:
$$\arccos(\sqrt{0})+a=\pi/2\\\arccos(\sqrt{1})+a=\pi/2$$
Which clearly won't work. The answer turned out to be
$$y=\frac{1}{2}\arccos(2x-1)\quad\quad\quad y=\pi -\frac{1}{2}\arccos(2x-1)$$
for the first and second questions.Now, this hinted me the following idea - for example - in the second case I should write:
$$a\cdot \arccos(bx+c)+d=\pi/2 \\ a\cdot \arccos(bx+c)+d=\pi$$
And find coefficients $a,b,c,d$ such that both equations are satisfied. The trouble is: I kinda understood how one can arrive at that the answers I mentioned: When $x=0$, we have $\arccos(-1)=\pi$ and when $x=1$, we have $\arccos(1)=0$ so we need to find:
$$\pi a + d = \pi/2 \\ 0a+d=\pi$$
With this, we find the function given in the answer. I've conjectured that we could also - for example - choose another polynomial function such that $x=0 \implies f(0)=-1$ and $x=1 \implies f(1)=0$ and then:
$$a\cdot \arccos(f(0))+d=\pi/2 \\ a\cdot \arccos(f(1))+d=\pi$$
Which then yields:
$$a\cdot \pi+d=\pi/2 \\ a\cdot \frac{\pi}{2}+d=\pi$$
And again, we could solve for $a,d$ and find it but this procedure does not work. I am confused: Why does that solution works but the one in my conjecture doesn't?