# For what reasons does the inverse trig functions have the images they have?

The trigonometric functions ($\sin, \cos, \tan,...$) aren't invertible in $R$ so we restrain it's domain to create inverses for them, the inverse trigonometric functions ($\arcsin, \arccos, \arctan,...$). I'm not sure if it's a convention but all the books i've read restrain the sine function to the interval $[-\pi/2, \pi/2]$ and the cosine function to the interval $[0, \pi]$. My question is:

For what reasons mathematicians chose these intervals?

Yes, it a convention. And a natural one. The longest possible length for such an interval os $\pi$ (after that, the functions aren't injective). And it's a natural choice to pick an interval starting at $0$ or containing it in its middle. The leaves us only with the choice of $\left[-\frac\pi2,\frac\pi2\right]$ for $\arcsin$ and $[0,\pi]$ for $\arccos$.
Well, we do need to restrain $\arcsin, \arccos$ to intervals on which $\sin, \cos$ are injective and upon which they achieve all possible values. As the maximums of these continuous functions are are $1$ and the minimums are $-1$ we must choose an intervals $[a,b]$ and $[c,d]$ where $\sin a = \pm 1; \sin b \mp 1$, $\cos a = \pm 1; \cos b = \mp 1$ and $b-a = d-c =$ half the period =$\pi$.
Those are $[a = n\pi - \frac \pi 2, b=n\pi +\frac \pi 2]$ and $[c = m\pi, d= (m+1)\pi]$.
So why the convention that we let $n = m = 0$? Well, is that not the most natural/simple choice?