Analyzing the ranges of inverse trigonometric functions.

Question

I was analyzing the ranges of inverse trigonometric functions.

1. $\sin^{-1}x$ has range $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$. So why this only?

Let's see what other options we have

$[0,\pi]$ - $\sin x$ is not onto in this domain

$\left[\dfrac{\pi}{2},\dfrac{3\pi}{2}\right]$ - Not sure but may be to preserve the symmetric nature of $\sin x$ about origin.

Hence the best suited interval will be $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ where $\sin x$ is invertible and symmetric about origin.

2. $\cos^{-1}x$ has range $[0,\pi]$. So why this only?

Let's see what other options we have

$\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$- $\cos x$ is not onto in this domain.

$\left[\dfrac{\pi}{2},\dfrac{3\pi}{2}\right]$- $\cos x$ is not onto in this domain.

$\left[\pi,2\pi\right]$ - May be to preserve the opposite nature of $\sin x$ and $\cos x$ like when the $\sin x$ is decreasing then $\cos x$ is increasing or vice-versa.

3. $\tan^{-1} x$ has range $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$. So why this only?

Let's see what other options we have

$\left[0,\pi\right]-\{\dfrac{\pi}{2}\}$- In this interval graph of $\tan x$ is not always increasing because of following reason:

$$\theta_1=\dfrac{\pi}{2}^{-},\theta_2=\dfrac{\pi}{2}^{+}$$

$$\theta_1<\theta_2$$ $$\tan(\theta_1)>\tan(\theta_2)$$

But as we have limited the domain of $\sin x$ to $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ and $\cos x$ to $\left[0,\pi\right]$, so $\tan x$ should be always increasing as $\sin x$ is always increasing in its range $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ and $\cos x$ is always decreasing in its range $\left[0,\pi\right]$

$\left[\dfrac{\pi}{2},\dfrac{3\pi}{2}\right]$- not sure but may be to preserve the symmetric nature of $\tan x$ graph about origin.

Others $3$ functions can be thought of in the similar way. I am not sure whether I am right but I have made an attempt to explain this stuff which one doesn't find in most of the text books.

Be free to give your opinions on this.

Answer 1

As you noticed the choice of the interval $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ is made in order to have bijective restriction for $\sin x: \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right] \to [-1,1]$ and then define the inverse function $\arcsin x:[-1,1]\to \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right] $.

Of course we could choose any other interval $\left[-\dfrac{\pi}{2}+k\pi,\dfrac{\pi}{2}+k\pi\right]$.

Similarly for $\cos x$ we need $\left[k\pi,\pi+k\pi\right]$ and for $\tan x$ we need $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right) $ in order to have an invertible restriction and define the corresponding inverse functions.

In that context, the symmetry issue is not so important but of course the choice for $\sin x$ and $\tan x$ allows to mantain this nice property. This is possible in this case since both functions are odd and therefore also the inverse can be defined with the same property, while for $\cos x$ this is not possible since the function is even and the interval $[0,\pi]$, in some sense, is the simpler choice.