Question

How do you determine if $f\left( x \right)={{\sin }^{-1}}x$ is an even or odd function?

Answer 1

We know that ${{\sin }^{-1}}x$ is the inverse function of the trigonometric function $\sin x.$ Also, $\sin x$ is an odd function. We prove that the inverse of an odd function is also an even function.

Complete step by step answer:
We are given with the function $f\left( x \right)={{\sin }^{-1}}x.$
We know that ${{\sin }^{-1}}x$ is an inverse trigonometric function.
Also, it is the inverse of the trigonometric function $\sin x.$
Besides, we are familiar with the fact that the function $\sin x$ is an odd function.
We have to show that if a function is odd, then its inverse function is also an odd function.
Suppose that $y=g\left( x \right)$ is an odd function.
Then we get the inverse of the given function as
$\Rightarrow x={{g}^{-1}}\left( y \right)$
So, from this, when we multiply the whole equation with $-1$ we get,
$\Rightarrow -x=-{{g}^{-1}}\left( y \right).......\left( 1 \right)$
Now, we recall the property of an odd function $f\left( x \right)=f\left( -x \right).$
Applying this property in the odd function we have defined, that is, in $y=g\left( x \right).$
We get,
$\Rightarrow g\left( -x \right)=-g\left( x \right)$
And from the definition of the function, we will get the following equation
$\Rightarrow g\left( -x \right)=-g\left( x \right)=-y$
That is,
$\Rightarrow g\left( -x \right)=-y.......\left( 2 \right)$
Also, remember,
$-g\left( x \right)=-y......\left( 3 \right)$
Now we are taking the value of the inverse function ${{g}^{-1}}$ at the points $g\left( -x \right)$ and $-g\left( x \right).$
Then we will get the following,
$\Rightarrow {{g}^{-1}}\left( g\left( -x \right) \right)={{g}^{-1}}\left( -g\left( x \right) \right).......\left( 4 \right)$ Since $g\left( -x \right)$ and $-g\left( x \right)$ are equal, the function values at these points are also equal.
We substitute the value of $-g\left( x \right)$ from the equation $\left( 3 \right)$ in the equation $\left( 4 \right).$
The equation $\left( 4 \right)$ becomes,
$\Rightarrow {{g}^{-1}}\left( g\left( -x \right) \right)={{g}^{-1}}\left( -y \right)$
In this equation, we apply the property of the operation, function composition $f\left( g\left( x \right) \right)=f\circ g\left( x \right)$ as follows,
$\Rightarrow {{g}^{-1}}\circ g\left( -x \right)={{g}^{-1}}\left( -y \right)$
We have the property, ${{g}^{-1}}\circ g\left( x \right)=x$
Then,
$\Rightarrow -x={{g}^{-1}}\left( -y \right)$
Let us recall what we have in the equation $\left( 1 \right),$ we get
$\Rightarrow -{{g}^{-1}}\left( y \right)={{g}^{-1}}\left( -y \right).......\left( 5 \right)$
Now we get what is required.
The equation $\left( 5 \right)$ shows that the inverse function $x={{g}^{-1}}\left( y \right)$ satisfies the property of odd functions.
That implies the inverse function is also odd.
Thus, we have shown the inverse of an odd function is also an odd function.
It is proved that the inverse function ${{\sin }^{-1}}x$ of the trigonometric function $\sin x$ is also an odd function.
Hence, $f\left( x \right)={{\sin }^{-1}}x$ is an odd function.

Note:
This is not applicable for an even function. Because the graphs of the even functions are symmetrical about $y$-axis, they are not injective. So, they do not have inverses.