What is the value of sin-1(-x) for all x belongs to [-1, 1]?

The correct option is -sin^-1(x).

To understand why this relationship holds, let us go through a step-by-step derivation using the properties of inverse trigonometric functions.

Let us define a variable:
$$y={\sin }^{-1}(-x)$$
where $x\in [-1,1]$. By definition of the principal value branch of the inverse sine function, the angle y must lie in the interval:
$$y\in [-\frac{\pi }{2},\frac{\pi }{2}]$$

By taking the sine of both sides of the equation, we can rewrite it in terms of the standard sine function:
$$\sin \left(y\right)=-x$$

Next, we multiply both sides of the equation by −1 to isolate x:
$$-\sin \left(y\right)=x$$

We know from trigonometric identities that the sine function is an odd function, meaning:
$$-\sin \left(y\right)=\sin (-y)$$
Substituting this identity back into our equation gives:
$$\sin (-y)=x$$

Since $y\in [-\frac{\pi }{2},\frac{\pi }{2}]$, it follows that its negative counterpart also lies in the same domain:
$$-y\in [-\frac{\pi }{2},\frac{\pi }{2}]$$
Therefore, we can safely take the inverse sine of both sides of the equation:
$$-y={\sin }^{-1}\left(x\right)$$

Multiplying both sides by −1 to solve for y, we get:
$$y=-{\sin }^{-1}\left(x\right)$$

Finally, replacing y with our original definition:
$${\sin }^{-1}(-x)=-{\sin }^{-1}\left(x\right)$$
This confirms that the inverse sine function is also an odd function over its defined domain.