Why is Sin(cos^-1(x)) is not defined on pi or any other x > pi/3.

Question

Why sin(sin^-1x) is not defined on every value of x

But in Sin(x), x takes all values. cos^-1(pi) = -1. But why is it not defined. I Want to understand how this function work.

Answer 1

While it is true that the $\sin$ function can take any value $x\in\mathbb{R}$ as input, this does not imply that the range of its inverse $\sin^{-1}$ is $\mathbb{R}$. In fact, it is defined to be $[-\pi/2,\pi/2]$ (if you take a look at the graph of $y=\sin(x)$, you can see why it makes sense to do so). Otherwise, we would encounter a case in which a single input maps to multiple outputs (violating the definition of a function).

For example, $\sin^{-1}(0)=0$ means that $\sin(0)=0$. This is true, but it is also true that $\sin(k\pi)=0$ for any $k\in\mathbb{Z}$.

As a result, when you take $\sin(\sin^{-1}(x))$, you're only passing values on the interval $[-\pi/2,\pi/2]$ into $\sin$, which means your output will be on the interval $[-1,1]$ (which you are confusing with $[-\pi/3,\pi/3]$).

This reasoning also applies to $\cos^{-1}$.