# Is arcsin(x) = csc(x) true?

Oct 21, 2015

No. This is confusing ${\sin}^{-} 1 \left(x\right)$ with ${\left(\sin \left(x\right)\right)}^{-} 1$.

#### Explanation:

$\arcsin \left(x\right) = {\sin}^{-} 1 \left(x\right)$ is the inverse function of the function $\sin \left(x\right)$

That is:

If $x \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$, then $\arcsin \left(\sin \left(x\right)\right) = x$

If $x \in \left[- 1 , 1\right]$ then $\sin \left(\arcsin \left(x\right)\right) = x$

On the other hand:

$\csc \left(x\right) = {\left(\sin \left(x\right)\right)}^{- 1} = \frac{1}{\sin} \left(x\right)$ is the reciprocal of the $\sin$ function.

I think some of the blame for this confusion has to lie with the common convention of writing ${\sin}^{2} \left(x\right)$ to mean $\sin {\left(x\right)}^{2}$. So when you have $\csc \left(x\right) = \frac{1}{\sin} \left(x\right) = \sin {\left(x\right)}^{- 1}$ you might think that we would also write that as ${\sin}^{- 1} \left(x\right)$, but that's reserved for $\arcsin \left(x\right)$.