# How do you find the exact value of csc^-1 0.5?

Jul 12, 2015

(a) If ${\csc}^{- 1} \left(0.5\right) = \sin \left(0.5\right)$ then it equals to $0.4794 \ldots$
(b) If ${f}^{- 1} \left(x\right)$ means inverse function, then there is no real value of ${\csc}^{- 1} \left(0.5\right)$.

#### Explanation:

Actually, it depends on what you mean by using the power of $- 1$.
(a) On one hand, ${f}^{- 1} \left(x\right)$ might mean $\frac{1}{f} \left(x\right)$.
(b) On the other hand, it might mean inverse function $g \left(x\right)$ that inverses the action of the function $f \left(x\right)$, that is $g \left(f \left(x\right)\right) = x$.

(a) If you mean that ${f}^{- 1} \left(x\right)$ is $\frac{1}{f} \left(x\right)$ then the following is true.

By definition, $\csc \left(x\right) = \frac{1}{\sin} \left(x\right)$.
Therefore, ${\csc}^{- 1} \left(x\right) = \sin \left(x\right)$

For $x = 0.5$:
${\csc}^{- 1} \left(0.5\right) = \sin \left(0.5\right)$ which you can find on any calculator as $0.4794 \ldots$

(b) If you mean that the power of $- 1$ is an inverse function (usually, prefixed arc- in trigonometry, like arcsin or arccsc) then the following is true.

${\csc}^{- 1} \left(0.5\right)$ is an angle $\phi$ (in radians), cosecant of which is equal to $0.5$.
That is, $\csc \left(\phi\right) = 0.5$.
From the definition of $\csc \left(\right)$ as $\frac{1}{\sin} \left(\right)$, we should find $\phi$ such that
$\frac{1}{\sin} \left(\phi\right) = 0.5$ or $\sin \left(\phi\right) = 2$.
But that equation has no solutions since $\sin \left(\phi\right) \le 1$ for any $\phi$.
So, if you consider this second meaning of the power of $- 1$, the expression ${\csc}^{- 1} \left(0.5\right)$ has no real value.