# Inverse trigonometric functions

1. Jun 23, 2015

I am familiar with the importance of the following inverse circular/hyperbolic functions:
$\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$, $\sinh^{-1}$, $\cosh^{-1}$, $\tanh^{-1}$.
However, I don't really get the point of $\csc^{-1}$, $\coth^{-1}$, and so on.
Given any equation of the form $f(x) = a$, where $f$ is the reciprocal of any circular/hyperbolic function, we can always write the equation as $\frac{1}{f(x)} = \frac{1}{a}$ before solving it with the aid of the six aforementioned inverse functions.
For instance, $\csc{x} = 2$, $0 < x < 2\pi$.
$\csc{x} = 2$
$\sin{x} = \frac{1}{2}$
$x = \frac{\pi}{6}, \frac{5\pi}{6}$
Do we use them to avoid cluttered notation in calculus?
For example, for $x \geq 1$:
$$\int \frac{1}{1 - x^2} dx = \coth^{-1}{x} + C = \frac{1}{2} \ln{\frac{x + 1}{x - 1}} + C$$

Last edited by a moderator: Jun 23, 2015
2. Jun 23, 2015

### Staff: Mentor

If the cosecent function is defined, it might be useful in some circumstances to have an inverse. Same for the hyperbolic cotangent.