# How do you evaluate  csc^-1 (cos(4/7))?

May 27, 2018

#### Explanation:

The $\csc$ function takes in any input that's not an integer multiple of $\pi$, and returns a number from (–oo, –1]uu[1,oo):

graph{cscx [-10, 10, -5, 5]}

Thus, the inverse function csc^(–1) can take in any input from (–oo, –1)uu(1,oo) and return, by definition, any number in [–pi/2, pi/2] except $0 :$

graph{x=cscy [-10, 10, -1.6, 1.6]}

In order for csc^(–1)(cos (4/7)) to be a real number, $\cos \left(\frac{4}{7}\right)$ needs to be in the domain of csc^(–1). Meaning, $\cos \left(\frac{4}{7}\right)$ needs to be any real number not between –1 and $1.$ But since $\frac{4}{7}$ is not an integer multiple of $\pi ,$ we know $\cos \left(\frac{4}{7}\right)$ actually is between –1 and $1.$

Thus, $\cos \left(\frac{4}{7}\right)$ is not in the domain of csc^(–1), and there is no value for csc^(–1)(cos (4/7)).

May 27, 2018

This expression is undefined for $\cos$ and $\csc$ considered as real-valued functions.

Note however that $\csc \left({\cos}^{- 1} \left(\frac{4}{7}\right)\right) = \frac{7 \sqrt{33}}{33}$

#### Explanation:

Note that as real valued functions:

$\cos \left(\theta\right) \in \left[- 1 , 1\right]$

$\csc \left(\theta\right) \in \left(- \infty , - 1\right] \cup \left[1 , \infty\right)$

Note that:

$0 < \frac{4}{7} < \frac{\pi}{2}$

and hence:

$0 < \cos \left(\frac{4}{7}\right) < 1$

So there is no real value of $\theta$ such that:

$\csc \left(\theta\right) = \cos \left(\frac{4}{7}\right)$

Footnote

I wonder whether the question should have actually been to find the value of $\csc \left({\cos}^{- 1} \left(\frac{4}{7}\right)\right)$, which we can do by considering a $4$, $\sqrt{33}$, $7$ right-angled triangle (since ${4}^{2} + {\left(\sqrt{33}\right)}^{2} = {7}^{2}$).

If so, then we find:

$\csc \left({\cos}^{- 1} \left(\frac{4}{7}\right)\right) = \frac{7}{\sqrt{33}} = \frac{7 \sqrt{33}}{33}$