Question

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

Answer 1

There is no answer.

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 (4/7)` needs to be in the domain of `csc^(–1).` Meaning, `cos(4/7)` needs to be any real number not between `–1` and `1.` But since `4/7` is not an integer multiple of `pi,` we know `cos(4/7)` actually is between `–1` and `1.`

Thus, `cos(4/7)` is not in the domain of `csc^(–1)`, and there is no value for `csc^(–1)(cos (4/7)).`

Answer 2

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

Note however that `csc(cos^(-1)(4/7)) = (7sqrt(33))/33`

Explanation:

Note that as real valued functions:

`cos(theta) in [-1, 1]`

`csc(theta) in (-oo, -1] uu [1, oo)`

Note that:

`0 < 4/7 < pi/2`

and hence:

`0 < cos(4/7) < 1`

So there is no real value of `theta` such that:

`csc(theta) = cos(4/7)`

Footnote

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

If so, then we find:

`csc(cos^(-1)(4/7)) = 7/sqrt(33) = (7sqrt(33))/33`