$arccsc(\csc(2\pi/3))$. What's the value?

Question

I tried to solve this problem using $f^{-1}(f(x))=x$. However, the range of Arccsc is greater or equal to $-1/2\pi$ and less or equal to $1/2 \pi$. And, apparently, $2\pi/3$ is not in the range. How do I solve this problem not using a calculator?

Answer 1

Let $y={\rm arccsc}(\csc(2\pi/3))$. This means $$\csc y=\csc(2\pi/3)\ ,\qquad -\pi/2\le y\le\pi/2\ .$$ Now $\csc t=\csc(\pi-t)$ provided $t$ is not a multiple of $\pi$, so this can be written $$\csc y=\csc(\pi/3)\ ,\qquad -\pi/2\le y\le\pi/2\ ,$$ and the answer is $y=\pi/3$.

Answer 2

The cosecant function is not 1-1, so you restrict it to $(-\pi/2, \pi/2)$ to make it 1-1 (this is the "principal branch"). You only get $f^{-1}\circ f(x)$ for $x\in (-\pi/2, \pi/2)$.

Answer 3

By definition the identity

$$ \sin ^{-1} x = \csc^{-1} {1/x} $$

we have

$$csc^{-1}(\csc(2\pi/3)) = \sin^{-1}(1/\csc(2\pi/3))=\sin^{-1}(\sin (2\pi/3))$$

$$=2 \pi/3, \pi- 2 \pi/3,... $$

i.e.,

$$ \pi/3, 2 \pi/3$$

plus for co-terminal angles add $2 \pi$

$$ 7 \pi/3, 8\pi/3$$

etc. lying in quadrants 1,2 only.