Is $\text{arccosec}(x) = \arcsin\left(\frac{1}{x}\right)$ for all $x \in ℝ?$

Question

Is $\text{arccosec}(x) = \arcsin\left(\frac{1}{x}\right)$ for all $x \in ℝ?$ I'm still really new to trigonometric inverses, so if the above was cleared up I'd be grateful. Thanks.

Answer 1

Yes if you were to take on conventional domain restrictions. Because $\text{arccsc} (x)=y_1$ will output a value $y_1$ such that $$\frac{1}{\sin y_1}=x\qquad (1)$$ And $\text{arcsin}(1/x)=y_2$ will output a value $y_2$ such that $$\sin y_2=\frac{1}{x}\Rightarrow\frac{1}{\sin y_2}=x\qquad (2)$$ And as you can see from $(1)$ and $(2)$, $y_1$ and $y_2$ are equal, because although $\sin(x)$ is periodic we restrict its domain for the inverses.

Conventional domain restrictions are as follows: $$\text{arccsc}(x):\mathbb{R}\backslash (-1,1)\rightarrow \mathbb{R}\Rightarrow\text{arccsc}(1/x):[-1,1]\rightarrow\mathbb{R}$$ $$\text{arcsin}(x):[-1,1]\rightarrow \mathbb{R}\Rightarrow\arcsin(1/x):\mathbb{R}\backslash(-1,1)\rightarrow \mathbb{R}$$ So we could also say $$\text{arccsc}\left(\frac{1}{x}\right)=\text{arcsin}(x)\qquad (3)$$ However the domain upon which both sides of $(3)$ are defined is the interval upon which both sides of your original statement are undefined.

Answer 2

Suppose $\alpha=\operatorname{arccsc} x$, which implies that $$ x=\csc\alpha=\frac{1}{\sin\alpha} $$ so $$ \sin\alpha=\frac{1}{x} $$ If you define the “arc cosecant” function to take its values in $(-\pi/2,0)\cup(0,\pi/2)$, then, yes, you have $$ \alpha=\arcsin\frac{1}{x} $$ and the relation holds for $x\le-1$ or $x\ge1$. The arc cosecant cannot be defined on $(-1,1)$, just like the arc sine is only defined on $[-1,1]$.

However, note that some authors define the arc cosecant to take its values in $(0,\pi/2]\cup(\pi,3\pi/2]$, in order to have an increasing function in each component of its domain. In this case, the above identity would be false.