Range of an inverse sine trigonometric function

Question

I am supposed to find the range of the function: $$f(x)=\frac{2}{\sin^{-1}x}$$

My approach: $$ \frac{-\pi }{2}\leq \sin^{-1}x\leq \frac{\pi }{2}$$ $$\frac{-2 }{\pi }\geq \frac{1}{\sin^{-1}x}\geq \frac{2}{\pi }$$ Going on solving further in a similar way, I'm getting the range as : $$\left[\frac{-4}{\pi },\frac{4}{\pi }\right]-{0}$$ But the correct answer is supposed to be: $$(-\infty , \frac{-4}{\pi }]\cup [\frac{-4}{\pi },\infty)$$ I would grateful if someone would point out the flaw in my approach.

Answer 1

The error is that you took reciprocals of the inequality $$ -\frac{\pi}{2}\leq\sin^{-1}(x)\leq\frac{\pi}{2} \, . $$ In general, if $a<b$, then it is not necessarily true that $$ \frac{1}{a}<\frac{1}{b} \, . $$ Sometimes it is true, e.g. if $a=-2$ and $b=5$. Sometimes it is not true, e.g. if $a=4$ and $b=5$. It's a mistake to assume that you can always apply the same 'rules' to inequalities as you can equations.

Anyway, there is a simpler method you can take to solve this problem:

1. Plot the graph of $\sin(x)$ for $-\dfrac{\pi}{2}\leq x\leq\dfrac{\pi}{2}$.
2. Reflect the graph in the line $y=x$ to get the graph of $\sin^{-1}(x)$.
3. Use this to plot the graph of $\dfrac{1}{\sin^{-1}(x)}$.
4. Then plot $\dfrac{2}{\sin^{-1}(x)}$.

Pay particular attention to what happens when $x$ is close to $0$.