Question

Find the principal value of ${{\sin }^{-1}}\left( \dfrac{-1}{2} \right)$ .

Answer 1

Hint: The solution in which the absolute value of the angle is the least is the principal solution. So, here we have ${{\sin }^{-1}}\left( \dfrac{-1}{2} \right)$ . So, first we have to assume $x={{\sin }^{-1}}\left( \dfrac{-1}{2} \right)$, then we can take sine on both sides. Then we can find the possible values of x that lie in the range of $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$ and conclude the principal value from it.

Complete step-by-step answer:

Every trigonometric function has their own range of principal value. It is not necessary for every function to have the same range of principal value.
According to the question, we have to find the principal value of ${{\sin }^{-1}}\left( \dfrac{-1}{2} \right)$.
As the given trigonometric function is in sine function, we need the range for principal value of sine function.
We know that, $-\dfrac{\pi }{2}\le {{\sin }^{-1}}x\le \dfrac{\pi }{2}$ .
Here, in question we have ${{\sin }^{-1}}\left( \dfrac{-1}{2} \right)$ .
Let us assume, $x={{\sin }^{-1}}\left( -\dfrac{1}{2} \right)$ .
Remove the inverse by taking sine in both of LHS as well as RHS in ${{\sin }^{-1}}\left( \dfrac{-1}{2} \right)$, we get,

& x={{\sin }^{-1}}\left( \dfrac{-1}{2} \right) \\\ & \Rightarrow \sin x=\dfrac{-1}{2} \\\ & \\\ \end{aligned}$$……………….(1) Keeping in mind, the range of principal value of sine function, we have $$\sin \left( -\dfrac{\pi }{6} \right)=-\dfrac{1}{2}$$………….(2) From equation (1) and equation (2), we get $$\begin{aligned} & \sin x=\sin \left( \dfrac{-\pi }{6} \right) \\\ & \Rightarrow x=-\dfrac{\pi }{6} \\\ \end{aligned}$$ We got $$x=-\dfrac{\pi }{6}$$ , and we know that $$-\dfrac{\pi }{6}$$ lies between $$-\dfrac{\pi }{2}$$ and $$\dfrac{\pi }{2}$$ . That is, $$-\dfrac{\pi }{2}\le -\dfrac{\pi }{6}\le \dfrac{\pi }{2}$$. Hence, the principal value of $${{\sin }^{-1}}\left( \dfrac{-1}{2} \right)$$ is $$-\dfrac{\pi }{6}$$. Note: In this type of question, one can make mistakes in principal range. Also we have, $$\sin \dfrac{7\pi }{6}=-\dfrac{1}{2}$$ , but $$\dfrac{7\pi }{6}$$ doesn’t lies in the range $$\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$$ . So, this approach is wrong. We have to get those values which should lie in the range $$\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$$.