Question

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

Answer 1

Hint: We will be using the concept of inverse trigonometric functions to solve the problem. We will first write $-\dfrac{1}{2}$ as sine of an angle then we will use the identity that for $x\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]{{\sin }^{-1}}\left( \sin x \right)=x$.

Complete step-by-step answer:
Now, we have to find the value of ${{\sin }^{-1}}\left( -\dfrac{1}{2} \right)$.
Now, we know that the value of $\sin \left( -\dfrac{\pi }{6} \right)=-\dfrac{1}{2}.........\left( 1 \right)$
We have taken $-\dfrac{1}{2}=\sin \left( -\dfrac{\pi }{6} \right)$ as in the view of the principal value convention, x is confined to be in $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$.
Now, we know that the graph of ${{\sin }^{-1}}\left( \sin x \right)$ is,

Now, we have to find the value of ${{\sin }^{-1}}\left( -\dfrac{1}{2} \right)$.
We will use the value of $-\dfrac{1}{2}$ from (1). So, we have,
${{\sin }^{-1}}\left( \sin \left( -\dfrac{\pi }{6} \right) \right)$
Also, we know that ${{\sin }^{-1}}\left( \sin x \right)=x$ for $x\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$. So, we have,
${{\sin }^{-1}}\left( \sin \left( -\dfrac{\pi }{6} \right) \right)=-\dfrac{\pi }{6}$

Note: To solve these type of question it is important to note that we have used a fact that ${{\sin }^{-1}}\left( \sin x \right)=x$ only for $x\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$. For another value of x the graph of ${{\sin }^{-1}}\left( \sin x \right)$ must be used to find the answer.