Question

The value of the expression $\sin \left( \dfrac{\pi }{3}-{{\sin }^{-1}}\left( \dfrac{-1}{2} \right) \right)$ is equal to

Answer 1

Hint: In the question, we have the inverse of the sine function. We know that the principal value of ${{\sin }^{-1}}\left( -\dfrac{1}{2} \right)$ , an inverse sine function has the range between $-\dfrac{\pi }{2}$ and $\dfrac{\pi }{2}$. After getting the principal value, just put that value in the given expression $\sin \left( \dfrac{\pi }{3}-{{\sin }^{-1}}\left( \dfrac{-1}{2} \right) \right)$ and then solve it further.

Complete step-by-step answer:

Let us assume, $y={{\sin }^{-1}}\left( \dfrac{-1}{2} \right)$………………(1)
We know the range of principal values of the sine function, $-\dfrac{\pi }{2}\le {{\sin }^{-1}}\theta \le \dfrac{\pi }{2}$ .
We have to remove the inverse function in equation (1). For that, we have to apply sine function in both LHS as well as RHS in equation (1).

& \sin y=-\dfrac{1}{2} \\\ & \sin y=\sin \left( -\dfrac{\pi }{6} \right) \\\ & y=-\dfrac{\pi }{6} \\\ \end{aligned}$$ Also, it satisfies the range of the inverse sine function. That is, $$-\dfrac{\pi }{2}\le \dfrac{\pi }{6}\le \dfrac{\pi }{2}$$. So, $$y=-\dfrac{\pi }{6}$$ is our principal value. Therefore, $${{\sin }^{-1}}\left( -\dfrac{1}{2} \right)=-\dfrac{\pi }{6}$$…………(2) Now, in the question we have $$\sin \left( \dfrac{\pi }{3}-{{\sin }^{-1}}\left( \dfrac{-1}{2} \right) \right)$$ . Using equation (2), we get $$\sin \dfrac{\pi }{2}$$ We know that $$\sin \dfrac{\pi }{2}$$ is equal to 1. Substituting the value of $$\sin \dfrac{\pi }{2}$$ in the above expression we get $$\begin{aligned} & \sin \left( \dfrac{\pi }{2} \right) \\\ & =1 \\\ \end{aligned}$$ Therefore, $$\sin \left( \dfrac{\pi }{3}-{{\sin }^{-1}}\left( \dfrac{-1}{2} \right) \right)=1$$ . Note: In this type of question, one can make a mistake in the principal value. Also, we have, $$\sin \dfrac{7\pi }{6}=-\dfrac{1}{2}$$ , but this doesn’t lie in the range $$\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$$ . So, this approach is wrong. We have to get those values of angles which should lie in the range $$\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$$ . Also, one can think of transforming the inverse sine function to inverse cosine function. But this will increase complexity. So, also don’t go for this approach.