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).
\[\begin{align}
  & \sin y=-\dfrac{1}{2} \\
 & \sin y=\sin \left( -\dfrac{\pi }{6} \right) \\
 & y=-\dfrac{\pi }{6} \\
\end{align}\]
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
\[\begin{align}
  & \sin \left( \dfrac{\pi }{2} \right) \\
 & =1 \\
\end{align}\]
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.