Question

How do you the inverse \[\sin (-1/2)\] or \[{{\sin }^{-1}}\left( {}^{-1}/{}_{2} \right)\] ?

Answer 1

Hint: Since we know that the range of the \[\sin \] function is \[\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\] and domain is \[\left[ -1,1 \right]\]. Let consider the value of this inverse be \[x\] then take the \[\sin \] function both sides and we also know the value of \[\sin \] function in the interval of \[\left[ 0,2\pi \right]\].

Complete step-by-step answer:
Let the value of the given inverse function be \[x\]
\[\Rightarrow x={{\sin }^{-1}}\left( {}^{-1}/{}_{2} \right)\]
Now taking \[\sin \] function both sides
\[\Rightarrow \sin x\text{ = }\sin ({{\sin }^{-1}}\left( {}^{-1}/{}_{2} \right))\]
And we also know that \[\sin ({{\sin }^{-1}}\theta )=\theta ,\text{ where }\theta \in \left[ -1,1 \right]\]
\[\Rightarrow \sin x={}^{-1}/{}_{2}\]
Since the \[\sin \] function is negative in third and fourth quadrant but the range of the \[\sin \] function is \[\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]\].
\[\Rightarrow \] The value of the \[x\] is \[{}^{-\pi }/{}_{6}\]
Thus, we have calculated
Hence the value of \[{{\sin }^{-1}}\left( {}^{-1}/{}_{2} \right)\] and that is \[{}^{-\pi }/{}_{6}\].

Note: First put down what the question has been given to us. To solve this type of questions we just remember the domain and the range of these inverse functions and apply the basics of trigonometric ratios and functions.