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

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

Explanation

Solution

Since we know that the range of the sin⁡\sin function is [−π2,π2]\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right] and domain is [−1,1]\left[ -1,1 \right]. Let consider the value of this inverse be xx then take the sin⁡\sin function both sides and we also know the value of sin⁡\sin function in the interval of [0,2π]\left[ 0,2\pi \right].

Complete step-by-step answer:
Let the value of the given inverse function be xx
⇒x=sin⁡−1(−1/2)\Rightarrow x={{\sin }^{-1}}\left( {}^{-1}/{}_{2} \right)
Now taking sin⁡\sin function both sides
⇒sin⁡x = sin⁡(sin⁡−1(−1/2))\Rightarrow \sin x\text{ = }\sin ({{\sin }^{-1}}\left( {}^{-1}/{}_{2} \right))
And we also know that sin⁡(sin⁡−1θ)=θ, where θ∈[−1,1]\sin ({{\sin }^{-1}}\theta )=\theta ,\text{ where }\theta \in \left[ -1,1 \right]
⇒sin⁡x=−1/2\Rightarrow \sin x={}^{-1}/{}_{2}
Since the sin⁡\sin function is negative in third and fourth quadrant but the range of the sin⁡\sin function is [−π2,π2]\left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right].
⇒\Rightarrow The value of the xx is −π/6{}^{-\pi }/{}_{6}
Thus, we have calculated
Hence the value of sin⁡−1(−1/2){{\sin }^{-1}}\left( {}^{-1}/{}_{2} \right) and that is −π/6{}^{-\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.