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Question: How do you simplify \[{{\sin }^{-1}}\left( \sin \left( \dfrac{2\pi }{3} \right) \right)\]?...

How do you simplify sin⁡−1(sin⁡(2π3)){{\sin }^{-1}}\left( \sin \left( \dfrac{2\pi }{3} \right) \right)?

Explanation

Solution

We have solved this problem using inverse trigonometric identities and rules. We know that sin⁡−1{{\sin }^{-1}} had range between [−π2,π2]\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right] .So we will covert our angle into the range of sin⁡−1{{\sin }^{-1}} then we will derive the solution using some trigonometric identities.

Complete step by step answer:
So first we have to check whether our angle given is in the range of sin⁡−1{{\sin }^{-1}} or not. If it is in the range then we can directly simplify otherwise we have to convert the angle to be in the range .
Now we have to check whether the angle given 2π3\dfrac{2\pi }{3} is in between range [−π2,π2]\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right].
Let y=sin⁡−1(sin⁡(2π3))y={{\sin }^{-1}}\left( \sin \left( \dfrac{2\pi }{3} \right) \right)
If we have a equation like x=sin⁡−1(sin⁡y)x={{\sin }^{-1}}\left( \sin y \right) then we can write it as
⇒sin⁡x=sin⁡y\Rightarrow \sin x=\sin y
From this we can write our equation as
⇒sin⁡y=sin⁡2π3\Rightarrow \sin y=\sin \dfrac{2\pi }{3}
Now we have to calculate the value of 2π3\dfrac{2\pi }{3}
⇒2π3=2×1803\Rightarrow \dfrac{2\pi }{3}=\dfrac{2\times 180}{3}
⇒120∘\Rightarrow {{120}^{\circ }}
So we the equation formed is
⇒sin⁡y=sin⁡(120∘)\Rightarrow \sin y=\sin \left( {{120}^{\circ }} \right)
As 120∘{{120}^{\circ }} is outside the range of sin⁡−1{{\sin }^{-1}} we have make it to come in between [−90∘,90∘]\left[ -{{90}^{\circ }},{{90}^{\circ }} \right].
Now we can rewrite the equation as follows
⇒sin⁡y=sin⁡(180∘−60∘)\Rightarrow \sin y=\sin \left( {{180}^{\circ }}-{{60}^{\circ }} \right)
From the formula we have sin⁡(180∘−θ)=sin⁡θ\sin \left( {{180}^{\circ }}-\theta \right)=\sin \theta
We can rewrite the equation as
⇒sin⁡y=sin⁡(60∘)\Rightarrow \sin y=\sin \left( {{60}^{\circ }} \right)
Now we have to convert the angle into the range required. Now we have to convert the degrees into radians.
To convert angle from degrees to radians we have to multiply it with π180\dfrac{\pi }{180}.
So after converting the angle our equation will be like
⇒sin⁡y=sin⁡(60∘×π180)\Rightarrow \sin y=\sin \left( {{60}^{\circ }}\times \dfrac{\pi }{180} \right)
⇒sin⁡y=sin⁡(π3)\Rightarrow \sin y=\sin \left( \dfrac{\pi }{3} \right)
Now we can have the angle π3\dfrac{\pi }{3} which lies between [−π2,π2]\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right].
Hence we can write
⇒y=π3\Rightarrow y=\dfrac{\pi }{3}
But our y is sin⁡−1(sin⁡(2π3)){{\sin }^{-1}}\left( \sin \left( \dfrac{2\pi }{3} \right) \right)
So we can substitute the above function in place of y then we will get
sin⁡−1(sin⁡(2π3))=π3{{\sin }^{-1}}\left( \sin \left( \dfrac{2\pi }{3} \right) \right)=\dfrac{\pi }{3}
So the value of the given function is π3\dfrac{\pi }{3}.

Note:
We can also solve it in another way. As the inverse sin is multivalued, to include the angle outside the range we will consider all its supplementary and coterminal angles. So that we can take the solution from those angles which will fit in the range.