Question

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

Answer 1

Hint: We have solved this problem using inverse trigonometric identities and rules. We know that \[{{\sin }^{-1}}\] had range between \[\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]\] .So we will covert our angle into the range of \[{{\sin }^{-1}}\] 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 \[{{\sin }^{-1}}\] 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 \[\dfrac{2\pi }{3}\] is in between range \[\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]\].
Let \[y={{\sin }^{-1}}\left( \sin \left( \dfrac{2\pi }{3} \right) \right)\]
If we have a equation like \[x={{\sin }^{-1}}\left( \sin y \right)\] then we can write it as
\[\Rightarrow \sin x=\sin y\]
From this we can write our equation as
\[\Rightarrow \sin y=\sin \dfrac{2\pi }{3}\]
Now we have to calculate the value of \[\dfrac{2\pi }{3}\]
\[\Rightarrow \dfrac{2\pi }{3}=\dfrac{2\times 180}{3}\]
\[\Rightarrow {{120}^{\circ }}\]
So we the equation formed is
\[\Rightarrow \sin y=\sin \left( {{120}^{\circ }} \right)\]
As \[{{120}^{\circ }}\] is outside the range of \[{{\sin }^{-1}}\] we have make it to come in between \[\left[ -{{90}^{\circ }},{{90}^{\circ }} \right]\].
Now we can rewrite the equation as follows
\[\Rightarrow \sin y=\sin \left( {{180}^{\circ }}-{{60}^{\circ }} \right)\]
From the formula we have \[\sin \left( {{180}^{\circ }}-\theta \right)=\sin \theta \]
We can rewrite the equation as
\[\Rightarrow \sin y=\sin \left( {{60}^{\circ }} \right)\]
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 \[\dfrac{\pi }{180}\].
So after converting the angle our equation will be like
\[\Rightarrow \sin y=\sin \left( {{60}^{\circ }}\times \dfrac{\pi }{180} \right)\]
\[\Rightarrow \sin y=\sin \left( \dfrac{\pi }{3} \right)\]
Now we can have the angle \[\dfrac{\pi }{3}\] which lies between \[\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]\].
Hence we can write
\[\Rightarrow y=\dfrac{\pi }{3}\]
But our y is \[{{\sin }^{-1}}\left( \sin \left( \dfrac{2\pi }{3} \right) \right)\]
So we can substitute the above function in place of y then we will get
\[{{\sin }^{-1}}\left( \sin \left( \dfrac{2\pi }{3} \right) \right)=\dfrac{\pi }{3}\]
So the value of the given function is \[\dfrac{\pi }{3}\].

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.