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Question: Find the principal value of \[{\sin ^{ - 1}}\left[ {\sin \left( {\dfrac{{2\pi }}{3}} \right)} \right...

Find the principal value of sin⁡−1[sin⁡(2π3)]{\sin ^{ - 1}}\left[ {\sin \left( {\dfrac{{2\pi }}{3}} \right)} \right]

Explanation

Solution

First of all we will assume the given quantity is xx and then find the value of xx in the range of sin⁡−1θ{\sin ^{ - 1}}\theta .
The range of sin⁡−1θ{\sin ^{ - 1}}\theta is [−π2,π2]\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right] which means the principal angle must lie in this range.

Complete answer: Let x=sin⁡−1[sin⁡(2π3)]x = {\sin ^{ - 1}}\left[ {\sin \left( {\dfrac{{2\pi }}{3}} \right)} \right]
Taking sin of both the sides we get:
sin⁡x=sin⁡[sin⁡−1[sin⁡(2π3)]]\sin x = \sin \left[ {{{\sin }^{ - 1}}\left[ {\sin \left( {\dfrac{{2\pi }}{3}} \right)} \right]} \right]
As we know that,
sin⁡(sin⁡−1θ)=θ\sin \left( {{{\sin }^{ - 1}}\theta } \right) = \theta
Therefore,
sin⁡x=sin⁡(2π3)\sin x = \sin \left( {\dfrac{{2\pi }}{3}} \right)………………………………(1)
According to above equation x=2π3x = \dfrac{{2\pi }}{3}
But the range of principal sin⁡−1θ{\sin ^{ - 1}}\theta is [−π2,π2]\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]
And 2π3>π2\dfrac{{2\pi }}{3} > \dfrac{\pi }{2} therefore, it does not lie in the range of principal value
Hence, x≠2π3x \ne \dfrac{{2\pi }}{3}
Now as we know that ,
2π3=π−π3\dfrac{{2\pi }}{3} = \pi - \dfrac{\pi }{3}
Hence, putting this value in equation (1) we get:
sin⁡x=sin⁡(π−π3)\sin x = \sin \left( {\pi - \dfrac{\pi }{3}} \right)
Also we know that,
sin⁡(π−θ)=sin⁡θ\sin \left( {\pi - \theta } \right) = \sin \theta
Therefore,
sin⁡x=sin⁡(π3)\sin x = \sin \left( {\dfrac{\pi }{3}} \right)
Which implies x=π3x = \dfrac{\pi }{3}
Now since −π2<π3<π2 - \dfrac{\pi }{2} < \dfrac{\pi }{3} < \dfrac{\pi }{2}
Therefore it lies in the range of principal value.

Hence the principal value of the given expression is π3\dfrac{\pi }{3}.

Note: The important formulas and identities used in this question are:
sin⁡(sin⁡−1θ)=θ\sin \left( {{{\sin }^{ - 1}}\theta } \right) = \theta
sin⁡(π−θ)=sin⁡θ\sin \left( {\pi - \theta } \right) = \sin \theta
Since the range of sin⁡−1x{\sin ^{ - 1}}xalways lie in the interval [−π2,π2]\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]
Therefore principal value should always lie in this interval, if it is not in this interval then it is not the principal value of the given expression.