Question

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

Answer 1

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

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

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

Note: The important formulas and identities used in this question are:
\[\sin \left( {{{\sin }^{ - 1}}\theta } \right) = \theta \]
\[\sin \left( {\pi - \theta } \right) = \sin \theta \]
Since the range of \[{\sin ^{ - 1}}x\]always lie in the interval \[\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]\]
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.