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Question: The principal value of \[{\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{5\pi }}{3}} \right)} \right)\] ...

The principal value of sin⁡−1(sin⁡(5π3)){\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{5\pi }}{3}} \right)} \right)
a) 5π3\dfrac{{5\pi }}{3}
b) −5π3 - \dfrac{{5\pi }}{3}
c) −π3 - \dfrac{\pi }{3}
d) −4π3 - \dfrac{{4\pi }}{3}

Explanation

Solution

Hint : Here the question is related to the inverse trigonometry topic. We have to determine the principal value of the given term. By considering the domain of the maximum value of the sine trigonometric ratios. Then by the trigonometric ratios for standard angles and ASTC rule we are determining the value.

Complete step-by-step answer :
Principal values for sine, cosine and tangent trigonometric ratio have one solution in this interval called the principal value of θ. It is in the first or fourth quadrant.
The principal value of sin⁡−1(x){\sin ^{ - 1}}\left( x \right) is [−π2,π2]\left[ {\dfrac{{ - \pi }}{2},\dfrac{\pi }{2}} \right]
Now we have to determine the principal value of sin⁡−1(sin⁡(5π3)){\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{5\pi }}{3}} \right)} \right), the 5π3∉[−π2,π2]\dfrac{{5\pi }}{3} \notin \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]. So we rewrite it as sin⁡−1(sin⁡(2π−π3)){\sin ^{ - 1}}\left( {\sin \left( {2\pi - \dfrac{\pi }{3}} \right)} \right)
So now the given angle will lie in the fourth quadrant. By the ASTC rule the sine trigonometric ratio is negative in the fourth quadrant. According to the trigonometric ratios for the allied angles, the above term is written as
⇒sin⁡−1(sin⁡(−π3))\Rightarrow {\sin ^{ - 1}}\left( {\sin \left( { - \dfrac{\pi }{3}} \right)} \right)
By the trigonometric ratios for allied angles we have sin⁡(−θ)=−sin⁡θ\sin ( - \theta ) = - \sin \theta . The above term is written as
⇒sin⁡−1(−sin⁡(π3))\Rightarrow {\sin ^{ - 1}}\left( { - \sin \left( {\dfrac{\pi }{3}} \right)} \right)
Take a negative sign outside
⇒−sin⁡−1(sin⁡(π3))\Rightarrow - {\sin ^{ - 1}}\left( {\sin \left( {\dfrac{\pi }{3}} \right)} \right)
The sine trigonometric ratio and sine inverse trigonometric ratio are inverse to each other and hence it will cancel.
⇒−π3\Rightarrow - \dfrac{\pi }{3}
Hence the option c) is the correct one.
So, the correct answer is “Option c”.

Note : The principal value will represent the range of a given trigonometric function. Every inverse trigonometric ratio has a range, when the given angle does not belong to the given domain then we are writing the angle in terms of sum or difference of the standard angles.