Question

The principal value of $ {\sin ^{ - 1}}[\sin (\dfrac{{2\pi }}{3})] $ is
 $ 1) \dfrac{{ - 2\pi }}{3} $
 $ 2) \dfrac{{2\pi }}{3} $
 $ 3) \dfrac{{4\pi }}{3} $
 $ 4) $ None of these

Answer 1

Hint: First, we need to analyze the given information so that we can able to solve the problem. Generally, in Mathematics, the trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified. Here we are asked to calculate the principal value of the function $ {\sin ^{ - 1}}[\sin (\dfrac{{2\pi }}{3})] $ .
Let us assume the given function to $ y $ which is $ y = {\sin ^{ - 1}}[\sin (\dfrac{{2\pi }}{3})] $
We need to apply the appropriate trigonometric identities to obtain the required answer.
Formula used:
 $ \sin (\pi - \theta ) = \sin \theta $
 $ \sin ({\sin ^{ - 1}}) = 1 $

Complete step by step answer:
Now we are going to use the inverse trigonometric functions so that we are able to solve them further.
Which is $ \sin ({\sin ^{ - 1}}) = 1 $
Let us start with the solution $ y = {\sin ^{ - 1}}[\sin (\dfrac{{2\pi }}{3})] $ and now multiply the sine values on both sides.
Thus, we get, $ \sin y = \sin ({\sin ^{ - 1}}[\sin (\dfrac{{2\pi }}{3})]) $
Now as per the sine inverse formula we get; $ \sin y = [\sin (\dfrac{{2\pi }}{3})] $ (since $ \sin ({\sin ^{ - 1}}) = 1 $ )
But, the range of the sine inverse is not between $ [ - \dfrac{\pi }{2},\dfrac{\pi }{2}] $ (because sine is not the $ 1 - 1 $ function and the given domain must be limited)
Therefore, from the above equation $ {\sin ^{ - 1}} $ is not possible principal value at $ [ - \dfrac{\pi }{2},\dfrac{\pi }{2}] $ and thus $ y = \dfrac{{2\pi }}{3} $ is not possible.
We know that $ \sin x $ is a positive value in the first quadrant and also in the second quadrant, but negative in the third and fourth quadrant.
Thus $ \sin (\pi - \theta ) = \sin \theta $
Now using the value of sine in the equation $ \sin y = [\sin (\dfrac{{2\pi }}{3})] $ and convert it into $ \sin y = [\sin (\dfrac{{2\pi }}{3})] \Rightarrow [\sin (\pi - \dfrac{{2\pi }}{3})] $ (the value of the sine is not changed as $ \sin (\pi - \theta ) = \sin \theta $ )
Hence solving the values, we get $ \sin y = [\sin (\pi - \dfrac{{2\pi }}{3})] \Rightarrow [\sin (\dfrac{\pi }{3})] $ and canceling we get $ y = \dfrac{\pi }{3} $
This value is in the range of $ [ - \dfrac{\pi }{2},\dfrac{\pi }{2}] $
Hence, the principal value of $ {\sin ^{ - 1}}[\sin (\dfrac{{2\pi }}{3})] $ $ = \dfrac{\pi }{3} $
Hence option $ 4) $ None of these is correct.
For option, $ 1)\dfrac{{ - 2\pi }}{3} $ there is no possibility because it will no in the range of $ [ - \dfrac{\pi }{2},\dfrac{\pi }{2}] $ the point be exceeding.
For option
For options $ 3)\dfrac{{4\pi }}{3} $ , $ 2)\dfrac{{2\pi }}{3} $ there is no possibility, because it will no in the range of $ [ - \dfrac{\pi }{2},\dfrac{\pi }{2}] $ the point be exceeding.

So, the correct answer is “Option 4”.

Note: The six basic trigonometric ratios are sine, cosine, tangent, cosecant, secant, and cotangent and all the fundamental trigonometric identities are derived from the six trigonometric ratios.
Remember the inverse trigonometric values which we used in the solution and properties of trigonometric functions like in the given question we need the property $ \sin x $ that is positive in the first two-quadrant and negative in the last two quadrants.