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# Evaluate the following(a) $\sin \left[ {\dfrac{\pi }{3} - {{\sin }^{ - 1}}\left( { - \dfrac{1}{2}} \right)} \right]$(b) $\sin \left[ {\dfrac{\pi }{2} - {{\sin }^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right)} \right]$

Last updated date: 25th Mar 2023
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Hint: Make use of the formula of inverse trigonometric functions and solve this.

Complete step by step solution:
(a) $\sin \left[ {\dfrac{\pi }{3} - {{\sin }^{ - 1}}\left( { - \dfrac{1}{2}} \right)} \right]$

To solve this let's make use of the formula of ${\sin ^{ - 1}}( - x) = - {\sin ^{ - 1}}x$
In this question ,we have $- {\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$ , so on comparing with
the formula we can write this as $- {\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)$=(-)(-)${\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)$ =${\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)$
So, now the equation will become $\sin \left( {\dfrac{\pi }{3} + {{\sin }^{ - 1}}\left( {\dfrac{1}{2}} \right)} \right)$
We know the value of ${\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{6}$
So, now the equation will become $\sin \left( {\dfrac{\pi }{3} + \dfrac{\pi }{6}} \right) = \sin \dfrac{\pi }{2} = 1$
So, therefore the value of $\sin \left[ {\dfrac{\pi }{3} - {{\sin }^{ - 1}}\left( { - \dfrac{1}{2}} \right)} \right]$=1

(b) $\sin \left[ {\dfrac{\pi }{2} - {{\sin }^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right)} \right]$
To solve this let's make use of the formula of ${\sin ^{ - 1}}( - x) = - {\sin ^{ - 1}}x$
In the question we have ${\sin ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right)$, so on comparing this with the formula, we can write this as ${\sin ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right) = ( - )( - ){\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)$
So, now we get the equation as $\sin \left[ {\dfrac{\pi }{2} + {{\sin }^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)} \right]$
We know that the value of ${\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) = \dfrac{\pi }{3}$
So, now we can write the equation as $\sin \left( {\dfrac{\pi }{2} + \dfrac{\pi }{3}} \right) = \cos \dfrac{\pi }{3} = \dfrac{1}{2}$
(Since the value of $\sin \left( {\dfrac{\pi }{2} + \theta } \right) = \cos \theta$ )
So, therefore the value of $\sin \left[ {\dfrac{\pi }{2} - {{\sin }^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right)} \right]$=$\dfrac{1}{2}$

Note: When we are solving these kind of problems make use of the appropriate formula of inverse trigonometric functions to solve, also take care of the sign associated with the functions.