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

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(a) \[\sin \left[ {\frac{\pi }{3} - {{\sin }^{ - 1}}\left( { - \frac{1}{2}} \right)} \right]\]
Hint-Make use of the formula of inverse trigonometric functions and solve this
The first part is
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( { - \frac{1}{2}} \right)$ , so on comparing with
 the formula we can write this as $ - {\sin ^{ - 1}}\left( { - \frac{1}{2}} \right)$=(-)(-)${\sin ^{ -
 1}}\left( {\frac{1}{2}} \right)$ =${\sin ^{ - 1}}\left( {\frac{1}{2}} \right)$
So, now the equation will become $\sin \left( {\frac{\pi }{3} + {{\sin }^{ - 1}}\left( {\frac{1}{2}}
 \right)} \right)$
We know the value of ${\sin ^{ - 1}}\left( {\frac{1}{2}} \right) = \frac{\pi }{6}$

So, now the equation will become $\sin \left( {\frac{\pi }{3} + \frac{\pi }{6}} \right) = \sin \frac{\pi }{2} = 1$
So, therefore the value of \[\sin \left[ {\frac{\pi }{3} - {{\sin }^{ - 1}}\left( { - \frac{1}{2}} \right)} \right]\]=1
Note :When solving these kind of problems make use of the appropriate formula of inverse trigonometric functions and also take care of the sign associated with it

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