Question

Evaluate the following

(a) \[\sin \left[ {\dfrac{\pi }{3} - {{\sin }^{ - 1}}\left( { - \dfrac{1}{2}} \right)} \right]\]

(b) \[\sin \left[ {\dfrac{\pi }{2} - {{\sin }^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right)} \right]\]

Answer 1

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.