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Question: Evaluate the following (a) \[\sin \left[ {\dfrac{\pi }{3} - {{\sin }^{ - 1}}\left( { - \dfrac{1}{2}...

Evaluate the following

(a) sin⁡[π3−sin⁡−1(−12)]\sin \left[ {\dfrac{\pi }{3} - {{\sin }^{ - 1}}\left( { - \dfrac{1}{2}} \right)} \right]

(b) sin⁡[π2−sin⁡−1(−32)]\sin \left[ {\dfrac{\pi }{2} - {{\sin }^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right)} \right]

Explanation

Solution

Make use of the formula of inverse trigonometric functions and solve this.

Complete step by step solution:

(a) sin⁡[π3−sin⁡−1(−12)]\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⁡−1x{\sin ^{ - 1}}( - x) = - {\sin ^{ - 1}}x

In this question ,we have −sin⁡−1(−12) - {\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right) , so on comparing with

the formula we can write this as −sin⁡−1(−12) - {\sin ^{ - 1}}\left( { - \dfrac{1}{2}} \right)=(-)(-)sin⁡−1(12){\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) =sin⁡−1(12){\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)

So, now the equation will become sin⁡(π3+sin⁡−1(12))\sin \left( {\dfrac{\pi }{3} + {{\sin }^{ - 1}}\left( {\dfrac{1}{2}} \right)} \right)

We know the value of sin⁡−1(12)=π6{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{6}

So, now the equation will become sin⁡(π3+π6)=sin⁡π2=1\sin \left( {\dfrac{\pi }{3} + \dfrac{\pi }{6}} \right) = \sin \dfrac{\pi }{2} = 1

So, therefore the value of sin⁡[π3−sin⁡−1(−12)]\sin \left[ {\dfrac{\pi }{3} - {{\sin }^{ - 1}}\left( { - \dfrac{1}{2}} \right)} \right]=1

(b) sin⁡[π2−sin⁡−1(−32)]\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⁡−1x{\sin ^{ - 1}}( - x) = - {\sin ^{ - 1}}x

In the question we have sin⁡−1(−32){\sin ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right), so on comparing this with the formula, we can write this as sin⁡−1(−32)=(−)(−)sin⁡−1(32){\sin ^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right) = ( - )( - ){\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)

So, now we get the equation as sin⁡[π2+sin⁡−1(32)]\sin \left[ {\dfrac{\pi }{2} + {{\sin }^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)} \right]

We know that the value of sin⁡−1(32)=π3{\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right) = \dfrac{\pi }{3}

So, now we can write the equation as sin⁡(π2+π3)=cos⁡π3=12\sin \left( {\dfrac{\pi }{2} + \dfrac{\pi }{3}} \right) = \cos \dfrac{\pi }{3} = \dfrac{1}{2}

(Since the value of sin⁡(π2+θ)=cos⁡θ\sin \left( {\dfrac{\pi }{2} + \theta } \right) = \cos \theta )

So, therefore the value of sin⁡[π2−sin⁡−1(−32)]\sin \left[ {\dfrac{\pi }{2} - {{\sin }^{ - 1}}\left( { - \dfrac{{\sqrt 3 }}{2}} \right)} \right]=12\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.