# 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

•

Total views: 309.6k

•

Views today: 3.89k

Answer

Verified

309.6k+ views

**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.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE