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Question: The solution of \[{\sin ^{ - 1}}x - {\sin ^{ - 1}}2x = \pm \dfrac{\pi }{3}\] is A.\[ \pm \dfrac{1...

The solution of sin⁡−1x−sin⁡−12x=±π3{\sin ^{ - 1}}x - {\sin ^{ - 1}}2x = \pm \dfrac{\pi }{3} is
A.±13 \pm \dfrac{1}{3}
B.±14 \pm \dfrac{1}{4}
C.±32 \pm \dfrac{{\sqrt 3 }}{2}
D.±12 \pm \dfrac{1}{2}

Explanation

Solution

Hint : In mathematics , the inverse trigonometric functions are the inverse functions of the trigonometric functions. Specifically, inverse trigonometric functions are the inverses of the sine, the cosine, the tangent, the cotangent, the secant, and the cosecant functions and are used to obtain an angle from any of the angle's trigonometric ratios.
Formulas used in the solution part are as follows :
sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin (\alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta
cos⁡(sin⁡−1θ)=1−θ2\cos \left( {{{\sin }^{ - 1}}\theta } \right) = \sqrt {1 - {\theta ^2}}
sin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡β\sin (\alpha - \beta ) = \sin \alpha \cos \beta - \cos \alpha \sin \beta

Complete step-by-step answer :
We are given the equation sin⁡−1x−sin⁡−12x=±π3{\sin ^{ - 1}}x - {\sin ^{ - 1}}2x = \pm \dfrac{\pi }{3}
We have sin⁡−1x−sin⁡−12x=π3{\sin ^{ - 1}}x - {\sin ^{ - 1}}2x = \dfrac{\pi }{3} or sin⁡−1x−sin⁡−12x=−π3{\sin ^{ - 1}}x - {\sin ^{ - 1}}2x = - \dfrac{\pi }{3}
By taking sin⁡−1x−sin⁡−12x=π3{\sin ^{ - 1}}x - {\sin ^{ - 1}}2x = \dfrac{\pi }{3}
We get sin⁡−1x=π3+sin⁡−12x{\sin ^{ - 1}}x = \dfrac{\pi }{3} + {\sin ^{ - 1}}2x
Now computing sin⁡\sin on both the sides we get
sin⁡(sin⁡−1x)=sin⁡(π3+sin⁡−12x)\sin \left( {{{\sin }^{ - 1}}x} \right) = \sin \left( {\dfrac{\pi }{3} + {{\sin }^{ - 1}}2x} \right)
We know that the trigonometric function and its inverse trigonometric part gets cancelled or compensated . therefore we get
x=sin⁡(π3+sin⁡−12x)x = \sin \left( {\dfrac{\pi }{3} + {{\sin }^{ - 1}}2x} \right)
Now using the identity sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin (\alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta
We get ,
x=sin⁡π3cos⁡(sin⁡−12x)+cos⁡π3sin⁡(sin⁡−12x)x = \sin \dfrac{\pi }{3}\cos \left( {{{\sin }^{ - 1}}2x} \right) + \cos \dfrac{\pi }{3}\sin \left( {{{\sin }^{ - 1}}2x} \right)
Which simplifies to
x=32cos⁡(sin⁡−12x)+12(2x)x = \dfrac{{\sqrt 3 }}{2}\cos \left( {{{\sin }^{ - 1}}2x} \right) + \dfrac{1}{2}(2x)
we know that cos⁡(sin⁡−1θ)=1−θ2\cos \left( {{{\sin }^{ - 1}}\theta } \right) = \sqrt {1 - {\theta ^2}}
hence we get ,
x=321−(2x)2+12(2x)x = \dfrac{{\sqrt 3 }}{2}\sqrt {1 - {{(2x)}^2}} + \dfrac{1}{2}(2x)
On further simplification we get ,
0=1−4x20 = 1 - 4{x^2}
Which gives us x=±12x = \pm \dfrac{1}{2}
Now taking sin⁡−1x−sin⁡−12x=−π3{\sin ^{ - 1}}x - {\sin ^{ - 1}}2x = - \dfrac{\pi }{3}
We get sin⁡−1x=−π3+sin⁡−12x{\sin ^{ - 1}}x = - \dfrac{\pi }{3} + {\sin ^{ - 1}}2x
Now computing sin⁡\sin on both the sides we get ,
sin⁡(sin⁡−1x)=sin⁡(sin⁡−12x−π3)\sin ({\sin ^{ - 1}}x) = \sin \left( {{{\sin }^{ - 1}}2x - \dfrac{\pi }{3}} \right)
We know that the trigonometric function and its inverse trigonometric part gets cancelled or compensated . therefore we get ,
x=sin⁡(sin⁡−12x−π3)x = \sin \left( {{{\sin }^{ - 1}}2x - \dfrac{\pi }{3}} \right)
Now using the identity sin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡β\sin (\alpha - \beta ) = \sin \alpha \cos \beta - \cos \alpha \sin \beta
We get ,
x=sin⁡(sin⁡−12x)cos⁡π3−cos⁡(sin⁡−12x)sin⁡π3x = \sin \left( {{{\sin }^{ - 1}}2x} \right)\cos \dfrac{\pi }{3} - \cos \left( {{{\sin }^{ - 1}}2x} \right)\sin \dfrac{\pi }{3}
Which simplifies to ,
x=(2x)12−cos⁡(sin⁡−12x)32x = (2x)\dfrac{1}{2} - \cos \left( {{{\sin }^{ - 1}}2x} \right)\dfrac{{\sqrt 3 }}{2}
we know that cos⁡(sin⁡−1θ)=1−θ2\cos \left( {{{\sin }^{ - 1}}\theta } \right) = \sqrt {1 - {\theta ^2}}
hence we get ,
x=(2x)12−321−(2x)2x = (2x)\dfrac{1}{2} - \dfrac{{\sqrt 3 }}{2}\sqrt {1 - {{(2x)}^2}}
On simplification we get ,
0=1−4x20 = 1 - 4{x^2}
Which gives us x=±12x = \pm \dfrac{1}{2}
Therefore option(4) is the correct answer.
So, the correct answer is “Option 4”.

Note: Inverse trigonometric functions are the inverse functions of the trigonometric functions. Specifically, inverse trigonometric functions are the inverses of the sine, the cosine, the tangent, the cotangent, the secant, and the cosecant functions and are used to obtain an angle from any of the angle's trigonometric ratios. keep in mind all the trigonometric identities.