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Question: Find value of \[x\] if \[{\sin ^{ - 1}}\dfrac{1}{3} + {\sin ^{ - 1}}\dfrac{2}{3} = {\sin ^{ - 1}}x\]...

Find value of xx if sin⁡−113+sin⁡−123=sin⁡−1x{\sin ^{ - 1}}\dfrac{1}{3} + {\sin ^{ - 1}}\dfrac{2}{3} = {\sin ^{ - 1}}x.

Explanation

Solution

First we have to solve the equation given in the problem in order to find the value of xx .
We have to solve the equation using the appropriate formula and then equate both sides of the equation to calculate the value of xx.

Formula used:
sin⁡−1x+sin⁡−1y=sin⁡−1(x1−y2+y1−x2){\sin ^{ - 1}}x + {\sin ^{ - 1}}y = {\sin ^{ - 1}}(x\sqrt {1 - {y^2}} + y\sqrt {1 - {x^2}} )

Complete step by step solution:
Let us note down the given equation,
sin⁡−113+sin⁡−123=sin⁡−1x{\sin ^{ - 1}}\dfrac{1}{3} + {\sin ^{ - 1}}\dfrac{2}{3} = {\sin ^{ - 1}}x
Let us solve the L.H.S. of the equation,
L.H.S. =sin⁡−113+sin⁡−123 = {\sin ^{ - 1}}\dfrac{1}{3} + {\sin ^{ - 1}}\dfrac{2}{3}
Use the formula,
sin⁡−1x+sin⁡−1y=sin⁡−1(x1−y2+y1−x2){\sin ^{ - 1}}x + {\sin ^{ - 1}}y = {\sin ^{ - 1}}(x\sqrt {1 - {y^2}} + y\sqrt {1 - {x^2}} )
Consider L.H.S. of the equation,
∴x=13,y=23\therefore x = \dfrac{1}{3},y = \dfrac{2}{3}
On using the formula for these values of xx and yy we get,
sin⁡−113+sin⁡−123=sin⁡−1(131−(23)2+231−(13)2){\sin ^{ - 1}}\dfrac{1}{3} + {\sin ^{ - 1}}\dfrac{2}{3} = {\sin ^{ - 1}}\left( {\dfrac{1}{3}\sqrt {1 - {{\left( {\dfrac{2}{3}} \right)}^2}} + \dfrac{2}{3}\sqrt {1 - {{\left( {\dfrac{1}{3}} \right)}^2}} } \right)
On taking squares of the fractions we get,
sin⁡−113+sin⁡−123=sin⁡−1(131−49+231−19){\sin ^{ - 1}}\dfrac{1}{3} + {\sin ^{ - 1}}\dfrac{2}{3} = {\sin ^{ - 1}}\left( {\dfrac{1}{3}\sqrt {1 - \dfrac{4}{9}} + \dfrac{2}{3}\sqrt {1 - \dfrac{1}{9}} } \right)
On making denominators of the fractions equal by cross-multiplying we get,
sin⁡−113+sin⁡−123=sin⁡−1(139−49+239−19){\sin ^{ - 1}}\dfrac{1}{3} + {\sin ^{ - 1}}\dfrac{2}{3} = {\sin ^{ - 1}}\left( {\dfrac{1}{3}\sqrt {\dfrac{{9 - 4}}{9}} + \dfrac{2}{3}\sqrt {\dfrac{{9 - 1}}{9}} } \right)
On performing the subtraction of the numerators of the fractions we get,
sin⁡−113+sin⁡−123=sin⁡−1(1359+2389){\sin ^{ - 1}}\dfrac{1}{3} + {\sin ^{ - 1}}\dfrac{2}{3} = {\sin ^{ - 1}}\left( {\dfrac{1}{3}\sqrt {\dfrac{5}{9}} + \dfrac{2}{3}\sqrt {\dfrac{8}{9}} } \right)
On taking the squares of whole square terms we get,
sin⁡−113+sin⁡−123=sin⁡−1(13×53+23×83){\sin ^{ - 1}}\dfrac{1}{3} + {\sin ^{ - 1}}\dfrac{2}{3} = {\sin ^{ - 1}}\left( {\dfrac{1}{3} \times \dfrac{{\sqrt 5 }}{3} + \dfrac{2}{3} \times \dfrac{{\sqrt 8 }}{3}} \right)
On performing multiplication of the fractions we get,
sin⁡−113+sin⁡−123=sin⁡−1(59+429){\sin ^{ - 1}}\dfrac{1}{3} + {\sin ^{ - 1}}\dfrac{2}{3} = {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 5 }}{9} + \dfrac{{4\sqrt 2 }}{9}} \right)
On performing addition of factors as denominator is common, we get,
sin⁡−113+sin⁡−123=sin⁡−1(5+429){\sin ^{ - 1}}\dfrac{1}{3} + {\sin ^{ - 1}}\dfrac{2}{3} = {\sin ^{ - 1}}\left( {\dfrac{{5 + 4\sqrt 2 }}{9}} \right)
This is the solution of the L.H.S.
But the equation given in the question is,
sin⁡−113+sin⁡−123=sin⁡−1x{\sin ^{ - 1}}\dfrac{1}{3} + {\sin ^{ - 1}}\dfrac{2}{3} = {\sin ^{ - 1}}x
On comparing calculated value of the equation and equation given in the problem we get,
sin⁡−1x=sin⁡−1(5+429){\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {\dfrac{{5 + 4\sqrt 2 }}{9}} \right)
On equating values on the both sides of the equation we get,
x=5+429x = \dfrac{{5 + 4\sqrt 2 }}{9}
This is the required solution.

Note: Inverse trigonometric functions are the inverse of the main trigonometric functions. Inverse trigonometric functions perform opposite functions of the main trigonometric functions. Hence sin⁡−1x{\sin ^{ - 1}}x performs the opposite function of sin⁡x\sin x . Inverse trigonometric functions are used to obtain angle from its trigonometric ratio. Inverse trigonometric functions can also be written using the arc keyword which means sin⁡−1x{\sin ^{ - 1}}x can also be written as arcsin⁡x\arcsin x .