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

Answer

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**Hint:**First we have to solve the equation given in the problem in order to find the value of \[x\] .

We have to solve the equation using the appropriate formula and then equate both sides of the equation to calculate the value of \[x\].

**Formula used:**

\[{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = {\sin ^{ - 1}}(x\sqrt {1 - {y^2}} + y\sqrt {1 - {x^2}} )\]

**Complete step by step solution:**

Let us note down the given equation,

\[{\sin ^{ - 1}}\dfrac{1}{3} + {\sin ^{ - 1}}\dfrac{2}{3} = {\sin ^{ - 1}}x\]

Let us solve the L.H.S. of the equation,

L.H.S. \[ = {\sin ^{ - 1}}\dfrac{1}{3} + {\sin ^{ - 1}}\dfrac{2}{3}\]

Use the formula,

\[{\sin ^{ - 1}}x + {\sin ^{ - 1}}y = {\sin ^{ - 1}}(x\sqrt {1 - {y^2}} + y\sqrt {1 - {x^2}} )\]

Consider L.H.S. of the equation,

\[\therefore x = \dfrac{1}{3},y = \dfrac{2}{3}\]

On using the formula for these values of $ x $ and $ y $ we get,

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

On taking squares of the fractions we get,

\[{\sin ^{ - 1}}\dfrac{1}{3} + {\sin ^{ - 1}}\dfrac{2}{3} = {\sin ^{ - 1}}\left( {\dfrac{1}{3}\sqrt {1 - \dfrac{4}{9}} + \dfrac{2}{3}\sqrt {1 - \dfrac{1}{9}} } \right)\]

On making denominators of the fractions equal by cross-multiplying we get,

\[{\sin ^{ - 1}}\dfrac{1}{3} + {\sin ^{ - 1}}\dfrac{2}{3} = {\sin ^{ - 1}}\left( {\dfrac{1}{3}\sqrt {\dfrac{{9 - 4}}{9}} + \dfrac{2}{3}\sqrt {\dfrac{{9 - 1}}{9}} } \right)\]

On performing the subtraction of the numerators of the fractions we get,

\[{\sin ^{ - 1}}\dfrac{1}{3} + {\sin ^{ - 1}}\dfrac{2}{3} = {\sin ^{ - 1}}\left( {\dfrac{1}{3}\sqrt {\dfrac{5}{9}} + \dfrac{2}{3}\sqrt {\dfrac{8}{9}} } \right)\]

On taking the squares of whole square terms we get,

\[{\sin ^{ - 1}}\dfrac{1}{3} + {\sin ^{ - 1}}\dfrac{2}{3} = {\sin ^{ - 1}}\left( {\dfrac{1}{3} \times \dfrac{{\sqrt 5 }}{3} + \dfrac{2}{3} \times \dfrac{{\sqrt 8 }}{3}} \right)\]

On performing multiplication of the fractions we get,

\[{\sin ^{ - 1}}\dfrac{1}{3} + {\sin ^{ - 1}}\dfrac{2}{3} = {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 5 }}{9} + \dfrac{{4\sqrt 2 }}{9}} \right)\]

On performing addition of factors as denominator is common, we get,

\[{\sin ^{ - 1}}\dfrac{1}{3} + {\sin ^{ - 1}}\dfrac{2}{3} = {\sin ^{ - 1}}\left( {\dfrac{{5 + 4\sqrt 2 }}{9}} \right)\]

This is the solution of the L.H.S.

But the equation given in the question is,

\[{\sin ^{ - 1}}\dfrac{1}{3} + {\sin ^{ - 1}}\dfrac{2}{3} = {\sin ^{ - 1}}x\]

On comparing calculated value of the equation and equation given in the problem we get,

\[{\sin ^{ - 1}}x = {\sin ^{ - 1}}\left( {\dfrac{{5 + 4\sqrt 2 }}{9}} \right)\]

On equating values on the both sides of the equation we get,

\[x = \dfrac{{5 + 4\sqrt 2 }}{9}\]

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 ^{ - 1}}x $ performs the opposite function of $ \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 ^{ - 1}}x $ can also be written as $ \arcsin x $ .

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