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Last updated date: 06th Dec 2023
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# Find value of $x$ if ${\sin ^{ - 1}}\dfrac{1}{3} + {\sin ^{ - 1}}\dfrac{2}{3} = {\sin ^{ - 1}}x$.

Answer
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280.5k+ views
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 $\arcsin x$ .