Question

If \[{\sin ^{ - 1}}(1 - x) - 2{\sin ^{ - 1}}x = \dfrac{\pi }{2},\] then what is the value of x?

Answer 1

Hint: In this question, we have to find the value of x from the given equation. For that we will use the concept of Inverse trigonometric functions. Simply transform the equation such that move the \[2{\sin ^{ - 1}}x\] to the R.H.S and then take sin of both the sides. After that use the identity $\sin \left( {\dfrac{\pi }{2} + \theta } \right) = \cos \theta $ on the R.H.S and $\sin \left( {{{\sin }^{ - 1}}\theta } \right) = \theta $ on the L.H.S . Now, use the property $\cos (2\theta ) = 1 - 2{\sin ^2}\theta $ on the R.H.S to simplify the equation and get the required value by simply solving the quadratic equation we get after that.

Complete step-by-step answer:
It is given that, \[{\sin ^{ - 1}}(1 - x) - 2{\sin ^{ - 1}}x = \dfrac{\pi }{2}\]
\[ \Rightarrow {\sin ^{ - 1}}(1 - x) = \dfrac{\pi }{2} + 2{\sin ^{ - 1}}x\]
Take sine of both sides
\[ \Rightarrow \sin \left( {{{\sin }^{ - 1}}(1 - x)} \right) = \sin \left( {\dfrac{\pi }{2} + 2{{\sin }^{ - 1}}x} \right)\]
\[ \Rightarrow (1 - x) = \sin \left( {\dfrac{\pi }{2} + 2{{\sin }^{ - 1}}x} \right)\]
We know that, $\sin \left( {\dfrac{\pi }{2} + \theta } \right) = \cos \theta $
\[ \Rightarrow (1 - x) = \cos \left( {2{{\sin }^{ - 1}}x} \right)\]
Now, we know that $\cos (2\theta ) = 1 - 2{\sin ^2}\theta $
\[ \Rightarrow (1 - x) = 1 - 2{\sin ^2}\left( {{{\sin }^{ - 1}}x} \right)\]
\[ \Rightarrow (1 - x) = 1 - 2\sin \left( {{{\sin }^{ - 1}}x} \right) \times \sin \left( {{{\sin }^{ - 1}}x} \right)\]
We know that $\sin \left( {{{\sin }^{ - 1}}\theta } \right) = \theta $ for $\theta \in [ - 1,1]$
\[ \Rightarrow (1 - x) = 1 - 2{x^2}\]
\[ \Rightarrow 1 - x - 1 + 2{x^2} = 0\]
\[ \Rightarrow 2{x^2} - x = 0\]
Taking x common
\[ \Rightarrow x(2x - 1) = 0\]
Now, either $x = 0$ or $(2x - 1) = 0$
When $(2x - 1) = 0$
$ \Rightarrow 2x = 1$
Divide both sides by 2
$ \Rightarrow x = \dfrac{1}{2}$
Now, when $x = \dfrac{1}{2};$ check that by substituting that in the equation \[{\sin ^{ - 1}}(1 - x) - 2{\sin ^{ - 1}}x = \dfrac{\pi }{2}\]
Substitute in the L.H.S and check whether it gets equal to R.H.S
\[ = {\sin ^{ - 1}}\left( {1 - \dfrac{1}{2}} \right) - 2{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)\]
\[ = {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) - 2{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)\]
Now, we know that \[{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{6}\]
\[ = \dfrac{\pi }{6} - \dfrac{{2\pi }}{6}\]
\[ = - \dfrac{\pi }{6}\] which is not equal to R.H.S
Now, when x = 0, substitute that in the equation \[{\sin ^{ - 1}}(1 - x) - 2{\sin ^{ - 1}}x = \dfrac{\pi }{2}\]
Substitute in the L.H.S and check whether it gets equal to R.H.S
\[ = {\sin ^{ - 1}}\left( {1 - 0} \right) - 2{\sin ^{ - 1}}\left( 0 \right)\]
\[ = {\sin ^{ - 1}}\left( 1 \right) - 2{\sin ^{ - 1}}\left( 0 \right)\]
We know that \[{\sin ^{ - 1}}\left( 1 \right) = \dfrac{\pi }{2},{\sin ^{ - 1}}(0) = 0\]
\[ = \dfrac{\pi }{2} - 2 \times 0\]
\[ = \dfrac{\pi }{2}\] which is equal to R.H.S
So, the value of x is 0 which satisfies the required equation.
$\therefore $ x = 0 is the required value of x.


Note- In such types of questions, just remember some simple trigonometric and inverse trigonometric functions, their values and their properties like $\sin \left( {\dfrac{\pi }{2} + \theta } \right) = \cos \theta $ , $\cos (2\theta ) = 1 - 2{\sin ^2}\theta $ , $\sin \left( {{{\sin }^{ - 1}}\theta } \right) = \theta $ , \[{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{6}\] , \[{\sin ^{ - 1}}\left( 1 \right) = \dfrac{\pi }{2}\] , \[{\sin ^{ - 1}}(0) = 0\] . Also, simplify the expression step by step using them to get the answer.