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

If sin⁡−1(1−x)−2sin⁡−1x=π2,{\sin ^{ - 1}}(1 - x) - 2{\sin ^{ - 1}}x = \dfrac{\pi }{2}, then what is the value of x?

Explanation

Solution

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 2sin⁡−1x2{\sin ^{ - 1}}x to the R.H.S and then take sin of both the sides. After that use the identity sin⁡(π2+θ)=cos⁡θ\sin \left( {\dfrac{\pi }{2} + \theta } \right) = \cos \theta on the R.H.S and sin⁡(sin⁡−1θ)=θ\sin \left( {{{\sin }^{ - 1}}\theta } \right) = \theta on the L.H.S . Now, use the property cos⁡(2θ)=1−2sin⁡2θ\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)−2sin⁡−1x=π2{\sin ^{ - 1}}(1 - x) - 2{\sin ^{ - 1}}x = \dfrac{\pi }{2}
⇒sin⁡−1(1−x)=π2+2sin⁡−1x\Rightarrow {\sin ^{ - 1}}(1 - x) = \dfrac{\pi }{2} + 2{\sin ^{ - 1}}x
Take sine of both sides
⇒sin⁡(sin⁡−1(1−x))=sin⁡(π2+2sin⁡−1x)\Rightarrow \sin \left( {{{\sin }^{ - 1}}(1 - x)} \right) = \sin \left( {\dfrac{\pi }{2} + 2{{\sin }^{ - 1}}x} \right)
⇒(1−x)=sin⁡(π2+2sin⁡−1x)\Rightarrow (1 - x) = \sin \left( {\dfrac{\pi }{2} + 2{{\sin }^{ - 1}}x} \right)
We know that, sin⁡(π2+θ)=cos⁡θ\sin \left( {\dfrac{\pi }{2} + \theta } \right) = \cos \theta
⇒(1−x)=cos⁡(2sin⁡−1x)\Rightarrow (1 - x) = \cos \left( {2{{\sin }^{ - 1}}x} \right)
Now, we know that cos⁡(2θ)=1−2sin⁡2θ\cos (2\theta ) = 1 - 2{\sin ^2}\theta
⇒(1−x)=1−2sin⁡2(sin⁡−1x)\Rightarrow (1 - x) = 1 - 2{\sin ^2}\left( {{{\sin }^{ - 1}}x} \right)
⇒(1−x)=1−2sin⁡(sin⁡−1x)×sin⁡(sin⁡−1x)\Rightarrow (1 - x) = 1 - 2\sin \left( {{{\sin }^{ - 1}}x} \right) \times \sin \left( {{{\sin }^{ - 1}}x} \right)
We know that sin⁡(sin⁡−1θ)=θ\sin \left( {{{\sin }^{ - 1}}\theta } \right) = \theta for θ∈[−1,1]\theta \in [ - 1,1]
⇒(1−x)=1−2x2\Rightarrow (1 - x) = 1 - 2{x^2}
⇒1−x−1+2x2=0\Rightarrow 1 - x - 1 + 2{x^2} = 0
⇒2x2−x=0\Rightarrow 2{x^2} - x = 0
Taking x common
⇒x(2x−1)=0\Rightarrow x(2x - 1) = 0
Now, either x=0x = 0 or (2x−1)=0(2x - 1) = 0
When (2x−1)=0(2x - 1) = 0
⇒2x=1\Rightarrow 2x = 1
Divide both sides by 2
⇒x=12\Rightarrow x = \dfrac{1}{2}
Now, when x=12;x = \dfrac{1}{2}; check that by substituting that in the equation sin⁡−1(1−x)−2sin⁡−1x=π2{\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(1−12)−2sin⁡−1(12)= {\sin ^{ - 1}}\left( {1 - \dfrac{1}{2}} \right) - 2{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)
=sin⁡−1(12)−2sin⁡−1(12)= {\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) - 2{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right)
Now, we know that sin⁡−1(12)=π6{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{6}
=π6−2π6= \dfrac{\pi }{6} - \dfrac{{2\pi }}{6}
=−π6= - \dfrac{\pi }{6} which is not equal to R.H.S
Now, when x = 0, substitute that in the equation sin⁡−1(1−x)−2sin⁡−1x=π2{\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(1−0)−2sin⁡−1(0)= {\sin ^{ - 1}}\left( {1 - 0} \right) - 2{\sin ^{ - 1}}\left( 0 \right)
=sin⁡−1(1)−2sin⁡−1(0)= {\sin ^{ - 1}}\left( 1 \right) - 2{\sin ^{ - 1}}\left( 0 \right)
We know that sin⁡−1(1)=π2,sin⁡−1(0)=0{\sin ^{ - 1}}\left( 1 \right) = \dfrac{\pi }{2},{\sin ^{ - 1}}(0) = 0
=π2−2×0= \dfrac{\pi }{2} - 2 \times 0
=π2= \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⁡(π2+θ)=cos⁡θ\sin \left( {\dfrac{\pi }{2} + \theta } \right) = \cos \theta , cos⁡(2θ)=1−2sin⁡2θ\cos (2\theta ) = 1 - 2{\sin ^2}\theta , sin⁡(sin⁡−1θ)=θ\sin \left( {{{\sin }^{ - 1}}\theta } \right) = \theta , sin⁡−1(12)=π6{\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right) = \dfrac{\pi }{6} , sin⁡−1(1)=π2{\sin ^{ - 1}}\left( 1 \right) = \dfrac{\pi }{2} , sin⁡−1(0)=0{\sin ^{ - 1}}(0) = 0 . Also, simplify the expression step by step using them to get the answer.