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Question: Prove that \[{\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right) = 2{\rm{ si}}{{\rm{n}}^{ - 1}}x,...

Prove that
sin⁡−1(2x1−x2)=2sin−1x,∣x∣≤112{\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right) = 2{\rm{ si}}{{\rm{n}}^{ - 1}}x,\left| x \right| \le \dfrac{1}{{12}}

Explanation

Solution

First write the whole equation which we have to prove. Now take the left hand side of the equation separately. Substitute Sina in place of x and use trigonometric identity you know to simplify the left hand side. Now find the value of a in terms of x. substitute it back. By using the formula of sin2x you can remove the sine term. After removing the sine term you get a direct expression in terms of a. Now By substituting the value of ‘a’ we can prove that left hand side and right hand are equal.

Complete step-by-step answer:
Given condition which we need to prove in the question:
sin−1(2x1−x2)=2sin−1xsi{n^{ - 1}}{\rm{ }}\left( {2x\sqrt {1 - {x^2}} } \right){\rm{ = 2 si}}{{\rm{n}}^{ - 1}}{\rm{ x}} ………………….(1)
By writing the right hand side separately, we get it as:
RHS=2sin−1xRHS{\rm{ = 2 si}}{{\rm{n}}^{ - 1}}{\rm{ x}} ………………….(2)
By writing the left hand side separately, we get it as:
LHS=sin−1(2x1−x2)LHS{\rm{ = si}}{{\rm{n}}^{ - 1}}{\rm{ }}\left( {2x\sqrt {1 - {x^2}} } \right) ………………….(3)
By taking an assumption as x = Sin a, we get it as:
x=sinax{\rm{ = sin a}} ………………….(4)
By applying Sin-1 on both sides of equation, we get it as:
sin−1(x)=sin−1(sina)si{n^{ - 1}}{\rm{ }}\left( x \right){\rm{ = si}}{{\rm{n}}^{ - 1}}{\rm{ }}\left( {sin{\rm{ a}}} \right)
By simplifying the above equation, we can write it as:
sin−1(x)=asi{n^{ - 1}}{\rm{ }}\left( x \right){\rm{ = a}}
By inverting the left, right sides of above equation, we get:
a=sin−1xa{\rm{ = si}}{{\rm{n}}^{ - 1}}x ………………….(5)
Now by substituting equation (4) in equation (3), we get it as:
LHS=sin−1(2(sina)1−(sina)2)LHS{\rm{ = si}}{{\rm{n}}^{ - 1}}\left( {2\left( {sin{\rm{ a}}} \right)\sqrt {1 - {{\left( {sin{\rm{ a}}} \right)}^2}} } \right)
By simplifying the term inside the square, we get it as:
LHS=sin−1(2sina1−sin2 a)LHS{\rm{ = si}}{{\rm{n}}^{ - 1}}\left( {2{\rm{ sin a}}\sqrt {1 - si{n^2}\,a} } \right)
By basic knowledge of trigonometry, we know the identity:
sin2θ+cos⁡2θ=1si{n^2}\theta {\rm{ + }}{\cos ^2}\theta {\rm{ = 1}}
By subtracting sin2θsi{n^2}\theta on both sides, we get it as:
sin2θ+cos⁡2θ−Sin2θ=1−Sin2θsi{n^2}\theta + {\cos ^2}\theta - Si{n^2}\theta = 1 - Si{n^2}\theta
By simplifying the above equation, we get relation as:
cos⁡2θ=1−Sin2θ{\cos ^2}\theta = 1 - Si{n^2}\theta
By substituting this inside the square root of LHS, we get
LHS=sin⁡−1(2sin⁡acos⁡2a)LHS = {\sin ^{ - 1}}\left( {2\sin a\sqrt {{{\cos }^2}a} } \right)
By removing square root, we can write the term as:
LHS=sin⁡−1(2sin⁡acos a)LHS = {\sin ^{ - 1}}\left( {2\sin a{\rm{ cos}}\,{\rm{a}}} \right)
By basic knowledge of trigonometry, we get the equation:
2sin⁡acosa=sin2a2\sin a{\rm{ cosa = sin2a}}
By substituting this question into the LHS, we get it as:
LHS=sin⁡−1(sin⁡2a)LHS = {\sin ^{ - 1}}\left( {\sin 2a} \right)
By applying basic property of inverse given by as follows:
sin⁡−1(sin⁡k)=k{\sin ^{ - 1}}\left( {\sin k} \right) = k
−π2<k<π2\dfrac{{ - \pi }}{2} < k < \dfrac{\pi }{2}
By substituting this, we get it as follows:
LHS=2aLHS = 2a
By substituting value from equation (5), we get:
LHS=2sin⁡−1xLHS = 2{\sin ^{ - 1}}x
By substituting the equation (2) here, we get it as:
LHS=RHSLHS = RHS
So, we proved LHS=RHSLHS = RHS . Hence, proved the given equation.

Note: Be careful with range of x. We apply the formula only if ∣x∣≤12\left| x \right| \le \dfrac{1}{{\sqrt 2 }} or else we cannot write the inverse formula which we used here. The idea of substituting sina must be used whenever you terms like 1−x2\sqrt {1 - {x^2}} which generally is cosa term which in turn makes the solution simple. As −π2<2a<π2\dfrac{{ - \pi }}{2} < 2a < \dfrac{\pi }{2} , a will lie between −π4<a<π4\dfrac{{ - \pi }}{4} < a < \dfrac{\pi }{4} . Thus sina which x will lie between −12≤x≤12∣x∣≤12.\dfrac{{ - 1}}{{\sqrt 2 }} \le x \le \dfrac{1}{{\sqrt 2 }}{\rm{ }}\left| x \right| \le \dfrac{1}{{\sqrt 2 }}. This is proof for ∣x∣≤12.\left| x \right| \le \dfrac{1}{{\sqrt 2 }}.