Question

Prove that
 \[{\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right) = 2{\rm{ si}}{{\rm{n}}^{ - 1}}x,\left| x \right| \le \dfrac{1}{{12}}\]

Answer 1

Hint: 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:
 \[si{n^{ - 1}}{\rm{ }}\left( {2x\sqrt {1 - {x^2}} } \right){\rm{ = 2 si}}{{\rm{n}}^{ - 1}}{\rm{ x}}\] ………………….(1)
By writing the right hand side separately, we get it as:
 \[RHS{\rm{ = 2 si}}{{\rm{n}}^{ - 1}}{\rm{ x}}\] ………………….(2)
By writing the left hand side separately, we get it as:
 \[LHS{\rm{ = si}}{{\rm{n}}^{ - 1}}{\rm{ }}\left( {2x\sqrt {1 - {x^2}} } \right)\] ………………….(3)
By taking an assumption as x = Sin a, we get it as:
 \[x{\rm{ = sin a}}\] ………………….(4)
By applying Sin-1 on both sides of equation, we get it as:
 \[si{n^{ - 1}}{\rm{ }}\left( x \right){\rm{ = si}}{{\rm{n}}^{ - 1}}{\rm{ }}\left( {sin{\rm{ a}}} \right)\]
By simplifying the above equation, we can write it as:
 \[si{n^{ - 1}}{\rm{ }}\left( x \right){\rm{ = a}}\]
By inverting the left, right sides of above equation, we get:
 \[a{\rm{ = si}}{{\rm{n}}^{ - 1}}x\] ………………….(5)
Now by substituting equation (4) in equation (3), we get it as:
 \[LHS{\rm{ = si}}{{\rm{n}}^{ - 1}}\left( {2\left( {sin{\rm{ a}}} \right)\sqrt {1 - {{\left( {sin{\rm{ a}}} \right)}^2}} } \right)\]
By simplifying the term inside the square, we get it as:
 \[LHS{\rm{ = si}}{{\rm{n}}^{ - 1}}\left( {2{\rm{ sin a}}\sqrt {1 - si{n^2}\,a} } \right)\]
By basic knowledge of trigonometry, we know the identity:
 \[si{n^2}\theta {\rm{ + }}{\cos ^2}\theta {\rm{ = 1}}\]
By subtracting \[si{n^2}\theta \] on both sides, we get it as:
 \[si{n^2}\theta + {\cos ^2}\theta - Si{n^2}\theta = 1 - Si{n^2}\theta \]
By simplifying the above equation, we get relation as:
 \[{\cos ^2}\theta = 1 - Si{n^2}\theta \]
By substituting this inside the square root of LHS, we get
 \[LHS = {\sin ^{ - 1}}\left( {2\sin a\sqrt {{{\cos }^2}a} } \right)\]
By removing square root, we can write the term as:
 \[LHS = {\sin ^{ - 1}}\left( {2\sin a{\rm{ cos}}\,{\rm{a}}} \right)\]
By basic knowledge of trigonometry, we get the equation:
 \[2\sin a{\rm{ cosa = sin2a}}\]
By substituting this question into the LHS, we get it as:
 \[LHS = {\sin ^{ - 1}}\left( {\sin 2a} \right)\]
By applying basic property of inverse given by as follows:
\[{\sin ^{ - 1}}\left( {\sin k} \right) = k\]
\[\dfrac{{ - \pi }}{2} < k < \dfrac{\pi }{2}\]
By substituting this, we get it as follows:
 \[LHS = 2a\]
By substituting value from equation (5), we get:
 \[LHS = 2{\sin ^{ - 1}}x\]
By substituting the equation (2) here, we get it as:
 \[LHS = RHS\]
So, we proved \[LHS = RHS\] . Hence, proved the given equation.

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