Question

Use Theorem 5 to prove the identity

\(2{\sin ^{ - 1}}x = {\cos ^{ - 1}}\left( {1 - 2{x^2}} \right)\), \(x \ge 0\).

Short answer

The identity \(2{\sin ^{ - 1}}x = {\cos ^{ - 1}}\left( {1 - 2{x^2}} \right)\) if \(x \ge 0\) is proved.

Step-by-step solution

Given data

The functions is \(f(x) = 2{\sin ^{ - 1}}x - {\cos ^{ - 1}}\left( {1 - 2{x^2}} \right)\).

Use theorem

"If \({f^\prime }(x) = 0\) for all \(x\) in an interval \((a,\;b)\), then \(f\) is constant on \((a,\;b)\)."

Differentiate \(f(x)\) with respect to \(x\)

Consider the function \(f(x) = 2{\sin ^{ - 1}}x - {\cos ^{ - 1}}\left( {1 - 2{x^2}} \right)\) as follows:

Differentiate \(f(x)\) with respect to \(x\) as follows:

\(\begin{aligned}{c}{f^\prime }(x) &= \frac{2}{{\sqrt {1 - {x^2}} }} - \left( {\frac{{ - 1}}{{\sqrt {1 - {{\left( {1 - 2{x^2}} \right)}^2}} }}} \right)\frac{d}{{dx}}\left( {1 - 2{x^2}} \right)\\ &= \frac{2}{{\sqrt {1 - {x^2}} }} + \frac{1}{{\sqrt {1 - 1 - 4{x^4} + 4{x^2}} }}( - 4x)\\ &= \frac{2}{{\sqrt {1 - {x^2}} }} - \frac{{4x}}{{\sqrt { - 4{x^4} + 4{x^2}} }}\\ &= \frac{2}{{\sqrt {1 - {x^2}} }} - \frac{{4x}}{{2x\sqrt {1 - {x^2}} }}\end{aligned}\)

Simplify further as follows:

\(\begin{aligned}{c}{f^\prime }(x) &= \frac{2}{{\sqrt {1 - {x^2}} }} - \frac{2}{{\sqrt {1 - {x^2}} }}\\ &= 0\end{aligned}\)

Hence by the above theorem for \(x \ge 0\), the function \(f(x)\) is constant as follows:

\(\begin{aligned}{c}f(x) &= C\\2{\sin ^{ - 1}}x - {\cos ^{ - 1}}\left( {1 - 2{x^2}} \right) &= C\end{aligned}\)

Evaluate the constant \(C\)

Evaluate the constant \(C\) by substituting \(x = 0\) as follows:

\(\begin{aligned}{c}2{\sin ^{ - 1}}(0) - {\cos ^{ - 1}}\left( {1 - 2{{(0)}^2}} \right) &= C\\2{\sin ^{ - 1}}(0) - {\cos ^{ - 1}}(1) &= C\\2(0) - 0 &= C\\C &= 0\end{aligned}\)

Therefore, for \(x \ge 0\), the function \(f(x)\) becomes as follows:

\(\begin{aligned}{c}2{\sin ^{ - 1}}x - {\cos ^{ - 1}}\left( {1 - 2{x^2}} \right) &= 0\\2{\sin ^{ - 1}}x &= {\cos ^{ - 1}}\left( {1 - 2{x^2}} \right)\end{aligned}\)

Thus, \(2{\sin ^{ - 1}}x = {\cos ^{ - 1}}\left( {1 - 2{x^2}} \right)\) if \(x \ge 0\).

Hence, the identity \(2{\sin ^{ - 1}}x = {\cos ^{ - 1}}\left( {1 - 2{x^2}} \right)\) if \(x \ge 0\) is proved.