Question

Use the method of Example 6 to prove the identity.

39. \(2{\sin ^{ - 1}}x = {\cos ^{ - 1}}\left( {1 - 2{x^2}} \right)\), \(x \ge 0\)

Short answer

The identity has been proved.

Step-by-step solution

Theorem 5

If \(f'\left( x \right) = 0\) for all \(x\) in an interval \(\left( {a,b} \right)\), then \(f\) is constanton \(\left( {a,b} \right)\).

Proof

Let a function \(f\left( x \right) = 2{\sin ^{ - 1}}x - {\cos ^{ - 1}}\left( {1 - 2{x^2}} \right)\).

Differentiate both the sides with respect to \(x\) as follows:

\(\begin{aligned}{c}f'\left( x \right) &= \frac{d}{{dx}}\left( {2{{\sin }^{ - 1}}x} \right) - \frac{d}{{dx}}\left( {{{\cos }^{ - 1}}\left( {1 - 2{x^2}} \right)} \right)\\ &= \frac{2}{{\sqrt {1 - {x^2}} }} - \left( { - \frac{1}{{\sqrt {1 - {{\left( {1 - 2{x^2}} \right)}^2}} }} \cdot \frac{d}{{dx}}\left( {1 - 2{x^2}} \right)} \right)\\ &= \frac{2}{{\sqrt {1 - {x^2}} }} + \left( {\frac{1}{{\sqrt {1 - 1 + 4{x^2} - 4{x^2}} }} \cdot \left( { - 4x} \right)} \right)\\ &= \frac{2}{{\sqrt {1 - {x^2}} }} - \left( {\frac{{4x}}{{\sqrt {4{x^2} - 4{x^2}} }}} \right)\\ &= \frac{2}{{\sqrt {1 - {x^2}} }} - \frac{2}{{\sqrt {1 - {x^2}} }}\\ &= 0\end{aligned}\)

Here, \(f'\left( x \right) = 0\). Therefore, by theorem 5, \(f\left( x \right) = c\) where \(c\) is a constant.

Find the value of c

Here, \(f\left( x \right) = 2{\sin ^{ - 1}}x - {\cos ^{ - 1}}\left( {1 - 2{x^2}} \right)\). Let \(x = 0.5\), then \(c = 2{\sin ^{ - 1}}0.5 - {\cos ^{ - 1}}\left( {1 - 2{{\left( {0.5} \right)}^2}} \right) = 2\left( {\frac{\pi }{6}} \right) - {\cos ^{ - 1}}\left( {1 - 0.5} \right)\).

That implies, \(\frac{\pi }{3} - \frac{\pi }{3} = c \Rightarrow c = 0\).

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

Hence, it is proved that \(2{\sin ^{ - 1}}x = {\cos ^{ - 1}}\left( {1 - 2{x^2}} \right)\).