2013 ACJC P1 Q12

Given that \(y={{({{\sin }^{-1}}x)}^{2}}\), show that

\[\left( 1-{{x}^{2}} \right){{\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)}^{2}}=4y\]

and \(\left( 1-{{x}^{2}} \right)\frac{{{\mathrm{d}}^{\text{2}}}y}{\mathrm{d}{{x}^{2}}}-x\frac{\mathrm{d}y}{\mathrm{d}x}=2\).[3]

By further differentiation of these results, find the Maclaurin series of \(y\) up to including the term in \({{x}^{4}}\).[3]

The series found above is to be used to estimate a value for \({{\pi }^{2}}\). Explain with the aid of a sketch why using \(x=\frac{1}{2}\) in the series gives a better approximation than \(x=1\), and hence find an approximation to \({{\pi }^{2}}\), giving your answer in fraction form. [4]