It is given that $y={{\sin }^{-1}}x$, $-1\le x\le 1$. Show that \(\frac{{{\mathrm{d}}^

Question

It is given that $y={{\sin }^{-1}}x$, $-1\le x\le 1$. Show that $\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}=x{{\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)}^{3}}$.[3] Using the result in part (a), find the Maclaurin expansion for $y$, up to and including the term in ${{x}^{3}}$.[3] By substituting $x=0.5$ into the expansion found in part (b), find an approximate value of $\pi $.[2]

Accepted Answer

(a) $\frac{\mathrm d^2y}{\mathrm dx^2}=x(\frac{\mathrm dy}{\mathrm dx})^3$. (b) $y=x+\frac16x^3+\cdots$. (c) $\pi\approx\frac{25}{8}=3.125$.

2023 CJC Promo Q5

2023 CJC Promo Q5

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