It is given that $y=\sqrt{2+{{\cos }^{2}}x}$. Show that $2y\frac{\mathrm{d}y}{\mathrm{d}x}=-\sin 2x$

Question

It is given that $y=\sqrt{2+{{\cos }^{2}}x}$. Show that $2y\frac{\mathrm{d}y}{\mathrm{d}x}=-\sin 2x$, [1] $y\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}+{{\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right)}^{2}}=-\cos 2x$. [1] Hence find the Maclaurin series of $y$, up to and including the term in ${{x}^{4}}$. [5] By substituting $x=\frac{\pi }{6}$, show that $\sqrt{11}\approx 2\sqrt{3}\left( 1-\frac{{{\pi }^{2}}}{216}+\frac{{{\pi }^{4}}}{31104} \right)$ [2]

Accepted Answer

$\sqrt{11} \approx 2\sqrt{3}\left(1 - \dfrac{\pi^2}{216} + \dfrac{\pi^4}{31104}\right)$

2022 RI P1 Q7

2022 RI P1 Q7

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