It is given that $\mathrm{f}\left( x \right)=\ln \left( \sin x+\cos x \right)$ for $-\frac{\pi }{4}

Question

It is given that $\mathrm{f}\left( x ight)=\ln \left( \sin x+\cos x ight)$ for $-\frac{\pi }{4} Show that $\mathrm{f}\,\mathrm{ }\!\!'\!\!\mathrm{ }\!\!'\!\!\mathrm{ }\left( x ight)+{{\left[ \mathrm{{f}'}(x) ight]}^{2}}+1=0$. [2] By further differentiation of the result in part (i), find the Maclaurin series for $\mathrm{f}\left( x ight)$ up to and including the term in ${{x}^{3}}$. [4] Using the expansion obtained in part (ii) and the standard series from the List of Formulae (MF26), find the series expansion of $\mathrm{g}\left( x ight)=\ln \left( 1+	an x ight)$ for $-\frac{\pi }{4} [4]

Accepted Answer

(i) $f''(0) = -2$; (ii) $\ln(\sin x + \cos x) = x - x^2 + \dfrac{2}{3}x^3 + \ldots$; (iii) $g(x) \approx x - \dfrac{1}{2}x^2 + \dfrac{2}{3}x^3$

2018 AJC P1 Q5

2018 AJC P1 Q5

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