(a)     Given that $y = \ln \cos x$ , show that the first two non-zero terms of the Maclaurin series for y are $- \frac{{{x^2}}}{2} - \frac{{{x^4}}}{{12}}$.

(b)     Use this series to find an approximation in terms of $\pi {\text{ for }}\ln 2$ .

(a)     $f(x) = \ln \cos x$

$f'(x) = \frac{{ - \sin x}}{{\cos x}} = - \tan x$     M1A1

$f''(x) = - {\sec ^2}x$     M1

$f'''(x) = - 2\sec x\sec x\tan x$     A1

${f^{iv}}(x) = - 2{\sec ^2}x({\sec ^2}x) - 2\tan x(2{\sec ^2}x\tan x)$

$= - 2{\sec ^4}x - 4{\sec ^2}x{\tan ^2}x$     A1

$f(x) = f(0) + xf'(0) + \frac{{{x^2}}}{{2!}}f''(0) + \frac{{{x^3}}}{{3!}}f'''(0) + \frac{{{x^4}}}{{4!}}{f^{iv}}(0) + …$

$f(0) = 0,$     M1

$f'(0) = 0,$

$f''(0) = - 1,$

$f'''(0) = 0,$

${f^{iv}}(0) = - 2,$     A1

Notes: Award the A1 if all the substitutions are correct.

Allow FT from their derivatives.

$\ln (\cos x) \approx  - \frac{{{x^2}}}{{2!}} - \frac{{2{x^4}}}{{4!}}$     A1

$= - \frac{{{x^2}}}{2} - \frac{{{x^4}}}{{12}}$     AG

[8 marks]

(b)     Some consideration of the manipulation of ln 2     (M1)

Attempt to find an angle     (M1)

EITHER

Taking $x = \frac{\pi }{3}$     A1

$\ln \frac{1}{2} \approx - \frac{{{{\left( {\frac{\pi }{3}} \right)}^2}}}{{2!}} - \frac{{2{{\left( {\frac{\pi }{3}} \right)}^4}}}{{4!}}$     A1

$- \ln 2 \approx - \frac{{\frac{{{\pi ^2}}}{9}}}{{2!}} - \frac{{2\frac{{{\pi ^4}}}{{81}}}}{{4!}}$     A1

$\ln 2 \approx \frac{{{\pi ^2}}}{{18}} + \frac{{{\pi ^4}}}{{972}} = \frac{{{\pi ^2}}}{9}\left( {\frac{1}{2} + \frac{{{\pi ^2}}}{{108}}} \right)$     A1

OR

Taking $x = \frac{\pi }{4}$     A1

$\ln \frac{1}{{\sqrt 2 }} \approx - \frac{{{{\left( {\frac{\pi }{4}} \right)}^2}}}{{2!}} - \frac{{2{{\left( {\frac{\pi }{4}} \right)}^4}}}{{4!}}$     A1

$- \frac{1}{2}\ln 2 \approx - \frac{{\frac{{{\pi ^2}}}{{16}}}}{{2!}} - \frac{{2\frac{{{\pi ^4}}}{{256}}}}{{4!}}$     A1

$\ln 2 \approx \frac{{{\pi ^2}}}{{16}} + \frac{{{\pi ^4}}}{{1536}} = \frac{{{\pi ^2}}}{8}\left( {\frac{1}{2} + \frac{{{\pi ^2}}}{{192}}} \right)$     A1

[6 marks]

Total [14 marks]

Some candidates had difficulty organizing the derivatives but most were successful in getting the series. Using the series to find the approximation for $\ln 2$ in terms of $\pi$ was another story and it was rare to see a good solution.