(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.