 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-1/section-b/3/a/iii/solution

The square is on the cosine: near zero,

ln (cos^{2} x) = 2 ln (cos x)

. Substituting

cos x = 1 - x^{2} /2 + x^{4} /24 + O (x^{6})

ln (cos x) = - \frac{x ^{2}}{2} + \frac{x ^{4}}{24} - \frac{1}{2} \frac{x ^{4}}{4} + O (x^{6}) = - \frac{x ^{2}}{2} - \frac{x ^{4}}{12} + O (x^{6}) .

Therefore

ln (cos^{2} x) = - x^{2} - \frac{x ^{4}}{6} + O (x^{6}) .

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