 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2005/paper-1/1/a/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

sin x = x - x^{3} /6 + O (x^{5})

gives

sin^{2} x = x^{2} (1 - x^{2} /3 + O (x^{4}))

. Taking its reciprocal and cancelling the singular term gives

\frac{1}{s i n ^{2} x} - \frac{1}{x ^{2}} = \frac{1}{x ^{2}} [1 + \frac{x ^{2}}{3} + O (x^{4}) - 1] = \frac{1}{3} + O (x^{2}) .

Thus the apparent singularity at zero is removable, and the first nonzero term is the constant

1/3

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