 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-2/18/c/ii/solution

Let

J = A / a

, so

J^{2} = - I

. The even and odd powers in the matrix exponential give

D = I k = 0 \sum \infty \frac{( - 1 ) ^{k} a ^{2 k}}{( 2 k )!} + J k = 0 \sum \infty \frac{( - 1 ) ^{k} a ^{2 k + 1}}{( 2 k + 1 )!} = I cos a + J sin a .

Here the two scalar sums are the Taylor series for cosine and sine. Consequently

D = (cos a - sin a sin a cos a), θ = a (mod 2 π) .

With the usual convention that positive angles turn anticlockwise, this rotation matrix rotates by

- a

; the parameter in the form printed in the paper is

θ = a

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