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

Table of contents
- Solution 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

.

 Ancestors (13)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version