Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 1 / 10E / e / Solution

From the addition formula and $\sin x_0=0$, $\cos x_0=-1$, so $2x_0$ is a period. Repeatedly subtract $2x_0$ from a positive period $2x_1$; the remainder in $[0,2x_0)$ is also a period, and positivity plus the addition identities force it to be $0$. Thus $x_1=n{x}_0$; the converse follows from periodicity and oddness.

Solved by gpt-5.6-sol high.