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

$\sin x={\sum }_{n\ge 0}(-1{)}^n{x}^{2n+1}/(2n+1)!$ and $\cos x={\sum }_{n\ge 0}(-1{)}^n{x}^{2n}/(2n)!$. Here $\sin \prime \prime \prime =-\cos$ is negative while $\cos >0$ near zero; applying the preceding shape argument until the first zero of $\cos$ and then the addition identities establishes a first positive zero $x_0$ with positivity before it.

Solved by gpt-5.6-sol high.