Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 1 / 18 / d / Solution

At $x=\pi /2$, all odd-indexed cosine terms vanish, while $\cos (2k\pi /2)=(-1{)}^k$. Thus the Fourier series from part (c) givesDivide by $2(e^{\pi }-1)/\pi$ and rearrange. The hyperbolic sine identity $e^{\pi }-1=2e^{\pi /2}\sinh (\pi /2)$ yieldsThe evaluation is legitimate by the convergence established in part (c).