Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2022 / Paper 2 / 1 / c / Solution

Put $Y(\theta )=y(\cos \theta )$, with $0<\theta <\pi$. The chain rule givesConsequently the original differential expression becomes $-Y^{\prime \prime }-2\cot \theta Y^{\prime }$. If $Y=F/\sin \theta$, direct differentiation givesTaking $F=\sin n\theta$ therefore yieldsThe apparent endpoint singularities are removable. Expansion at $\theta =0,\pi$ givesThus both endpoint regularity conditions hold. This is the Chebyshev polynomial of the second kind $U_{n-1}(x)$.