Past exam of the mathematics course of the University of Cambridge / 2006 / ia / Paper 1 / 8A / i / Solution

For every integer $m$, De Moivre's theorem states $(\cos \theta +i\sin \theta {)}^m=\cos (m\theta )+i\sin (m\theta )$. Taking the real part for $m=5$ givesReplace ${\sin }^2\theta$ by $1-{\cos }^2\theta$ and expand:The resulting polynomial is the fifth Chebyshev polynomial.