Past exam of the mathematical methods I course of the University of Oxford / 2015 resit / 2 / Solution

De Moivre's theorem states $(\cos x+i\sin x{)}^n=\cos (nx)+i\sin (nx)$ for integer $n$. For $n=3$, expanding the cube and taking its real part givesRearranging gives the required identity withNo division by $\cos x$ is needed, so the proof also covers its zeros.