Past exam of the mathematical methods I course of the University of Oxford / 2017 resit / 1 / a / Solution

By De Moivre's theorem, $(\cos \theta +i\sin \theta {)}^3=\cos 3\theta +i\sin 3\theta$. Expanding the cube and equating its real part givesThe last form uses ${\sin }^2\theta +{\cos }^2\theta =1$; comparing the imaginary parts similarly gives the corresponding sine identity.