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

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.

A root of unity satisfies $|z{|}^3=1$, hence has modulus one. Writing $z=e^{i\varphi }$, its argument satisfies $3\varphi =2\pi k$ for some integer $k$. The three distinct complex roots are thereforeThey are equally spaced on the unit circle. Taking $k=0,1,2$ exhausts them, since adding three to $k$ repeats a root and a nonzero polynomial of degree three has at most three distinct roots.

Factor out the same half-angle phase from the numerator and denominator:Here Euler's formula converts the bracketed sums to cosine and sine. Both sides require $\theta \notin 2\pi \mathbb{Z}$; at the excluded angles the original denominator vanishes.