Chapter 2.16, Problem 11P

Prove that

$\begin{array}{l}\cos \theta +\cos 3\theta +\cos 5\theta +\mathrm{...}+\cos (2n-1)\theta =\frac{\sin 2n\theta }{2\sin \theta }\\ \sin \theta +\sin 3\theta +\sin 5\theta +\mathrm{...}+\sin (2n-1)\theta =\frac{{\sin }^{2}2n\theta }{\sin \theta }\end{array}$

Hint: Use Euler’s formula and the geometric progression formula.