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Question Number 87497 by Rio Michael last updated on 04/Apr/20

A complex number z is defined by z = \frac{1}{2} (\cos θ + i \sin θ), such that

z^{n} = \frac{1}{2^{n}} (\cos n θ + i \sin n θ)

Using De {Moivre}^{'} s theorem, or otherwise, show that

(i) \underset{r = 0}{\sum^{\infty}} \frac{1}{4^{r}} \sin 2 r θ is a convergent geometic progression .

(ii) \underset{r = 0}{\sum^{\infty}} \frac{1}{4^{r}} \sin 2 r = \frac{14 \sin 2 θ}{17 - 16 \cos 2 θ}