All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$\mathrm{sin}(r+\frac{1}{2})\theta - \mathrm{sin}(r-\frac{1}{2})\theta$

$= \mathrm{sin}r\theta \mathrm{cos}\frac{\theta}{2} + \mathrm{cos}r\theta \mathrm{sin}\frac{\theta}{2} -[\mathrm{sin}r\theta \mathrm{cos}\frac{\theta}{2} - \mathrm{cos}r\theta \mathrm{sin}\frac{\theta}{2}]$

$= 2\mathrm{cos}r\theta \mathrm{sin}\frac{\theta}{2}$

(ii)

$\sum_{r=1}^{n} \mathrm{cos}r\theta$

$= \sum_{r=1}^{n} \frac{\mathrm{sin}(r+\frac{1}{2})\theta - \mathrm{sin}(r-\frac{1}{2})\theta}{2\mathrm{sin}\frac{\theta}{2}}$

$= \frac{1}{2\mathrm{sin}\frac{\theta}{2}} \sum_{r=1}^{n} \mathrm{sin}(r+\frac{1}{2})\theta - \mathrm{sin}(r-\frac{1}{2})\theta$

$= \frac{1}{2\mathrm{sin}\frac{\theta}{2}} [ \mathrm{sin}(\frac{3\theta}{2}) - \mathrm{sin}(\frac{\theta}{2})]$

$+ \mathrm{sin}(\frac{5\theta}{2}) - \mathrm{sin}(\frac{3\theta}{2})$

$+ \ldots$

$+ \mathrm{sin}(n-\frac{1}{2})\theta - \mathrm{sin}(n-\frac{3}{2})\theta$

$+ \mathrm{sin}(n+\frac{1}{2})\theta - \mathrm{sin}(n-\frac{1}{2})\theta$

$= \frac{1}{2\mathrm{sin}\frac{\theta}{2}} (\mathrm{sin}(n+\frac{1}{2})\theta - \mathrm{sin}(\frac{\theta}{2}))$

(iii)

Let P(n) be the statement $\sum_{r=1}^n \mathrm{sin}r\theta = \frac{\mathrm{cos}\frac{\theta}{2} - \mathrm{cos}(n+\frac{1}{2})\theta}{2\mathrm{sin}\frac{\theta}{2}}, n \in \mathbb{Z}^+$

When $n = 1$,

$\mathrm{LHS}~=~ \mathrm{sin}\theta$

$\mathrm{RHS}~=~ \frac{\mathrm{cos}\frac{\theta}{2} - \mathrm{cos}(1+\frac{1}{2})\theta}{2\mathrm{sin}\frac{\theta}{2}} = sin\theta$

Assume P(k) is true for some positive integers k,

$\sum_{r=1}^k \mathrm{sin}r\theta = \frac{\mathrm{cos}\frac{\theta}{2} - \mathrm{cos}(k+\frac{1}{2})\theta}{2\mathrm{sin}\frac{\theta}{2}}$

To prove P(k+1) is true,

$\sum_{r=1}^{k+1} \mathrm{sin}r\theta = \frac{\mathrm{cos}\frac{\theta}{2} - \mathrm{cos}(k+\frac{3}{2})\theta}{2\mathrm{sin}\frac{\theta}{2}}$

$\mathrm{LHS}~=~ \sum_{r=1}^{k+1} \mathrm{sin}r\theta$

$= \sum_{r=1}^{k} \mathrm{sin}r\theta + \mathrm{sin}(k+1)\theta$

$= \frac{\mathrm{cos}\frac{\theta}{2} - \mathrm{cos}(k+\frac{1}{2})\theta}{2\mathrm{sin}\frac{\theta}{2}} + \mathrm{sin}(k+1)\theta$

$= \frac{\mathrm{cos}\frac{\theta}{2} - \mathrm{cos}(k+\frac{1}{2})\theta + 2\mathrm{sin}(k+1)\theta \mathrm{sin}\frac{\theta}{2}}{2\mathrm{sin}\frac{\theta}{2}}$

$= \frac{\mathrm{cos}\frac{\theta}{2} - \mathrm{cos}(k+\frac{1}{2})\theta - \mathrm{cos}(k+\frac{3}{2})\theta + \mathrm{cos}(k+\frac{1}{2})\theta}{2\mathrm{sin}\frac{\theta}{2}}$

$= \frac{\mathrm{cos}\frac{\theta}{2} - \mathrm{cos}(k+\frac{3}{2})\theta}{2\mathrm{sin}\frac{\theta}{2}}$

$\therefore$, P(k+1) is true.

Since P(1) is true, and P(k) is true $\Rightarrow$ P(k+1) is true, by mathematical induction, P(n) is true for all positive integers n.