Explanation

(c)
$\because\ \ \ x + \frac{1}{x} = 2\sin\alpha$
$\Rightarrow x^{2} - 2x\sin\alpha + 1 = 0$
$\therefore\ \ \ \ x = \frac{2\sin\alpha \pm \sqrt{4\sin^{2}{\alpha - 4}}}{2}$
$\Rightarrow \ \ \ \ \ x = \sin{\alpha \pm i\ \cos\alpha}$
Similarly, $y = \cos{\beta \pm i\cos\beta}$
$\therefore\ \ \ \ xy = \left( \sin{\alpha \pm i\cos\alpha} \right)(\cos\beta \pm i\sin\beta)$
$= \sin{(\beta - \alpha) \pm i\cos{(\beta - \alpha)}}$
$xy = \pm i\lbrack\cos{(\beta - \alpha) - i\sin{(\beta - \alpha)}}\rbrack$
And $\frac{1}{xy} = \pm \frac{1}{i}\lbrack\cos(\beta - \alpha) + i\sin{(\beta - \alpha)}\rbrack$
Now, $(xy)^{3} + \frac{1}{(xy)^{3}} = \pm i^{3}\left\lbrack \cos{3\ (\beta - \alpha)} - i\sin{3(\beta - \alpha)} \right\rbrack$
$\pm \frac{1}{i^{3}}\lbrack\cos{3(\beta - \alpha)} + i\sin{3(\beta - \alpha)}\rbrack$
$= \pm i\ \lbrack\cos{3(\beta - \alpha)} - i\sin{3(\beta - \alpha)}\rbrack$
$\pm \frac{1}{i}\lbrack\cos{3(\beta - \alpha) + i\sin{3(\beta - \alpha)}}\rbrack$
$= \pm \frac{1}{i}\lbrack\left\{ \cos{3(\beta - \alpha) - i\sin{3(\beta - \alpha)}} \right\} - \{\cos{3(\beta - \alpha) + i\sin{3(\beta - \alpha)}}\}\rbrack$
$= \pm \frac{1}{i}\left( - 2i\sin{3(\beta - \alpha)} \right) = 2\sin{3(\beta - \alpha)}$