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If $\alpha ,\,\,\beta ,\gamma ,\,\,\delta $ are the smallest positive angles in ascending order of magnitude which have their sines equal to the positive quantity $k$, then the value of $4\,\sin \frac{\alpha }{2} + 3\,\sin \frac{\beta }{2} + 2\,\sin \frac{\gamma }{2} + \sin \frac{\delta }{2}$ is equal to
$2\,\sqrt {1 - k} $
$\frac{1}{2}\sqrt {1 + k} $
$2\,\sqrt {1 + k} $
None of these
Solution
(c) Given $\alpha < \beta < \gamma < \delta $ and $\sin \alpha = \sin \beta = \sin \gamma = \sin \delta = k$. Also $\alpha ,\beta ,\gamma ,\delta $ are smallest positive angles satisfying above two conditions.
$\therefore$ We can take $\beta = \pi – \alpha ,\gamma = 2\pi + \alpha ,\delta = 3\pi – \alpha $.
Given expression
$ = 4\sin \frac{\alpha }{2} + 3\sin \left( {\frac{\pi }{2} – \frac{\alpha }{2}} \right) + 2\sin \left( {\pi + \frac{\alpha }{2}} \right) + \sin \left( {\frac{{3\pi }}{2} – \frac{\alpha }{2}} \right)$
$ = 4\sin \frac{\alpha }{2} + 3\cos \frac{\alpha }{2} – 2\sin \frac{\alpha }{2} – \cos \frac{\alpha }{2} = 2\left( {\sin \frac{\alpha }{2} + \cos \frac{\alpha }{2}} \right)$
$ = 2\sqrt {{{\left( {\sin \frac{1}{2}\alpha + \cos \frac{1}{2}\alpha } \right)}^2}} = 2\sqrt {1 + \sin \alpha } = 2\sqrt {1 + k} $.
