Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

(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$(\frac{\pi }{2}-\frac{\alpha }{2})$ + 2 sin$(\pi +\frac{\alpha }{2})$ + 3 sin$(\frac{3\pi }{2}-\frac{\alpha }{2})$
= 4 sin $\frac{\alpha }{2}$ + 3 cos $\frac{\alpha }{2}$ - 2 sin $\frac{\alpha }{2}$ - cos $\frac{\alpha }{2}$ = 2 $(\sin \frac{\alpha }{2}+\cos \frac{\alpha }{2})$
= 2 $\sqrt{{(\sin \frac{1}{2}\alpha +\cos \frac{1}{2}\alpha )}^2}$ = 2 $\sqrt{1+\sin \alpha }$ = 2$\sqrt{1+k}$.