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 $4sin\frac{\alpha }{2}+3sin\frac{\beta }{2}+2sin\frac{\gamma }{2}+sin\frac{\delta }{2}$ is equal to

• A. $2\sqrt{1-k}$
• B. $\frac{1}{2}\sqrt{1+k}$
• C. $2\sqrt{1+k}$
• D. $Noneofthese$

Answer 1

The correct option is C $2\sqrt{1+k}$

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