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. $2\sqrt{1-k}$
• B. $2\sqrt{1+k}$
• C. $\frac{\sqrt{1+k}}{2}$
• D. none of these

Answer 1

Answer: (B)

$2\sqrt{1+k}$

Given $\sin \alpha =\sin \beta =\sin \gamma =\sin \delta =k>0$

and $\alpha <\beta <\gamma <\delta$

$\alpha$ is acute

$\beta =\pi -\alpha$

$\gamma =2\pi +\alpha$

$\delta =3\pi -\alpha$

Hence,

$4\sin \left(\frac{\alpha }{2}\right)+3\sin \left(\frac{\beta }{2}\right)+2\sin \left(\frac{\gamma }{2}\right)+\sin \left(\frac{\delta }{2}\right)$

$=4\sin \left(\frac{\alpha }{2}\right)+3\sin \left(\frac{\pi -\alpha }{2}\right)+2\sin \left(\frac{2\pi +\alpha }{2}\right)+\sin \left(\frac{3\pi -\alpha }{2}\right)$

$=4\sin \left(\frac{\alpha }{2}\right)+3\sin (\frac{\pi }{2}-\frac{\alpha }{2})+2\sin (\pi +\frac{\alpha }{2})+\sin (\frac{3\pi }{2}-\frac{\alpha }{2})$

$=4\sin \left(\frac{\alpha }{2}\right)+3\cos \left(\frac{\alpha }{2}\right)-2\sin \left(\frac{\alpha }{2}\right)-\cos \left(\frac{\alpha }{2}\right)$

$=2(\sin \left(\frac{\alpha }{2}\right)+\cos \left(\frac{\alpha }{2}\right))$

$=2\left(\sqrt{{(\sin \left(\frac{\alpha }{2}\right)+\cos \left(\frac{\alpha }{2}\right))}^2}\right)$

$=2\left(\sqrt{1+\sin \alpha }\right)=2\sqrt{1+k}$