Question

lf $\alpha ,\beta ,\gamma ,\delta$ are the smallest positive angles in ascending order of magnitude which have their sines equal to the positive quantity $K$ then value of $4\sin \left(\frac{\alpha }{2}\right)+3\sin \left(\frac{\beta }{2}\right)+2\sin \left(\frac{\gamma }{2}\right)+\sin \left(\frac{\delta }{2}\right)$ equals

• A. $2\sqrt{1-K}$
• B. $2\sqrt{1+K}$
• C. $2\sqrt{K}$
• D. $\sqrt{K+1}$

Answer 1

The correct option is B $2\sqrt{1+K}$

$\sin \alpha =\sin \beta =\sin \gamma =\sin \delta =k$

$\beta =\pi -\alpha \Rightarrow \frac{\beta }{2}=\frac{\pi }{2}-\frac{\alpha }{2}$

$\gamma =2\pi +\alpha \Rightarrow \frac{\gamma }{2}=\pi +\frac{\alpha }{2}$

$\delta =3\pi -\alpha \Rightarrow \frac{\delta }{2}=\frac{3\pi }{2}-\frac{\alpha }{2}$

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

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

$=4\sin \frac{\alpha }{2}+3\cos \frac{\alpha }{2}-2\sin \frac{\alpha }{2}-\cos \frac{\alpha }{2}$

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

$=2\sqrt{{(\sin \frac{\alpha }{2}+\cos \frac{\alpha }{2})}^2}=2\sqrt{1+\sin \alpha }=2\sqrt{1+K}$