Question

If ${\theta }_1,{\theta }_2,{\theta }_3,{\theta }_4$ are the smallest positive angles in ascending order of magnitude which have their sines equal to a positive number $\lambda ,$ then the value of
$4\sin \frac{{\theta }_1}{2}+3\sin \frac{{\theta }_2}{2}+2\sin \frac{{\theta }_3}{2}+\sin \frac{{\theta }_4}{2}$ is equal to

• A. $2\sqrt{\lambda }$
• B. $2\sqrt{1+\lambda }$
• C. $2\sqrt{1-\lambda }$
• D. None of these

Answer 1

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

${\theta }_1<{\theta }_2<{\theta }_3<{\theta }_4$ are in ascending order $\cdots \left(1\right)$

$\sin {\theta }_1=\sin {\theta }_2=\sin {\theta }_3=\sin {\theta }_4=\lambda \cdots \left(2\right)$

$\therefore {\theta }_2=\pi -{\theta }_1,{\theta }_3=2\pi +{\theta }_1,{\theta }_4=3\pi -{\theta }_1$

We have chosen the values to satisfy the conditions $\left(1\right)$ and $\left(2\right)$

$4\sin \frac{{\theta }_1}{2}+3\sin \frac{{\theta }_2}{2}+2\sin \frac{{\theta }_3}{2}+\sin \frac{{\theta }_4}{2}=4\sin \frac{{\theta }_1}{2}+3\sin \frac{\pi -{\theta }_1}{2}+2\sin \frac{2\pi +{\theta }_1}{2}+\sin \frac{3\pi -{\theta }_1}{2}$

$=4\sin \frac{{\theta }_1}{2}+3\cos \frac{{\theta }_1}{2}-2\sin \frac{{\theta }_1}{2}-\cos \frac{\theta }{2}$

$=2(\sin \frac{{\theta }_1}{2}+\cos \frac{{\theta }_1}{2})=2\sqrt{1+\sin {\theta }_1}=2\sqrt{1+\lambda }$