The value of sin

$\sin \frac{\pi}{18}+\sin \frac{\pi}{9}+\sin \frac{2 \pi}{9}+\sin \frac{5 \pi}{18}$

$=\sin \frac{\pi}{18}+\sin \frac{5 \pi}{18}+\sin \frac{\pi}{9}+\sin \frac{2 \pi}{9}$

$=2 \sin \left(\frac{\frac{5 \pi}{18}+\frac{\pi}{18}}{2}\right) \cos \left(\frac{\frac{5 \pi}{18}-\frac{\pi}{18}}{2}\right)+2 \sin \left[\left(\frac{2 \pi}{9}+\frac{\pi}{9}\right) / 2\right] \cos \left(\frac{\frac{2 \pi}{9}-\frac{\pi}{9}}{2}\right)$

(using identity : $\left.-\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)\right)$

$=2 \sin \frac{3 \pi}{18} \cos \frac{2 \pi}{18}+2 \sin \frac{3 \pi}{18} \cos \left(\frac{\pi}{18}\right)$

$=2 \sin \frac{3 \pi}{18}\left(\cos \frac{2 \pi}{18}+\cos \frac{\pi}{18}\right)$

$=2 \sin \frac{\pi}{6}\left(\cos \frac{\pi}{9}+\cos \frac{\pi}{18}\right)$

$=2 \times \frac{1}{2}\left(\cos \frac{\pi}{9}+\cos \frac{\pi}{18}\right)$     $\left(\because \sin \frac{\pi}{6}=\frac{1}{2}\right)$

$=\cos \frac{\pi}{9}+\cos \frac{\pi}{18}$

$=\sin \left(\frac{\pi}{2}-\frac{\pi}{9}\right)+\sin \left(\frac{\pi}{2}-\frac{\pi}{18}\right)$    $\left(\because \sin \left(\frac{\pi}{2}-\theta\right)=\cos \theta\right)$

$=\sin \left(\frac{9 \pi-2 \pi}{18}\right)+\sin \left(\frac{9 \pi-\pi}{18}\right)$

$=\sin \frac{7 \pi}{18}+\sin \left(\frac{8 \pi}{18}\right)$

$=\sin \frac{7 \pi}{18}+\sin \frac{4 \pi}{9}$

Hence, the correct answer is option A.