Question

The value of $\sin \frac{\mathrm{\pi }}{18}+\sin \frac{\mathrm{\pi }}{9}+\sin \frac{2\mathrm{\pi }}{9}+\sin \frac{5\mathrm{\pi }}{18}$ is given by

(a) $\sin \frac{7\mathrm{\pi }}{18}+\sin \frac{4\mathrm{\pi }}{9}$

(b) 1

(c) $\cos \frac{\mathrm{\pi }}{6}+\cos \frac{3\mathrm{\pi }}{7}$

(d) $\cos \frac{\mathrm{\pi }}{9}+\sin \frac{\mathrm{\pi }}{9}$

Answer 1

​$$\begin{gathered} \sin \frac{\mathrm{\pi }}{18}+\sin \frac{\mathrm{\pi }}{9}+\sin \frac{2\mathrm{\pi }}{9}+\sin \frac{5\mathrm{\pi }}{18} \\ =\sin \frac{\pi }{18}+\sin \frac{5\pi }{18}+\sin \frac{\pi }{9}+\sin \frac{2\pi }{9} \\ =2\sin \left(\frac{{\displaystyle \frac{5\pi }{18}}+{\displaystyle \frac{\pi }{18}}}{2}\right)\cos \left(\frac{{\displaystyle \frac{5\pi }{18}}-{\displaystyle \frac{\pi }{18}}}{2}\right)+2\sin \left[\left(\frac{2\pi }{9}+\frac{\pi }{9}\right)/2\right]\cos \left(\frac{{\displaystyle \frac{2\pi }{9}}-{\displaystyle \frac{\pi }{9}}}{2}\right) \\ \left(\mathrm{using}\mathrm{identity}:-\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)\right) \\ =2\sin \frac{3\pi }{18}\cos \frac{2\pi }{18}+2\sin \frac{3\pi }{18}\cos \left(\frac{\pi }{18}\right) \\ =2\sin \frac{3\pi }{18}\left(\cos \frac{2\pi }{18}+\cos \frac{\pi }{18}\right) \\ =2\sin \frac{\pi }{6}\left(\cos \frac{\pi }{9}+\cos \frac{\pi }{18}\right) \\ =2\times \frac{1}{2}\left(\cos \frac{\pi }{9}+\cos \frac{\pi }{18}\right)\left(\because \sin \frac{\pi }{6}=\frac{1}{2}\right) \\ =\cos \frac{\pi }{9}+\cos \frac{\pi }{18} \\ =\sin \left(\frac{\pi }{2}-\frac{\pi }{9}\right)+\sin \left(\frac{\pi }{2}-\frac{\pi }{18}\right)\left(\because \sin \left(\frac{\pi }{2}-\theta \right)=\cos \theta \right) \\ =\sin \left(\frac{9\pi -2\pi }{18}\right)+\sin \left(\frac{9\pi -\pi }{18}\right) \\ =\sin \frac{7\pi }{18}+\sin \left(\frac{8\pi }{18}\right) \\ =\sin \frac{7\pi }{18}+\sin \frac{4\pi }{9} \end{gathered}$$

Hence, the correct answer is option A.