Question

The value of $\sin \frac{2\pi }{7}+\sin \frac{4\pi }{7}+\sin \frac{8\pi }{7}$ is

• A. 1
• B. $\frac{\sqrt{7}}{2}$
• C. $\frac{3\sqrt{3}}{4}$
• D. $\frac{\sqrt{15}}{4}$

Answer 1

The correct option is B $\frac{\sqrt{7}}{2}$

Let $A=\frac{2\pi }{7}$
$\Rightarrow 7A=2\pi$
$\sin 4A=\sin (7A-3A)=\sin (2\pi -3A)$
$\Rightarrow \sin 4A=-\sin 3A$
$\Rightarrow 2\sin 2A\cos 2A=4{\sin }^{3}A-3\sin A$
$\Rightarrow 4\sin A\cos A(1-2{\sin }^{2}A)=\sin A(4{\sin }^{2}A-3)$
$\Rightarrow 4\cos A(1-2{\sin }^{2}A)=4{\sin }^{2}A-3$
Square both the sides, we get
$16(1-{\sin }^{2}A)(1-2{\sin }^{2}A{)}^2=(4{\sin }^{2}A-3{)}^2$
$=64{\sin }^{6}A-112{\sin }^{4}A+56{\sin }^{2}A-7=0$
It is cubic in ${\sin }^{2}A$ with roots:
${\sin }^2\left(\frac{2\pi }{7}\right),{\sin }^2\left(\frac{4\pi }{7}\right),{\sin }^2\left(\frac{8\pi }{7}\right)$
Sum of roots, ${\sin }^2\left(\frac{2\pi }{7}\right)+{\sin }^2\left(\frac{4\pi }{7}\right)+{\sin }^2\left(\frac{8\pi }{7}\right)=\frac{7}{4}$
Also, from trigonometric identities, we can prove that:
$\sin \left(\frac{2\pi }{7}\right)\sin \left(\frac{4\pi }{7}\right)+\sin \left(\frac{4\pi }{7}\right)\sin \left(\frac{8\pi }{7}\right)+\sin \left(\frac{8\pi }{7}\right)\sin \left(\frac{2\pi }{7}\right)=0$
$\Rightarrow {(\sin \left(\frac{2\pi }{7}\right)+\sin \left(\frac{4\pi }{7}\right)+\sin \left(\frac{8\pi }{7}\right))}^2=\frac{7}{4}$
$\Rightarrow \sin \left(\frac{2\pi }{7}\right)+\sin \left(\frac{4\pi }{7}\right)+\sin \left(\frac{8\pi }{7}\right)=\frac{\sqrt{7}}{2}$