Find (trigonometry)

\cos \frac{2 \pi}{7}, \cos \frac{4 \pi}{7}, \cos \frac{6\pi}{7} are roots of the cubic 8t^3 +4t^2-4t-1 = 0 and hence

\cos \frac{2 \pi}{7}+ \cos \frac{4 \pi}{7}+ \cos \frac{6\pi}{7} =-\frac{1}{2}

Now if the given expression is S, then

S^2 = \left(\sin \frac{2 \pi}{7} + \sin \frac{4 \pi}{7} - \sin \frac{ \pi}{7} \right)^2 \\ \\ = \left[\sin^2 \frac{2\pi}{7} + \sin^2 \frac{4\pi}{7} + \sin^2 \frac{\pi}{7}\right] + 2 \sin \frac{2\pi}{7}\sin \frac{4\pi}{7}-2 \sin \frac{2\pi}{7}\sin \frac{\pi}{7}-2 \sin \frac{4\pi}{7}\sin \frac{\pi}{7}

\left[\sin^2 \frac{2\pi}{7} + \sin^2 \frac{4\pi}{7} + \sin^2 \frac{\pi}{7}\right] = \frac{3}{2} - \frac{\cos \frac{4\pi}{7} + \cos \frac{8\pi}{7} + \cos \frac{2\pi}{7} }{2} \\ \\ = \frac{3}{2} - \frac{\cos \frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{6\pi}{7} }{2} = \frac{3}{2} + \frac{1}{4} = \frac{7}{4}

Also

2 \sin \frac{2 \pi}{7} \sin \frac{4 \pi}{7} - 2 \sin \frac{ \pi}{7} \sin \frac{2 \pi}{7} - 2 \sin \frac{ \pi}{7} \sin \frac{4 \pi}{7} \\ \\ = \left[\cos \frac{2\pi}{7} -\cos \frac{6\pi}{7}\right] - \left[\cos \frac{\pi}{7} -\cos \frac{3\pi}{7}\right] - \left[\cos \frac{3\pi}{7} -\cos \frac{5\pi}{7}\right] \\ \\ = \cos \frac{2\pi}{7} + \cos \frac{5\pi}{7} - \cos \frac{\pi}{7} - \cos \frac{6\pi}{7} = 0

Hence S^2 = \frac{7}{4} \Rightarrow S = \frac{\sqrt 7}{2}

(since S>0)

gr8 thnk u for wonderful solutions [1]

is my solution incorrect ?