Tính $\sin \dfrac{{2\pi }}{7} + \sin \dfrac{{4\pi }}{7} + \sin \dfrac{{6\pi }}{7}$

Ta có:

$\begin{array}{l}\sin \dfrac{\pi }{7}\left( {\sin \dfrac{{2\pi }}{7} + \sin \dfrac{{4\pi }}{7} + \sin \dfrac{{6\pi }}{7}} \right) = \sin \dfrac{\pi }{7}\sin \dfrac{{2\pi }}{7} + \sin \dfrac{\pi }{7}\sin \dfrac{{4\pi }}{7} + \sin \dfrac{\pi }{7}\sin \dfrac{{6\pi }}{7}\\ = \dfrac{1}{2}\left( {\cos \dfrac{\pi }{7} - \cos \dfrac{{3\pi }}{7}} \right) + \dfrac{1}{2}\left( {\cos \dfrac{{3\pi }}{7} - \cos \dfrac{{5\pi }}{7}} \right) + \dfrac{1}{2}\left( {\cos \dfrac{{5\pi }}{7} - \cos \dfrac{{7\pi }}{7}} \right)\\ = \dfrac{1}{2}\cos \dfrac{\pi }{7} + \dfrac{1}{2} = {\cos ^2}\dfrac{\pi }{{14}}\\\sin \dfrac{\pi }{7} = 2\sin \dfrac{\pi }{{14}}\cos \dfrac{\pi }{{14}} \Rightarrow \left( {\sin \dfrac{{2\pi }}{7} + \sin \dfrac{{4\pi }}{7} + \sin \dfrac{{6\pi }}{7}} \right) = \dfrac{1}{2}\cot \dfrac{\pi }{{14}}\end{array}$