Rezolvare:
Folosim:
[tex]\boldsymbol{ \red{ 2 \sin x \cos y = \sin(x + y) + \sin(x - y)}}[/tex]
Înmulțim cu
[tex]\dfrac{2 \sin\dfrac{\pi}{7} }{2 \sin\dfrac{\pi}{7}}\bigg(\cos\dfrac{2\pi}{7} + \cos\dfrac{4\pi}{7} + \cos\dfrac{6\pi}{7}\bigg) =[/tex]
[tex]= \dfrac{1 }{2 \sin\dfrac{\pi}{7}}\bigg(2 \sin\dfrac{\pi}{7}\cos\dfrac{2\pi}{7} + 2 \sin\dfrac{\pi}{7}\cos\dfrac{4\pi}{7} + 2 \sin\dfrac{\pi}{7}\cos\dfrac{6\pi}{7}\bigg) \\[/tex]
[tex]2 \sin\dfrac{\pi}{7}\cos\dfrac{2\pi}{7} = \sin\dfrac{\pi+2\pi}{7} + \sin\dfrac{\pi - 2\pi}{7} = \sin\dfrac{3\pi}{7} + \sin\dfrac{-\pi}{7} = \sin\dfrac{3\pi}{7} - \sin\dfrac{\pi}{7}[/tex]
[tex]2 \sin\dfrac{\pi}{7}\cos\dfrac{4\pi}{7} = \sin\dfrac{\pi+4\pi}{7} + \sin\dfrac{\pi - 4\pi}{7} = \sin\dfrac{5\pi}{7} + \sin\dfrac{-3\pi}{7} = \sin\dfrac{5\pi}{7} - \sin\dfrac{3\pi}{7}[/tex]
[tex]2 \sin\dfrac{\pi}{7}\cos\dfrac{6\pi}{7} = \sin\dfrac{\pi+6\pi}{7} + \sin\dfrac{\pi - 6\pi}{7} = \sin\dfrac{7\pi}{7} + \sin\dfrac{-5\pi}{7} = \sin\dfrac{7\pi}{7} - \sin\dfrac{5\pi}{7}[/tex]
Expresia devine:
[tex]= \dfrac{1 }{2 \sin\dfrac{\pi}{7}}\bigg( \sin\dfrac{3\pi}{7} - \sin\dfrac{\pi}{7} + \sin\dfrac{5\pi}{7} - \sin\dfrac{3\pi}{7} + \sin \pi - \sin\dfrac{5\pi}{7}\bigg) \\[/tex]
Reducem termenii asemenea (sin π = 0)
[tex]= \dfrac{1 }{2 \sin\dfrac{\pi}{7}}\bigg( - \sin\dfrac{\pi}{7} \bigg) = \bf- \dfrac{1}{2}[/tex]