The value of cos( 2π 7)+ cos( 4π 7)+ cos( 6π 7) is equal to

Given, $\mathrm{cos}\frac{2\pi }{7}+\mathrm{cos}\frac{4\pi }{7}+\mathrm{cos}\frac{6\pi }{7}$cos72π​+cos74π​+cos76π​

$$=\frac{1}{2\mathrm{sin}(\frac{\pi }{7})}{2\mathrm{cos}(\frac{2\pi }{7})\mathrm{sin}(\frac{\pi }{7})+2\mathrm{cos}(\frac{4\pi }{7})\mathrm{sin}(\frac{\pi }{7})+2\mathrm{cos}(\frac{6\pi }{7})\mathrm{sin}(\frac{\pi }{7})}$$=2sin(7π​)1​2cos(72π​)sin(7π​)+2cos(74π​)sin(7π​)+2cos(76π​)sin(7π​)

$$=\frac{1}{2\mathrm{sin}\frac{\pi }{7}}{\mathrm{sin}\frac{3\pi }{7}-\mathrm{sin}\frac{\pi }{7}+\mathrm{sin}\frac{5\pi }{7}-\mathrm{sin}\frac{3\pi }{7}+\mathrm{sin}\frac{7\pi }{7}-\mathrm{sin}\frac{5\pi }{7}}$$=2sin7π​1​sin73π​−sin7π​+sin75π​−sin73π​+sin77π​−sin75π​

Since, $2\mathrm{cos}A\mathrm{sin}B=\mathrm{sin}(A+B)-\mathrm{sin}(A-B)$2cosAsinB=sin(A+B)−sin(A−B)

$$=\frac{1}{2\mathrm{sin}\frac{\pi }{7}}{-\mathrm{sin}\frac{\pi }{7}+sin\pi }$$=2sin7π​1​−sin7π​+sinπ

$$=\frac{-1}{2},(\because sin\pi =0)$$=2−1​,(∵sinπ=0)