求值：cos2π7+cos4π7+cos6π7．

Question

求值：cos(2π)/(7)+cos(4π)/(7)+cos(6π)/(7)．

Answer 1

原式=(sin(π)/(7)(cos(2π)/(7)+cos(4π)/(7)+cos(6π)/(7)))/(sin(π)/(7))
=(sin(π)/(7)cos(2π)/(7)+sin(π)/(7)cos(4π)/(7)+sin(π)/(7)cos(6π)/(7))/(sin(π)/(7))
=((1)/(2)(sin(3π)/(7)−sin(π)/(7))+(1)/(2)(sin(5π)/(7)−sin(3π)/(7))+(1)/(2)(sin(7π)/(7)−sin(5π)/(7)))/(sin(π)/(7))
=-(1)/(2)．