Question

The value of cos 0°+ cos pi/7+cos2pi/7+cos 3pi/7+..........+cos6pi/7 is equals to =》》

Answer 1

Step-by-step explanation:

Refer to the attachment..

Answer 2

Step-by-step explanation:

cos(x) = -cos(π - x)

So

cos(π/7) = -cos(6π/7) ==> cos(π/7) + cos(6π/7) = 0

cos(2π/7) = -cos(5π/7) ==> cos(2π/7) + cos(5π/7) = 0

cos(3π/7) = -cos(4π/7) ==> cos(3π/7) + cos(4π/7) = 0

Thus

cos(π/7) + cos(2π/7) + cos(3π/7) + cos(4π/7) + cos(5π/7) + cos(6π/7) + cos(7π/7)

= cos(π/7) + cos(2π/7) + cos(3π/7) + cos(4π/7) + cos(5π/7) + cos(6π/7) + cos(π)

= cos(π/7) + cos(2π/7) + cos(3π/7) + cos(4π/7) + cos(5π/7) + cos(6π/7) - 1

= [ cos(π/7) + cos(6π/7) ] + [ cos(2π/7) + cos(5π/7) ] + [ cos(3π/7) + cos(4π/7) ] - 1

= 0 + 0 + 0 - 1

= -1