IF A=PAI
2COSA/13.COS9A/13+COS3A/13+COS5A/13=??
Your notation seems to be rather confusing - it is not clear what do you mean. Try to write the formulae in a more usual way in your questions. I only guess it may be
2 * cos(π/13) * cos(9π/13) + cos(3π/13) + cos(5π/13) ?
If it is true, then by using the formulae
cos(θ) * cos(φ) = (cos(θ+φ) + cos(θ-φ)) / 2
cos(θ) + cos(φ) = 2 * cos((θ+φ)/2) + cos((θ-φ)/2)
(see the source) and the well-known formula cos(π/2) = 0 we get
2 * cos(π/13) * cos(9π/13) + cos(3π/13) + cos(5π/13) =
cos(10π/13) + cos(8π/13) + cos(3π/13) + cos(5π/13) =
[cos(10π/13) + cos(3π/13)] + [cos(8π/13) +cos(5π/13)] =
2*cos(π/2)*cos(7π/26) + 2*cos(π/2)*cos(3π/26) = 0
Comments
Your notation seems to be rather confusing - it is not clear what do you mean. Try to write the formulae in a more usual way in your questions. I only guess it may be
2 * cos(π/13) * cos(9π/13) + cos(3π/13) + cos(5π/13) ?
If it is true, then by using the formulae
cos(θ) * cos(φ) = (cos(θ+φ) + cos(θ-φ)) / 2
cos(θ) + cos(φ) = 2 * cos((θ+φ)/2) + cos((θ-φ)/2)
(see the source) and the well-known formula cos(π/2) = 0 we get
2 * cos(π/13) * cos(9π/13) + cos(3π/13) + cos(5π/13) =
cos(10π/13) + cos(8π/13) + cos(3π/13) + cos(5π/13) =
[cos(10π/13) + cos(3π/13)] + [cos(8π/13) +cos(5π/13)] =
2*cos(π/2)*cos(7π/26) + 2*cos(π/2)*cos(3π/26) = 0