proof

cos(10)+cos(82)+cos(154)+cos(226)+cos(298)=0

Before beginning this proof, we need to note several things...

(1)  cos A + cos B =   2cos[(A + B)/2]*cos[(A - B)/2]

(2)  2cos(36)  =  [1 +sqrt(5)] / 2 =  Phi

(3) 2cos(108)  = - [sqrt(5) - 1]/ 2 =   -phi

So we have:

cos(226) + cos(154)  = 2cos(380/2) *cos(72/2)  = 2cos(190)cos(36)  = cos(190)[2cos(36)] = cos(190)*Phi = cos(180 + 10)*Phi  = [cos(180)cos(10) - sin(180)sin(10)]*Phi  = -cos(10)*Phi

cos (298) + cos (82) =  2cos(380/2)*cos(216/2) = 2cos(190)cos(108) = cos(190)[2cos(108)] =

cos(190)(-phi)  = -cos(190)*phi   ....amd we have shown that cos(190) = -cos(10).....so.....-[cos(190)*phi ]  = -[-cos(10)]*phi  =  cos(10)*phi

So....

cos(10)+cos(82)+cos(154)+cos(226)+cos(298) =

cos(10) + [cos(226) + cos(154)] + [cos(298) + cos(82)] =

cos (10 ) - cos(10)*Phi + cos(10)*phi  =   [factor out cos(10) ]

cos(10) [ 1 - Phi + phi]  =

cos(10) [ 1 + phi - Phi]       and by definition,  1 + phi = Phi   ....so....

cos(10) [ Phi - Phi]  = cos(10) * 0   =  0