Infatti:
cos : [cos2b + cos2(120° + b) + cos2(120° - b )] =
= : [cos2b + (cos120°cosb - sin120°sinb)2 + (cos120°cosb + sin120°sinb)2] =
osservo che:cos120° = cos(90° + 30°) = cos( + ) = -sin = -
sin120° = sin(90° + 30°) = sin( + ) = cos =
avrò quindi:
= : [cos2b + (-cosb - sinb)2 + (-cosb + sinb)2] =
= : [cos2b + cos2b + 3sin2b/4 + sinbcosb + cos2b + 3sin2b/4 - sinbcosb] =
= : [3cos2b/2 + 3sin2b/2] =
= : 3[cos2b + sin2b]/2 =