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 =