If #cos(alpha-beta) +cos(beta-gamma) +cos(gamma-alpha)=-frac{3}2#,can you prove that #cos alpha +cos beta +cos gamma=0# and #sin alpha +sin beta +sin gamma =0#?
1 Answer
We have:
# cos(alpha-beta) +cos(beta-gamma) +cos(gamma-alpha) =-3/2 #
Using the cosine sum of angle formula:
# cos(A-B) -= cosAcosB + sinAsinB #
we have:
# cosalphacosbeta + sinalphasinbeta + cosbetacosgamma + sinbetasingamma + cosgammacosalpha + singammasinalpha = -3/2 #
# :. 2cosalphacosbeta + 2sinalphasinbeta + 2cosbetacosgamma + 2sinbetasingamma + 2cosgammacosalpha + 2singammasinalpha +3 = 0 #
And using the identity:
# sin^2A + cos^A -= 1 #
we can write:
# 2cosalphacosbeta + 2sinalphasinbeta + 2cosbetacosgamma + 2sinbetasingamma + 2cosgammacosalpha + 2singammasinalpha + sin^2alpha + cos^2alpha + sin^2beta + cos^2beta + sin^2gamma + cos^2gamma = 0 #
And we can rearrange and collect terms:
# (2cosalphacosbeta + 2cosbetacosgamma + 2cosgammacosalpha + cos^2alpha + cos^2beta + cos^2gamma) + (2sinalphasinbeta + 2sinbetasingamma + 2singammasinalpha + sin^2alpha + sin^2beta + sin^2gamma) = 0 #
Which factorises:
# (cosalpha + cosbeta + cosgamma)^2 + (sinalpha + sinbeta + singamma)^2 = 0 #
The squared terms must be greater or equal to zero, thus for the sum to be zero we have:
# cosalpha + cosbeta + cosgamma = 0 #
# sinalpha + sinbeta + singamma = 0 \ \ \ # QED