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
Steve M
Mar 11, 2018

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

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