#cos(a-b)+cos(b-c)+cos(c-a)=-3/2#
#=>cosacosb+sinasinb+cosbcosc+sinbsinc+cosc cosa+sincsina=-3/2#
#=>2cosacosb+2sinasinb+2cosbcosc+2sinbsinc+2cosc cosa+2sincsina+3=0#
#=>2cosacosb+2sinasinb+2cosbcosc+2sinbsinc+2cosc cosa+2sincsina+sin^2a+cos^2a+sin^2b+cos^2b+sin^2c+cos^2c=0#
#=>(cosa+cosb+cosc)^2+(sin a+sin b+sinc)^2=0 #
Sum of two squared quantities being zero each of them must be zero.
Hence #cosa+cosb+cosc=0......[1]#
and
#sin a+sin b+sinc=0 .......[2]#
Now from [1] and [2] we get
#(cosa+cosb)^2+(sin+sinb)^2=(-cosc)^2+(-sinc)^2#
#=>cos^2a+cos^2b+2cosacosb+sin^2a+sin^2b+2sinasinb=cos^2c+sin^2c#
#=>2+2cos(a-b)=1#
#=>2cos(a-b)=1-2=-1#
#=>cos(a-b)=-1/2#
Similarly we can have
#cos(b-c)=-1/2 and cos(c-a)=-1/2#
Hence it is proved that
#cos(a-b)=cos(b-c)=cos(c-a)=-1/2#
