#sin(120^@+alpha)=sin{180^@-(60^@-alpha)}#,
# =sin(60^@-alpha)#,
#=sin60^@cosalpha-cos60^@sinalpha#.
# rArr sin(120^@+alpha)=1/2(sqrt3cosalpha-sinalpha)......(1)#.
Similarly, #sin(120^@-alpha)=sin(60^@+alpha)#.
#:. sin(120^@-alpha)=1/2(sqrt3cosalpha+sinalpha)......(2)#.
#"From "(1) and (2)#,
#sin^2alpha+sin^2(120^@+alpha)+sin^2(120^@-alpha)#,
#=sin^2alpha+1/4(sqrt3cosalpha-sinalpha)^2+1/4(sqrt3cosalpha+sinalpha)^2#,
#=sin^2alpha+1/4{2(3cos^2alpha+sin^2alpha)}..........[because, (a+b)^2+(a-b)^2=2(a^2+b^2)]#,
#=sin^2alpha+3/2cos^2alpha+1/2sin^2alpha#,
#=3/2sin^2alpha+3/2cos^2alpha#,
#=3/2#, as desired!
Spread the Joy of Maths.!