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if cos(y-z) + cos(z-x)+ cos(x-y) = -3/2, prove that cosx + cosy +cosz = 0 = sinx + siny+ sinz

if cos(y-z) + cos(z-x)+ cos(x-y) = -3/2, prove that cosx + cosy +cosz = 0 = sinx + siny+ sinz

Grade:11

1 Answers

Aarushi Ahlawat
41 Points
5 years ago

cos (x-y) + cos (y-z) + cos (z-x) = -3/2
=3+2(cos (x-y) + cos (y-z) + cos (z-x))=0
=1+1+1+2(cos (x-y) + cos (y-z) + cos (z-x))=0
=(sin2x + cos2x) + (sin2y + cos2y) + (sin2z + cos2z) + 2(cosx cosy + sinx siny + cosy cosz + siny sinz + cosz cosx + sinz sinx)=0
=(sin2x + sin2y + sin2z+2(sinx siny + siny sinz + sinz sinx))+(cos2x + cos2y + cos2z +2(cosx cosy + cosy cosz + cosz cosx))=0
=(sinx + siny + sinz)2 + (cosx + cosy + cosz)2=0
only and only iff
sinx + siny + sinz=0
cosx + cosy + cosz=0

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