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sinx+siny+sinz=0=cosx+cosy+cosz then prove that cos(x+y)+cos(y+z)+cos(z+x)=0
cos(θ − x) + cos(θ − y) + cos(θ − z)= cos θ cos x + sin θ sin x + cos θ cos y + sin θ sin y + cos θ cos z + sin θ sin z= cos θ[cos x + cos y + cos z] + sin θ[sin x + sin y + sin z]= 0Setting θ = x + y + z , we get cos(x + y) + cos(y + z) + cos(x + z) = 0.
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