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If cos(y-z)+cos(z-x)+cos (x-y)=-3/2 then prove that cosx + cos y + cos z =0 and sin x+ sin y+sinz=0?
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If cos(y-z)+cos(z-x)+cos (x-y)=-3/2 then prove that cosx + cos y + cos...

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If cos(y-z)+cos(z-x)+cos (x-y)=-3/2 then prove that cosx + cos y + cos...
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If cos(y-z)+cos(z-x)+cos (x-y)=-3/2 then prove that cosx + cos y + cos...
Solution:

Given:
cos(y-z) + cos(z-x) + cos(x-y) = -3/2

Proof:

Using trigonometric identities:
cos(y-z) + cos(z-x) + cos(x-y) = -3/2
cos(y)cos(z) + sin(y)sin(z) + cos(z)cos(x) + sin(z)sin(x) + cos(x)cos(y) + sin(x)sin(y) = -3/2

Re-arranging terms:
cos(x)(cos(y) + cos(z)) + sin(x)(sin(y) + sin(z)) = -3/2

Comparing with the given equations:
cos(x) = cos(y) + cos(z)
sin(x) = sin(y) + sin(z)

Adding the two equations:
cos(x) + sin(x) = cos(y) + cos(z) + sin(y) + sin(z)
cos(x) + sin(x) = 0

Therefore:
cos x + cos y + cos z = 0
sin x + sin y + sin z = 0

Conclusion:
Hence, it is proved that cos x + cos y + cos z = 0 and sin x + sin y + sin z = 0.
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