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If $\sin(A+B) + \sin(B+C) + \cos(C-A) = \frac{3}{2}$ show that, $$1.\sin A + \cos B + \sin C = 0$$ $$2. \cos A + \sin B +\cos C = 0$$

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    @Debanjan: What have you tried?2010-11-17
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    @ Chandru1 : I have the solution for this one ... this is from my text book actually,I like the way they solved which is like breaking the LHS into sum of product of $\sin$ and $\cos$ and then writing the 3 as $ \sin ^2 + \cos ^2 $ form and then by rearranging you ultimately get $ (\sin A + \cos B + \sin C)^2 + (\cos A + \sin B +\cos C )^2 = 0$ which implies the proof.2010-11-17
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    I post it here in hope that you would show me another way of solving it ;)2010-11-17
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    @Debanjan: Please include whatever you have said in the question. So that people may not ask you the question which i asked2010-11-17
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    Is $A+B+C=\pi$?2010-11-17
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    @Américo Tavares:Nopes.2010-11-17
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    @Deb: Consider: $B=0, A=C$, such that $\sin(A) = 1/4$2010-11-17
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    @ Moron : I don't understand :|2010-11-17
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    @deb: Putting those values in shows that eqn 1 and eqn 2 need not be true. See my answer.2010-11-17
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    @Debanjan: I spent an *hour* struggling with this problem.2010-11-17

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