If $\sin(A+B) + \sin(B+C) + \cos(C-A) = \frac{3}{2}$ show that, $$1.\sin A + \cos B + \sin C = 0$$ $$2. \cos A + \sin B +\cos C = 0$$
Proving trignometrical identities:
1
$\begingroup$
trigonometry
-
0@Debanjan: What have you tried? – 2010-11-17
-
1@ 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
-
0I post it here in hope that you would show me another way of solving it ;) – 2010-11-17
-
1@Debanjan: Please include whatever you have said in the question. So that people may not ask you the question which i asked – 2010-11-17
-
1Is $A+B+C=\pi$? – 2010-11-17
-
0@Américo Tavares:Nopes. – 2010-11-17
-
0@Deb: Consider: $B=0, A=C$, such that $\sin(A) = 1/4$ – 2010-11-17
-
0@ Moron : I don't understand :| – 2010-11-17
-
0@deb: Putting those values in shows that eqn 1 and eqn 2 need not be true. See my answer. – 2010-11-17
-
0@Debanjan: I spent an *hour* struggling with this problem. – 2010-11-17