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:
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0@Debanjan: What have you tried? – 2010-11-17
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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
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0I post it here in hope that you would show me another way of solving it ;) – 2010-11-17
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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
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1Is $A+B+C=\pi$? – 2010-11-17
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0@Américo Tavares:Nopes. – 2010-11-17
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0@Deb: Consider: $B=0, A=C$, such that $\sin(A) = 1/4$ – 2010-11-17
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0@ Moron : I don't understand :| – 2010-11-17
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0@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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0@Debanjan: I spent an *hour* struggling with this problem. – 2010-11-17
1 Answers
The problem seems to be missing some assumptions, as noted by Americo.
For instance, If $B = 0$ and $A=C$ are acute angles, such that $\sin A = 1/4$ the we have that
$\sin(A+B) + \sin(B+C) + \cos(A-C) = 3/2$, but none of
$\sin A + \cos B + \sin C$ or $\cos A + \sin B + \cos C$ are $0$.
In any case, this looks like a perfect problem for using complex numbers.
If $B' = \pi/2 - B$ and
$z_1 = \cos A + i \sin A$
$z_2 = \cos B' + i \sin B'$
$z_3 = \cos C + i \sin C$
The given identity is $\cos (A-B') + \cos (C - B') + \cos (A-C) = 3/2$
i.e.
$$\frac{z_1}{z_2} +\frac{z_2}{z_1} +\frac{z_3}{z_2} +\frac{z_2}{z_3} +\frac{z_1}{z_3} +\frac{z_3}{z_1} = 3$$
The two identities
$\sin A + \cos B + \sin C = 0$
$\cos A + \sin B +\cos C = 0$
are equivalent to showing that $z_1 + z_2 + z_3 = 0$.
Eq 1, says that the imaginary part of $z_1 + z_2 + z_3$ is $0$ and Eq 2 says that the real part is $0$.
Hope that helps.
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1Hey, is there any remote chance so that I could 'learn' your name (if it is not a national secret) ? – 2010-11-17
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1@Deb: I could tell you, but then I would have to kill you... ;-) – 2010-11-17
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0Well I can pretty much sure that you are someone who has some good understanding of mathematics and Algorithms ?! And it will not surprise me if you have at-least a yellow handle in topcoder ;-) – 2010-11-18
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0And nice explanation, however if they gave this one in exam I am supposed to follow the method mentioned there. – 2010-11-18
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1@Deb: No, no topcoder handle. btw, what was the correct problem? Perhaps you can edit the question with the solution given... – 2010-11-18