# Proving trignometrical identities:

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? – anonymous Nov 17 '10 at 15:51
@ 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. – Quixotic Nov 17 '10 at 16:00
I post it here in hope that you would show me another way of solving it ;) – Quixotic Nov 17 '10 at 16:00
@Debanjan: Please include whatever you have said in the question. So that people may not ask you the question which i asked – anonymous Nov 17 '10 at 16:03
Is $A+B+C=\pi$? – Américo Tavares Nov 17 '10 at 18:13

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

1. $\sin A + \cos B + \sin C = 0$

2. $\cos A + \sin B +\cos C = 0$

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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Hey, is there any remote chance so that I could 'learn' your name (if it is not a national secret) ? – Quixotic Nov 17 '10 at 19:19
@Deb: I could tell you, but then I would have to kill you... ;-) – Aryabhata Nov 17 '10 at 22:04
Well 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 ;-) – Quixotic Nov 18 '10 at 9:42
And nice explanation, however if they gave this one in exam I am supposed to follow the method mentioned there. – Quixotic Nov 18 '10 at 9:44
@Deb: No, no topcoder handle. btw, what was the correct problem? Perhaps you can edit the question with the solution given... – Aryabhata Nov 18 '10 at 15:16