1
$\begingroup$

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

  • 0
    @Debanjan: I spent an *hour* struggling with this problem.2010-11-17

1 Answers 1

5

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.

  • 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