If $\frac{\cos x}{\cos y}+\frac{\sin x}{\sin y}=-1$, then what's the value of $4\left(\frac{\cos^3y}{\cos x}+\frac{\sin^3y}{\sin x}\right)$? I was trying to solve this problem:

If 
  $$\frac{\cos x}{\cos y}+\frac{\sin x}{\sin y}=-1$$ 
  then what is the value of 
  $$S=4\left(\frac{\cos^3y}{\cos x}+\frac{\sin^3y}{\sin x}\right)$$

I tried to solve this problem using a change of variables
$\cos^3y=a\cos x$ and $\sin^3y=b \sin x$.
So: 
$$\frac{S}{4}=a+b$$
$$\frac{\cos^6y}{a^2}+\frac{\sin^6y}{b^2}=1$$
and 
$$\frac{\cos^2y}{a} + \frac{\sin^2y}{b}=-1$$
I tried to manipulate these equations, but it gives no result 

http://comentario.fariasbrito.com.br/vest/index.php?vid=68 There you go (it's the seventh question).
It's a pretty interesting problem, but I think when you're searching problems from IME or ITA, you should first check the sites of courses that are about them if the problems aren't from before 2000 (like the one I sent and Poliedro).
@edit: A person told me that I should outline the solution here, so here I go:
$$S = \frac{(3\sin (x+y) + \sin(x-3y))}{\sin x \cos x}$$
(as found by D.R.)
Then defining $a = 3\sin (x+y) + \sin(x-3y)$, $b = \sin x \cos x$ and seeing that $\frac{\cos x}{\cos y} + \frac{\sin x}{\sin y} = -1 \Rightarrow \sin y \cdot \cos x + \sin x \cdot \cos y = -\sin y \cdot \cos y \Rightarrow 2\sin(x+y) +\sin 2y = 0 = c$ we get:
$$S = \frac ab$$
$$a-4b-2c = a-4b = 3\sin (x+y) + \sin(x-3y) - 2\sin 2x - 4\sin (x+y) - 2\sin2y = -\sin (x+y) + \sin(x-3y) - 2\sin 2x - 2\sin2y = 2\sin (-2y)\cdot \cos (x-y) - 2\cdot 2\sin (x+y) \cdot \cos(x-y)$$
Since $2\sin(x+y) = -\sin 2y$ we get:
$$a - 4b = -2\sin 2y \cdot \cos (x-y) + 2\sin 2y \cdot \cos (x-y) = 0 \Rightarrow a = 4b$$
Thus: $$S = \frac {4b}{b} \Rightarrow S = 4$$
In the link there's another solution that is bigger, but I think it should be more acessible to people that don't want the trick of using $a, b$ and $c$.
