<s>[INST] 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 
 [/INST] I don't know if this helps, but this is what I've discovered so far
The first equation can be written as
$$\cos x \sin y + \sin x \cos y = -\cos y \sin y$$
or using the double angle formula
$$\sin (x+y) = -\cos y \sin y$$
By the triple angle formulas for sine and cosine, we have that 
$$\sin^3(y)=\frac{3\sin(y)-\sin(3y)}{4}, \, \cos^3(y) = \frac{3\cos(y) + \cos(3y)}{4}$$
Plugging this in and simplifying the fractions, we get
$$S=\frac{3\cos(y)\sin(x) + \cos(3y)\sin(x) +3\sin(y)\cos (x)-\sin(3y)\cos(x)}{\cos x \sin x}$$
Rearranged, this is
$$S=\frac{3\cos(y)\sin(x) +3\sin(y)\cos (x) + \cos(3y)\sin(x) -\sin(3y)\cos(x)}{\cos x \sin x}$$
And using the sum and difference of angle formulas for sine, we get
$$S=\frac{3\sin(x+y) + \sin(x-3y)}{\cos x \sin x}$$
EDIT: Using the graphing software desmos and plugging in test values for $x,y$ that satisfy the first equation, we get $\fbox{4 }$. I guess I'll leave it up to someone smarter to prove why is works, instead of just showing that it does.

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