Tricky trig problem If $\sin x \cos x + \sin y \cos y + \sin x \sin y + \cos x \cos y = 1$ and $\cos (x-y)$ is the smallest possible, what is the value of $2x-y$, expressed in degrees, that is closest 360º? (AMC 2012, Senior)
I tried using a number of trigonometric identities, going backwards from $\cos (x-y)$, but didn't really get anywhere. Other than that, I'm not too sure of how to approach this question.

First step is to rewrite the equation as :
$$ \sin x \cos x + \sin y \cos y + cos (x-y) = 1$$
$$ \frac{1}{2}(\sin(2x)+\sin(2y)) + \cos (x-y) = 1$$
$$ \cos (x-y) (\sin(x+y) + 1) = 1$$
Since $\cos (x-y)$ must be smallest possible, $\sin (x+y)$ must be as big as possible. Which means that $\sin (x+y) = 1$, and then $\cos (x-y)$ must be equal to $\frac {1}{2}$. Therefore,
$$ x+y = \frac{\pi}{2}$$
$$ x-y = \frac{\pi}{3}$$
$$2x-y = \frac{3\pi}{4}$$
