# 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, $$x+y = \frac{\pi}{2}$$ $$x-y = \frac{\pi}{3}$$ $$2x-y = \frac{3\pi}{4}$$

The equation $\sin (x) \sin (y)+\cos (x) \cos (y)+\sin (x) \cos (x)+\sin (y) \cos (y)=1$ can be written in this way look here

$\cos(x-y) (\sin (x+y)+1)=1$

As they say that $\cos(x-y)$ is the minimum, then $\sin (x+y)+1$ is the maximum, which means that $\sin (x+y)=1$ and $\cos(x-y)=\dfrac{1}{2}$

Then we solve the systems $$\left\{x-y=\frac{\pi }{3}+2 \pi k,x+y=\pi h+\frac{\pi }{2}\right\}$$ and $$\left\{x-y=-\frac{\pi }{3}+2 \pi k,x+y=\pi h+\frac{\pi }{2}\right\}$$ which give infinitely many solutions $$x= \frac{1}{12} (6 \pi h+12 \pi k+5 \pi ),y= \frac{1}{12} (6 \pi h-12 \pi k+\pi )$$ and $$x\to \frac{1}{12} (6 \pi h+12 \pi k+\pi ),y\to \frac{1}{12} (6 \pi h-12 \pi k+5 \pi );\;k,\;h\in\mathbb{Z}$$ The value of $2x-y$ closest to 360° is $\dfrac{7\pi}{4}=315^{\circ}$

Maybe the following will help. $$1=\sin x \cos x + \sin y \cos y + \sin x \sin y + \cos x \cos y$$ $$=\frac{1}{2}(\sin2x+\sin2y)+\cos(x-y)=\cos(x-y)[1+\sin(x+y)]$$ I think we need something else in the given.