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.
 A: 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,
$$ x+y = \frac{\pi}{2}$$
$$ x-y = \frac{\pi}{3}$$
$$2x-y = \frac{3\pi}{4}$$
A: 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 $\dfrac{7\pi}{4}=315^{\circ}$
A: 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.
