1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $\dfrac{7\pi}{4}=315^{\circ}$

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.