How many ordered pairs of $(x, y$) exist in the following case?
$\cos(x+y) = 1/e$
$\cos(x-y) =1$
where both $x, y$ are less than or equal to pi.
How many ordered pairs of $(x, y$) exist in the following case?
$\cos(x+y) = 1/e$
$\cos(x-y) =1$
where both $x, y$ are less than or equal to pi.
Infinitely many, unless $x$ and $y$ are also both greater than zero (or some other lower bound). If $x$ and $y$ are both greater than zero, you can consider the second line first, which implies that $x-y=n2\pi$ for some integer $n$. Using $0\le x \le \pi$ and $0\le y \le \pi$, we have $-\pi \le x - y \le \pi$ and therefore the only possible $n$ is $n=0$. Hence $x=y$, and you can consider the first equation which looks for a value of $x$ such that $\cos 2x = 1/e$. Noting that, as $x$ takes possible values from $0$ to $\pi$, $\cos 2x$ covers the unit circle exactly one time, we note that it will intersect a vertical line placed at $1/e$ exactly twice. Thus there are two possible values of $x$, and hence two possible values of coordinates $(x,y)$ (where $x=y$) such that both equations hold.