1
$\begingroup$

The problem is finding the real-number $ (x, y) $ of this simultaneous equation $$\begin{cases} (\cos{x})^2+(\sin{y})^2=(x-\dfrac{\pi}{2})^2 \\ (\cos{y})^2+(\sin{x})^2=(y-\dfrac{\pi}{2})^2\end{cases}$$ By adding two equations, I can assume $ (x, y) $is on the circle which is centered at $(\dfrac{\pi}{2},\dfrac{\pi}{2})$ with a radius $\sqrt{2}$, and obviously have the solution when $x=y$. But I can't show the uniqueness of the solution, or find another one. Can anybody help me to solve this problem?

$\endgroup$
0
$\begingroup$

You can solve the circle equation for $y$ and plug it into the first equation, which then gives you

$$ f(x) = (\cos(x))^2 + (\sin(\pi/2 \pm \sqrt{2 - (x-\pi/2)^2 )})^2 - (x-\pi/2)^2 $$ and you are looking for $f(x) = 0$. Here is a plot.

enter image description here

Indeed there are the only solutions $x = y = \pi/2 \pm 1 \simeq \{0.57 ; 2.57\}$. I wouldn't know an easy way of how to prove that formally though. You would need a curve discussion on $f(x)$.

$\endgroup$

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.