plot of $\sin(x) + \sin(y)= \cos(x) + \cos(y)$ I was playing arround with implicit plots of the form $f(x,y) = g(x,y)$, and I noticed that if you plot in the plane the following equation: $\sin(x) + \sin(y)= \cos(x) + \cos(y)$ you get the following graph:

My question is why does this trigonometric functions give us this squares spanning the entire plane?
 A: Using Prosthaphaeresis Formulas
$$2\sin\dfrac{x+y}2\cos\dfrac{x-y}2=\cos\dfrac{x+y}2\cos\dfrac{x-y}2$$
If $\cos\dfrac{x-y}2=0\implies\dfrac{x-y}2=(2n+1)\dfrac\pi2, x-y=(2n+1)\pi$
else $\sin\dfrac{x+y}2=\cos\dfrac{x+y}2\iff\tan\dfrac{x+y}2=1\implies\dfrac{x+y}2=m\pi+\dfrac\pi4\iff x+y=\dfrac{(4m+1)\pi}2$
So we are getting continuous perpendicular & equidistant straight lines.
In the first case, the distance between two consecutive lines is $$\dfrac{2(m+1)+1-(2m+1)}{\sqrt2}\cdot\pi$$
and in the second, $$\dfrac{2\pi}{\sqrt2}$$
So,  we get infinite number of squares with each side $=\sqrt2\pi$
A: $$\sin(x)+\sin(y)=\cos(x)+\cos(y)\iff$$
$$\sin(x)-\cos(x)=\cos(y)-\sin(y)\iff$$
$$\sqrt{2}\sin(x-\frac{\pi}{4})=\sqrt{2}\sin(\frac{\pi}{4}-y)\iff$$
$$x=-y+\frac{\pi}{2}+2k\pi$$
or
$$x=y+\pi+2k\pi$$
thus, there are two kind of lines :
increasing lines with equation
$$y=x+(2k+1)\pi$$
and decreasing ones
$$y=-x+(\frac 12+2k)\pi$$
where $ k\in \Bbb Z.$
A: \begin{align*}
\sin(x)-\cos(x)&=\cos(y)-\sin(y)\Rightarrow \sin\left(x-\frac{\pi}4\right)=\sin\left(\frac{\pi}4-y\right)\\
&\Rightarrow x-\frac{\pi}4=2n\pi+\left(\frac{\pi}4-y\right)\\
&=x+y=2n\pi+\frac{\pi}2
\end{align*}
Repeat for $x-\frac{\pi}4=n\pi-\left(\frac{\pi}4-y\right)$
Family of straight lines :)
A: $$\sin(x)-\cos(x)=\cos(y)-\sin(y)$$
$$(\sin(x)-\cos(x))/\sqrt 2=(\cos(y)-\sin(y))/\sqrt 2$$
$$ \sin (x-\pi/4) = \sin (\pi/4-y) $$
Shall explain for two principal inverse sine functions
$$x- \pi/4= \pi/4-y \rightarrow x+y = \pi/2$$
$$x- \pi/4= \pi-[\pi/4-y] \rightarrow y=x-\pi$$
You see these two straight lines in your plot around the origin.
Other co-terminal inverse angle with periods of $ 2 k \pi$.
