plot of $\sin(x) + \sin(y)= \cos(x) + \cos(y)$

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 $=\sqrt2\pi$

$$\sin(x)+\sin(y)=\cos(x)+\cos(y)\iff$$

$$\sin(x)-\cos(x)=\cos(y)-\sin(y)\iff$$

$$\sqrt{2}\sin(x-\frac{\pi}{4})=\sqrt{2}\sin(\frac{\pi}{4}-y)\iff$$

$$x=-y+\frac{\pi}{2}+2k\pi$$ 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.$

\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 :)