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

$$sin(x)-cos(x)=cos(y)-sin(y)$$ $$(sin(x)-cos(x))/sqrt 2=(cos(y)-sin(y))/sqrt 2$$

$$ sin (x-pi/4) = sin (pi/4-y) $$

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$.

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

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

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

$$x=-y+frac{pi}{2}+2kpi$$ or $$x=y+pi+2kpi$$ 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 $ kin Bbb Z.$

begin{align*} sin(x)-cos(x)&=cos(y)-sin(y)Rightarrow sinleft(x-frac{pi}4right)=sinleft(frac{pi}4-yright)\ &Rightarrow x-frac{pi}4=2npi+left(frac{pi}4-yright)\ &=x+y=2npi+frac{pi}2 end{align*} Repeat for $x-frac{pi}4=npi-left(frac{pi}4-yright)$

Family of straight lines :)

Using Prosthaphaeresis Formulas

$$2sindfrac{x+y}2cosdfrac{x-y}2=cosdfrac{x+y}2cosdfrac{x-y}2$$

If $cosdfrac{x-y}2=0impliesdfrac{x-y}2=(2n+1)dfracpi2, x-y=(2n+1)pi$

else $sindfrac{x+y}2=cosdfrac{x+y}2ifftandfrac{x+y}2=1impliesdfrac{x+y}2=mpi+dfracpi4iff x+y=dfrac{(4m+1)pi}2$

So we are getting continuous perpendicular & equidistant straight lines.

In the first case, the distance between two consecutive lines is $$dfrac{2(m+1)+1-(2m+1)}{sqrt2}cdotpi$$

and in the second, $$dfrac{2pi}{sqrt2}$$

So, we get infinite number of squares with each side $=sqrt2pi$