Sineful mistake

$sinx + siny = sin(x + y)$

$\Rightarrow 2sin\Big(\frac{x + y}{2}\Big)cos\Big(\frac{x - y}{2}\Big) - 2sin\Big(\frac{x + y}{2}\Big)cos\Big(\frac{x + y}{2}\Big) = 0$

$\Rightarrow sin\Big(\frac{x + y}{2}\Big)\Bigg(cos\Big(\frac{x - y}{2}\Big) - cos\Big(\frac{x + y}{2}\Big)\Bigg) = 0$

$\Rightarrow sin\Big(\frac{x + y}{2}\Big) sin\Big(\frac{x}{2}\Big)sin\Big(\frac{y}{2}\Big) = 0$

Case - 1 : $x = y = 0^{\circ}$ , or $x = y = 360^{\circ}$ , $2$ pairs.

Case - 2 : $x = 0^{\circ} , y \neq 0^{\circ}$ , or $x = 360^{\circ} , y \neq 360^{\circ}$ $2 \times 360 = 720$ pairs .

Case - 3 : $y = 0^{\circ} , x \neq {0^{\circ} , 360^{\circ}}$ , or $y = 360^{\circ} , x \neq {0^{\circ} , 360^{\circ}}$ , $2 \times 359 = 718$ pairs.

Case - 4 : $x,y \neq 0^{\circ}, 360^{\circ} , x + y = 360^{\circ}, 359$ pairs.

Summing up , total number of ordered pairs = $2 + 720 + 718 + 359$ = $\fbox{1799}$

The equation is equivalent to

$M-M_0 \in \mathbb{R}$

where $M$ and $M_0$ are the "midpoints" defined by

$M = ($ cis $x ^\circ +$ cis $y ^\circ )/2$

$M_0 = (1+$ cis $( x + y) ^\circ )/2$

(where cis $\theta$ means $\cos \theta + i \sin \theta$ , or $e^{i\theta}$ assuming $\theta$ in radians.)

But $M$ and $M_0$ are both on the line through $0$ and cis $\left ( \frac{ x + y}{2}\right) ^\circ$ (possibly on the "negative" side of it). So if $x+y$ is a multiple of $360$ then $M-M_0 \in \mathbb{R}$ holds automatically. If not, $M-M_0 \in \mathbb{R}$
becomes $M=M_0$ . But two distinct chords of a circle cannot have the same midpoint, unless they're diameters. That is to say, $M=M_0$ if and only if $1$ is the same as either cis $x ^\circ$ or cis $y ^\circ$ .

(In the special case that the chords are diameters, cis $x ^\circ = -$ cis $y ^\circ$ and

$-1 =$ cis $(x + y) ^\circ = - ($ cis $x ^\circ)^2$

so cis $x ^\circ = \pm 1$ , and we again have $1 \in \{$ cis $x ^\circ$ , cis $y ^\circ \}$ . We should also consider the possibility of a degenerate "chord", i.e. that the "endpoints" are actually the same point. But in that case we would have to have $1 =$ cis $x ^\circ =$ cis $y ^\circ$ .)

So what we have shown is that $M-M_0 \in \mathbb{R}$ if and only if $\{x,y,x+y\}$ contains at least one multiple of $360$ . (This can also be shown with a nice trigonometric/algebraic argument, but I thought it might interesting to post a complex/geometric method.)

Now consider the requirement that $x,y \in \{0,1,...,360\}$ . If $x \in \{0,360\}$ , then no additional restriction on $y$ is needed; if $x \in \{1,...,359\}$ , then $y \in \{0,360,360-x\}$ (which are in fact three distinct numbers). So the number of pairs is: $2(361)+359(3)=1799$ and our final answer is $799$ .

sin x + sin y = sin (x+y) 2 sin (x+y)/2 cos (x-y)/2 = 2 sin (x+y)/2 cos (x+y)/2 sin (x+y)/2 (cos (x-y)/2 - cos (x+y)/2 ) = 0 sin (x+y)/2 (-2sin x/2 sin y/2) = 0 Hence sin (x+y)/2 = 0 and (x+y)/2 = 0 or 180 and x+y = 0 or 360 (361 solutions) sin x/2 = 0 and x = 0 or 360 and y any integer in the range (361 x 2 = 722 solutions) sin y/2 = 0 and y= 0 or 360 and y any integer in the range (722 solutions) However, do note that the roots (0,0) and (360, 360) have been repeated twice and the roots (0, 360) and (360, 0) have been repeated thirce. So we have 2(2-1+3-1) = 6 repeated roots. Hence N = 361+ 722 +722 - 6 = 1799. Last 3 digits of N = 799

sin x + sin y = sin (x+y) 2 sin (x+y)/2 cos (x-y)/2 = 2 sin (x+y)/2 cos (x+y)/2 sin (x+y)/2 (cos (x-y)/2 - cos (x+y)/2 ) = 0 sin (x+y)/2 (-2sin x/2 sin y/2) = 0 Hence sin (x+y)/2 = 0 and (x+y)/2 = 0 or 180 and x+y = 0 or 360 (361 solutions) sin x/2 = 0 and x = 0 or 360 and y any integer in the range (361 x 2 = 722 solutions) sin y/2 = 0 and y= 0 or 360 and y any integer in the range (722 solutions) However, do note that the roots (0,0) and (360, 360) have been repeated twice and the roots (0, 360) and (360, 0) have been repeated thirce. So we have 2(2-1+3-1) = 6 repeated roots. Hence N = 361+ 722 +722 - 6 = 1799. Last 3 digits of N = 799

$sin x + sin y$ = $sin (x+y)$ ;
$2sin ((x+y)/2)cos ((x-y)/2)$ = $2sin ((x+y)/2)cos ((x+y)/2)$ ; $sin((x+y)/2) [cos((x-y)/2) - cos ((x+y)/2)] = 0$
$sin((x+y)/2) [2sin(x/2)sin(y/2)] = 0$ ;
$sin((x+y)/2) = 0$ ; or $sin(x/2) = 0$ ; or $sin(y/2) = 0$ ;
$x+y =0,360,720$ ; or $x = 0,360$ ; or $y = 0,360$ ;

  When x = 0,  y can take 361 values   ;                                                                                            

  When x = 360, y can take 361 values ;                                                                          

 When y = 0  ,  x can take 361 values (already counted 0 and 360)  ;                     

 When y = 360 , x can take 361 values (already counted 0 and 360) ;                            

 x+y=0   is possible for x=0 , y=0 (already counted ) ;   

 x+y=720  is possible for x=360 , y=360 (already counted ) ;  

 x+y=360 , (x,y) = (0,360) , (1,359), ... , (359,1), (360,0) . { already counted (0,360) and (360,0) }  ;
     Hence 359 possible pairs.

 Total number of ordered pairs of integers (x,y) = 2*361 + 3*359 
                               = 1799