If `sinx+siny=a` and `cosx+cosy=b` how do you find `x,y` ?

Given

`sinx+siny=a......(1)`

`cosx+cosy=b.....(2)`

Squaring and adding (1) and (2)

`cos^2x+sin^2x+cos^2y+sin^2y+2cosxcosy+2sinxsiny=a^2+b^2`

`=>2+2cos(x-y)=a^2+b^2`

`=>2cos(x-y)=a^2+b^2-2`

`=>x-y=cos^-1((a^2+b^2-2)/2)......(3)`

Dividing (1) by (2)

`(sinx+siny)/(cosx+cosy)=a/b`

`=>(2sin((x+y)/2)cos((x-y)/2))/ (2cos((x+y)/2)cos((x-y)/2))=a/b`

`=>tan((x+y)/2)=a/b`

`=>(x+y)/2=tan^-1(a/b)`

`=>(x+y)=2tan^-1(a/b).......(4)`

Adding (3) & (4)

`=>2x=cos^-1((a^2+b^2-2)/2)+2tan^-1(a/b)`

`=>x`
`=tan^-1(a/b) +1/2cos^-1((a^2+b^2-2)/2)`

Subtracting (3) from (4)

`=>2y`
`=2tan^-1(a/b)-cos^-1((a^2+b^2-2)/2)`

`=>y`
`=tan^-1(a/b)-1/2cos^-1((a^2+b^2-2)/2)`

Calling `u = sin x` and `v = sin y` we have

`{ (u + v = a), (sqrt(1-u^2)+sqrt(1-v^2) = b) :}`

or

`{ (u + v = a), (sqrt(1-(a-v)^2)+sqrt(1-v^2) = b) :}`

squaring the second equation

`1-a^2-v^2-2av = b^2+1-v^2-2bsqrt(1-v^2)`

or

`2bsqrt(1-v^2) = b^2+a^2-2av`

squaring again

`4b^2(1-v^2)=(b^2+a^2)^2+4a^2v^2-4(a^2+b^2)av`

and

`4(a^2+b^2)v^2-4(a^2+b^2)av+(b^2+a^2)^2-4b^2=0`

Solving for `v` we obtain

`v = 1/2 (a pm abs(b)sqrt( 4 -(a^2 + b^2)))`

and

`u = a - v = 1/2 (a -(pm abs(b)sqrt( 4 -(a^2 + b^2))))`

Finally

`y = arcsin(1/2 (a + abs(b)sqrt( 4 -(a^2 + b^2))))` and
`x = arcsin(1/2 (a - abs(b)sqrt( 4 -(a^2 + b^2))))`