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))))#