Question

Given `{(sinx+siny=a),(cosx+cosy=b):}` calculate `sin((x-y)/2)=` ?

Answer 1

`sin(u+v)+sin(u-v)=2sin(u)cos(v) = a`
`cos(u+v)+cos(u-v)=2cos(u)cos(v) = b`

so

`4sin^2(u)cos^2(v) = a^2`
`4cos^2(u)cos^2(v) = b^2`

adding

`4cos^2(v) = a^2+b^2` or

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

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

but `u+v=x` and `u-v=y` so `v = (x-y)/2` and finally

`sin((x-y)/2)= pm 1/2sqrt(4-(a^2+b^2))`

Answer 2

Given

`sinx+siny=a........[1]`

`cosx+cosy=b.......[2]`

Squaring and adding [1] and [2] we get

`(sinx+siny)^2+(cosx+cosy)^2 =a^2+b^2`

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

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

`=>4-2+2(cos(x-y))=a^2+b^2`

`=>4-2[1-cos(x-y)]=a^2+b^2`

`=>4-2*2sin^2((x-y)/2)=a^2+b^2`

`=>sin^2((x-y)/2)=(4-a^2-b^2)/4`

`=>sin((x-y)/2)=pmsqrt((4-a^2-b^2)/4)=pm(sqrt(4-a^2-b^2))/2`