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

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Answer 1 · 2016-12-21T15:30:46

#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 · 2016-12-24T03:15:56

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#