If, #Cosx+Cosy=a# and #Sinx+Siny=b# So Prove That ?? #Sin2x+Sin2y = 2ab{1-(2/(a^2+b^2))}#

1 Answer
May 31, 2018

Given that,

#rarrcosx+coy=a#....[1]

#rarrsinx+siny=b#....[2]

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

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

#rarr2+2(cosxcosy+sinxsiny)=a^2+b^2#

#rarr2(1+cos(x-y))=a^2+b^2#

#rarrcos(x-y)=(a^2+b^2)/2-1#

Dividing equation [1] by [2], we get,

#rarr(cosx+cosy)/(sinx+siny)=a/b#

#rarr(2cos((x+y)/2)cos((x-y)/2))/(2sin((x+y)/2)cos((x-y)/2))=a/b#

#rarrcot((x+y)/2)=a/b#

#rarrtan((x+y)/2)=b/a#

#rarr(x+y)/2=tan^(-1)(b/a)#

#rarrx+y=2tan^(-1)(b/a)#

As, #2tan^(-1)x=sin^(-1)((2x)/(1+x^2))#,we have,

#rarrx+y=sin^(-1)((2*(b/a))/(1+(b/a)^2))=sin^(-1)((2ab)/(a^2+b^2))#

#rarrsin(x+y)=(2ab)/(a^2+b^2)#

Now,

#LHS=sin2x+sin2y#

#=2sin(x+y)*cos(x-y)#

#=2[(2ab)/(a^2+b^2)][(a^2+b^2)/2-1]#

#=2ab[2/(a^2+b^2)*(a^2+b^2)/2-2/(a^2+b^2)]#

#=2ab[1-2/(a^2+b^2)]=RHS#