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