Question

sinx + siny =a and cosx +cost =b then prove that sin(x+y) =2ab/a^2+b^2​

Answer 1

We, know that sin(x+y) = sinxcosy + sinycosx

First, multiply the 2 equations,

ab = (sinx + siny)(cosx + cosy)

    = sinxcosx + (sinxcosy + sinycosx) + sinycosy

    = ½sin(2x) + sin(x+y) + ½sin(2y)

{ since,  sin(2x) = 2 sinxcosx  }

    = sin(x+y) + ½[sin(2x) + sin(2y)]    

{ since , sinx + siny = 2sin[(x+y}/2]cos[(x-y)/2]  }

    = sin(x+y) + sin(x+y)cos(x-y)

    = sin(x+y) [1 + cos(x-y)]  ---------> 1

 

a² = (sinx + siny)²   = sin²x + 2sinxsiny  + sin²y

b² = (cosx + cosy)² = cos²x + 2cosxcosy + cos²y

a² + b² = 2 + 2 cos(x-y)

           = 2 [1 + cos(x-y)] ----------> 2

From (1) and (2)

sin(x+y) = 2ab/(a² + b²)

Hence proved.