## Guest

Ajay
209 Points
6 years ago
Hi Riya
You can use the formula for Sum and Difference for Sin to get the required solution. Have  look below and hope you understand the derivation
${ sin }^{ 2 }A-{ sin }^{ 2 }B\quad =\quad (sinA+sinB)(sinA-sinB)\\ Using\quad formula\quad for\quad sum\quad and\quad difference\quad for\quad sin\\ =\quad \left( 2cos\frac { (A+B) }{ 2 } sin\frac { (A-B) }{ 2 } \right) \left( 2sin\frac { (A+B) }{ 2 } cos\frac { (A-B) }{ 2 } \right) \\ Rearranging\quad \\ =\quad \left( 2sin\frac { (A+B) }{ 2 } cos\frac { (A+B) }{ 2 } \right) \left( 2sin\frac { (A-B) }{ 2 } cos\frac { (A-B) }{ 2 } \right) \\ We\quad know\quad \quad 2sinxcosx\quad =\quad sin2x\\ =sin(A+B)sin(A-B)$

riya
10 Points
6 years ago
Thank you .I got it now. I was not equating the second last LHS to sin2x but now I got it thanks a lot .I need to concentrate more.