Question

Show that :- Sin²A + Sin²B + 2sinA SinB Cos(A+B) = Sin²(A+B)

Answer 1

Applying 2sin(A)*cos(B) = sin(A + B) + sin(A - B), Right side is:

= sin²A + sin²(A - B) - sin(A - B){sin(A + B) + sin(A - B)}

= sin²A + sin²(A - B) - sin(A + B)*sin(A - B) - sin²(A - B)

= sin²A - {sin(A + B)*sin(A - B)}

= sin²A - sin²A + sin²B [Since sin(A + B)*sin(A - B) = sin²A - sin²B]

= sin²B = Left side [Proved]

Proof for sin(A + B)*sin(A - B) = sin²A - sin²B:

sin(A + B)*sin(A - B) = {sin(A)*cos(B) + cos(A)*sin(B)}*{sin(A)*cos(B) - cos(A)*sin(B)}

= sin²A*cos²B - cos²A*sin²B [Applying (a + b)(a - b) = a² - b²]

= sin²A(1 - sin²B) - (1 - sin²A)sin²B

= sin²A - sin²A*sin²B - sin²B + sin²A*sin²B

= sin²A - sin²B