Prove that:
sin2B = sin2A + sin2(A-B) – 2sinA cosB sin(A-B)
RHS = sin2A + sin2(A -B) – 2 sinA cosB sin(A -B)
= sin2A + sin(A -B) [sin(A –B) – 2 sinA cosB]
We know that sin(A –B) = sinA cosB – cosA sinB
= sin2A + sin(A -B) [sinA cosB – cosA sinB – 2 sinA cosB]
= sin2A + sin(A -B) [-sinA cosB – cosA sinB]
= sin2A - sin(A -B) [sinA cosB + cosA sinB]
We know that sin(A +B) = sinA cosB + cosA sinB
= sin2A – sin(A –B) sin(A +B)
= sin2A – sin2A + sin2B
= sin2B = LHS
Hence proved.