Q. 16 D4.2( 16 Votes )

# 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.

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