Prove that: sin2B = sin2A + sin2(A-B) – 2sinA cosB sin(A-B)

1.	Prove that: sin2B = sin2A + sin2(A-B) – 2sinA cosB sin(A-B)
Answer» 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

Answer»

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

Discussion