Question

Prove that: 

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

Answer 1

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