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Question

Prove that:
Sin A Sin(A+2B) - Sin B Sin(B+2A) = Sin(A-B)Sin (A+B)


Solution

sina sin(a + 2b) - sinbsin(b + 2a)

⇒ 2 [ sina sin(a + 2b) - sinbsin(b + 2a) ] /2
∴ [ multiply and divide by "2" ]

⇒[ cos(a + 2b - a) - cos(a + 2b + a) - {cos(b + 2a - b) - cos(b + 2a + b) ] /2
∴ [ 2sinAsinB = cos(A - B) - cos(A + B) ]

⇒[ cos(2b) - cos(2a + 2b) - cos(2a) + cos(2a + 2b) ] / 2

⇒[ cos2b - cos2a ] /2
∴ [ cosA - cosB = -2sin(A + B)/2 * sin(A - B)/2 ]

⇒[ -2sin(2a + 2b)/2 * sin(2b - 2a)/2 ] /2

⇒[ -2sin(a + b) * (-)sin(a - b) ] /2

⇒sin(a + b) * sin(a - b)

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