Answer:
Given:
cos3A = 3sinA - 4sin³A
To prove:
sinAsin(60°-A)sin(60°+A) = 1/4 sin3A
Proof:
LHS = sinAsin(60°-A)sin(60°+A)
= sinA[sin(60°-A)sin(60°+A)]
= sinA[2sin30°cosA]
= sinA[2(1/2)cosA]
= sinAcosA
= 1/2sin2A
= 1/2[2sinAcosA]
= 1/2[cos(3A-A) - cos(3A+A)]
= 1/2[cos2A - cos4A]
= 1/2[1 - 2cos2Acos2A]
= 1/2[1 - 2(1/2)^2cos2A]
= 1/2[1 - 2(1/4)cos2A]
= 1/2[1 - 1/2cos2A]
= 1/2[1/2(1 - cos2A)]
= 1/4[1 - cos2A]
= 1/4[2sin^2A]
= 1/4sin^3A
= RHS
Hence proved.