Question: prove sin2A*cosA + cos2A*sinA = sin4A*cosA - cos4A*sinA
Divide both sides by sinA.cosA, giving
sin2A/sinA + cos2A/cosA = sin4A/sinA - cos4A/cosA
lhs
2cosA + (2cos^2A - 1)/cosA
2cosA + 2cosA - secA
4cosA - secA
rhs
2.sin2A.cos2A/sinA - (1 - 2sin^2(2A))/cosA
4sinA.cosA.cos2A/sinA - secA + 2.(2sinA.cosA)^2/cosA
4cosA.cos2A - secA + 8.sin^2A,cosA
4.cosA(cos2A + 2sin^2A) - secA
4.cosA(1 - 2sin^2A + 2sin^2A) - secA
4.cosA(1) - secA = lhs
i.e. lhs = rhs
Q.E.D