prove sin2A*cosA + cos2A*sinA = sin4A*cosA - cos4A*sinA

sin2A = 2sinA*cosA cos2A = cosA*cosA - sinA*sinA = 1 - 2sinA*sinA = 2cosA*cosA - 1 tan2A = 2tanA/1 - tanA*tanA
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1 Answer

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

by Level 11 User (81.5k points)

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