To prove: cos 3A = 4cos^3A - 3cosA
PROOF: cos 3A = cos( A +2A) = cosA cos2A - sinA sin2A
[using cos(C+D) = cosCcosD -sinCsinD]
Put cos2A = 2cos^2 A -1 & sin 2A = 2sinAcosA
cos3A = cosA{2cos^2 A - 1} - sinA {2sinAcosA}
cos3A = 2cos^3 A - cos A -2cosA sin^2 A
=> 2cos^3 A - cos A -2cosA(1 - cos^2 A)
=> 2cos^3 A - cos A -2cosA + 2cos^3 A
=> 4cos^3 A - 3cosA answer
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Verified answer
To prove: cos 3A = 4cos^3A - 3cosA
PROOF: cos 3A = cos( A +2A) = cosA cos2A - sinA sin2A
[using cos(C+D) = cosCcosD -sinCsinD]
Put cos2A = 2cos^2 A -1 & sin 2A = 2sinAcosA
cos3A = cosA{2cos^2 A - 1} - sinA {2sinAcosA}
cos3A = 2cos^3 A - cos A -2cosA sin^2 A
=> 2cos^3 A - cos A -2cosA(1 - cos^2 A)
=> 2cos^3 A - cos A -2cosA + 2cos^3 A
=> 4cos^3 A - 3cosA answer