I assume the final A is indeed A. sin4A=4sinAcosAcos2A
sin4A=sin(2A+2A)=2sin2Acos2A
sin2A=2sinAcosA
sin4A=2(2sinAcosA)cos2A=4sinAcosAcos2A
Johnathan B.
asked • 10/24/14Verifying trig. Equations
sin4A=4sinAcosAcos2a
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Johnathan B.
10/24/14