prove that 4cosA (60°-A) cos (60°-A) =cos3A
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cos(A-B)cos(A+B) = cos2B- sin2A ==> L.H.S = 4cosA (cos2A - sin260) = 4cosA (cos2A - 3/4)
=4cos3A -3cosA = cos3A =R.H.S .Hence proved
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