cosa.cos2A.cos4Acos8A=sin16A÷16SINA
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Gaurav Seth 6 years, 3 months ago
CosAcos2Acos4Acos8A
=1/2sinA[(2sinAcosA)cos2Acos4Acos8A]
=1/2sinA(sin2Acos2Acos4Acos8A)
=1/4sinA[(2sin2Acos2A)cos4Acos8A]
=1/4sinA(sin4Acos4Acos8A)
=1/8sinA[(2sin4Acos4A)cos8A]
=1/8sinA(sin8Acos8A)
=1/16sinA(2sin8Acos8A)
=1/16sinA(sin16A)
=sin16A/16sinA (Proved)
3Thank You