prove that ...
cosA* cos2A*cos4A = sin8A/8sinA
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Answer:
Step-by-step explanation:
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)
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