Math, asked by Niscesscyru6, 1 year ago

Prove that : cosA cos2A cos4A cos8A =sin16A / 16sinA

Answers

Answered by ARoy

170

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)

Answered by safwanmuhammedsm

0

Answer:

Proved

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

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