Math, asked by bhavyayadav3986, 10 months ago

PROVE
CosAcos2Acos4Acos8A=sin16A\16sinA

Answers

Answered by rutujagajhans

6

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)

thank you!!!!

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