Question

can anyone answer this:cos4x=1-8sin^2xcos^x spammers will be reported....​

Answer 1

cos4x

= cos2(2x) =1-2sin2(2x)

=1-2(2sinx . cosx)2

{ sin2x = 2sinx. cosx}

=1-2(4sin2x.cos2x)

=1-8sin2x.cos2x

LHS=RHS

Answer 2

you can solve

cos4x as

cos2(2x) =1-2sin2(2x)

=1-2(2sinx . cosx)2 { sin2x = 2sinx. cosx}

=1-2(4sin2x.cos2x)

=1-8sin2x.cos2x

LHS=RHS

Step-by-step explanation:

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