Question

cos4x=1-8sin2x*cos2x

Answer 1

i might think that  it is

                                 cos4x=1-8sin²x.cos²x

LHS,

  cos4x

=  cos2{2x}

 = 1-2sin²2x  ......................................{cos2x=1-2sin²x}

 =1-2{sin2x}²

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

= 1-2{4sin²x.cos²x}

= 1-8sin²x.cos²x          

proved..................