 # Prove that cos4x=1-8sin^2xcos^2x 11 years ago

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

10 years ago

cos4x = cos2(2x)

2cos^2(2x) - 1

2(cos2x)^2 -1

2[(2cosx -1 )]^2 -1

2[(4cos^2(x)+1-4cox)] -1

8cos^2(x) - 8cosx +1

approve my answer by clicking yes

4 years ago
cos4x=cos2(2x)=cos^2(2x)-sin^2(2x)
=[cos^2(x)-sin^2(x)]^2-(2sinx cosx)^2
=[cos^4(x)+sin^4(x)-2cos^(x)sin^2(x)]-[4sin^2(x)cos^2(x)]
=cos^4(x)+sin^4(x)-6cos^2(x)sin^2(x)
=[cos^4(x)+sin^4(x)+2cos^2(x)cos^2(x)]-[8cos^2(x)sin^2(x)]
=[cos^2(x)+sin^2(x)]^2 - 8cos^2(x)sin^2(x)
= (1)^2 - 8sin^2(x)cos^2(x)
= 1- 8sin^2(x)cos^2(x)
HENCE PROOVED
USED FORMULAE
cos2x=cos^2(x)-sin^2(x)
Sin2x=2sinx cosx
(a+b)^2=a^2+b^2+2ab
2 years ago
You’ll have to go for half angle method :

cos4x=cos2(2x)=cos^2(2x)-sin^2(2x)
=[cos^2(x)-sin^2(x)]^2-(2sinx cosx)^2
=[cos^4(x)+sin^4(x)-2cos^(x)sin^2(x)]-[4sin^2(x)cos^2(x)]
=cos^4(x)+sin^4(x)-6cos^2(x)sin^2(x)
=[cos^4(x)+sin^4(x)+2cos^2(x)cos^2(x)]-[8cos^2(x)sin^2(x)]
=[cos^2(x)+sin^2(x)]^2 - 8cos^2(x)sin^2(x)
= (1)^2 - 8sin^2(x)cos^2(x)
= 1- 8sin^2(x)cos^2(x)
HENCE PROOVED
USED FORMULAE
cos2x=cos^2(x)-sin^2(x)
Sin2x=2sinx cosx
(a+b)^2=a^2+b^2+2ab