Verify Cos^2x-cos^4x=sin^2x-sin^4x?

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Answer 1 · 2018-05-10T17:32:35

Please see the proof below

We need

#cos^2x+sin^2x=1#

Therefore,

#LHS=cos^2x-cos^4x#

#=cos^2x(1-cos^2x)#

#=cos^2xsin^2x#

#=(1-sin^2x)sin^2x#

#=sin^2x-sin^4x#

#=RHS#

#QED#

Answer 2 · 2018-05-10T17:34:17

Kindly refer to Explanation.

#cos^2x-cos^4x#,

#=(1-cos^2x)cos^2x#,

#=(sin^2x)(1-sin^2x)#,

#=sin^2x-sin^4x#, as desired!

Aliter :

We have, #cos^4x-sin^4x#,

#=(cos^2x+sin^2x)(cos^2x-sin^2x)#,

#=cos^2x-sin^2x, i.e., #

# cos^4x-sin^4x=cos^2x-sin^2x," equivalently, "#

# sin^2x-sin^4x=cos^2x-cos^4x#.

Answer 3 · 2018-05-10T17:34:27

Read below

#cos^2x=1-sin^2x#
#cos^4x=(1-sin^2x)^2=1-2sin^2x+sin^4x#

#cos^2x-cos^4x=1-sin^2x-1+2sin^2x-sin^4x#
#=(1-1)+ (2sin^2x-sin^2(x))-sin^4x#