How do I simplify #sin^4x-2sin^2x+1#?

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Answer 1 · 2018-03-12T17:34:20

It is a perfect square:

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

Answer 2 · 2018-03-12T17:37:25

Let, #sin ^2 x=a^2# so, # sin^4 x=(sin^2x)^2 =(a^2)^2=a^4#

So,the given expression becomes,

#a^4 -2a^2 +1#

#=(a^2)^2 - 2 a^2 *1 +(1)^2#

#=(a^2 -1)^2#

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

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

#==cos^4x#

Answer 3 · 2018-03-12T17:37:41

# cos^4x#.

#sin^4x-2sin^2x+1#,

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

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

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

#={-cos^2x}^2#,

#=cos^4x#.