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

Answer 1

Daniel Acosta

It is a perfect square:

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

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Answer 2

Dylan Arnold

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#

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Answer 3

James Adkins

# 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#.

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Answer 4

Aaliyah Abdullah

To simplify ( \sin^4(x) - 2\sin^2(x) + 1 ), you can use the identity ( \sin^2(x) = 1 - \cos^2(x) ).

Replace ( \sin^2(x) ) with ( 1 - \cos^2(x) ):

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

Expand and simplify:

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

Combine like terms:

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

This is the simplified form of ( \sin^4(x) - 2\sin^2(x) + 1 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.