How do you factor the expression and use the fundamental identities to simplify $1-2cos^2x+cos^4x$ ?

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Answer 1 · 2017-10-31T10:43:14

The answer is $=sin^4x$

We need

$(a-b)^2=a^2-2ab+b^2$

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

Therefore,

$1-2cos^2x+cos^4x=(1-cos^2x)-cos^2x+cos^4x$

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

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

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

$=sin^4x$

Answer 2 · 2017-10-31T14:02:50

$color(blue)(sin^4x)$

This makes use of the Pythagorean identity:

$sin^2x + sin^2x = 1$

$1-2cos^2x+cos^4x$

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

$cos^2x= 1-sin^2x$

$:.$

$1 -2(1-sin^2x)+1-2sin^2x+sin^4x$

$1 -2+2sin^2x+1-2sin^2x+sin^4x$

$1 -2+2sin^2x+1-2sin^2x+sin^4x=color(blue)(sin^4x)$

How do you factor the expression and use the fundamental identities to simplify 1-2cos^2x+cos^4x1-2cos^2x+cos^4x?