Simplify $sin^4x-cos^4x+cos^2x$ ?

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Answer 1 · 2017-04-17T06:55:18

The answer is $=sin^2x$

We need

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

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

$a^4-b^4=(a^2+b^2)(a^2-b^2)$

The expression is

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

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

$=(sin^2x-cancelcos^2x)+cancelcos^2x$

$=sin^2x$

Answer 2 · 2017-04-17T06:55:48

$sin^4x-cos^4x+cos^2x=sin^2x$

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

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

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

= $sin^2x$

Simplify sin^4x-cos^4x+cos^2xsin^4x-cos^4x+cos^2x?