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

Question

13588 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-17T06:55:18

The answer is $= {sin}^{2} x$ $=sin^2x$

We need

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

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

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

The expression is

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

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

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

$=sin^2x$

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^2xsin4x−cos4x+cos2xsin^4x-cos^4x+cos^2x?