How do you prove the identity $({sin}^{4}) x - ({cos}^{4}) x$ $(sin^4)x-(cos^4)x$ ?

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Answer 1 · 2015-09-03T22:15:53

You probably mean this one:

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

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

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

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

...using the difference of squares identity:

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

Now ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x = 1$ , so

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

$= {sin}^{2} x - {cos}^{2} x$ $= sin^2x-cos^2x$

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

$= - cos 2 x$ $= -cos 2x$

...using the double angle formula for $cos$ $cos$

How do you prove the identity (sin4)x−(cos4)x(sin^4)x-(cos^4)x?