How do you prove ${sin}^{4} x - {cos}^{4} x = {sin}^{2} x - {cos}^{2} x$ $sin^4x-cos^4x=sin^2x-cos^2x$ ?

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How do you prove ${sin}^{4} x - {cos}^{4} x = {sin}^{2} x - {cos}^{2} x$ $sin^4x-cos^4x=sin^2x-cos^2x$ ?

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Answer 1 · 2015-05-07T19:29:57

Use the identities:

$a^{2} - b^{2} = (a - b) (a + b))$ $a^2 - b^2 = (a - b)(a + b))$
${cos}^{2} x + {sin}^{2} x = 1$ $cos^2 x + sin^2 x = 1$

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

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How do you prove ${sin}^{4} x - {cos}^{4} x = {sin}^{2} x - {cos}^{2} x$ $sin^4x-cos^4x=sin^2x-cos^2x$ ?

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How do you prove sin4x−cos4x=sin2x−cos2xsin^4x-cos^4x=sin^2x-cos^2x?

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How do you prove ${sin}^{4} x - {cos}^{4} x = {sin}^{2} x - {cos}^{2} x$ $sin^4x-cos^4x=sin^2x-cos^2x$ ?