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

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Answer 1 · 2015-05-22T06:48:45

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

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

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

$= ({cos}^{2} x - {sin}^{2} x) \times 1 = ({cos}^{2} x - {sin}^{2} x)$ $= (cos^2x-sin^2x)xx1 = (cos^2x-sin^2x)$

based on the identities:

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

${cos}^{2} x + {sin}^{2} x = 1$ $cos^2x+sin^2x = 1$ from Pythagoras (right angled triangle with hypotenuse of length 1).

How do you verify cos4x−sin4x=cos2x−sin2xcos^4x - sin^4x = cos^2x - sin^2x?