How do you prove ${cos}^{4} x - {sin}^{4} x = 1 - 2 {sin}^{2} x$ $cos^4x - sin^4x = 1 - 2sin^2x$ ?

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Answer 1 · 2016-04-18T21:47:28

Start by factoring the left side as a difference of squares.

${cos}^{4} x - {sin}^{4} x = 1 - 2 {sin}^{2} x$ $cos^4x - sin^4x = 1 - 2sin^2x$

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

Now, applying the pythagorean identity ${cos}^{2} x + {sin}^{2} x = 1$ $cos^2x + sin^2x = 1$ :

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

Rearranging the previously stated pythagorean identity to solve for sin:

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

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

Substituting:

$1 (1 - {sin}^{2} x - {sin}^{2} x) =$ $1(1 - sin^2x - sin^2x) =$

$1 - 2 {sin}^{2} x = 1 - 2 {sin}^{2} x \to$ $1 - 2sin^2x = 1 - 2sin^2x ->$ identity proved

Hopefully this helps!

How do you prove cos^4x - sin^4x = 1 - 2sin^2xcos4x−sin4x=1−2sin2xcos^4x - sin^4x = 1 - 2sin^2x?