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

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Answer 1 · 2016-04-18T21:47:28

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

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

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

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

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

Rearranging the previously stated pythagorean identity to solve for sin:

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

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

Substituting:

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

$1 - 2sin^2x = 1 - 2sin^2x ->$ identity proved

Hopefully this helps!

How do you prove cos^4x - sin^4x = 1 - 2sin^2xcos^4x - sin^4x = 1 - 2sin^2x?