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

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Answer 1 · 2015-09-03T22:15:53

You probably mean this one:

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

$sin^4x-cos^4x$

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

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

...using the difference of squares identity:

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

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

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

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

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

$= -cos 2x$

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

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