Question #62e00

Question

2145 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-26T16:39:51

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

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

I love these questions.

First you need to identify that this is the difference of two squares.

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

Now we use the identity $sin^2x+cos^2x=1$ and in the rearranged form $cos^2x=1-sin^2x$

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

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