Question

Please help me simplify this: (sin^4x -cos^4x)/(sin^2x -cos^2x)

Answer 1

\[\frac{ \sin ^{4x}-\cos ^{4x} }{ \sin ^{2x}-\cos ^{2x} }\]

Answer 2

is that correct?

Answer 3

yes

Answer 4

oh wait no

Answer 5

\[\Large \frac{ \sin ^{4}x-\cos ^{4}x }{ \sin ^{2}x-\cos ^{2}x }\]

Answer 6

yes that is what it is teren

Answer 7

There's something to love about the difference of two squares... or two quartics, as long as the exponents are even.

Answer 8

so use difference of 2 squares on the top and bottom?

Answer 9

Yeah, but, pro-tip, don't mind the bottom. Focus on the top, factor away...

Answer 10

alright thank you very much

Answer 11

You left me hanging :P
Do me a favour... when you're done, post your answer. I'm curious :)

Answer 12

will do, and keep an eye out for the ton of identiy problems i'm stuck on

Answer 13

i got 1 as my answer

Answer 14

Very good.

Answer 15

thanks a lot! :)

Answer 16

No problem. Heck, you did most of the work yourself ^.^