Question

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

Answer 1

Start by factoring the left side. Use the difference of squares idea, like in quadratic equations.

Answer 2

like this (sin^2x+cos^2x)(sin^2+cos^2x)?

Answer 3

Close, it's + and -.

Answer 4

and what do I do afterwards?

Answer 5

do i turn sin^2x+cos^2x into a 1?

Answer 6

yes. good thinking !

Answer 7

ok so i turned cos^2x into 1-2sin^2x but i'm left with -1-2sin^2x

Answer 8

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

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

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

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

-cos^2x=1-3sin^2x


This is what you are left with so far...

Answer 9

yeah so -1-2sin^2x =/= 1-2sin^2x , what do I do now?

Answer 10

where do you get that from ?

Answer 11

@Enlightened , can you work both sides at the same time, or you only can work with 1 side ?

Answer 12

1 side? cos^2x = 1-2sin^2x but how do I get rid of the negative?

Answer 13

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

-cos^2x=1-sin^2x THE COSINE IS NEGATIVE.

Answer 14

-1+sin^2x=1-sin^2x is certainly NOT always the case. I hate verifying, without being able to use both sides....

Answer 15

well it's the side I have to prove, i don't think you can use both sides.