Question

Prove:

sin^4x – cos^4x = sin^2x – cos^2x

Step by step please.
prove it as simply as you can :)

Answer 1

\[\sin^4(x) - \cos^4(x) = \left(\sin^2(x)\right)^2 - \left(\cos^2(x)\right)^2\]Now use an identityt.

Answer 2

Identity*

Answer 3

\[x^2 - y^2 = (x + y)(x - y)\]

Answer 4

Things will begin to unfold.

Answer 5

\[\Large \sin^4x-\cos^4x\]
\[\Large (\sin^2x)^2-(\cos^2x)^2\]
you would have to know difference of two squares:

\[a^2-b^2=(a+b)(a-b)\]
\[\Large (\sin^2x+\cos^2x)(\sin^2x-\cos^x)\]
you would have to know that:
\[\sin^2x+\cos^2x=1\]
\[\Large 1(\sin^2x-\cos^2x)~~=~~\sin^2x-\cos^2x\]

Answer 6

i meant to write cos^2x in my second step

Answer 7

Thank you :)

Answer 8

Isn't that exactly what I was proving too?