Does not Compute

[Walter]

Can someone explain to me how the two fractions are equal? I don't see how (sin^2x)^2 - (cos^2x)^2 = (sin^2x + cos^2x) (sin^2x - cos^2x).

 

[pka]

Can someone explain to me how the two fractions are equal? I don't see how (sin^2x)^2 - (cos^2x)^2 = (sin^2x + cos^2x) (sin^2x - cos^2x).
View attachment 1789

Are you really telling the world that you do not understand that:
\(\displaystyle a^4-b^4=(a^2+b^2)(a^2-b^2)~?\)

 

[Walter]

Not the world, per se.

It's a perfect square, I see it now...

 

[Walter]

No, it is NOT a perfect square. It is a difference of squares just as the problem indicated.

Thanks for clearing that up, I haven't taken math in a semester and I'm trying to refresh myself by getting all my stupid questions out of the way over the internet.