Expressing trig identities

Express:

(cosx - sinx)² in terms of sin2x
I expanded the brackets and got the answer to just be cos2x, I cant seem to think how I could express it in terms of sin2x.

cos⁴x - sin⁴x in terms of cos2x
I got the answer to be cos²(2x) but the book says the answer is just cos2x

Not quite correct. Expansion is $\cos^2x - 2\cos x \sin x + \sin^2 x$

Secondly, note that $\cos^4 x - \sin^4 x \neq (\cos^2 x - \sin^2 x )^2$. Rather, $\cos^4 x - \sin^4 x=(\cos^2x - \sin^2 x )(\cos^2 x + \sin^2 x)$

so (cos²x - sin²x)(cos²x + sin²x) = (cos(2x))(cos(2x)) = cos²(2x)?

How do you do that fancy text again?

Express:

(cosx - sinx)² in terms of sin2x

cos⁴x - sin⁴x in terms of cos2x

a) You'd get cos^2(x) + sin^2(x) - 2sin(x)cos(x). What do you notice about cos^2(x) + sin^2(x) and the -2sin(x)cos(x). What identities are these?

b) cos⁴x - sin⁴x = [cos^2(x) + sin^2(x)][cos^2(x)-sin^2(x]. Look at the 2 brackets. You know the first equals 1 and the second equals cos(2x). So 1cos(2x) = ... ?