Let A=(cosA -sinA)

Question

(sinA cosA)
show that
A^2=(cos2A sin2A)
(sin2A cos2A)

Answer 1

A^2 = (cosA-sinA)^2 (sinAcosA)^2
= (cos^2A-2sinAcosA+sin^2A)(sin^2Acos^2A)
= (1-2sinAcosA)(1/4)(2sinAcosA)^2
= (1/4)(1-sin2A)(sin2A)(sin2A)

Hmmm. What if A = -pi/4?
(1/√2 + 1/√2)(-1/√2 * 1/√2) =? (0)(0)
-1/√2 =? 0
Nope. I don't think it's true. Typo?