(sinθ-cosθ)^2
= sin^2 θ-2sinθcosθ-cos^2 θ
is this all?? what else can i do?
Not sure what you are trying to do but
$\displaystyle (\sin\theta-\cos\theta)^2 = (\sin\theta-\cos\theta)(\sin\theta-\cos\theta)= \sin^2\theta-2\sin\theta\cos\theta+\cos^2\theta $ $\displaystyle \neq \sin^2\theta-2\sin\theta\cos\theta-\cos^2\theta
$
Also $\displaystyle \cos^2\theta = 1-\sin^2\theta$
and $\displaystyle 2\sin\theta\cos\theta = \sin 2\theta$
Good luck!
two identities involved here ...
(1) the basic Pythagorean identity ...
$\displaystyle \sin^2{x} + \cos^2{x} = 1$
(2) the double angle identity for sine ...
$\displaystyle \sin(2x) = 2\sin{x}\cos{x}$
...
$\displaystyle \sin^2{\theta} - 2\sin{\theta}\cos{\theta} + \cos^2{\theta} =$
$\displaystyle 1 - 2\sin{\theta}\cos{\theta} =$
$\displaystyle 1 - \sin(2\theta)$