How do you prove (sinx+cosx)^2 + (sinx-cosx)^2 = 2?

${\sin}^{2} \left(x\right) + 2 \sin \left(x\right) \cos \left(x\right) + {\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right) - 2 \sin \left(x\right) \cos \left(x\right) + {\cos}^{2} \left(x\right) = 2$
simplifying the $2 \sin \left(x\right) \cos \left(x\right)$ and considering that ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$ you get:
${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 2$
$1 + 1 = 2$