Square, Then Double

Squaring a sine and a cosine together gives the two squares, which make 1, and a cross term 2 sin x cos x, which is sin 2x. Take ${\left(\sin\left(x\right)-\cos\left(x\right)\right)}^{2}$. Expand it: ${\sin\left(x\right)}^{2}-2\sin\left(x\right)\cos\left(x\right)+{\cos\left(x\right)}^{2}$. Fold the cross term into $-\sin\left(2x\right)$ and the squares into 1, and the line reads $1-\sin\left(2x\right)$.

That says something about sizes: a square is never negative, so 1 - sin 2x is never negative, which is why sin 2x is never more than 1.