If A = 30°; show that :

(sin A - cos A)² = 1 - sin 2 A

(sin A - cos A)² = 1 - sin 2 A

$\text{L.H.S.} = \text{(sin A - cos A)}^2\\[1em] = \text{(sin 30° - cos 30°)}^2\\[1em] = \Big(\dfrac{1}{2} - \dfrac{\sqrt3}{2}\Big)^2\\[1em] = \Big(\dfrac{1 - \sqrt3}{2}\Big)^2\\[1em] = \dfrac{(1 - \sqrt3)^2}{2^2}\\[1em] = \dfrac{1 + 3 - 2\sqrt3}{4}\\[1em] = \dfrac{4 - 2\sqrt3}{4}\\[1em] = \dfrac{2(2 - 1\sqrt3)}{4}\\[1em] = \dfrac{2 - \sqrt3}{2}\\[1em]$

$\text{R.H.S.} = \text{1 - sin 2 A}\\[1em] = \text{1 - sin (2 x 30°)}\\[1em] = \text{1 - sin 60°}\\[1em] = 1 - \dfrac{\sqrt3}{2}\\[1em] = \dfrac{2}{2} - \dfrac{\sqrt3}{2}\\[1em] = \dfrac{2 - \sqrt3}{2}$

∴ L.H.S. = R.H.S.

Hence, (sin A - cos A)² = 1 - sin 2 A.