4. Calculate $\sin 2x$, if $\sin x - \cos x = \frac{1}{2}$.
4. We write the relation for $\sin 2 x=2 \sin x \cos x$ and $(\sin x-\cos x)^{2}=\sin ^{2} x-2 \sin x \cos x+\cos ^{2} x$,
where we substitute $\sin x-\cos x=\frac{1}{2}$. We get the equation $\left(\frac{1}{2}\right)^{2}=1-2 \sin x \cos x$. We calculate
$2 \sin x \cos x=\frac{3}{4}=\sin 2 x$.
Writing $\sin 2 x=2 \sin x \cos x$ ..... 1 point
Squaring $(\sin x-\cos x)^{2}=\sin ^{2} x-2 \sin x \cos x+\cos ^{2} x$ ..... 1 point
Using $\sin ^{2} x+\cos ^{2} x=1$ ..... 1 point
Writing the equation $\left(\frac{1}{2}\right)^{2}=1-2 \sin x \cos x$ ..... 1 point
Calculating $2 \sin x \cos x=1-\left(\frac{1}{2}\right)^{2}=\frac{3}{4}$ ..... 1 point
Writing the solution $\sin 2 x=\frac{3}{4}$ ..... 1 point