<s>[INST] If $\sin(x) - \cos(x) = 1/3$ then determine $\sin(x)\cos(x)$ 
If
  $$\sin(x) - \cos(x) = \frac{1}{3}$$
  then determine
  $$\sin(x)\cos(x)$$

I know that the expected solution is squaring both sides of equation and solving it this way:
\begin{gather}
\sin^2(x)+\cos^2(x)= 1 \\[4px]
(\sin(x) - \cos(x))^2 = \left(\frac{1}{3}\right)^2 \\[4px]
\sin^2(x) - 2\sin(x)\cos(x) + \cos^2(x) =\frac{1}{9} \\[4px]
-2\sin(x)\cos(x)=\frac{1}{9} -\sin^2(x)-\cos^2(x) \\[4px]
2\sin(x)\cos(x)=-\frac{1}{9} +\sin^2(x)+\cos^2(x) \\[4px]
2\sin(x)\cos(x)=-\frac{1}{9} +1\\[4px]
2\sin(x)\cos(x)=\frac{8}{9} \\[4px]
\sin(x)\cos(x)=\frac{4}{9}
\end{gather}
But assume I haven't noticed that I can solve it by squaring both sides in the first place. I can't figure it out how to solve it any other way.
 [/INST] You can notice that
$$
\sin x-\cos x=\sqrt{2}\sin(x-\pi/4)
$$
Set $y=x-\pi/4$; then $\sin y=\frac{1}{3\sqrt{2}}$ and
\begin{align}
\sin x\cos x
&=\sin(y+\pi/4)\cos(y+\pi/4) \\[4px]
&=\left(\frac{1}{\sqrt{2}}\cos y+\frac{1}{\sqrt{2}}\sin y\right)
  \left(\frac{1}{\sqrt{2}}\cos y-\frac{1}{\sqrt{2}}\sin y\right) \\[4px]
&=\frac{1}{2}(\cos^2y-\sin^2y) \\[4px]
&=\frac{1}{2}(1-2\sin^2y) \\[4px]
&=\frac{1}{2}\left(1-\frac{2}{(3\sqrt{2})^2}\right)
\end{align}
</s>