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.

You can use $$\sin x = \frac{2t}{1+t^2}; \cos x = \frac {1-t^2}{1+t^2}$$ 
where $t=\tan (x/2)$, but you just have to know that some value of $t$ works.
Then you get a quadratic to solve for $t$. And once you know $t$ you can solve the problem.
