<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] Note that
$$\cos(x + \pi/4) = \cos x \cos(\pi/4) - \sin x \sin(\pi/4) = \frac{1}{\sqrt{2}}(\cos x - \sin x) = -\frac{1}{3\sqrt{2}}.$$
Now, using a double-angle formula:
$$\cos(2x + \pi/2) = 2 \cos^2(x + \pi/4) - 1 = 2 \left( -\frac{1}{3\sqrt{2}}\right)^2 - 1 = -\frac{8}{9}.$$
On the other hand,
$$\cos(2x + \pi/2) = \cos(2x) \cos(\pi/2) - \sin(2x) \sin(\pi/2) = -\sin(2x) = -2 \sin x \cos x.$$
Equating the two, $-2 \sin x \cos x = -\frac{8}{9}$.

In general, this method of using a sum-of-angles formula in reverse is very often useful in analyzing expressions of the form $A \cos x + B \sin x$ when $A$ and $B$ are constants, and is a nice tool to keep available in your toolbox.
</s>