<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] To save typing, let $s = \sin x$ and $c = \cos x$.  Then we have two equations in two variables:
$$
\begin{align}
s^2 + c^2 &= 1\\
s - c &= \frac13.
\end{align}
$$
A widely applicable approach in such a situation is to solve one equation for one of the variables, and use that result to substitute into the other.  Here, the second equation can easily be solved for $s$, yielding $s = c + \frac13$.  Substituting for $s$ in the first equation gives a quadratic equation in $c$:
$$
2c^2 + \frac23 c - \frac89 = 0.
$$
The roots of this quadratic are
$$
c = \frac{-1+\sqrt{17}}{6}, \frac{-1-\sqrt{17}}{6}.
$$
Since $s = c + \frac13$, the corresponding values for $s$ are
$$
s = \frac{1+\sqrt{17}}{6}, \frac{1-\sqrt{17}}{6}.
$$
Multiplying either pair of corresponding values gives the result $sc = \frac49$.
This method is much more laborious than the original solution, but requires no cleverness or flash of insight.
</s>