<s>[INST] Triangle proof using law of sines 
In triangle $ABC$, suppose that angle $C$ is twice angle $A$. Use the law of sines to show that $ab= c^2 - a^2$.

 [/INST] A common way to invoke the Law of Sines is to note what its ratios equal:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = d$$
where $d$ is the circum-diameter of $\triangle ABC$. Thus, 
$$a = d \sin A \qquad b = d \sin B \qquad c = d \sin C$$
Consequently, the identity in question, upon division by $d^2$, is equivalent to
$$\sin A \sin B \stackrel{?}{=} \sin^2 C - \sin^2 A$$
Now, $C=2A$ implies $B=\pi-3A$, so that $\sin B = \sin 3A$, and the identity to be proven becomes
$$\begin{align}
\sin A \sin 3A &\stackrel{?}{=} \sin^2 2A - \sin^2 A \\[6pt]
&= \left( \sin 2A + \sin A \right) \left( \sin 2 A - \sin A \right) \\[6pt]
&= 2 \sin\frac{3A}{2}\cos\frac{A}{2} \cdot 2 \sin\frac{A}{2} \cos\frac{3A}{2} \\[6pt]
&= 2 \sin\frac{A}{2}\cos\frac{A}{2} \cdot 2 \sin\frac{3A}{2} \cos\frac{3A}{2} \\[6pt]
&= \sin A \; \sin 3A
\end{align}$$
In the above, I use the sum-to-product identity $\sin\theta \pm \sin\phi = 2 \sin\frac{\theta\pm \phi}{2} \cos\frac{\theta\mp \phi}{2}$, which is familiar to me. Someone familiar with multiple-angle identities may have proceeded thusly:
$$\begin{align}
\sin A \sin 3A &\stackrel{?}{=} \sin^2 2A - \sin^2 A \\[6pt]
&= 4\sin^2 A\cos^2 A - \sin^2 A \\[6pt]
&= \sin A \cdot \sin A \left( 4 \cos^2 A - 1 \right) \\[6pt]
&= \sin A \cdot \sin A \left( 3 - 4\sin^2 A \right) \\[6pt]
&= \sin A \; \sin 3A
\end{align}$$
</s>