First use $\sin(180^{\circ}-x)=\sin x$ so :
$$a=\sin 20^{\circ} \sin 40^{\circ} \sin 70^{\circ} $$
$$a=\sin 20^{\circ} \sin 40^{\circ} \cos 20^{\circ}$$
Now use $\sin x \cos x=\frac{\sin(2x)}{2}$ :
$$a=\frac{\sin^2 40^{\circ}}{2}$$ and this hasn't a nice form .
If you have angle $\theta$ in quadrant $1$, you can find its "corresponding" angle in quadrant $2$ by $(\pi - \theta)$, in quadrant $3$ by $(\pi+\theta)$, and in quadrant $4$ by $(2\pi-\theta)$. For example, $\frac{\pi}{4}$ corresponds to $\frac{3\pi}{4}$, $\frac{5\pi}{4}$, and $\frac{7\pi}{8}$ in quadrants $2$, $3$, and $4$, respectively. (That's how I always think of them at least.)
Also, recall sine functions correspond to the height of the right triangle ($y$-axis), so they are positive in quadrants $1$ and $2$. Cosine functions correspond to base of the right triangle ($x$-axis), so they are positive in quadrants $2$ and $4$. (Tangent functions can be found through sine and cosine functions.)
You can use the following identities (which are derived from the aforementioned facts).
$$\sin\bigg(\frac{\pi}{2}+\theta\bigg) = \cos\theta \quad \sin\bigg(\frac{\pi}{2}-\theta\bigg) = \cos\theta$$
$$\cos\bigg(\frac{\pi}{2}+\theta\bigg) = -\sin\theta \quad \cos\bigg(\frac{\pi}{2}-\theta\bigg) = \sin\theta$$
$$\tan\bigg(\frac{\pi}{2}+\theta\bigg) = -\cot\theta \quad \tan\bigg(\frac{\pi}{2}-\theta\bigg) = \cot\theta$$
$$\sin\bigg(\pi+\theta\bigg) = -\sin\theta \quad \sin\bigg(\pi-\theta\bigg) = \sin\theta$$
$$\cos\bigg(\pi+\theta\bigg) = -\cos\theta \quad \cos\bigg(\pi-\theta\bigg) = -\cos\theta$$
$$\tan\bigg(\pi+\theta\bigg) = \tan\theta \quad \tan\bigg(\pi-\theta\bigg) = -\tan\theta$$
$$\sin\bigg(\frac{3\pi}{2}+\theta\bigg) = -\cos\theta \quad \sin\bigg(\frac{3\pi}{2}-\theta\bigg) = -\cos\theta$$
$$\cos\bigg(\frac{3\pi}{2}+\theta\bigg) = \sin\theta \quad \cos\bigg(\frac{3\pi}{2}-\theta\bigg) = -\sin\theta$$
$$\tan\bigg(\frac{3\pi}{2}+\theta\bigg) = -\cot\theta \quad \tan\bigg(\frac{3\pi}{2}-\theta\bigg) = \cot\theta$$
$$\sin\bigg(2\pi+\theta\bigg) = \sin\theta \quad \sin\bigg(2\pi-\theta\bigg) = -\sin\theta$$
$$\cos\bigg(2\pi+\theta\bigg) = \cos\theta \quad \cos\bigg(2\pi-\theta\bigg) = \cos\theta$$
$$\tan\bigg(2\pi+\theta\bigg) = \tan\theta \quad \tan\bigg(2\pi-\theta\bigg) = -\tan\theta$$
I certainly wouldn't recommend memorizing these though since knowing how the unit circle works basically means you know them already.
For example, in an equation you reach $$\cos \theta = -\frac{\sqrt{3}}{2}$$
You already know that $\cos {\frac{\pi}{6}} = \frac{\sqrt{3}}{2}$ and you also know cosine is negative in quadrants $2$ and $3$, so all you need to do is find the corresponding angle for ${\frac{\pi}{6}}$ in those quadrants.
$$\text{Quadrant II} \implies \theta = \pi-{\frac{\pi}{6}} = \frac{5\pi}{6}$$
$$\text{Quadrant III} \implies \theta = \pi+{\frac{\pi}{6}} = \frac{7\pi}{6}$$
This might take a bit of practice, but once you get this whole "corresponding" angle concept, it all becomes simple. Perhaps you can start by trying to visualize this by solving equations with a unit circle. You'll eventually get the hang of it.
Best Answer
Rather than manipulating to get things in terms of $12^\circ$, we'll try to get things in terms of $18^\circ$, because we (hopefully) know that $\sin(18^\circ)=\frac{\sqrt{5}-1}{4}$. Now, let's do it. First factor out $4\sin(24^\circ)$ and use sum-to-product on the remaining factor to get $$8\sin(24^\circ)(\sin(18^\circ)\cos(6^\circ)).$$ Then use product-to-sum on the first and third terms to finish. I leave the details to you, as there is value in knowing how to do these manipulations on your own.
EDIT: In general, if there are trigonometric expressions involving multiples of $6^\circ$, there is a good chance the solution will involve $\sin(18^\circ)=\frac{\sqrt 5 -1}{4}$ and/or $\cos(36^\circ)=\frac{\sqrt 5+1}{4}$. If you'd like another example for more practice, try to show that $$\sin(6^\circ)\sin(12^\circ)\sin(24^\circ)\sin(42^\circ)+\sin(12^\circ)\sin(24^\circ)\sin(42^\circ)=\frac{1}{16}.$$ (Source: ARML 2019)