I am looking for the exact value of $a = \sin(160^\circ)\sin(140^\circ)\sin(110^\circ)$.
The hard way would be to compute each factor, wich is doable, and basically amounts to compute trigonometric values for $20^\circ=\frac{\pi}{9}$. But this computation involves the resolution of a cubic equation, which I would like to avoid (that's for a preCalculus class that didn't know about methods cubic equations).
I am looking for a more elegant way using trigonometric formulas like the Product-to-Sum formula and similar ones. Does anyone have an idea on this?
The anecdote is that I know such a method exists (a friend showed it to me a while back) but I cannot remember it.
Best Answer
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 .