How to calculate this $\sin\frac{\pi}{9}\sin\frac{2\pi}{9}\sin\frac{4\pi}{9}$? I'm stuck with the expression
$$\sin\frac{\pi}{9}\sin\frac{2\pi}{9}\sin\frac{4\pi}{9}.$$
I have no idea how to begin, please give me a hint!
(The answer should be $\sqrt3/8$.)

Remember that the polynomial $P(x)=3x-4x^3$ has the property $P(\sin t)=\sin 3t$. Therefore $x_1=\sin \pi/9$,
$x_2=\sin 2\pi/9$ and $x_3=-\sin4\pi/9$ are all solutions of the equation
$P(x_i)=\sin \pi/3=\sqrt3/2$. In other words they are zeros of $P(x)-\sqrt3/2$. The leading coefficient of $P$ is $-4$, so
$$
-4x^3+3x-\sqrt3/2=P(x)-\sqrt3/2=-4(x-x_1)(x-x_2)(x-x_3).
$$
Expand the product on the right hand side, and compare the constant terms.
