1. Given the value of $\sin \alpha$. What is the maximum number of values that a) $\sin \frac{\alpha}{2}$, b)* $\sin \frac{\alpha}{3}$ can take?
1. a) First, let's show that $\sin \frac{\alpha}{2}$ cannot take more than 4 values. Indeed, if $\sin \alpha = \sin \beta$, then $\beta = \alpha + 2\pi k$ or $\beta = \pi - \alpha + 2\pi k$ (where $k$ is an integer). Accordingly, $\frac{\beta}{2} = \frac{\alpha}{2} + \pi k$ or $\frac{\beta}{2} = \frac{\pi - \alpha}{2} + \pi k$. These values correspond to points on the unit circle: $\frac{\alpha}{2}, \frac{\alpha}{2} + \pi$, $\frac{\pi}{2} - \frac{\alpha}{2}, \frac{3\pi}{2} - \frac{\alpha}{2}$, and only these. Some of these points may coincide, but in any case, there are no more than four points.

It remains to provide an example where the sine values at these four points are pairwise distinct. For example, let $\sin \alpha = \frac{\sqrt{3}}{2}$, then the points on the circle corresponding to the angle $\frac{\alpha}{2}$ are $\frac{\pi}{6}, \frac{\pi}{3}, \frac{7\pi}{6}, \frac{4\pi}{3}$. The sine takes the following values at these points: $\frac{1}{2}, \frac{\sqrt{3}}{2}, -\frac{1}{2}$, and $-\frac{\sqrt{3}}{2}$.

b) If $\sin \alpha = 0$, then $\alpha = k\pi$ (where $k$ is an integer). Therefore, $\sin \frac{\alpha}{3}$ can equal $\sin 0 = 0$, $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$, and $\sin \frac{4\pi}{3} = -\frac{\sqrt{3}}{2}$.

Let's show that $\sin \frac{\alpha}{3}$ cannot take more than three values. We can reason as in part "a". We will provide a different proof. Let $\sin \frac{\alpha}{3} = t$. Using the triple angle formula for sine:

$$
\sin \alpha = \sin \left(3 \cdot \frac{\alpha}{3}\right) = -4t^3 + 3t
$$

It remains to note that a polynomial of degree three has no more than three roots (see fact 20).

Comments. $1^{\circ}$. It is known that if $n$ is odd, then $\sin n\varphi$ is a polynomial of degree $n$ in $\sin \varphi$, and $\cos n\varphi$ is a polynomial of degree $n$ in $\cos \varphi$ for any $n$. See the comment to problem 2 for 11th grade of the 1996 competition.

$2^{\circ}$. If $\sin n\varphi$ is given, then the maximum number of values that $\sin \varphi$ can take is $n$ for odd $n$, and $2n$ for even $n$. Prove this yourself.

For a given $\cos n\varphi$, $\cos \varphi$ can take no more than $n$ distinct values (regardless of the parity of $n$).