Substitute values of theta so that the values of 3 * theta come out to multiples of pi/6, or pi/4.

For example if you use multiples of 1/3 * pi/4 = pi / 12 your values of 3 * theta will come out in multiples of pi/4.

You will then see that between theta = 0 and theta = pi/3, the value of sin(3 * theta) goes from 0 to 1 and back to 0. Between theta = pi/3 and theta = 2 pi/3 the value goes from 0 to -1 and back to 0.

Between theta = 2 pi/3 and theta = pi, the value of sin(3 * theta) goes from 0 to 1 and back to 0. Between theta = pi and theta = 4 pi/3 the value goes from 0 to -1 and back to 0.

Between theta = 4 pi/3 and theta = 5 pi/3, the value of sin(3 * theta) goes from 0 to 1 and back to 0. Between theta = 5 pi/3 and theta = 2 pi the value goes from 0 to -1 and back to 0.

So the value of 6 sin(theta) would go from 0 to 6 to 0 to -6 to 0 between theta = 0 and theta = 2 pi/3, then again between theta = 2 pi/3 and theta = 4 pi/3, then again between theta = 4 pi/3 and theta = 2 pi.

This will eventually form a 3-petaled 'rose'.

It turns out that this aspect of the dot product is not covered in your text, so this problem would not be required.

So you don't have to know the following, but here it is in case you want to learn it or refer to it later:

The answer is that the component of v parallel to any unit vector u is (v dot u) * u. A unit vector parallel to w is w / || w ||, so the component of v parallel to w is

v_parallel = (v dot (w / || w || ) * w / || w || = (v dot w) * w / || w ||^2.

Since v_parallel + v_perpendicular = v, the component of v perpendicular to w is

v_perpendicular = v - v_parallel = v - (v dot w) * w / || w ||^2.

If repeated letters are allowed, then the same letter can be used as many times as desired. So for example, aba would be allowed, as would aab, or even aaa.

Thus for each of the three chosen letters, there are 8 choices, and the total number of possible 3-letter codes would therefore be 8 * 8 * 8 = 512.