Solution:

Answer: 14

The existence of the Chebyshev polynomials, which express $\cos (n \theta)$ as a polynomial in $\cos (\theta)$, imply that Bob draws a blue line between $\cos (\theta)$ and each other vertex, and also between $\cos (2 \theta)$ and $\cos (4 \theta)$, between $\cos (2 \theta)$ and $\cos (6 \theta)$, and between $\cos (3 \theta)$ and $\cos (6 \theta)$ (by substituting $\theta' = 2 \theta$ or $3 \theta$ as necessary). We now show that Roberta draws a red line through each other pair of vertices.

Let $m$ and $n$ be positive integers. Notice that $\cos (n \theta)$ is a periodic function with period $\frac{2 \pi}{n}$, and $\cos (m \theta)$ is periodic with period $\frac{2 \pi}{m}$. Thus, any polynomial in $\cos (m \theta)$ is also periodic of period $\frac{2 \pi}{m}$. This may not be the minimum period of the polynomial, however, so the minimum period is $\frac{2 \pi}{m k}$ for some $k$. Therefore, if $\cos (n \theta)$ can be expressed as a polynomial in $\cos (m \theta)$ then $\frac{2 \pi}{n} = \frac{2 \pi}{m k}$ for some $k$, so $m \mid n$. This shows that there is a blue line between two vertices $\cos (a \theta)$ and $\cos (b \theta)$ if and only if one of $a$ or $b$ divides the other.

Drawing the graph, one can easily count that there are 3 triangles with all blue edges, 3 triangles with all red edges, and $\binom{6}{3} = 20$ triangles total. Thus there are $20 - 3 - 3 = 14$ triangles having at least one red and at least one blue edge.