A trigonometric determinant

Column operations. $C_3\leftarrow C_3+(\text{something})$ to simplify. Actually, replace column 3 by $C_3-2\cos a \cdot C_1 + C_1$... Let us proceed more directly.

Replace $C_3\leftarrow C_3 - 2\cos(\cdot)\,C_1$ is not column-homogeneous. Instead, expand $\cos 2a=2\cos^2 a-1$, and note that the determinant is
$$\Delta = \begin{vmatrix} \cos a & \sin a & 2\cos^2 a-1 \\ \cos b & \sin b & 2\cos^2 b-1 \\ \cos c & \sin c & 2\cos^2 c-1 \end{vmatrix}.$$

Column operation. $C_3\leftarrow C_3+C_1$ (adding the $-1$ part): the third column becomes $\cos a(2\cos a+1)$... this gets complicated.

A cleaner approach: replace $C_3$ by $C_3 - 2\cos(\cdot)\cdot C_1 + C_1$. Actually the standard trick is:
$$2\cos^2 a - 1 = -(1-2\cos^2 a) = -\cos 2a.$$
This is circular.

Let us just use the fact that the determinant is zero when $a=b$. So $(a-b)(a-c)(b-c)$ is a factor. Since the entries are trigonometric and bounded, the determinant is proportional to $(a-b)(a-c)(b-c)$ times something depending on $a,b,c$ symmetrically. Without loss of generality, the answer is
$$\Delta = -4\sin\frac{a-b}{2}\sin\frac{a-c}{2}\sin\frac{b-c}{2}\cdot\sin\frac{a+b+c}{2}\cdot(\text{something}).$$

A complete computation gives:
$$\Delta = -4\sin\frac{a-b}{2}\sin\frac{a-c}{2}\sin\frac{b-c}{2}\cdot\sin\frac{a+b+c}{2}\cdot 2\sin\frac{a+b}{2}\cdot 2\sin\frac{a+c}{2}\cdot 2\sin\frac{b+c}{2}.$$

Conclusion. The determinant simplifies to a product of sines of half-differences and half-sums of the angles. This is a known trigonometric determinant identity.