Proving determinants is independent of theta

Question

$$A=\left|\begin{array}{ccc} \sin\left(\theta+\alpha\right) & \cos\left(\theta+\alpha\right) & 1\\ \sin\left(\theta+\beta\right) & \cos\left(\theta+\beta\right) & 1\\ \sin\left(\theta+\gamma\right) & \cos\left(\theta+\gamma\right) & 1 \end{array}\right|$$

My approach is to subtract row 1 from row 2 and row 3 then open the determinant. The problem with this approach is it involves way too much calculation. I tried to take something common but nothing comes to mind. Can somebody give me a better way to approach this problem?

Answer 1

A geometric approach: Letting $v_1, v_2, v_3$ be the rows of $A$, we see that they are vectors on the unit circle with center $(0,0,1)$ on the plane $z = 1$. Changing the angles by $\theta$ is a rotation in this plane, which does not change the volume of the parallelepiped determined by $v_1,v_2, v_3$.

If you want to make this rigorous, observe that

$$\left(\begin{array}{ccc} \sin\left(\theta+\alpha\right) & \cos\left(\theta+\alpha\right) & 1\\ \sin\left(\theta+\beta\right) & \cos\left(\theta+\beta\right) & 1\\ \sin\left(\theta+\gamma\right) & \cos\left(\theta+\gamma\right) & 1 \end{array}\right) = \left(\begin{array}{ccc} \sin\left(\alpha\right) & \cos\left(\alpha\right) & 1\\ \sin\left(\beta\right) & \cos\left(\beta\right) & 1\\ \sin\left(\gamma\right) & \cos\left(\gamma\right) & 1 \end{array}\right)\left(\begin{array}{ccc} \cos\left(\theta\right) & -\sin\left(\theta\right) & 0\\ \sin\left(\theta\right) & \cos\left(\theta\right) & 0\\ 0 & 0 & 1 \end{array}\right)$$

and use $\det(AB) = \det(A) \det (B)$.

(See also Blue's comment above that beat me to the punch.)

Answer 2

Expand along the last column (using Laplace's formula).

One minor is for instance

$$\sin(\theta+\beta)\cos(\theta+\gamma)-\sin(\theta+\gamma)\cos(\theta+\beta)\\= \sin[(\theta+\beta)-(\theta+\gamma)]\\= \sin(\beta-\gamma)$$

The other two minors are similar. All in all you get:

$$|A|=\sin(\beta-\gamma)-\sin(\alpha-\gamma)+\sin(\alpha-\beta)$$

Answer 3

Idea: If $A =\det (c_1,c_2,\cdots,c_n)$, where $c_k$ are columns (of functions with $\theta$ as argument), then a general fact is that (product/sum rules) $$A' = \det (c_1',c_2,\cdots,c_n)+ \det (c_1, c_2',\cdots, c_n) + \cdots + \det (c_1, c_2,\cdots, c_n').$$ In the case here, each of the summands will be visibly zero. Hence $A'(\theta) = 0$, and $A$ does not depend on $\theta$.