Trigonometric determinants I've recently obtained my University entrance papers from 1967 (yes,52 years ago!) and I found the question below difficult. I presume the answer is a symmetric expression in the differences between alpha,beta and gamma.Am I missing some obvious trick? Any help would be appreciated.
Simplify and evaluate the determinant

and show that its value is independent of theta.
 A: Since row and column operations don't affect the determinant, we may add the third column to the first column to obtain a first column of ones then by RREF obtain as first column $(1,0,0)$ and finally use Laplace expansion on the first column to obtain
$$\det(A)=\left(\sin(\theta+\beta)\cos(\theta+\beta)-\sin(\theta+\alpha)\cos(\theta+\alpha)\right)\left(\cos^2(\theta+\gamma)-\cos^2(\theta+\alpha)\right)+$$
$$-\left(\sin(\theta+\gamma)\cos(\theta+\gamma)-\sin(\theta+\alpha)\cos(\theta+\alpha)\right)\left(\cos^2(\theta+\beta)-\cos^2(\theta+\alpha)\right)$$
and by Product-to-sum and sum-to-product identities
$$\sin(\theta+\beta)\cos(\theta+\beta)-\sin(\theta+\alpha)\cos(\theta+\alpha)=\frac12\sin(2\theta+2\beta)-\frac12\sin(2\theta+2\alpha)=$$
$$=-\sin(\alpha-\beta)\cos(2\theta+\alpha+\beta)$$
and
$$\cos^2(\theta+\gamma)-\cos^2(\theta+\alpha)=\sin(\alpha-\gamma)\sin(2\theta+\alpha+\gamma)$$
and similarly
$$\sin(\theta+\gamma)\cos(\theta+\gamma)-\sin(\theta+\alpha)\cos(\theta+\alpha)=-\sin(\alpha-\gamma)\cos(2\theta+\alpha+\gamma)$$
and
$$\cos^2(\theta+\beta)-\cos^2(\theta+\alpha)=\sin(\alpha-\beta)\sin(2\theta+\alpha+\beta)$$
therefore
$$\det(A)=\sin(\alpha-\beta)\sin(\alpha-\gamma)$$$$\left(\cos(2\theta+\alpha+\gamma)\sin(2\theta+\alpha+\beta)-\cos(2\theta+\alpha+\beta)\sin(2\theta+\alpha+\gamma)\right)$$
and since
$$\cos(2\theta+\alpha+\gamma)\sin(2\theta+\alpha+\beta)=\frac12\sin(4\theta+2\alpha+\beta +\gamma)-\frac12\sin(\gamma-\beta)$$
$$\cos(2\theta+\alpha+\beta)\sin(2\theta+\alpha+\gamma)=\frac12\sin(4\theta+2\alpha+\beta +\gamma)-\frac12\sin(\beta-\gamma)$$
finally we obtain
$$\det(A)=\sin(\alpha-\beta)\sin(\alpha-\gamma)\sin(\beta-\gamma)$$
A: To make it a bit better readable, you may substitute


*

*$u = \theta + \alpha, v= \theta + \beta, w = \theta + \gamma$
So, the first row of your determinant looks like
$$\begin{pmatrix} \sin^2 u & \sin u \cos u & \cos^2 u
\end{pmatrix}$$
Now, you may use 


*

*$\cos^2 u - \sin^2 u = \cos 2u$ and 

*$2 \sin u \cos u = \sin 2u$ and 

*$\sin^2 u= \frac{1}{2}(1-\cos 2u)$
Applying this to the columns of the determinant you get (here only for the first row)
$$\begin{pmatrix} \frac{1}{2}(1-\cos 2u) & \frac 12\sin 2u & \cos 2u
\end{pmatrix}$$
So, your determinant looks now like this
$$\left|
\begin{pmatrix} 
\frac{1}{2} & \frac 12\sin 2u & \cos 2u \\
\frac{1}{2} & \frac 12\sin 2v & \cos 2v \\
\frac{1}{2} & \frac 12\sin 2w & \cos 2w \\
\end{pmatrix}\right| = \frac 14 \left|
\begin{pmatrix} 
1 & \sin 2u & \cos 2u \\
1 & \sin 2v & \cos 2v \\
1 & \sin 2w & \cos 2w \\
\end{pmatrix}\right|$$
Now, expand along the first column and use 


*

*$\sin a\cos b- \cos a \sin b = \sin (a-b)$
$$\sin(2(v-w)) - \sin(2(u-w)) + \sin(2(u-v))$$
This also shows the independence from $\theta$.
