<s>[INST] Determinant of a matrix with trigonometric functions 
Prove:
det
  $\begin{pmatrix}
  \cos(a-b) & \cos(b-c) & \cos(c-a) \\
  \cos(a+b) & \cos(b+c) & \cos(c+a) \\
  \sin(a+b) & \sin(b+c) & \sin(c+a)
\end{pmatrix}=-2\sin(a-b)\sin(b-c)\sin(c-a) 
$.

I know that
$det\begin{pmatrix}
  \cos(a-b) & \cos(b-c) & \cos(c-a) \\
  \cos(a+b) & \cos(b+c) & \cos(c+a) \\
  \sin(a+b) & \sin(b+c) & \sin(c+a)
\end{pmatrix}=2\cdot det\begin{pmatrix}
  \sin a\sin b & \sin b\sin c & \sin c\sin a \\
  \cos a\cos b & \cos b\cos c & \cos c\cos a \\
  \sin(a+b) & \sin(b+c) & \sin(c+a)
\end{pmatrix} 
$
Inspired by seeing the answers of Determinant of matrix with trigonometric functions, I try to tackle this problem likewise,but fail.
 [/INST] $$det=\sum_{cyc}(\cos(a-b)\cos(b+c)\sin(a+c)-\cos(a-b)\sin(b+c)\cos(c+a))=$$
$$=\sum_{cyc}\cos(a-b)\sin(a+c-b-c)=\sum_{cyc}\cos(a-b)\sin(a-b).$$
In another hand,
$$-2\sin(c-a)\sin(b-c)\sin(a-b)=-(\cos(2c-a-b)-\cos(a-b))\sin(a-b)=$$
$$=\cos(a-b)\sin(a-b)-\frac{1}{2}(\sin(2c-2b)+\sin(2a-2c))=\sum_{cyc}\cos(a-b)\sin(a-b)$$
and we are done.
</s>