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.
$$det=\sum_{cyc}(\cos(a-b)\cos(b+c)\sin(a+c)-\cos(a-b)\sin(b+c)\cos(c+a))=$$ $$=\sum_{cyc}\cos(a-b)\sin(a+c-b-c)=\sum_{cyc}\cos(a-b)\sin(a-b).$$ 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.