Ireland_2017

The first identity follows since $$\begin{array}{rl}(x-w)(y-z)+(y-w)(z-x)+(z-w)(x-y)& =(xy-xz-wy+wz)+(yz-yx-wz+wx)+(zx-zy-wx+wy)\\ & =(xy+wx+yz+wx+zx+wy)-(xx+wy+yx+wz+zy+wx)\\ & =0.\end{array}$$

The second because in the first place $$2\sin A\sin B=\cos (A-B)-\cos (A+B),$$ and so $$\begin{array}{rl}2\sin (x-w)\sin (y-z)& =\cos ((x-w)-(y-z))-\cos ((x-w)+(y-z))\\ & =\cos (x-y+z-w)-\cos (x-y-z-w),\end{array}$$ $$\begin{array}{rl}2\sin (y-w)\sin (z-x)& =\cos ((y-w)-(z-x))-\cos ((y-w)+(z-x))\\ & =\cos (x+y-z-w)-\cos (x-y-z+w),\text{ and}\end{array}$$ $$\begin{array}{rl}2\sin (z-w)\sin (x-y)& =\cos ((z-w)-(x-y))-\cos ((z-w)+(x-y))\\ & =\cos (x-y-z+w)-\cos (x-y+z-w).\end{array}$$ The result follows by adding these together.