Irish Mathematical Olympiad

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

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