Five. (20 points) Given
$$
\frac{\cos x+\cos y+\cos z}{\cos (x+y+z)}=\frac{\sin x+\sin y+\sin z}{\sin (x+y+z)}=a \text {. }
$$

Find the value of $\cos (y+z)+\cos (z+x)+\cos (x+y)$.
Let $S=x+y+z$.
Also, $\mathrm{e}^{\mathrm{i} x}=\cos x+\mathrm{i} \sin x, \mathrm{e}^{\mathrm{i} y}=\cos y+\mathrm{i} \sin y$, $\mathrm{e}^{\mathrm{i} z}=\cos z+\mathrm{i} \sin z$,
then
$$
\begin{array}{l}
\mathrm{e}^{\mathrm{i} x}+\mathrm{e}^{\mathrm{i} y}+\mathrm{e}^{\mathrm{i} z} \\
=(\cos x+\cos y+\cos z)+\mathrm{i}(\sin x+\sin y+\sin z) \\
=a \cos (x+y+z)+\mathrm{i} a \sin (x+y+z) \\
=a \mathrm{e}^{\mathrm{i}(x+y+z)}=a \mathrm{e}^{\mathrm{i}S} .
\end{array}
$$

Similarly, $\mathrm{e}^{-\mathrm{i} x}+\mathrm{e}^{-\mathrm{i} y}+\mathrm{e}^{-\mathrm{i} z}=a \mathrm{e}^{-\mathrm{i} S}$.
Thus, $\mathrm{e}^{\mathrm{i}(y+x)}+\mathrm{e}^{\mathrm{i}(x+z)}+\mathrm{e}^{\mathrm{i}(y+z)}=\mathrm{e}^{\mathrm{i}(S-x)}+\mathrm{e}^{\mathrm{i}(S-y)}+\mathrm{e}^{\mathrm{i}(S-z)}$
$$
=\mathrm{e}^{\mathrm{i} S}\left(\mathrm{e}^{-\mathrm{i} x}+\mathrm{e}^{-\mathrm{i} y}+\mathrm{e}^{-\mathrm{i} z}\right)=\mathrm{e}^{\mathrm{i} S}\left(a \mathrm{e}^{-\mathrm{i} S}\right)=a \mathrm{e}^{\mathrm{i} S-\mathrm{i} S}=a \text {. }
$$

Therefore, $[\cos (x+y)+\cos (x+z)+\cos (y+z)]+$
$$
\mathrm{i}[\sin (x+y)+\sin (x+z)+\sin (y+z)]=a \text {. }
$$

Hence, $\cos (x+y)+\cos (x+z)+\cos (y+z)=a$,
and $\sin (x+y)+\sin (x+z)+\sin (y+z)=0$.