= Solution
Let $k_S=\omega/c_S$ and $k_P=\omega/c_P$. <Snell law for elastic waves> fixes the reflected waves' common tangential wavenumber:
$$
k_S\sin\theta=k_P\sin\phi,
\qquad
\sin\phi=\frac{c_P}{c_S}\sin\theta.
$$
Choose the reflected <SV-wave> and converted reflected <P wave> as
$$
\mathbf u_{RS}
=\operatorname{Re}\left\{
R(\cos\theta,0,\sin\theta)
e^{ik_S(x\sin\theta-z\cos\theta)-i\omega t}
\right\},
$$
$$
\mathbf u_{RP}
=\operatorname{Re}\left\{
T(\sin\phi,0,-\cos\phi)
e^{ik_P(x\sin\phi-z\cos\phi)-i\omega t}
\right\}.
$$
The polarizations are respectively perpendicular and parallel to their <wavevector>[wavevectors]. Phase matching makes all three exponential factors equal at $z=0$. The <rigid boundary condition for an elastic wave> $\mathbf u_I+\mathbf u_{RS}+\mathbf u_{RP}=0$ there gives
$$
\cos\theta(1+R)+T\sin\phi=0,
$$
$$
-\sin\theta+R\sin\theta-T\cos\phi=0.
$$
Solving this two-by-two <linear system> yields
$$
\boxed{R=-\frac{\cos(\theta+\phi)}{\cos(\theta-\phi)}},
\qquad
\boxed{T=-\frac{\sin2\theta}{\cos(\theta-\phi)}}.
$$
These fields give the <Reflection of an SV-wave from a rigid plane>.
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