= Solution
Let $\omega=kc_P$, and let $\psi$ be the transmission angle. Since $\hat c_P=2c_P$, tangential phase matching gives
$$
k\sin\phi=\hat k\sin\psi,
\qquad
\hat k=\frac k2,
\qquad
\boxed{\sin\psi=2\sin\phi.}
$$
The assumption of no evanescence requires $\sin\phi\leq1/2$.
With a common suppressed factor $e^{-i\omega t}$, take
$$
\begin{aligned}
\mathbf u_I&=(\sin\phi,0,\cos\phi)
e^{ik(x\sin\phi+z\cos\phi)},\\
\mathbf u_R&=R(\sin\phi,0,-\cos\phi)
e^{ik(x\sin\phi-z\cos\phi)},\\
\mathbf u_T&=T(\sin\psi,0,\cos\psi)
e^{i\hat k(x\sin\psi+z\cos\psi)}.
\end{aligned}
$$
For an inviscid liquid, continuity of normal displacement and normal stress gives
$$
\cos\phi(1-R)=T\cos\psi,
\qquad
Z(1+R)=\hat ZT,
$$
where $Z=\lambda/c_P$ and $\hat Z=\hat\lambda/\hat c_P$. Solving,
$$
\boxed{
R=\frac{\hat Z\cos\phi-Z\cos\psi}
{\hat Z\cos\phi+Z\cos\psi},
\qquad
T=\frac{2Z\cos\phi}
{\hat Z\cos\phi+Z\cos\psi}.}
$$
The condition $|R|=|T|$ gives
$$
|\hat Z\cos\phi-Z\cos\psi|=2Z\cos\phi.
$$
The sign that could instead produce $Z\cos\psi=(\hat Z+2Z)\cos\phi$ leads to $\sin^2\phi>1$ and is inadmissible. Hence
$$
Z\cos\psi=(\hat Z-2Z)\cos\phi.
$$
Using $\cos^2\psi=1-4\sin^2\phi$ and squaring gives
$$
\boxed{
\sin^2\phi
=\frac{3Z^2-4Z\hat Z+\hat Z^2}
{\hat Z(\hat Z-4Z)}.}
$$
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