= Solution
Taking the <Hermitian adjoint> gives $\varphi_{\mathbf k}=c_k\psi_{\mathbf k}+s_k\psi_{-\mathbf k}^\dagger$. For nonzero <wavevectors>, the <canonical commutation relations> imply
$$
\begin{aligned}
[\varphi_{\mathbf k},\varphi_{\mathbf q}^\dagger]
&=c_kc_q[\psi_{\mathbf k},\psi_{\mathbf q}^\dagger]
+s_ks_q[\psi_{-\mathbf k}^\dagger,\psi_{-\mathbf q}]\\
&=(c_kc_q-s_ks_q)\delta_{\mathbf k,\mathbf q}.
\end{aligned}
$$
When the <Kronecker delta> is nonzero, $k=q$, so the <hyperbolic Pythagorean identity> $c_k^2-s_k^2=1$ gives
$$
\boxed{[\varphi_{\mathbf k},\varphi_{\mathbf q}^\dagger]=\delta_{\mathbf k,\mathbf q}.}
$$
For the two <annihilation operators>, the only potentially nonzero terms are
$$
\begin{aligned}
[\varphi_{\mathbf k},\varphi_{\mathbf q}]
&=c_ks_q[\psi_{\mathbf k},\psi_{-\mathbf q}^\dagger]
+s_kc_q[\psi_{-\mathbf k}^\dagger,\psi_{\mathbf q}]\\
&=(c_ks_q-s_kc_q)\delta_{\mathbf k,-\mathbf q}=0.
\end{aligned}
$$
Here $\mathbf q=-\mathbf k$ implies equal magnitudes and hence equal $c,s$, because $\theta_k$ depends only on $|\mathbf k|$. Thus
$$
\boxed{[\varphi_{\mathbf k},\varphi_{\mathbf q}]=0.}
$$
The corresponding <commutators> of two <creation operators> vanish by the <Hermitian adjoint>. This verifies that the <Bogoliubov transformation> preserves the full <bosonic> <canonical commutation relations>.
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