Calculate the von Neumann Entropy of a two-qubit entangled state

After working through an exercise I got a confusion answer/solution that either may be because I’ve made a mistake or I’m not understanding von Neumann Entropy.

I have the two qubit system
$$ | psi rangle = cos(theta) rangle 1 rangle | 0 rangle + sin(theta) | 0 rangle | 1 rangle $$
And I want to calculate the von Neumann entropy

$$ S = – Tr ( rho_A ln rho_A ) $$

I got $rho_A$ as

$$ rho_A = begin{pmatrix}
cos^2 (theta) & 0 0 & sin^2 (theta)
end{pmatrix} $$

Thus the von Neumann entropy as

$$ S = – Tr ( begin{pmatrix}
cos^2 (theta) & 0 0 & sin^2 (theta)
end{pmatrix} ln begin{pmatrix}
cos^2 (theta) & 0 0 & sin^2 (theta)
end{pmatrix} ) $$
The natural log of the matrix is
$$ ln rho_A = PDP^{-1} $$
$$ = begin{pmatrix}
1 & 0 0 & 1
end{pmatrix} begin{pmatrix}
ln(cos^2 (theta)) & 0 0 & ln(sin^2 (theta))
end{pmatrix} begin{pmatrix}
1 & 0 0 & 1
end{pmatrix} $$
$$ = begin{pmatrix}
ln(cos^2 (theta)) & 0 0 & ln(sin^2 (theta))
end{pmatrix} $$
$$ S = – Tr ( begin{pmatrix}
cos^2 (theta) cdot ln(cos^2 (theta)) & 0 0 & sin^2 (theta) cdot ln(sin^2 (theta))
end{pmatrix} ) $$

Perhaps I am not understanding but this answer does not make any sense? The trace of that matrix cannot be simplified further and I don’t get an actual answer that doesn’t depend on $theta$. Also in this case the values of $theta$ which make $rho_A$ a pure state is any $theta$ and the same can be said for values of $theta$ which make is a maximally mixed state?