= Solution
For the indicated function the cosine substitution gives
$$
\widetilde f_0(\theta)=\sqrt{1-\cos^2\theta}=|\sin\theta|.
$$
The absolute-value function is <Lipschitz continuous> with constant one, and so is sine. Their composition therefore satisfies
$$
\big||\sin(\theta+h)|-|\sin\theta|\big|\le |h|,
\qquad \omega(\widetilde f_0,\delta)\le\delta.
$$
Use the sharper intermediate estimate from the preceding part, rather than the ordinary interval <modulus of continuity>:
$$
\boxed{E_n^{\mathrm{alg}}(f_0)\le C/n=O(n^{-1})}.
$$
The usual <inverse theorem for trigonometric approximation> cannot hold verbatim with the ordinary interval <modulus of continuity>. It would imply
$$
\omega(f_0,1/n)\le\frac{C'}n\sum_{\nu=0}^n E_\nu^{\mathrm{alg}}(f_0)
=O\!\left(\frac{\log(n+1)}n\right).
$$
But comparison with the endpoint $1$ gives
$$
\omega(f_0,1/n)\ge |f_0(1-1/n)-f_0(1)|
=\sqrt{\frac2n-\frac1{n^2}},
$$
which is of order $n^{-1/2}$ and contradicts that bound as $n\to\infty$. This is an <endpoint obstruction to an algebraic inverse approximation theorem>. The algebraic approximation rate measures smoothness after cosine substitution; cosine compresses distances quadratically near the endpoints. A valid algebraic inverse theorem must account for that endpoint geometry rather than using the unchanged periodic formulation.
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