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The first Jackson theorem for periodic approximation applied to $|\sin \theta |$ gives algebraic best error $O(n^{-1})$ for $\sqrt{1-x^2}$. Its ordinary interval modulus of continuity at $1/n$ is at least $\sqrt{2/n-1/n^2}$. Substituting that error rate into the unmodified inverse theorem for trigonometric approximation would instead bound the modulus by $O(\log (n+1)/n)$, a contradiction. Algebraic inverse estimates must incorporate the endpoint compression of cosine substitution for polynomial approximation.

The first Jackson theorem for periodic approximation asserts that a universal constant $C$ satisfiesfor every continuous $2\pi$-periodic function $g$, where the infimum is over degree-at-most-$n$ trigonometric polynomials.

Apply it to the even function $g(\theta )=f(\cos \theta )$. If $q_n$ is a trigonometric polynomial approximating $g$, its even parthas no larger error, because $g(-\theta )=g(\theta )$ and the triangle inequality bounds each averaged error by $\Vert g-q_n{\Vert }_{\infty }$. Every even trigonometric polynomial has the formHere the Chebyshev polynomials $T_j$ have algebraic degree $j$, so $p_n$ has degree at most $n$. Surjectivity of cosine givesConversely, every algebraic polynomial of degree at most $n$ yields such an even trigonometric polynomial. Taking infima therefore proves the exact identity $E_n^{\mathrm{alg}}(f)=E_n^{\mathrm{trig}}(\tilde{f})$, not merely an inequality. Combine this with the periodic theorem and the preceding modulus of continuity estimate:Symmetrization, the degree correspondence and the equality of norms justify every step in transferring the Jackson-type estimate.