 Solution

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.