= Solution
Take $x_j=-1+j/4$, $0\le j\le8$. These nine points satisfy $\cos(4\pi x_j)=(-1)^j$. Thus the zero <polynomial> has nine alternating error extrema of magnitude one and is minimax among <polynomials> of degree at most seven by the <Chebyshev alternation theorem>. Hence \b[every such <polynomial> has uniform error at least one].
One can also prove the bound directly. If the error were strictly below one everywhere, $g(x_j)$ would have sign $(-1)^j$. The <intermediate value theorem> would give a distinct zero in each of the eight intervening intervals, contradicting the degree bound. Since the <continuous> error attains its maximum on the <compact> interval, there is an $x$ with $|f(x)-g(x)|\ge1$.
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