We prove that the function g(x)=1/(1-cos(x))\documentclass[12pt]{minimal}
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\begin{document}$$g(x)= 1 / ( 1 - \cos(x) )$$\end{document} is completely monotonic on (0,π]\documentclass[12pt]{minimal}
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\begin{document}$$(0,\pi]$$\end{document} and absolutely monotonic on [π,2π)\documentclass[12pt]{minimal}
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\begin{document}$$[\pi, 2\pi)$$\end{document}, and we determine the best possible bounds λn\documentclass[12pt]{minimal}
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\begin{document}$$\lambda_n$$\end{document} and μn\documentclass[12pt]{minimal}
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\begin{document}$$\mu_n$$\end{document} such that the inequalities
λn≤g(n)(x)+g(n)(y)-g(n)(x+y)(n≥0even)\documentclass[12pt]{minimal}
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\begin{document}$$
\lambda_n \leq g^{(n)}(x)+g^{(n)}(y)-g^{(n)}(x+y) \quad (n \geq 0 \ \mbox{even})
$$\end{document}
and
μn≤g(n)(x+y)-g(n)(x)-g(n)(y)(n≥1odd)\documentclass[12pt]{minimal}
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\begin{document}$$
\mu_n \leq g^{(n)}(x+y)-g^{(n)}(x)-g^{(n)}(y) \quad (n \geq 1 \ \mbox{odd})
$$\end{document}
hold for all x,y∈(0,π)\documentclass[12pt]{minimal}
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\begin{document}$$x,y\in (0,\pi)$$\end{document} with x+y≤π\documentclass[12pt]{minimal}
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\begin{document}$$x+y\leq \pi$$\end{document}.