\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
We first consider the event \(Y \leq t\).
\begin {align*} Y \leq t & \iff \max \left \{X_1, X_2, \ldots , X_n\right \} \leq t \\ & \iff X_1, X_2, \ldots , X_n \leq t \\ & \iff X_1 \leq t, X_2 \leq t, \cdots , X_n \leq t. \end {align*}
Hence, \begin {align*} \Prob (Y \leq t) & = \Prob (X_1 \leq t, X_2 \leq t, \cdots , X_n \leq t) \\ & = \Prob (X_1 \leq t) \Prob (X_2 \leq t) \cdot \Prob (X_n \leq t) \\ & = \left [\Prob (X_1 \leq t)\right ]^n \end {align*}
as desired.
We first find the cumulative distribution function of \(X\), \(F\). For \(0 \leq x \leq \pi \), \begin {align*} F(x) & = \int _{0}^{x} f(t) \Diff t \\ & = \frac {1}{2} \int _{0}^{x} \sin t \Diff t \\ & = - \frac {1}{2} \left [\cos t\right ]_{0}^{x} \\ & = \frac {1}{2} \left (1 - \cos x\right ). \end {align*}
Now, let \(G\) be the cumulative distribution function of \(Y\). We have \(0 \leq Y \leq \pi \). For \(0 \leq y \leq \pi \), \begin {align*} G(y) & = \Prob (Y \leq y) \\ & = \left [\Prob (X_1 \leq y)\right ]^n \\ & = \left [F(y)\right ]^n \\ & = \left [\frac {1}{2} \left (1 - \cos y\right )\right ]^n & = \frac {1}{2^n} \left (1 - \cos y\right )^n. \end {align*}
Hence, the probability density function of \(Y\), \(g\), is given by \begin {align*} g(y) & = G'(y) \\ & = \frac {1}{2^n} \cdot n \cdot \sin x \cdot \left (1 - \cos X\right )^{n - 1} \\ & = \frac {n \sin x \left (1 - \cos X\right )^{n - 1}}{2^n} \end {align*}
for \(0 \leq t \leq \pi \), and \(0\) otherwise.
\(m(n)\) is such that \begin {align*} G(m(n)) & = \frac {1}{2} \\ \frac {1}{2^n} \left (1 - \cos m(n)\right )^n & = \frac {1}{2} \\ \left (1 - \cos m(n)\right )^n & = 2^{n - 1} \\ 1 - \cos m(n) & = 2^{\frac {n - 1}{n}} \\ \cos m(n) & = 1 - 2^{1 - \frac {1}{n}} \\ m(n) & = \arccos \left (1 - 2^{1 - \frac {1}{n}}\right ). \end {align*}
As \(n\) increases, \(\frac {1}{n}\) decreases, \(1 - \frac {1}{n}\) increases, \(2^{1 - \frac {1}{n}}\) increases, \(1 - 2^{1 - \frac {1}{n}}\) increases, and so \(m(n)\) increases. \(m(n) \to \pi \) as \(n \to \infty \).
By definition, we have \begin {align*} \mu (n) & = \Expt (Y) \\ & = \int _{0}^{\pi } \frac {n}{2^n} x \sin x \left (1 - \cos x\right )^{n - 1} \Diff x \\ & = \frac {1}{2^n} \int _{0}^{\pi } x \cdot n \sin x \left (1 - \cos x\right )^{n - 1} \Diff x \\ & = \frac {1}{2^n} \int _{0}^{\pi } x \cdot \left (1 - \cos x\right )^n \Diff x \\ & = \frac {1}{2^n} \left [x \left (1 - \cos x \right )^{n}\right ]_{0}^{\pi } - \frac {1}{2^n} \int _{0}^{\pi } \left (1 - \cos x\right )^{n} \Diff x \\ & = \frac {1}{2^n} \left [\pi \cdot \left (1 + 1 \right )^n - 0 \cdot \left (1 - 1\right )^n\right ] - \frac {1}{2^n} \int _{0}^{\pi } \left (1 - \cos x\right )^n \Diff x \\ & = \frac {1}{2^n} \cdot \pi \cdot 2^n - \frac {1}{2^n} \int _{0}^{\pi } \left (1 - \cos x\right )^n \Diff x \\ & = \pi - \frac {1}{2^n} \int _{0}^{\pi } \left (1 - \cos x \right )^n \Diff x. \end {align*}
By taking difference of two consecutive terms of \(\mu (n)\), we have \begin {align*} \mu (n + 1) - \mu (n) & = \left [\pi - \frac {1}{2^{n + 1}} \int _{0}^{\pi } \left (1 - \cos x \right )^{n + 1} \Diff x\right ] - \left [\pi - \frac {1}{2^{}} \int _{0}^{\pi } \left (1 - \cos x \right )^{n} \Diff x\right ] \\ & = \frac {1}{2^n} \int _{0}^{\pi } \left (1 - \cos x\right )^n \Diff x - \frac {1}{2^{n + 1}} \int _{0}^{\pi } \left (1 - \cos x\right )^{n + 1} \Diff x \\ & = \frac {1}{2^{n + 1}} \int _{0}^{\pi } \left [2 \left (1 - \cos x\right )^{n} - \left (1 - \cos x\right )^{n + 1}\right ] \Diff x \\ & = \frac {1}{2^{n + 1}} \int _{0}^{\pi } \left (1 - \cos x\right )^{n} \left [2 - \left (1 - \cos x\right )\right ] \Diff x \\ & = \frac {1}{2^{n + 1}} \int _{0}^{\pi } \left (1 - \cos x\right )^{n} \left (1 + \cos x\right ) \Diff x. \end {align*}
For \(0 < x < \pi \), we have \(0 < \cos x < 1\), and so the integrand is positive on the interval.
Hence, \(\mu (n + 1) - \mu (n) > 0\), and \(\mu (n + 1) > \mu (n)\), and hence \(\mu (n)\) increases with \(n\).
On one hand, we have \[ m(2) = \arccos \left (1 - 2^{1 - \frac {1}{2}}\right ) = \arccos \left (1 - \sqrt {2}\right ). \]
On the other hand, \begin {align*} \mu (2) & = \pi - \frac {1}{4} \int _{0}^{\pi } \left (1 - \cos x\right )^2 \Diff x \\ & = \pi - \frac {1}{4} \int _{0}^{\pi } \left (1 - 2 \cos x + \cos ^2 x\right ) \Diff x \\ & = \pi - \frac {1}{4} \int _{0}^{\pi } \left (1 - 2 \cos x + \frac {\cos 2x + 1}{2}\right ) \Diff x \\ & = \pi - \frac {1}{4} \int _{0}^{\pi } \left (\frac {3}{2} - 2 \cos x + \frac {1}{2} \cos 2x\right ) \Diff x \\ & = \pi - \frac {1}{4} \left (\frac {3}{2} x - 2 \sin x + \frac {1}{4} \sin 2x\right )_{0}^{\pi } \\ & = \pi - \frac {1}{4} \left [\frac {3}{2} \left (\pi - x\right ) - 2 \left (\sin \pi - \sin 0\right ) + \frac {1}{4} \left (\sin 2\pi - \sin 0\right )\right ] \\ & = \pi - \frac {1}{4} \cdot \frac {3}{2} \pi \\ & = \frac {5}{8} \pi . \end {align*}
We want to show that \[ \left (0 < \frac {1}{2}\pi < \right ) \frac {5}{8} \pi < \arccos \left (1 - \sqrt {2}\right ) \left (< \pi \right ), \] and this is equivalent to showing that \[ \cos \frac {5}{8}\pi > 1 - \sqrt {2}. \]
We first notice that \(\cos \frac {5}{8}\pi = \cos \left (\frac {1}{2}\pi + \frac {1}{8}\pi \right )\), and notice that \(\cos \left (\frac {1}{8}\pi \right )\) is such that \[ 2 \cos ^2 \left (\frac {1}{8} \pi \right ) - 1 = \cos \left (2 \cdot \frac {1}{8}\pi \right ) = \cos \frac {\pi }{4} = \frac {1}{\sqrt {2}}, \] and hence \[ 2 \cos ^2 \frac {\pi }{8} = 1 + \frac {1}{\sqrt {2}} = \frac {2 + \sqrt {2}}{2}, \] meaning \[ \cos \frac {\pi }{8} = \sqrt {\frac {2 + \sqrt {2}}{4}} = \frac {\sqrt {2 + \sqrt {2}}}{2}. \]
Therefore, \[ \sin ^2 \frac {\pi }{8} = 1 - \frac {2 + \sqrt {2}}{4} = \frac {2 - \sqrt {2}}{4} \] and hence \[ \sin \frac {\pi }{8} = \frac {\sqrt {2 - \sqrt {2}}}{2}. \]
Hence, \begin {align*} \cos \frac {5}{8}\pi & = \cos \left (\frac {1}{2} \pi + \frac {1}{8} \pi \right ) \\ & = \cos \frac {1}{2}\pi \cos \frac {1}{8}\pi - \sin \frac {1}{2} \pi \sin \frac {1}{8} \pi \\ & = 0 - \sin \frac {1}{8} \pi \\ & = -\frac {\sqrt {2 - \sqrt {2}}}{2}. \end {align*}
Finally, we have the following being equivalent: \begin {align*} \cos \frac {5}{8}\pi & > 1 - \sqrt {2} \\ (0>) - \frac {\sqrt {2 - \sqrt {2}}}{2} & > 1 - \sqrt {2} \\ \sqrt {2} - 1 & > \frac {\sqrt {2 - \sqrt {2}}}{2} \\ 2 + 1 - 2 \sqrt {2} & > \frac {2 - \sqrt {2}}{4} \\ 12 - 8 \sqrt {2} & > 2 - \sqrt {2} \\ 7 \sqrt {2} & < 10 \\ 49 \cdot 2 = 98 & < 100 \end {align*}
is true, and hence \(\mu (2) < m(2)\) as desired.