Problem source

By simplifying $\sin(r+\frac12)x - \sin(r-\frac12)x$ or otherwise show that, for $\sin\frac12 x \ne0$,
  \[
  \cos x + \cos 2x +\cdots + \cos nx = \frac{\sin(n+\frac12)x -
    \sin\frac12 x}{2\sin\frac12x}\,.
  \]
  The functions $S_n$, for $n=1, 2, \dots$, are defined  by
  \[
  S_n(x) = \sum_{r=1}^n \frac 1 r \sin rx \qquad (0\le
  x \le \pi).
  \]
  \begin{questionparts}
  \item Find the stationary points of $S_2(x)$ for $0\le x\le\pi$, and sketch this function.
  \item Show that if $S_n(x)$ has a stationary point at $x=x_0$, where $0< x_0 < \pi$, then
    \[
    \sin nx_0 = (1-\cos nx_0) \tan\tfrac12 x_0
    \]
    and hence that $S_n(x_0) \ge S_{n-1}(x_0)$.  Deduce that if $S_{n-1}(x) > 0$ for all $x$ in the interval $0 < x < \pi$, then $S_{n}(x) > 0$ for all $x$ in this interval.
  \item Prove that $S_n(x)\ge0$ for $n\ge1$ and $0\le x\le\pi$.
  \end{questionparts}

Solution source

\begin{align*}
&& \sin(r + \tfrac12)x - \sin(r - \tfrac12) x &= \sin rx \cos \tfrac12x + \cos r x\sin\tfrac12x - \sin r x \cos \tfrac12 x + \cos rx \sin \tfrac12 x \\
&&&= 2\cos r x \sin\tfrac12 x \\
\\
&& S &= \cos x + \cos 2x + \cdots + \cos n x \\
&& 2\sin \tfrac12 x S &= \sin(1 + \tfrac12)x - \sin \tfrac12 x + \\
&&&\quad+ \sin(2+\tfrac12)x - \sin(2-  \tfrac12)x + \\
&&&\quad+ \sin(3+\tfrac12)x - \sin(3 - \tfrac12)x + \\
&&& \quad + \cdots + \\
&&&\quad + \sin(n+\tfrac12)x - \sin(n-\tfrac12)x \\
&&&=\sin(n+\tfrac12)x - \sin\tfrac12 x \\
\Rightarrow && S &= \frac{\sin(n+\tfrac12)x - \sin\tfrac12 x}{2 \sin \tfrac12 x}
\end{align*}

\begin{questionparts}
\item $\,$ \begin{align*}
&& S_2(x) &= \sin x + \tfrac12 \sin 2 x \\
&& S'_2(x) &= \cos x + \cos 2x \\
&&&= \cos x + 2\cos^2 x - 1 \\
&&&= (2\cos x -1)(\cos x + 1) \\
\end{align*}

Therefore the turning points are $\cos x= \frac12 \Rightarrow x = \frac{\pi}{3}$ and $\cos x = -1 \Rightarrow x = \pi$



\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){sin(deg(#1))+0.5*sin(2*deg(#1))};
    \def\xl{-0.5}; 
    \def\xu{3.4};
    \def\yl{-.25}; 
    \def\yu{2};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the reusable styles to keep code clean
    \tikzset{
        x=\xscale cm, y=\yscale cm,
        axis/.style={thick, draw=black!80, -{Stealth[scale=1.2]}},
        grid/.style={thin, dashed, gray!30},
        curveA/.style={very thick, color=cyan!70!black, smooth},
        curveB/.style={very thick, color=orange!90!black, smooth},
        dot/.style={circle, fill=black, inner sep=1.2pt},
        labelbox/.style={fill=white, inner sep=2pt, rounded corners=2pt} % Protects text from lines
    }

    \foreach \x/\xtext in {
        {0}/0, 
        {pi/4}/\frac{\pi}{4}, 
        {pi/2}/\frac{\pi}{2}, 
        {3*pi/4}/\frac{3\pi}{4}, 
        {pi}/\pi
    } {
        % Draw tick marks slightly above and below the x-axis, then attach the label
        \draw[thick, black!80] ({\x}, 0.05) -- ({\x}, -0.05) node[below, labelbox] {$\xtext$};
    }
    
    % Draw background grid
    \draw[grid] (\xl,\yl) grid[xstep={pi/4},ystep=0.5] (\xu,\yu);
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right, black] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above, black] {$y$};
    
    % Define the bounding region with clip
    \begin{scope}
        \clip (\xl,\yl) rectangle (\xu,\yu);

        
        \draw[curveA, domain=0:\xu, samples=150] 
            plot ({\x},{\functionf(\x)});

    \end{scope}
    
    % Annotate Function Names
    \node[curveA, labelbox] at ({(\xl+\xu)/2}, {1}) {$y = S_2(x)$};

    

    \end{tikzpicture}
\end{center}

\item Suppose $S_n(x)$ has a stationary point at $x_0$, then $$ therefore

\begin{align*}
&&0 &= S_n'(x_0) \\
&&&= \cos x_0 + \cos 2x_0 + \cdots + \cos n x_0 \\
&&&= \frac{\sin(n+\tfrac12)x_0 - \sin \tfrac12x_0}{2 \sin \tfrac12 x_0} \\
\Rightarrow &&\sin\tfrac12 x_0&= \sin nx_0 \cos \tfrac12 x_0 + \cos nx_0 \sin \tfrac12x_0  \\
\Rightarrow && \sin nx_0 &= (1-\cos nx_0)\tan \tfrac12 x_0
\end{align*}

Therefore $S_n(x_0) -S_{n-1}(x_0) = \tfrac1n \sin n x_0 = \tfrac1n \underbrace{(1-\cos nx_0)}_{\geq 0}\underbrace{\tan\tfrac12 x_0}_{\geq 0} \geq 0$. Therefore if $S_{n-1}(x) > 0$ for all $x$ on $0 < x < \pi$ then since $S_n(x) > S_{n-1}(x)$ at the turning points and since they agree at the end points, it must be larger at all points inbetween.

\item Notice that $S_1(x) = \sin x \geq 0$ for all $x \in [0,1]$ and by our previous argument we can show $S_n > S_{n-1}$ inside the interval and equal on the boundary we must have $S_n(x) \geq 0$ for $x \in [0, \pi]$
\end{questionparts}