Problem source

The unit circle is the circle with radius 1 and centre the origin, $O$.
$N$ and $P$ are distinct points on the unit circle. $N$ has coordinates $(-1, 0)$, and $P$ has coordinates $(\cos\theta, \sin\theta)$, where $-\pi < \theta < \pi$. The line $NP$ intersects the $y$-axis at $Q$, which has coordinates $(0, q)$.
\begin{questionparts}
\item Show that $q = \tan\frac{1}{2}\theta$.
\item In this part, $q \neq 1$.
\begin{enumerate}
\item Let $\mathrm{f}_1(q) = \dfrac{1+q}{1-q}$. Show that $\mathrm{f}_1(q) = \tan\frac{1}{2}\!\left(\theta + \frac{1}{2}\pi\right)$.
\item Let $Q_1$ be the point with coordinates $(0, \mathrm{f}_1(q))$ and $P_1$ be the point of intersection (other than $N$) of the line $NQ_1$ and the unit circle. Describe geometrically the relationship between $P$ and $P_1$.
\end{enumerate}
\item 
\begin{enumerate}
\item $P_2$ is the image of $P$ under an anti-clockwise rotation about $O$ through angle $\frac{1}{3}\pi$. The line $NP_2$ intersects the $y$-axis at the point $Q_2$ with co-ordinates $(0, \mathrm{f}_2(q))$. Find $\mathrm{f}_2(q)$ in terms of $q$, for $q \neq \sqrt{3}$.
\item In this part, $q \neq -1$. Let $\mathrm{f}_3(q) = \dfrac{1-q}{1+q}$, let $Q_3$ be the point with coordinates $(0, \mathrm{f}_3(q))$ and let $P_3$ be the point of intersection (other than $N$) of the line $NQ_3$ and the unit circle. Describe geometrically the relationship between $P$ and $P_3$.
\item In this part, $0 < q < 1$. Let $\mathrm{f}_4(q) = \mathrm{f}_2^{-1}\!\Big(\mathrm{f}_3\!\big(\mathrm{f}_2(q)\big)\Big)$, let $Q_4$ be the point with coordinates $(0, \mathrm{f}_4(q))$ and let $P_4$ be the point of intersection (other than $N$) of the line $NQ_4$ and the unit circle. Describe geometrically the relationship between $P$ and $P_4$.
\end{enumerate}
\end{questionparts}

Solution source

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){exp(1/(#1)*ln(#1))};
    \def\xl{-1.6}; 
    \def\xu{1.6};
    \def\yl{-1.6}; 
    \def\yu{1.6};
    
    % 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},
        curveC/.style={very thick, color=green!90!black, smooth},
        curveBlack/.style={very thick, color=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
    }
    
    % Draw background grid
    \draw[grid] (\xl,\yl) (\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] (0,0) circle (1);

        \coordinate (O) at (0,0);
        \coordinate (X) at (1,0);
        \coordinate (P) at  ({cos(50)}, {sin(50)}); 
        \coordinate (N) at  (-1, 0); 

        \filldraw (P) circle (1.5pt) node[right] {$P = (\cos \theta, \sin \theta)$};
        \filldraw (N) circle (1.5pt) node[below left] {$N$};

        \draw[curveB] (P) -- (N); 

        \draw[dashed] (O) -- (P);

        \pic [draw, angle radius=.6cm, angle eccentricity=1.5, "$\theta$"] {angle = X--O--P};
        \pic [draw, angle radius=.6cm, angle eccentricity=1.5, "$\frac12\theta$"] {angle = O--N--P};

    \end{scope}
    
    % Annotate Function Names
    % \node[curveB, labelbox] at (1.85, -1.1) {$x = \frac{4y^2+1}{3}$};
    

    \end{tikzpicture}
\end{center}
\begin{questionparts}

\item $\,$ \begin{align*}
&& \frac{y-0}{x-(-1)} &= \frac{\sin \theta }{\cos \theta + 1} \\
\Rightarrow && y_0 &= \frac{\sin \theta}{\cos \theta + 1} \\
&&&= \frac{\frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}+1} \\
&&&= t = \tan \tfrac{\theta}{2}
\end{align*}
Alternatively, it is straightforward to see from the angles.

\item \begin{enumerate}
\item $f_1(q) = \frac{1+q}{1-q}$ so \begin{align*}
&& f_1(\tan\tfrac12\theta) &= \frac{1+\tan\tfrac12\theta}{1-\tan\tfrac12\theta} \\
&&&= \frac{\cos \tfrac12 \theta + \sin \tfrac12 \theta}{\cos \tfrac12 \theta  - \sin \tfrac12 \theta} \\
&&&= \frac{\sin(\tfrac14 \pi + \tfrac12 \theta)}{\cos(\tfrac14 \pi + \tfrac12 \theta)} \\
&&&= \tan \tfrac12(\theta + \tfrac{\pi}{2})
\end{align*}

\item $Q_1$ is the point $(0, f_1(q))$ so $P_1$ will be the point $(\cos (\theta + \tfrac{\pi}{2}), \sin (\theta + \tfrac{\pi}{2}))$ which is a rotation anticlockwise by $\frac{\pi}{2}$
\end{enumerate}


\item \begin{enumerate}
\item $P_2 = (\cos(\theta + \tfrac{\pi}{3}), \sin( \theta + \tfrac{\pi}{3})$ and so $f_2(q) = \tan (\tfrac12(\theta + \tfrac{\pi}{3}))$ so
\begin{align*}
&& f_2(q) &= \tan (\tfrac12(\theta + \tfrac{\pi}{3})) \\
&&&= \frac{q + \tan \frac{\pi}{3}}{1 - \tan \frac{\pi}{3} \cdot q} \\
&&&= \frac{q + \frac{1}{\sqrt3}}{1 - \frac{q}{\sqrt{3}}} \\
&&&= \frac{\sqrt3 q + 1}{\sqrt3-q}
\end{align*}

\item Since $q \to -q$ reflects $(0,q)$ in the $x$-axis, $f_3(q) = f_1(-q)$ so $P_3$ is the reflection of $P_1$ so it's rotation by $\frac{\pi}{2}$ followed by reflection in the $x$-axis, which is reflection in $y=x$. [ie $\theta \to -\theta + \frac{\pi}{2} \to \frac{\pi}{2}-\theta$]

\item We are rotating by $\frac{\pi}{3}$ then reflecting in $y=x$ and then rotating by $-\frac{\pi}{3}$, ie $\theta \to \theta + \frac{\pi}{3} \to \frac{\pi}{6}-\theta  \to -\theta -\frac{\pi}{6} $
\end{enumerate}

\end{questionparts}