\(% 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}}\)

[next] [prev] [prev-tail] [tail] [up]

2024.2.3 Question 3

  1. The line \(NP\) has gradient \[ m_{NP} = \frac {\sin \theta - 0}{\cos \theta - (-1)} = \frac {\sin \theta }{\cos \theta + 1}, \] and hence it has equation \[ l_{NP}: y = \frac {\sin \theta }{\cos \theta + 1} \cdot (x + 1). \]

    When \(x = 0\), we have \begin {align*} q & = \frac {\sin \theta }{\cos \theta + 1} \\ & = \frac {2 \sin \frac {\theta }{2} \cos \frac {\theta }{2}}{2 \cos ^2 \frac {\theta }{2} - 1 + 1} \\ & = \frac {\sin \frac {\theta }{2}}{\cos \frac {\theta }{2}} \\ & = \tan \frac {\theta }{2}. \end {align*}

    1. \begin {align*} \RHS & = \tan \frac {1}{2} \left (\theta + \frac {1}{2} \pi \right ) \\ & = \tan \left (\frac {\theta }{2} + \frac {\pi }{4}\right ) \\ & = \frac {\tan \frac {\theta }{2} + \tan \frac {\pi }{4}}{1 - \tan \frac {\theta }{2} \tan \frac {\pi }{4}} \\ & = \frac {q + 1}{1 - q} \\ & = f_1(q), \end {align*}

      as desired.

    2. Let the coordinates of \(P_1\) be \((\cos \phi , \sin \phi )\), and hence we must have \begin {align*} f_1(q) & = \tan \frac {1}{2} \phi \\ \tan \frac {1}{2} \left (\theta + \frac {1}{2}\pi \right ) & = \tan \frac {1}{2} \phi \\ \phi & = \theta + \frac {1}{2}\pi , \end {align*}

      and so \(P_1\) is the image of \(P\) being rotated through an angle of \(\pi \) counterclockwise about the origin.

    1. The coordinates of \(P_2\) are \(\left (\cos \left (\theta + \frac {1}{3}\pi \right ), \sin \left (\theta + \frac {1}{3}\pi \right )\right )\), and hence we must have that \begin {align*} f_3(q) & = \tan \frac {1}{2} \left (\theta + \frac {1}{3}\pi \right ) \\ & = \tan \left (\frac {\theta }{2} + \frac {\pi }{6}\right ) \\ & = \frac {\tan \frac {\theta }{2} + \tan \frac {\pi }{6}}{1 - \tan \frac {\theta }{2} \tan \frac {\pi }{6}} \\ & = \frac {q + \frac {1}{\sqrt {3}}}{1 - q \cdot \frac {1}{\sqrt {3}}} \\ & = \frac {1 + \sqrt {3}q}{\sqrt {3} - q}. \end {align*}
    2. Notice that \(f_3(q) = f_1(-q) = \tan \frac {1}{2} \left (- \theta + \frac {1}{2}\pi \right )\), and so the coordinates of \(P_3\) must be \[ \left (\cos \left (\frac {1}{2}\pi - \theta \right ), \sin \left (\frac {1}{2}\pi - \theta \right )\right ), \] which is \(P_3 (\sin \theta , \cos \theta )\), a reflection of \(P\) in the line \(y = x\).

    3. \(P_4\) must be the image of \(P\) under the following transformations:

      • Rotation counterclockwise by \(\frac {1}{3}\pi \) about the origin \(O\);
      • Reflection in the line \(y = x\);
      • Rotation clockwise by \(\frac {1}{3}\pi \) about the origin \(O\).

      This is precisely the reflection in which the axis after the second step is \(y = x\). Hence, the axis of this reflection has an angle of \(\frac {1}{4} \pi - \frac {1}{3}\pi = \frac {1}{12}\pi \) with the positive \(x\)-axis.

      \(P_4\) is the image of \(P\) reflected in the line which makes an angle of \(-\frac {\pi }{12}\) with the positive \(x\)-axis, passing through the origin.

[next] [prev] [prev-tail] [front] [up]