From f6326f1c24bebb2a4042ee0dc9d9ac489f56bc8e Mon Sep 17 00:00:00 2001
From: Andrew Lorimer <andrew@lorimer.id.au>
Date: Tue, 28 May 2019 14:44:26 +1000
Subject: [PATCH] [spec] minor editors; add inverse sin&cos graphs

---
 spec/spec-collated.pdf |  Bin 452801 -> 464037 bytes
 spec/spec-collated.tex | 1720 ++++++++++++++++++++--------------------
 2 files changed, 872 insertions(+), 848 deletions(-)

diff --git a/spec/spec-collated.pdf b/spec/spec-collated.pdf
index 6db681c27b7f3febdd4796c27e81035d35425d8e..6e6dc208768da9be5b522be916a5644b69d5ccf9 100644
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diff --git a/spec/spec-collated.tex b/spec/spec-collated.tex
index b669b55..e2ccd0e 100644
--- a/spec/spec-collated.tex
+++ b/spec/spec-collated.tex
@@ -41,164 +41,167 @@
 
   \section{Complex numbers}
 
-    \[\mathbb{C}=\{a+bi:a,b\in\mathbb{R}\}\]
+  \[\mathbb{C}=\{a+bi:a,b\in\mathbb{R}\}\]
 
-    \begin{align*}
-      \text{Cartesian form: } & a+bi\\
-      \text{Polar form: } & r\operatorname{cis}\theta
-    \end{align*}
+  \begin{align*}
+    \text{Cartesian form: } & a+bi\\
+    \text{Polar form: } & r\operatorname{cis}\theta
+  \end{align*}
 
-    \subsection*{Operations}
+  \subsection*{Operations}
 
-\definecolor{shade1}{HTML}{ffffff}
-\definecolor{shade2}{HTML}{e6f2ff}
+  \definecolor{shade1}{HTML}{ffffff}
+  \definecolor{shade2}{HTML}{e6f2ff}
   \definecolor{shade3}{HTML}{cce2ff}
-      \begin{tabularx}{\columnwidth}{r|X|X}
-        & \textbf{Cartesian} & \textbf{Polar} \\
-        \hline
-        \(z_1 \pm z_2\) & \((a \pm c)(b \pm d)i\) & convert to \(a+bi\)\\
-        \hline
-        \(+k \times z\) & \multirow{2}{*}{\(ka \pm kbi\)} & \(kr\operatorname{cis} \theta\)\\
-        \cline{1-1}\cline{3-3}
-        \(-k \times z\) & & \(kr \operatorname{cis}(\theta\pm \pi)\)\\
-        \hline
-        \(z_1 \cdot z_2\) & \(ac-bd+(ad+bc)i\) & \(r_1r_2 \operatorname{cis}(\theta_1 + \theta_2)\)\\
-        \hline
-        \(z_1 \div z_2\) & \((z_1 \overline{z_2}) \div |z_2|^2\) & \(\left(\frac{r_1}{r_2}\right) \operatorname{cis}(\theta_1 - \theta_2)\)
-      \end{tabularx}
-
-      \subsubsection*{Scalar multiplication in polar form}
-      
-        For \(k \in \mathbb{R}^+\):
-        \[k\left(r \operatorname{cis}\theta\right)=kr \operatorname{cis}\theta\]
-
-        \noindent For \(k \in \mathbb{R}^-\):
-        \[k\left(r \operatorname{cis}\theta\right)=kr \operatorname{cis}\left(\begin{cases}\theta - \pi & |0<\operatorname{Arg}(z)\le \pi \\ \theta + \pi & |-\pi<\operatorname{Arg}(z)\le 0\end{cases}\right)\]
+  \begin{tabularx}{\columnwidth}{r|X|X}
+    & \textbf{Cartesian} & \textbf{Polar} \\
+    \hline
+    \(z_1 \pm z_2\) & \((a \pm c)(b \pm d)i\) & convert to \(a+bi\)\\
+    \hline
+    \(+k \times z\) & \multirow{2}{*}{\(ka \pm kbi\)} & \(kr\operatorname{cis} \theta\)\\
+    \cline{1-1}\cline{3-3}
+    \(-k \times z\) & & \(kr \operatorname{cis}(\theta\pm \pi)\)\\
+    \hline
+    \(z_1 \cdot z_2\) & \(ac-bd+(ad+bc)i\) & \(r_1r_2 \operatorname{cis}(\theta_1 + \theta_2)\)\\
+    \hline
+    \(z_1 \div z_2\) & \((z_1 \overline{z_2}) \div |z_2|^2\) & \(\left(\frac{r_1}{r_2}\right) \operatorname{cis}(\theta_1 - \theta_2)\)
+  \end{tabularx}
+
+  \subsubsection*{Scalar multiplication in polar form}
+
+  For \(k \in \mathbb{R}^+\):
+  \[k\left(r \operatorname{cis}\theta\right)=kr \operatorname{cis}\theta\]
+
+  \noindent For \(k \in \mathbb{R}^-\):
+  \[k\left(r \operatorname{cis}\theta\right)=kr \operatorname{cis}\left(\begin{cases}\theta - \pi & |0<\operatorname{Arg}(z)\le \pi \\ \theta + \pi & |-\pi<\operatorname{Arg}(z)\le 0\end{cases}\right)\]
 
     \subsection*{Conjugate}
 
-      \begin{align*}
-        \overline{z} &= a \mp bi\\
-        &= r \operatorname{cis}(-\theta)
-      \end{align*}
+    \begin{align*}
+      \overline{z} &= a \mp bi\\
+      &= r \operatorname{cis}(-\theta)
+    \end{align*}
 
-      \noindent \colorbox{cas}{On CAS: \texttt{conjg(a+bi)}}
+    \noindent \colorbox{cas}{On CAS: \texttt{conjg(a+bi)}}
 
-      \subsubsection*{Properties}
+    \subsubsection*{Properties}
 
-        \begin{align*}
-          \overline{z_1 \pm z_2} &= \overline{z_1}\pm\overline{z_2}\\
-          \overline{z_1 \cdot z_2} &= \overline{z_1}\cdot\overline{z_2}\\
-          \overline{kz} &= k\overline{z} \quad | \quad k \in \mathbb{R}\\
-          z\overline{z} &= (a+bi)(a-bi)\\
-          &= a^2 + b^2\\
-          &= |z|^2
-        \end{align*}
+    \begin{align*}
+      \overline{z_1 \pm z_2} &= \overline{z_1}\pm\overline{z_2}\\
+      \overline{z_1 \cdot z_2} &= \overline{z_1}\cdot\overline{z_2}\\
+      \overline{kz} &= k\overline{z} \quad | \quad k \in \mathbb{R}\\
+      z\overline{z} &= (a+bi)(a-bi)\\
+      &= a^2 + b^2\\
+      &= |z|^2
+    \end{align*}
 
     \subsection*{Modulus}
 
-      \[|z|=|\vec{Oz}|=\sqrt{a^2 + b^2}\]
+    \[|z|=|\vec{Oz}|=\sqrt{a^2 + b^2}\]
 
-      \subsubsection*{Properties}
+    \subsubsection*{Properties}
 
-        \begin{align*}
-          |z_1z_2|&=|z_1||z_2|\\
-          \left|\frac{z_1}{z_2}\right|&=\frac{|z_1|}{|z_2|}\\
-          |z_1+z_2|&\le|z_1|+|z_2|
-        \end{align*}
+    \begin{align*}
+      |z_1z_2|&=|z_1||z_2|\\
+      \left|\frac{z_1}{z_2}\right|&=\frac{|z_1|}{|z_2|}\\
+      |z_1+z_2|&\le|z_1|+|z_2|
+    \end{align*}
 
     \subsection*{Multiplicative inverse}
 
-      \begin{align*}
-        z^{-1}&=\frac{a-bi}{a^2+b^2}\\
-        &=\frac{\overline{z}}{|z|^2}a\\
-        &=r \operatorname{cis}(-\theta)
-      \end{align*}
+    \begin{align*}
+      z^{-1}&=\frac{a-bi}{a^2+b^2}\\
+      &=\frac{\overline{z}}{|z|^2}a\\
+      &=r \operatorname{cis}(-\theta)
+    \end{align*}
 
     \subsection*{Dividing over \(\mathbb{C}\)}
 
-      \begin{align*}
-        \frac{z_1}{z_2}&=z_1z_2^{-1}\\
-        &=\frac{z_1\overline{z_2}}{|z_2|^2}\\
-        &=\frac{(a+bi)(c-di)}{c^2+d^2}\\
-        & \qquad \text{(rationalise denominator)}
-      \end{align*}
+    \begin{align*}
+      \frac{z_1}{z_2}&=z_1z_2^{-1}\\
+      &=\frac{z_1\overline{z_2}}{|z_2|^2}\\
+      &=\frac{(a+bi)(c-di)}{c^2+d^2}\\
+      & \qquad \text{(rationalise denominator)}
+    \end{align*}
 
     \subsection*{Polar form}
 
-      \begin{align*}
-        z&=r\operatorname{cis}\theta\\
-        &=r(\cos \theta + i \sin \theta)
-      \end{align*}
+    \begin{align*}
+      z&=r\operatorname{cis}\theta\\
+      &=r(\cos \theta + i \sin \theta)
+    \end{align*}
 
-      \begin{itemize}
-        \item{\(r=|z|=\sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}\)}
-        \item{\(\theta = \operatorname{arg}(z)\) \quad \colorbox{cas}{On CAS: \texttt{arg(a+bi)}}}
-        \item{\(\operatorname{Arg}(z) \in (-\pi,\pi)\) \quad \bf{(principal argument)}}
-        \item{\colorbox{cas}{Convert on CAS:}\\ \verb|compToTrig(a+bi)| \(\iff\) \verb|cExpand{r·cisX}|}
-        \item{Multiple representations:\\\(r\operatorname{cis}\theta=r\operatorname{cis}(\theta+2n\pi)\) with \(n \in \mathbb{Z}\) revolutions}
-        \item{\(\operatorname{cis}\pi=-1,\qquad \operatorname{cis}0=1\)}
-      \end{itemize}
+    \begin{itemize}
+      \item{\(r=|z|=\sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}\)}
+      \item{\(\theta = \operatorname{arg}(z)\) \quad \colorbox{cas}{On CAS: \texttt{arg(a+bi)}}}
+      \item{\(\operatorname{Arg}(z) \in (-\pi,\pi)\) \quad \bf{(principal argument)}}
+      \item{\colorbox{cas}{Convert on CAS:}\\ \verb|compToTrig(a+bi)| \(\iff\) \verb|cExpand{r·cisX}|}
+      \item{Multiple representations:\\\(r\operatorname{cis}\theta=r\operatorname{cis}(\theta+2n\pi)\) with \(n \in \mathbb{Z}\) revolutions}
+      \item{\(\operatorname{cis}\pi=-1,\qquad \operatorname{cis}0=1\)}
+    \end{itemize}
 
     \subsection*{de Moivres' theorem}
 
     \[(r \operatorname{cis} \theta)^n = r^n \operatorname{cis}(n\theta) \text{ where } n \in \mathbb{Z}\]
 
     \subsection*{Complex polynomials}
-    
-      Include \(\pm\) for all solutions, incl. imaginary
 
-      \begin{tabularx}{\columnwidth}{ R{0.55} X  }
-        \hline
-        Sum of squares & \(\begin{aligned} 
+    Include \(\pm\) for all solutions, incl. imaginary
+
+    \begin{tabularx}{\columnwidth}{ R{0.55} X  }
+      \hline
+      Sum of squares & \(\begin{aligned} 
         z^2 + a^2 &= z^2-(ai)^2\\
-        &= (z+ai)(z-ai) \end{aligned}\) \\
-        \hline
-        Sum of cubes & \(a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)\)\\
-        \hline
-        Division & \(P(z)=D(z)Q(z)+R(z)\) \\
-        \hline
-        Remainder theorem & Let \(\alpha \in \mathbb{C}\). Remainder of \(P(z) \div (z-\alpha)\) is \(P(\alpha)\)\\
-        \hline
-        Factor theorem & \(z-\alpha\) is a factor of \(P(z) \iff P(\alpha)=0\) for \(\alpha \in \mathbb{C}\)\\
-        \hline
-        Conjugate root theorem & \(P(z)=0 \text{ at } z=a\pm bi\) (\(\implies\) both \(z_1\) and \(\overline{z_1}\) are solutions)
-      \end{tabularx}
+      &= (z+ai)(z-ai) \end{aligned}\) \\
+      \hline
+      Sum of cubes & \(a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)\)\\
+      \hline
+      Division & \(P(z)=D(z)Q(z)+R(z)\) \\
+      \hline
+      Remainder theorem & Let \(\alpha \in \mathbb{C}\). Remainder of \(P(z) \div (z-\alpha)\) is \(P(\alpha)\)\\
+      \hline
+      Factor theorem & \(z-\alpha\) is a factor of \(P(z) \iff P(\alpha)=0\) for \(\alpha \in \mathbb{C}\)\\
+      \hline
+      Conjugate root theorem & \(P(z)=0 \text{ at } z=a\pm bi\) (\(\implies\) both \(z_1\) and \(\overline{z_1}\) are solutions)\\
+      \hline
+    \end{tabularx}
 
-    \subsection*{Roots}
+    \subsection*{\(n\)th roots}
 
-      \(n\)th roots of \(z=r\operatorname{cis}\theta\) are:
+    \(n\)th roots of \(z=r\operatorname{cis}\theta\) are:
 
-      \[z = r^{\frac{1}{n}} \operatorname{cis}\left(\frac{\theta+2k\pi}{n}\right)\]
+    \[z = r^{\frac{1}{n}} \operatorname{cis}\left(\frac{\theta+2k\pi}{n}\right)\]
 
-      \begin{itemize}
+    \begin{itemize}
 
-        \item{Same modulus for all solutions}
-        \item{Arguments are separated by \(\frac{2\pi}{n}\)}
-        \item{Solutions of \(z^n=a\) where \(a \in \mathbb{C}\) lie on the circle \(x^2+y^2=\left(|a|^{\frac{1}{n}}\right)^2\) \quad (intervals of \(\frac{2\pi}{n}\))}
-      \end{itemize}
+      \item{Same modulus for all solutions}
+      \item{Arguments separated by \(\frac{2\pi}{n} \therefore\) there are \(n\) roots}
+      \item{If one square root is \(a+bi\), the other is \(-a-bi\)}
+      \item{Give one implicit \(n\)th root \(z_1\), function is \(z=z_1^n\)}
+      \item{Solutions of \(z^n=a\) where \(a \in \mathbb{C}\) lie on the circle \(x^2+y^2=\left(|a|^{\frac{1}{n}}\right)^2\) \quad (intervals of \(\frac{2\pi}{n}\))}
+    \end{itemize}
 
-      \noindent For \(0=az^2+bz+c\), use quadratic formula:
+    \noindent For \(0=az^2+bz+c\), use quadratic formula:
 
-      \[z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
+    \[z=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
 
     \subsection*{Fundamental theorem of algebra}
 
-      A polynomial of degree \(n\) can be factorised into \(n\) linear factors in \(\mathbb{C}\):
+    A polynomial of degree \(n\) can be factorised into \(n\) linear factors in \(\mathbb{C}\):
 
-        \[\implies P(z)=a_n(z-\alpha_1)(z-\alpha_2)(z-\alpha_3)\dots(z-\alpha_n)\]
-        \[\text{ where } \alpha_1,\alpha_2,\alpha_3,\dots,\alpha_n \in \mathbb{C}\]
+    \[\implies P(z)=a_n(z-\alpha_1)(z-\alpha_2)(z-\alpha_3)\dots(z-\alpha_n)\]
+    \[\text{ where } \alpha_1,\alpha_2,\alpha_3,\dots,\alpha_n \in \mathbb{C}\]
 
     \subsection*{Argand planes}
-    
-      \begin{center}\begin{tikzpicture}[scale=2]
-        \draw [->] (-0.2,0) -- (1.5,0) node [right]  {$\operatorname{Re}(z)$};
-        \draw [->] (0,-0.2) -- (0,1.5) node [above] {$\operatorname{Im}(z)$};
-        \coordinate (P) at (1,1);
-        \coordinate (a) at (1,0);
-        \coordinate (b) at (0,1);
-        \coordinate (O) at (0,0);
-        \draw (0,0) -- (P) node[pos=0.5, above left]{\(r\)} node[pos=1, right]{\(\begin{aligned}z&=a+bi\\&=r\operatorname{cis}\theta\end{aligned}\)};
+
+    \begin{center}\begin{tikzpicture}[scale=2]
+      \draw [->] (-0.2,0) -- (1.5,0) node [right]  {$\operatorname{Re}(z)$};
+      \draw [->] (0,-0.2) -- (0,1.5) node [above] {$\operatorname{Im}(z)$};
+      \coordinate (P) at (1,1);
+      \coordinate (a) at (1,0);
+      \coordinate (b) at (0,1);
+      \coordinate (O) at (0,0);
+      \draw (0,0) -- (P) node[pos=0.5, above left]{\(r\)} node[pos=1, right]{\(\begin{aligned}z&=a+bi\\&=r\operatorname{cis}\theta\end{aligned}\)};
         \draw [gray, dashed] (1,1) -- (1,0) node[black, pos=1, below]{\(a\)};
         \draw [gray, dashed] (1,1) -- (0,1) node[black, pos=1, left]{\(b\)};
         \begin{scope}
@@ -207,880 +210,901 @@
           \node at ($(O)+(20:3mm)$) {$\theta$};
         \end{scope}
         \filldraw (P) circle (0.5pt);
-      \end{tikzpicture}\end{center}
+    \end{tikzpicture}\end{center}
 
-      \begin{itemize}
-        \item{Multiplication by \(i \implies\) CCW rotation of \(\frac{\pi}{2}\)}
-        \item{Addition: \(z_1 + z_2 \equiv\) \overrightharp{\(Oz_1\)} + \overrightharp{\(Oz_2\)}}
-      \end{itemize}
+    \begin{itemize}
+      \item{Multiplication by \(i \implies\) CCW rotation of \(\frac{\pi}{2}\)}
+      \item{Addition: \(z_1 + z_2 \equiv\) \overrightharp{\(Oz_1\)} + \overrightharp{\(Oz_2\)}}
+    \end{itemize}
 
     \subsection*{Sketching complex graphs}
-      
-      \subsubsection*{Linear}
 
-        \begin{itemize}
-          \item{\(\operatorname{Re}(z)=c\) or \(\operatorname{Im}(z)=c\) (perpendicular bisector)}
-          \item{\(\operatorname{Im}(z)=m\operatorname{Re}(z)\)}
-          \item{\(|z+a|=|z+b| \implies 2(a-b)x=b^2-a^2\)\\Geometric: equidistant from \(a,b\)}
-        \end{itemize}
-
-      \subsubsection*{Circles}
-
-        \begin{itemize}
-          \item \(|z-z_1|^2=c^2|z_2+2|^2\)
-          \item \(|z-(a+bi)|=c \implies (x-a)^2+_(y-b)^2=c^2\)
-        \end{itemize}
-
-      \noindent \textbf{Loci} \qquad \(\operatorname{Arg}(z)<\theta\)
-
-        \begin{center}\begin{tikzpicture}[scale=2,mydot/.style={circle, fill=white, draw, outer sep=0pt, inner sep=1.5pt}]
-          \draw [->] (0,0) -- (1,0) node [right]  {$\operatorname{Re}(z)$};
-          \draw [->] (0,-0.5) -- (0,1) node [above] {$\operatorname{Im}(z)$};
-          \draw [<-, dashed, thick, blue] (-1,0) -- (0,0);
-          \draw [->, thick, blue] (0,0) -- (1,1);
-          \fill [gray, opacity=0.2, domain=-1:1, variable=\x] (-1,-0.5) -- (-1,0) -- (0, 0) -- (1,1) -- (1,-0.5) -- cycle;
-          \begin{scope}
-            \path[clip] (0,0) -- (1,1) -- (1,0);
-            \fill[red, opacity=0.5, draw=black] (0,0) circle (2mm);
-            \node at ($(0,0)+(20:3mm)$) {$\frac{\pi}{4}$};
-          \end{scope}
-          \node [font=\footnotesize] at (0.5,-0.25) {\(\operatorname{Arg}(z)\le\frac{\pi}{4}\)};
-          \node [blue, mydot] {};
-        \end{tikzpicture}\end{center}
-
-      \noindent \textbf{Rays} \qquad \(\operatorname{Arg}(z-b)=\theta\)
-
-        \begin{center}\begin{tikzpicture}[scale=2,mydot/.style={circle, fill=white, draw, outer sep=0pt, inner sep=1.5pt}]
-          \draw [->] (-0.75,0) -- (1.5,0) node [right]  {$\operatorname{Re}(z)$};
-          \draw [->] (0,-1) -- (0,1) node [above] {$\operatorname{Im}(z)$};
-          \draw [->, thick, brown] (-0.25,0) -- (-0.75,-1);
-          \node [above, font=\footnotesize] at (-0.25,0) {\(\frac{1}{4}\)};
-          \begin{scope}
-            \path[clip] (-0.25,0) -- (-0.75,-1) -- (0,0);
-            \fill[orange, opacity=0.5, draw=black] (-0.25,0) circle (2mm);
-          \end{scope}
-          \node at (-0.08,-0.3) {\(\frac{\pi}{8}\)};
-          \node [font=\footnotesize, left] at (-0.75,-1) {\(\operatorname{Arg}(z+\frac{1}{4})=\frac{\pi}{8}\)};
-          \node [brown, mydot] at (-0.25,0) {};
-          \draw [<->, thick, green] (0,-1) -- (1.5,0.5) node [pos=0.25, black, font=\footnotesize, right] {\(|z-2|=|z-(1+i)|\)};
-          \node [left, font=\footnotesize] at (0,-1) {\(-1\)};
-          \node [below, font=\footnotesize] at (1,0) {\(1\)};
-        \end{tikzpicture}\end{center}
+    \subsubsection*{Linear}
+
+    \begin{itemize}
+      \item{\(\operatorname{Re}(z)=c\) or \(\operatorname{Im}(z)=c\) (perpendicular bisector)}
+      \item{\(\operatorname{Im}(z)=m\operatorname{Re}(z)\)}
+      \item{\(|z+a|=|z+b| \implies 2(a-b)x=b^2-a^2\)\\Geometric: equidistant from \(a,b\)}
+    \end{itemize}
+
+    \subsubsection*{Circles}
+
+    \begin{itemize}
+      \item \(|z-z_1|^2=c^2|z_2+2|^2\)
+      \item \(|z-(a+bi)|=c \implies (x-a)^2+_(y-b)^2=c^2\)
+    \end{itemize}
+
+    \noindent \textbf{Loci} \qquad \(\operatorname{Arg}(z)<\theta\)
+
+    \begin{center}\begin{tikzpicture}[scale=2,mydot/.style={circle, fill=white, draw, outer sep=0pt, inner sep=1.5pt}]
+      \draw [->] (0,0) -- (1,0) node [right]  {$\operatorname{Re}(z)$};
+      \draw [->] (0,-0.5) -- (0,1) node [above] {$\operatorname{Im}(z)$};
+      \draw [<-, dashed, thick, blue] (-1,0) -- (0,0);
+      \draw [->, thick, blue] (0,0) -- (1,1);
+      \fill [gray, opacity=0.2, domain=-1:1, variable=\x] (-1,-0.5) -- (-1,0) -- (0, 0) -- (1,1) -- (1,-0.5) -- cycle;
+      \begin{scope}
+        \path[clip] (0,0) -- (1,1) -- (1,0);
+        \fill[red, opacity=0.5, draw=black] (0,0) circle (2mm);
+        \node at ($(0,0)+(20:3mm)$) {$\frac{\pi}{4}$};
+      \end{scope}
+      \node [font=\footnotesize] at (0.5,-0.25) {\(\operatorname{Arg}(z)\le\frac{\pi}{4}\)};
+      \node [blue, mydot] {};
+    \end{tikzpicture}\end{center}
+
+    \noindent \textbf{Rays} \qquad \(\operatorname{Arg}(z-b)=\theta\)
+
+    \begin{center}\begin{tikzpicture}[scale=2,mydot/.style={circle, fill=white, draw, outer sep=0pt, inner sep=1.5pt}]
+      \draw [->] (-0.75,0) -- (1.5,0) node [right]  {$\operatorname{Re}(z)$};
+      \draw [->] (0,-1) -- (0,1) node [above] {$\operatorname{Im}(z)$};
+      \draw [->, thick, brown] (-0.25,0) -- (-0.75,-1);
+      \node [above, font=\footnotesize] at (-0.25,0) {\(\frac{1}{4}\)};
+      \begin{scope}
+        \path[clip] (-0.25,0) -- (-0.75,-1) -- (0,0);
+        \fill[orange, opacity=0.5, draw=black] (-0.25,0) circle (2mm);
+      \end{scope}
+      \node at (-0.08,-0.3) {\(\frac{\pi}{8}\)};
+      \node [font=\footnotesize, left] at (-0.75,-1) {\(\operatorname{Arg}(z+\frac{1}{4})=\frac{\pi}{8}\)};
+      \node [brown, mydot] at (-0.25,0) {};
+      \draw [<->, thick, green] (0,-1) -- (1.5,0.5) node [pos=0.25, black, font=\footnotesize, right] {\(|z-2|=|z-(1+i)|\)};
+      \node [left, font=\footnotesize] at (0,-1) {\(-1\)};
+      \node [below, font=\footnotesize] at (1,0) {\(1\)};
+    \end{tikzpicture}\end{center}
 
     \section{Vectors}
-\begin{center}\begin{tikzpicture}
-  \draw [->] (-0.5,0) -- (3,0) node [right]  {\(x\)};
-          \draw [->] (0,-0.5) -- (0,3) node [above] {\(y\)};
-          \draw [orange, ->, thick] (0.5,0.5) -- (2.5,2.5) node [pos=0.5, above] {\(\vec{u}\)};
-         \begin{scope}[very thick, every node/.style={sloped,allow upside down}]
+    \begin{center}\begin{tikzpicture}
+      \draw [->] (-0.5,0) -- (3,0) node [right]  {\(x\)};
+      \draw [->] (0,-0.5) -- (0,3) node [above] {\(y\)};
+      \draw [orange, ->, thick] (0.5,0.5) -- (2.5,2.5) node [pos=0.5, above] {\(\vec{u}\)};
+      \begin{scope}[very thick, every node/.style={sloped,allow upside down}]
         \draw [gray, dashed, thick] (0.5,0.5) -- (2.5,0.5) node [pos=0.5] {\midarrow} node[black, pos=0.5, below]{\(x\vec{i}\)};
         \draw [gray, dashed, thick] (2.5,0.5) -- (2.5,2.5) node [pos=0.5] {\midarrow};
-         \end{scope}
-        \node[black, right] at (2.5,1.5) {\(y\vec{j}\)};
-\end{tikzpicture}\end{center}
-\subsection*{Column notation}
+      \end{scope}
+      \node[black, right] at (2.5,1.5) {\(y\vec{j}\)};
+    \end{tikzpicture}\end{center}
+    \subsection*{Column notation}
 
-\[\begin{bmatrix}x\\ y \end{bmatrix} \iff x\boldsymbol{i} + y\boldsymbol{j}\]
-\(\begin{bmatrix}x_2-x_1\\ y_2-y_1 \end{bmatrix}\) \quad between \(A(x_1,y_1), \> B(x_2,y_2)\)
+    \[\begin{bmatrix}x\\ y \end{bmatrix} \iff x\boldsymbol{i} + y\boldsymbol{j}\]
+      \(\begin{bmatrix}x_2-x_1\\ y_2-y_1 \end{bmatrix}\) \quad between \(A(x_1,y_1), \> B(x_2,y_2)\)
 
-\subsection*{Scalar multiplication}
+        \subsection*{Scalar multiplication}
 
-\[k\cdot (x\boldsymbol{i}+y\boldsymbol{j})=kx\boldsymbol{i}+ky\boldsymbol{j}\]
+        \[k\cdot (x\boldsymbol{i}+y\boldsymbol{j})=kx\boldsymbol{i}+ky\boldsymbol{j}\]
 
-\noindent For \(k \in \mathbb{R}^-\), direction is reversed
+        \noindent For \(k \in \mathbb{R}^-\), direction is reversed
 
-\subsection*{Vector addition}
-\begin{center}\begin{tikzpicture}[scale=1]
+        \subsection*{Vector addition}
+        \begin{center}\begin{tikzpicture}[scale=1]
           \coordinate (A) at (0,0);
           \coordinate (B) at (2,2);
           \draw [->, thick, red] (0,0) -- (2,2) node [pos=0.5, below right] {\(\vec{u}=2\vec{i}+2\vec{j}\)};
           \draw [->, thick, blue] (2,2) -- (1,4) node [pos=0.5, above right] {\(\vec{v}=-\vec{i}+2\vec{j}\)};
           \draw [->, thick, orange] (0,0) -- (1,4) node [pos=0.5, left] {\(\vec{u}+\vec{v}=\vec{i}+4\vec{j}\)};
-\end{tikzpicture}\end{center}
-
-\[(x\boldsymbol{i}+y\boldsymbol{j}) \pm (a\boldsymbol{i}+b\boldsymbol{j})=(x \pm a)\boldsymbol{i}+(y \pm b)\boldsymbol{j}\]
-
-\begin{itemize}
-  \item Draw each vector head to tail then join lines
-  \item Addition is commutative (parallelogram)
-  \item \(\boldsymbol{u}-\boldsymbol{v}=\boldsymbol{u}+(-\boldsymbol{v}) \implies \overrightharp{AB}=\boldsymbol{b}-\boldsymbol{a}\)
-\end{itemize}
-
-\subsection*{Magnitude}
-
-\[|(x\boldsymbol{i} + y\boldsymbol{j})|=\sqrt{x^2+y^2}\]
-
-\subsection*{Parallel vectors}
-
-\[\boldsymbol{u} || \boldsymbol{v} \iff \boldsymbol{u} = k \boldsymbol{v} \text{ where } k \in \mathbb{R} \setminus \{0\}\]
-
-For parallel vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\):\\
-\[\boldsymbol{a \cdot b}=\begin{cases}
-|\boldsymbol{a}||\boldsymbol{b}| \hspace{2.8em} \text{if same direction}\\
--|\boldsymbol{a}||\boldsymbol{b}| \hspace{2em} \text{if opposite directions}
-\end{cases}\]
-%\includegraphics[width=0.2,height=\textheight]{graphics/parallelogram-vectors.jpg}
-%\includegraphics[width=1]{graphics/vector-subtraction.jpg}
-
-\subsection*{Perpendicular vectors}
-
-\[\boldsymbol{a} \perp \boldsymbol{b} \iff \boldsymbol{a} \cdot \boldsymbol{b} = 0\ \quad \text{(since \(\cos 90 = 0\))}\]
-
-\subsection*{Unit vector \(|\hat{\boldsymbol{a}}|=1\)}
-\[\begin{split}\hat{\boldsymbol{a}} & = {\frac{1}{|\boldsymbol{a}|}}\boldsymbol{a} \\ & = \boldsymbol{a} \cdot {|\boldsymbol{a}|}\end{split}\]
-
-  \subsection*{Scalar product \(\boldsymbol{a} \cdot \boldsymbol{b}\)}
-
-
-\begin{center}\begin{tikzpicture}[scale=2]
-  \draw [->] (0,0) -- (1,0.5) node [pos=0.5, above left] {\(\boldsymbol{b}\)};
-  \draw [->] (0,0) -- (1,0) node [pos=0.5, below] {\(\boldsymbol{a}\)};
-          \begin{scope}
-            \path[clip] (1,0.5) -- (1,0) -- (0,0);
-            \fill[orange, opacity=0.5, draw=black] (0,0) circle (2mm);
-            \node at ($(0,0)+(15:4mm)$) {\(\theta\)};
-          \end{scope}
-\end{tikzpicture}\end{center}
-\begin{align*}\boldsymbol{a} \cdot \boldsymbol{b} &= a_1 b_1 + a_2 b_2 \\  &= |\boldsymbol{a}| |\boldsymbol{b}| \cos \theta \\ &\quad (\> 0 \le \theta \le \pi) \text{ - from cosine rule}\end{align*}
-\noindent\colorbox{cas}{On CAS: \texttt{dotP({[}a\ b\ c{]},\ {[}d\ e\ f{]})}}
-
-\subsubsection*{Properties}
-
-\begin{enumerate}
-\item
-  \(k(\boldsymbol{a\cdot b})=(k\boldsymbol{a})\cdot \boldsymbol{b}=\boldsymbol{a}\cdot (k\boldsymbol{b})\)
-\item
-  \(\boldsymbol{a \cdot 0}=0\)
-\item
-  \(\boldsymbol{a} \cdot (\boldsymbol{b} + \boldsymbol{c})=\boldsymbol{a} \cdot \boldsymbol{b} + \boldsymbol{a} \cdot \boldsymbol{c}\)
-\item
-  \(\boldsymbol{i \cdot i} = \boldsymbol{j \cdot j} = \boldsymbol{k \cdot k}= 1\)
-\item
-  \(\boldsymbol{a} \cdot \boldsymbol{b} = 0 \quad \implies \quad \boldsymbol{a} \perp \boldsymbol{b}\)
-\item
-  \(\boldsymbol{a \cdot a} = |\boldsymbol{a}|^2 = a^2\)
-\end{enumerate}
-
-\subsection*{Angle between vectors}
-
-\[\cos \theta = \frac{\boldsymbol{a} \cdot \boldsymbol{b}}{|\boldsymbol{a}| |\boldsymbol{b}|} = \frac{a_1 b_1 + a_2 b_2}{|\boldsymbol{a}| |\boldsymbol{b}|}\]
-
-\noindent \colorbox{cas}{On CAS:} \texttt{angle([a b c], [a b c])}
-
-(Action \(\rightarrow\) Vector \(\rightarrow\)Angle)
-
-\subsection*{Angle between vector and axis}
-
-\noindent For\(\boldsymbol{a} = a_1 \boldsymbol{i} + a_2 \boldsymbol{j} + a_3 \boldsymbol{k}\)
-which makes angles \(\alpha, \beta, \gamma\) with positive side of
-\(x, y, z\) axes:
-\[\cos \alpha = \frac{a_1}{|\boldsymbol{a}|}, \quad \cos \beta = \frac{a_2}{|\boldsymbol{a}|}, \quad \cos \gamma = \frac{a_3}{|\boldsymbol{a}|}\]
-
-\noindent \colorbox{cas}{On CAS:} \texttt{angle({[}a\ b\ c{]},\ {[}1\ 0\ 0{]})}\\for angle
-between \(a\boldsymbol{i} + b\boldsymbol{j} + c\boldsymbol{k}\) and
-\(x\)-axis
-
-\subsection*{Projections \& resolutes}
-
-\begin{tikzpicture}[scale=3]
-  \draw [->, purple] (0,0) -- (1,0.5) node [pos=0.5, above left] {\(\boldsymbol{a}\)};
-  \draw [->, orange] (0,0) -- (1,0) node [pos=0.5, below] {\(\boldsymbol{u}\)};
-  \draw [->, blue] (1,0) -- (2,0) node [pos=0.5, below] {\(\boldsymbol{b}\)};
-          \begin{scope}
-            \path[clip] (1,0.5) -- (1,0) -- (0,0);
-            \fill[orange, opacity=0.5, draw=black] (0,0) circle (2mm);
-            \node at ($(0,0)+(15:4mm)$) {\(\theta\)};
-          \end{scope}
-         \begin{scope}[very thick, every node/.style={sloped,allow upside down}]
-        \draw [gray, dashed, thick] (1,0) -- (1,0.5) node [pos=0.5] {\midarrow} node[black, pos=0.5, right, rotate=-90]{\(\boldsymbol{w}\)};
-          \end{scope}
-\draw (0,0) coordinate (O)
-  (1,0) coordinate (A)
-  (1,0.5) coordinate (B)
-  pic [draw,red,angle radius=2mm] {right angle = O--A--B};
-\end{tikzpicture}
-
-\subsubsection*{\(\parallel\boldsymbol{b}\) (vector projection/resolute)}
-
-\begin{align*}
-  \boldsymbol{u} & = \frac{\boldsymbol{a}\cdot\boldsymbol{b}}{|\boldsymbol{b}|^2}\boldsymbol{b} \\
-  & = \left(\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{|\boldsymbol{b}|}\right)\left(\frac{\boldsymbol{b}}{|\boldsymbol{b}|}\right) \\
-  & = (\boldsymbol{a} \cdot \hat{\boldsymbol{b}})\hat{\boldsymbol{b}}
-\end{align*}
-
-\subsubsection*{\(\perp\boldsymbol{b}\) (perpendicular projection)}
-\[\boldsymbol{w} = \boldsymbol{a} - \boldsymbol{u}\]
+        \end{tikzpicture}\end{center}
 
-\subsubsection*{\(|\boldsymbol{u}|\) (scalar resolute)}
-\begin{align*}
-  r_s &= |\boldsymbol{u}|\\
-  &= \boldsymbol{a} \cdot \hat{\boldsymbol{b}}\\
-  &=\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{|\boldsymbol{b}|}
-\end{align*}
+        \[(x\boldsymbol{i}+y\boldsymbol{j}) \pm (a\boldsymbol{i}+b\boldsymbol{j})=(x \pm a)\boldsymbol{i}+(y \pm b)\boldsymbol{j}\]
 
-\subsubsection*{Rectangular (\(\parallel,\perp\)) components}
+        \begin{itemize}
+          \item Draw each vector head to tail then join lines
+          \item Addition is commutative (parallelogram)
+          \item \(\boldsymbol{u}-\boldsymbol{v}=\boldsymbol{u}+(-\boldsymbol{v}) \implies \overrightharp{AB}=\boldsymbol{b}-\boldsymbol{a}\)
+        \end{itemize}
 
-\[\boldsymbol{a}=\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{\boldsymbol{b}\cdot\boldsymbol{b}}\boldsymbol{b}+\left(\boldsymbol{a}-\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{\boldsymbol{b}\cdot\boldsymbol{b}}\boldsymbol{b}\right)\]
+        \subsection*{Magnitude}
 
+        \[|(x\boldsymbol{i} + y\boldsymbol{j})|=\sqrt{x^2+y^2}\]
+
+        \subsection*{Parallel vectors}
 
-\subsection*{Vector proofs}
+        \[\boldsymbol{u} || \boldsymbol{v} \iff \boldsymbol{u} = k \boldsymbol{v} \text{ where } k \in \mathbb{R} \setminus \{0\}\]
+
+        For parallel vectors \(\boldsymbol{a}\) and \(\boldsymbol{b}\):\\
+        \[\boldsymbol{a \cdot b}=\begin{cases}
+          |\boldsymbol{a}||\boldsymbol{b}| \hspace{2.8em} \text{if same direction}\\
+          -|\boldsymbol{a}||\boldsymbol{b}| \hspace{2em} \text{if opposite directions}
+        \end{cases}\]
+        %\includegraphics[width=0.2,height=\textheight]{graphics/parallelogram-vectors.jpg}
+        %\includegraphics[width=1]{graphics/vector-subtraction.jpg}
+
+        \subsection*{Perpendicular vectors}
 
-\textbf{Concurrent:} intersection of \(\ge\) 3 lines
+        \[\boldsymbol{a} \perp \boldsymbol{b} \iff \boldsymbol{a} \cdot \boldsymbol{b} = 0\ \quad \text{(since \(\cos 90 = 0\))}\]
 
-\begin{tikzpicture}
-  \draw [blue] (0,0) -- (1,1);
-  \draw [red] (1,0) -- (0,1);
-  \draw [brown] (0.4,0) -- (0.6,1);
-        \filldraw (0.5,0.5) circle (2pt);
-\end{tikzpicture}
+        \subsection*{Unit vector \(|\hat{\boldsymbol{a}}|=1\)}
+        \[\begin{split}\hat{\boldsymbol{a}} & = {\frac{1}{|\boldsymbol{a}|}}\boldsymbol{a} \\ & = \boldsymbol{a} \cdot {|\boldsymbol{a}|}\end{split}\]
 
-\subsubsection*{Collinear points}
+          \subsection*{Scalar product \(\boldsymbol{a} \cdot \boldsymbol{b}\)}
 
-\(\ge\) 3 points lie on the same line
 
-\begin{tikzpicture}
-  \draw [purple] (0,0) -- (4,1);
-  \filldraw (2,0.5) circle (2pt) node [above] {\(C\)};
-  \filldraw (1,0.25) circle (2pt) node [above] {\(A\)};
-  \filldraw (3,0.75) circle (2pt) node [above] {\(B\)};
-  \coordinate (O) at (2.8,-0.2);
-  \node at (O) [below] {\(O\)}; 
-         \begin{scope}[->, orange, thick] 
-           \draw (O) -- (2,0.5) node [pos=0.5, above, font=\footnotesize, black] {\(\boldsymbol{c}\)};
-           \draw (O) -- (1,0.25) node [pos=0.5, below, font=\footnotesize, black] {\(\boldsymbol{a}\)};
-           \draw (O) -- (3,0.75) node [pos=0.5, right, font=\footnotesize, black] {\(\boldsymbol{b}\)};
-         \end{scope}
-\end{tikzpicture}
+          \begin{center}\begin{tikzpicture}[scale=2]
+            \draw [->] (0,0) -- (1,0.5) node [pos=0.5, above left] {\(\boldsymbol{b}\)};
+            \draw [->] (0,0) -- (1,0) node [pos=0.5, below] {\(\boldsymbol{a}\)};
+            \begin{scope}
+              \path[clip] (1,0.5) -- (1,0) -- (0,0);
+              \fill[orange, opacity=0.5, draw=black] (0,0) circle (2mm);
+              \node at ($(0,0)+(15:4mm)$) {\(\theta\)};
+            \end{scope}
+          \end{tikzpicture}\end{center}
+          \begin{align*}\boldsymbol{a} \cdot \boldsymbol{b} &= a_1 b_1 + a_2 b_2 \\  &= |\boldsymbol{a}| |\boldsymbol{b}| \cos \theta \\ &\quad (\> 0 \le \theta \le \pi) \text{ - from cosine rule}\end{align*}
+            \noindent\colorbox{cas}{On CAS: \texttt{dotP({[}a\ b\ c{]},\ {[}d\ e\ f{]})}}
 
-\begin{align*}
-  \text{e.g. Prove that}\\
-  \overrightharp{AC}=m\overrightharp{AB} \iff \boldsymbol{c}&=(1-m)\boldsymbol{a}+m\boldsymbol{b}\\
-  \implies \boldsymbol{c} &= \overrightharp{OA} + \overrightharp{AC}\\
-  &= \overrightharp{OA} + m\overrightharp{AB}\\
-  &=\boldsymbol{a}+m(\boldsymbol{b}-\boldsymbol{a})\\
-  &=\boldsymbol{a}+m\boldsymbol{b}-m\boldsymbol{a}\\
-  &=(1-m)\boldsymbol{a}+m{b}
-\end{align*}
-\begin{align*}
-  \text{Also, } \implies \overrightharp{OC} &= \lambda \vec{OA} + \mu \overrightharp{OB} \\
-  \text{where } \lambda + \mu &= 1\\
-  \text{If } C \text{ lies along } \overrightharp{AB}, & \implies 0 < \mu < 1
-\end{align*}
-
-
-\subsubsection*{Parallelograms}
-
-\begin{center}\begin{tikzpicture}
-  \coordinate (O) at (0,0) node [below left] {\(O\)};
-  \coordinate (A) at (4,0);
-  \coordinate (B) at (6,2);
-  \coordinate (C) at (2,2);
-  \coordinate (D) at (6,0);
-
-    \draw[postaction={decorate}, decoration={markings, mark=at position 0.6 with {\arrow{>>}}}] (O)--(A) node [below left] {\(A\)};
-    \draw[postaction={decorate}, decoration={markings,mark=at position 0.5 with {\arrow{>}}}] (A)--(B) node [above right] {\(B\)};
-    \draw[postaction={decorate}, decoration={markings, mark=at position 0.6 with {\arrow{>>}}}] (B)--(C) node [above left] {\(C\)};
-    \draw[postaction={decorate}, decoration={markings,mark=at position 0.5 with {\arrow{>}}}] (C)--(O);
-
-    \draw [gray, dashed] (O) -- (B) node [pos=0.75] {\(\diagdown\diagdown\)} node [pos=0.25] {\(\diagdown\diagdown\)};
-    \draw [gray, dashed] (A) -- (C) node [pos=0.75] {\(\diagup\)} node [pos=0.25] {\(\diagup\)};
-          \begin{scope}
-            \path[clip] (C) -- (A) -- (O);
-            \fill[orange, opacity=0.5, draw=black] (0,0) circle (4mm);
-            \node at ($(0,0)+(20:8mm)$) {\(\theta\)};
-          \end{scope}
-          \draw [gray, thick, dotted] (B) -- (D) node [pos=0.5, right, black, font=\footnotesize] {\(|\boldsymbol{c}|\sin\theta\)} (A) -- (D) node [pos=0.5, below, black, font=\footnotesize] {\(|\boldsymbol{c}|\cos\theta\)};
-  \draw pic [draw,thick,red,angle radius=2mm] {right angle=O--D--B};
-\end{tikzpicture}\end{center}
+            \subsubsection*{Properties}
 
-\begin{itemize}
-  \item
-    Diagonals \(\overrightharp{OB}, \overrightharp{AC}\) bisect each other
-  \item
-    If diagonals are equal length, it is a rectangle
-  \item
-    \(|\overrightharp{OB}|^2 + |\overrightharp{CA}|^2 = |\overrightharp{OA}|^2 + |\overrightharp{AB}|^2 + |\overrightharp{CB}|^2 + |\overrightharp{OC}|^2\)
-  \item
-    Area \(=\boldsymbol{c} \cdot \boldsymbol{a}\)
-\end{itemize}
-
-  \subsubsection*{Useful vector properties}
+            \begin{enumerate}
+              \item
+                \(k(\boldsymbol{a\cdot b})=(k\boldsymbol{a})\cdot \boldsymbol{b}=\boldsymbol{a}\cdot (k\boldsymbol{b})\)
+              \item
+                \(\boldsymbol{a \cdot 0}=0\)
+              \item
+                \(\boldsymbol{a} \cdot (\boldsymbol{b} + \boldsymbol{c})=\boldsymbol{a} \cdot \boldsymbol{b} + \boldsymbol{a} \cdot \boldsymbol{c}\)
+              \item
+                \(\boldsymbol{i \cdot i} = \boldsymbol{j \cdot j} = \boldsymbol{k \cdot k}= 1\)
+              \item
+                \(\boldsymbol{a} \cdot \boldsymbol{b} = 0 \quad \implies \quad \boldsymbol{a} \perp \boldsymbol{b}\)
+              \item
+                \(\boldsymbol{a \cdot a} = |\boldsymbol{a}|^2 = a^2\)
+            \end{enumerate}
+
+            \subsection*{Angle between vectors}
+
+            \[\cos \theta = \frac{\boldsymbol{a} \cdot \boldsymbol{b}}{|\boldsymbol{a}| |\boldsymbol{b}|} = \frac{a_1 b_1 + a_2 b_2}{|\boldsymbol{a}| |\boldsymbol{b}|}\]
+
+            \noindent \colorbox{cas}{On CAS:} \texttt{angle([a b c], [a b c])}
+
+            (Action \(\rightarrow\) Vector \(\rightarrow\)Angle)
+
+            \subsection*{Angle between vector and axis}
+
+            \noindent For\(\boldsymbol{a} = a_1 \boldsymbol{i} + a_2 \boldsymbol{j} + a_3 \boldsymbol{k}\)
+            which makes angles \(\alpha, \beta, \gamma\) with positive side of
+            \(x, y, z\) axes:
+            \[\cos \alpha = \frac{a_1}{|\boldsymbol{a}|}, \quad \cos \beta = \frac{a_2}{|\boldsymbol{a}|}, \quad \cos \gamma = \frac{a_3}{|\boldsymbol{a}|}\]
+
+            \noindent \colorbox{cas}{On CAS:} \texttt{angle({[}a\ b\ c{]},\ {[}1\ 0\ 0{]})}\\for angle
+            between \(a\boldsymbol{i} + b\boldsymbol{j} + c\boldsymbol{k}\) and
+            \(x\)-axis
+
+            \subsection*{Projections \& resolutes}
+
+            \begin{tikzpicture}[scale=3]
+              \draw [->, purple] (0,0) -- (1,0.5) node [pos=0.5, above left] {\(\boldsymbol{a}\)};
+              \draw [->, orange] (0,0) -- (1,0) node [pos=0.5, below] {\(\boldsymbol{u}\)};
+              \draw [->, blue] (1,0) -- (2,0) node [pos=0.5, below] {\(\boldsymbol{b}\)};
+              \begin{scope}
+                \path[clip] (1,0.5) -- (1,0) -- (0,0);
+                \fill[orange, opacity=0.5, draw=black] (0,0) circle (2mm);
+                \node at ($(0,0)+(15:4mm)$) {\(\theta\)};
+              \end{scope}
+              \begin{scope}[very thick, every node/.style={sloped,allow upside down}]
+                \draw [gray, dashed, thick] (1,0) -- (1,0.5) node [pos=0.5] {\midarrow} node[black, pos=0.5, right, rotate=-90]{\(\boldsymbol{w}\)};
+              \end{scope}
+              \draw (0,0) coordinate (O)
+              (1,0) coordinate (A)
+              (1,0.5) coordinate (B)
+              pic [draw,red,angle radius=2mm] {right angle = O--A--B};
+            \end{tikzpicture}
+
+            \subsubsection*{\(\parallel\boldsymbol{b}\) (vector projection/resolute)}
+
+            \begin{align*}
+              \boldsymbol{u} & = \frac{\boldsymbol{a}\cdot\boldsymbol{b}}{|\boldsymbol{b}|^2}\boldsymbol{b} \\
+              & = \left(\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{|\boldsymbol{b}|}\right)\left(\frac{\boldsymbol{b}}{|\boldsymbol{b}|}\right) \\
+              & = (\boldsymbol{a} \cdot \hat{\boldsymbol{b}})\hat{\boldsymbol{b}}
+            \end{align*}
+
+            \subsubsection*{\(\perp\boldsymbol{b}\) (perpendicular projection)}
+            \[\boldsymbol{w} = \boldsymbol{a} - \boldsymbol{u}\]
+
+            \subsubsection*{\(|\boldsymbol{u}|\) (scalar projection/resolute)}
+            \begin{align*}
+              s &= |\boldsymbol{u}|\\
+              &= \boldsymbol{a} \cdot \hat{\boldsymbol{b}}\\
+              &=\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{|\boldsymbol{b}|}\\
+              &= |\boldsymbol{a}| \cos \theta
+            \end{align*}
+
+            \subsubsection*{Rectangular (\(\parallel,\perp\)) components}
+
+            \[\boldsymbol{a}=\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{\boldsymbol{b}\cdot\boldsymbol{b}}\boldsymbol{b}+\left(\boldsymbol{a}-\frac{\boldsymbol{a}\cdot\boldsymbol{b}}{\boldsymbol{b}\cdot\boldsymbol{b}}\boldsymbol{b}\right)\]
+
+
+            \subsection*{Vector proofs}
+
+            \textbf{Concurrent:} intersection of \(\ge\) 3 lines
+
+            \begin{tikzpicture}
+              \draw [blue] (0,0) -- (1,1);
+              \draw [red] (1,0) -- (0,1);
+              \draw [brown] (0.4,0) -- (0.6,1);
+              \filldraw (0.5,0.5) circle (2pt);
+            \end{tikzpicture}
+
+            \subsubsection*{Collinear points}
+
+            \(\ge\) 3 points lie on the same line
+
+            \begin{tikzpicture}
+              \draw [purple] (0,0) -- (4,1);
+              \filldraw (2,0.5) circle (2pt) node [above] {\(C\)};
+              \filldraw (1,0.25) circle (2pt) node [above] {\(A\)};
+              \filldraw (3,0.75) circle (2pt) node [above] {\(B\)};
+              \coordinate (O) at (2.8,-0.2);
+              \node at (O) [below] {\(O\)}; 
+              \begin{scope}[->, orange, thick] 
+                \draw (O) -- (2,0.5) node [pos=0.5, above, font=\footnotesize, black] {\(\boldsymbol{c}\)};
+                \draw (O) -- (1,0.25) node [pos=0.5, below, font=\footnotesize, black] {\(\boldsymbol{a}\)};
+                \draw (O) -- (3,0.75) node [pos=0.5, right, font=\footnotesize, black] {\(\boldsymbol{b}\)};
+              \end{scope}
+            \end{tikzpicture}
+
+            \begin{align*}
+              \text{e.g. Prove that}\\
+              \overrightharp{AC}=m\overrightharp{AB} \iff \boldsymbol{c}&=(1-m)\boldsymbol{a}+m\boldsymbol{b}\\
+              \implies \boldsymbol{c} &= \overrightharp{OA} + \overrightharp{AC}\\
+              &= \overrightharp{OA} + m\overrightharp{AB}\\
+              &=\boldsymbol{a}+m(\boldsymbol{b}-\boldsymbol{a})\\
+              &=\boldsymbol{a}+m\boldsymbol{b}-m\boldsymbol{a}\\
+              &=(1-m)\boldsymbol{a}+m{b}
+            \end{align*}
+            \begin{align*}
+              \text{Also, } \implies \overrightharp{OC} &= \lambda \vec{OA} + \mu \overrightharp{OB} \\
+              \text{where } \lambda + \mu &= 1\\
+              \text{If } C \text{ lies along } \overrightharp{AB}, & \implies 0 < \mu < 1
+            \end{align*}
+
+
+            \subsubsection*{Parallelograms}
+
+            \begin{center}\begin{tikzpicture}
+              \coordinate (O) at (0,0) node [below left] {\(O\)};
+              \coordinate (A) at (4,0);
+              \coordinate (B) at (6,2);
+              \coordinate (C) at (2,2);
+              \coordinate (D) at (6,0);
+
+              \draw[postaction={decorate}, decoration={markings, mark=at position 0.6 with {\arrow{>>}}}] (O)--(A) node [below left] {\(A\)};
+              \draw[postaction={decorate}, decoration={markings,mark=at position 0.5 with {\arrow{>}}}] (A)--(B) node [above right] {\(B\)};
+              \draw[postaction={decorate}, decoration={markings, mark=at position 0.6 with {\arrow{>>}}}] (B)--(C) node [above left] {\(C\)};
+              \draw[postaction={decorate}, decoration={markings,mark=at position 0.5 with {\arrow{>}}}] (C)--(O);
+
+              \draw [gray, dashed] (O) -- (B) node [pos=0.75] {\(\diagdown\diagdown\)} node [pos=0.25] {\(\diagdown\diagdown\)};
+              \draw [gray, dashed] (A) -- (C) node [pos=0.75] {\(\diagup\)} node [pos=0.25] {\(\diagup\)};
+              \begin{scope}
+                \path[clip] (C) -- (A) -- (O);
+                \fill[orange, opacity=0.5, draw=black] (0,0) circle (4mm);
+                \node at ($(0,0)+(20:8mm)$) {\(\theta\)};
+              \end{scope}
+              \draw [gray, thick, dotted] (B) -- (D) node [pos=0.5, right, black, font=\footnotesize] {\(|\boldsymbol{c}|\sin\theta\)} (A) -- (D) node [pos=0.5, below, black, font=\footnotesize] {\(|\boldsymbol{c}|\cos\theta\)};
+              \draw pic [draw,thick,red,angle radius=2mm] {right angle=O--D--B};
+            \end{tikzpicture}\end{center}
+
+            \begin{itemize}
+              \item
+                Diagonals \(\overrightharp{OB}, \overrightharp{AC}\) bisect each other
+              \item
+                If diagonals are equal length, it is a rectangle
+              \item
+                \(|\overrightharp{OB}|^2 + |\overrightharp{CA}|^2 = |\overrightharp{OA}|^2 + |\overrightharp{AB}|^2 + |\overrightharp{CB}|^2 + |\overrightharp{OC}|^2\)
+              \item
+                Area \(=\boldsymbol{c} \cdot \boldsymbol{a}\)
+            \end{itemize}
+
+            \subsubsection*{Useful vector properties}
+
+            \begin{itemize}
+              \item
+                \(\boldsymbol{a} \parallel \boldsymbol{b} \implies \boldsymbol{b}=k\boldsymbol{a}\) for some
+                \(k \in \mathbb{R} \setminus \{0\}\)
+              \item
+                If \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are parallel with at
+                least one point in common, then they lie on the same straight line
+              \item
+                \(\boldsymbol{a} \perp \boldsymbol{b} \iff \boldsymbol{a} \cdot \boldsymbol{b}=0\)
+              \item
+                \(\boldsymbol{a} \cdot \boldsymbol{a} = |\boldsymbol{a}|^2\)
+            \end{itemize}
+
+            \subsection*{Linear dependence}
+
+            \(\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}\) are linearly dependent if they are \(\nparallel\) and:
+            \begin{align*}
+              0&=k\boldsymbol{a}+l\boldsymbol{b}+m\boldsymbol{c}\\
+              \therefore \boldsymbol{c} &= m\boldsymbol{a} + n\boldsymbol{b} \quad \text{(simultaneous)}
+            \end{align*}
+
+            \noindent \(\boldsymbol{a}, \boldsymbol{b},\) and \(\boldsymbol{c}\) are linearly
+            independent if no vector in the set is expressible as a linear
+            combination of other vectors in set, or if they are parallel.
+
+            \subsection*{Three-dimensional vectors}
+
+            Right-hand rule for axes: \(z\) is up or out of page.
+
+            \tdplotsetmaincoords{60}{120} 
+            \begin{center}\begin{tikzpicture} [scale=3, tdplot_main_coords, axis/.style={->,thick}, 
+              vector/.style={-stealth,red,very thick}, 
+              vector guide/.style={dashed,gray,thick}]
+
+              %standard tikz coordinate definition using x, y, z coords
+              \coordinate (O) at (0,0,0);
+
+              %tikz-3dplot coordinate definition using x, y, z coords
+
+              \pgfmathsetmacro{\ax}{1}
+              \pgfmathsetmacro{\ay}{1}
+              \pgfmathsetmacro{\az}{1}
+
+              \coordinate (P) at (\ax,\ay,\az);
+
+              %draw axes
+              \draw[axis] (0,0,0) -- (1,0,0) node[anchor=north east]{$x$};
+              \draw[axis] (0,0,0) -- (0,1,0) node[anchor=north west]{$y$};
+              \draw[axis] (0,0,0) -- (0,0,1) node[anchor=south]{$z$};
+
+              %draw a vector from O to P
+              \draw[vector] (O) -- (P);
+
+              %draw guide lines to components
+              \draw[vector guide]         (O) -- (\ax,\ay,0);
+              \draw[vector guide] (\ax,\ay,0) -- (P);
+              \draw[vector guide]         (P) -- (0,0,\az);
+              \draw[vector guide] (\ax,\ay,0) -- (0,\ay,0);
+              \draw[vector guide] (\ax,\ay,0) -- (0,\ay,0);
+              \draw[vector guide] (\ax,\ay,0) -- (\ax,0,0);
+              \node[tdplot_main_coords,above right]
+              at (\ax,\ay,\az){(\ax, \ay, \az)};
+            \end{tikzpicture}\end{center}
+
+            \subsection*{Parametric vectors}
+
+            Parametric equation of line through point \((x_0, y_0, z_0)\) and
+            parallel to \(a\boldsymbol{i} + b\boldsymbol{j} + c\boldsymbol{k}\) is:
+
+            \[\begin{cases}x = x_o + a \cdot t \\ y = y_0 + b \cdot t \\ z = z_0 + c \cdot t\end{cases}\]
+
+              \section{Circular functions}
+
+              \(\sin(bx)\) or \(\cos(bx)\): period \(=\frac{2\pi}{b}\)
+
+              \noindent \(\tan(nx)\): period \(=\frac{\pi}{n}\)\\
+              \indent\indent\indent asymptotes at \(x=\frac{(2k+1)\pi}{2n} \> \vert \> k \in \mathbb{Z}\)
+
+              \subsection*{Reciprocal functions}
+
+              \subsubsection*{Cosecant}
+
+              \[\operatorname{cosec} \theta = \frac{1}{\sin \theta} \> \vert \> \sin \theta \ne 0\]
+
+              \begin{itemize}
+                \item
+                  \textbf{Domain} \(= \mathbb{R} \setminus {n\pi : n \in \mathbb{Z}}\)
+                \item
+                  \textbf{Range} \(= \mathbb{R} \setminus (-1, 1)\)
+                \item
+                  \textbf{Turning points} at
+                  \(\theta = \frac{(2n + 1)\pi}{2} \> \vert \> n \in \mathbb{Z}\)
+                \item
+                  \textbf{Asymptotes} at \(\theta = n\pi \> \vert \> n \in \mathbb{Z}\)
+              \end{itemize}
 
-\begin{itemize}
-\item
-  \(\boldsymbol{a} \parallel \boldsymbol{b} \implies \boldsymbol{b}=k\boldsymbol{a}\) for some
-  \(k \in \mathbb{R} \setminus \{0\}\)
-\item
-  If \(\boldsymbol{a}\) and \(\boldsymbol{b}\) are parallel with at
-  least one point in common, then they lie on the same straight line
-\item
-  \(\boldsymbol{a} \perp \boldsymbol{b} \iff \boldsymbol{a} \cdot \boldsymbol{b}=0\)
-\item
-  \(\boldsymbol{a} \cdot \boldsymbol{a} = |\boldsymbol{a}|^2\)
-\end{itemize}
-
-\subsection*{Linear dependence}
+              \subsubsection*{Secant}
+           \begin{center}\includegraphics[width=0.7\columnwidth]{graphics/sec.png}\end{center}
 
-\(\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c}\) are linearly dependent if they are \(\nparallel\) and:
-\begin{align*}
-  0&=k\boldsymbol{a}+l\boldsymbol{b}+m\boldsymbol{c}\\
-  \therefore \boldsymbol{c} &= m\boldsymbol{a} + n\boldsymbol{b} \quad \text{(simultaneous)}
-\end{align*}
+                \[\operatorname{sec} \theta = \frac{1}{\cos \theta} \> \vert \> \cos \theta \ne 0\]
 
-\noindent \(\boldsymbol{a}, \boldsymbol{b},\) and \(\boldsymbol{c}\) are linearly
-independent if no vector in the set is expressible as a linear
-combination of other vectors in set, or if they are parallel.
+                \begin{itemize}
 
-\subsection*{Three-dimensional vectors}
+                  \item
+                    \textbf{Domain}
+                    \(= \mathbb{R} \setminus \frac{(2n + 1) \pi}{2} : n \in \mathbb{Z}\}\)
+                  \item
+                    \textbf{Range} \(= \mathbb{R} \setminus (-1, 1)\)
+                  \item
+                    \textbf{Turning points} at
+                    \(\theta = n\pi \> \vert \> n \in \mathbb{Z}\)
+                  \item
+                    \textbf{Asymptotes} at
+                    \(\theta = \frac{(2n + 1) \pi}{2} \> \vert \> n \in \mathbb{Z}\)
+                \end{itemize}
 
-Right-hand rule for axes: \(z\) is up or out of page.
+                \subsubsection*{Cotangent}
 
-\tdplotsetmaincoords{60}{120} 
-\begin{center}\begin{tikzpicture} [scale=3, tdplot_main_coords, axis/.style={->,thick}, 
-vector/.style={-stealth,red,very thick}, 
-vector guide/.style={dashed,gray,thick}]
+                \begin{center}\includegraphics[width=0.7\columnwidth]{graphics/cot.png}\end{center}
 
-%standard tikz coordinate definition using x, y, z coords
-\coordinate (O) at (0,0,0);
+                  \[\operatorname{cot} \theta = {{\cos \theta} \over {\sin \theta}} \> \vert \> \sin \theta \ne 0\]
 
-%tikz-3dplot coordinate definition using x, y, z coords
+                  \begin{itemize}
 
-\pgfmathsetmacro{\ax}{1}
-\pgfmathsetmacro{\ay}{1}
-\pgfmathsetmacro{\az}{1}
+                    \item
+                      \textbf{Domain} \(= \mathbb{R} \setminus \{n \pi: n \in \mathbb{Z}\}\)
+                    \item
+                      \textbf{Range} \(= \mathbb{R}\)
+                    \item
+                      \textbf{Asymptotes} at \(\theta = n\pi \> \vert \> n \in \mathbb{Z}\)
+                  \end{itemize}
 
-\coordinate (P) at (\ax,\ay,\az);
+                  \subsubsection*{Symmetry properties}
 
-%draw axes
-\draw[axis] (0,0,0) -- (1,0,0) node[anchor=north east]{$x$};
-\draw[axis] (0,0,0) -- (0,1,0) node[anchor=north west]{$y$};
-\draw[axis] (0,0,0) -- (0,0,1) node[anchor=south]{$z$};
+                  \[\begin{split}
+                    \operatorname{sec} (\pi \pm x) & = -\operatorname{sec} x \\
+                    \operatorname{sec} (-x) & = \operatorname{sec} x \\
+                    \operatorname{cosec} (\pi \pm x) & = \mp \operatorname{cosec} x \\
+                    \operatorname{cosec} (-x) & = - \operatorname{cosec} x \\
+                    \operatorname{cot} (\pi \pm x) & = \pm \operatorname{cot} x \\
+                    \operatorname{cot} (-x) & = - \operatorname{cot} x
+                  \end{split}\]
 
-%draw a vector from O to P
-\draw[vector] (O) -- (P);
+                  \subsubsection*{Complementary properties}
 
-%draw guide lines to components
-\draw[vector guide]         (O) -- (\ax,\ay,0);
-\draw[vector guide] (\ax,\ay,0) -- (P);
-\draw[vector guide]         (P) -- (0,0,\az);
-\draw[vector guide] (\ax,\ay,0) -- (0,\ay,0);
-\draw[vector guide] (\ax,\ay,0) -- (0,\ay,0);
-\draw[vector guide] (\ax,\ay,0) -- (\ax,0,0);
-\node[tdplot_main_coords,above right]
-at (\ax,\ay,\az){(\ax, \ay, \az)};
-\end{tikzpicture}\end{center}
+                  \[\begin{split}
+                    \operatorname{sec} \left({\pi \over 2} - x\right) & = \operatorname{cosec} x \\
+                    \operatorname{cosec} \left({\pi \over 2} - x\right) & = \operatorname{sec} x \\
+                    \operatorname{cot} \left({\pi \over 2} - x\right) & = \tan x \\
+                    \tan \left({\pi \over 2} - x\right) & = \operatorname{cot} x
+                  \end{split}\]
 
-\subsection*{Parametric vectors}
+                  \subsubsection*{Pythagorean identities}
 
-Parametric equation of line through point \((x_0, y_0, z_0)\) and
-parallel to \(a\boldsymbol{i} + b\boldsymbol{j} + c\boldsymbol{k}\) is:
+                  \[\begin{split}
+                    1 + \operatorname{cot}^2 x & = \operatorname{cosec}^2 x, \quad \text{where } \sin x \ne 0 \\
+                    1 + \tan^2 x & = \operatorname{sec}^2 x, \quad \text{where } \cos x \ne 0
+                  \end{split}\]
 
-\[\begin{cases}x = x_o + a \cdot t \\ y = y_0 + b \cdot t \\ z = z_0 + c \cdot t\end{cases}\]
+                  \subsection*{Compound angle formulas}
 
-\section{Circular functions}
+                  \[\cos(x \pm y) = \cos x + \cos y \mp \sin x \sin y\]
+                  \[\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y\]
+                  \[\tan(x \pm y) = {{\tan x \pm \tan y} \over {1 \mp \tan x \tan y}}\]
 
-\(\sin(bx)\) or \(\cos(bx)\): period \(=\frac{2\pi}{b}\)
+                  \subsection*{Double angle formulas}
 
-\noindent \(\tan(nx)\): period \(=\frac{\pi}{n}\)\\
-\indent\indent\indent asymptotes at \(x=\frac{(2k+1)\pi}{2n} \> \vert \> k \in \mathbb{Z}\)
+                  \[\begin{split}
+                    \cos 2x &= \cos^2 x - \sin^2 x \\
+                    & = 1 - 2\sin^2 x \\
+                    & = 2 \cos^2 x -1
+                  \end{split}\]
 
-\subsection*{Reciprocal functions}
+                  \[\sin 2x = 2 \sin x \cos x\]
 
-\subsubsection*{Cosecant}
+                  \[\tan 2x = {{2 \tan x} \over {1 - \tan^2 x}}\]
 
-\[\operatorname{cosec} \theta = \frac{1}{\sin \theta} \> \vert \> \sin \theta \ne 0\]
-
-\begin{itemize}
-\item
-  \textbf{Domain} \(= \mathbb{R} \setminus {n\pi : n \in \mathbb{Z}}\)
-\item
-  \textbf{Range} \(= \mathbb{R} \setminus (-1, 1)\)
-\item
-  \textbf{Turning points} at
-    \(\theta = \frac{(2n + 1)\pi}{2} \> \vert \> n \in \mathbb{Z}\)
-\item
-  \textbf{Asymptotes} at \(\theta = n\pi \> \vert \> n \in \mathbb{Z}\)
-\end{itemize}
+                  \subsection*{Inverse circular functions}
 
-\subsubsection*{Secant}
+                  \pgfplotsset{every axis/.append style={
+                    axis x line=middle,    % put the x axis in the middle
+                    axis y line=middle,    % put the y axis in the middle
+                    axis line style={<->}, % arrows on the axis
+                    xlabel={$x$},          % default put x on x-axis
+                    ylabel={$y$},          % default put y on y-axis
+                    }}
+
+% arrows as stealth fighters
+\tikzset{>=stealth}
 
+\begin{tikzpicture}
+  \begin{axis}[domain = -1:1, samples = 500]
+    \addplot[color = red]  {rad(asin(x))} node [pos=0.25, below right] {\(\sin^{-1}x\)};
+    \addplot[color = blue] {rad(acos(x))} node [pos=0.25, below left] {\(\cos^{-1}x\)};
+  \end{axis}
+\end{tikzpicture}
 
-\begin{center}\includegraphics[width=0.7\columnwidth]{graphics/sec.png}\end{center}
-
-\[\operatorname{sec} \theta = \frac{1}{\cos \theta} \> \vert \> \cos \theta \ne 0\]
-
-\begin{itemize}
+                  Inverse functions: \(f(f^{-1}(x)) = x\) (restrict domain)
+
+                  \[\sin^{-1}: [-1, 1] \rightarrow \mathbb{R}, \quad \sin^{-1} x = y\]
+                  \hfill where \(\sin y = x, \> y \in [{-\pi \over 2}, {\pi \over 2}]\)
+
+                  \[\cos^{-1}: [-1,1] \rightarrow \mathbb{R}, \quad \cos^{-1} x = y\]
+                  \hfill where \(\cos y = x, \> y \in [0, \pi]\)
+
+                  \[\tan^{-1}: \mathbb{R} \rightarrow \mathbb{R}, \quad \tan^{-1} x = y\]
+                  \hfill where \(\tan y = x, \> y \in \left(-{\pi \over 2}, {\pi \over 2}\right)\)
 
-\item
-  \textbf{Domain}
-    \(= \mathbb{R} \setminus \frac{(2n + 1) \pi}{2} : n \in \mathbb{Z}\}\)
-\item
-  \textbf{Range} \(= \mathbb{R} \setminus (-1, 1)\)
-\item
-  \textbf{Turning points} at
-  \(\theta = n\pi \> \vert \> n \in \mathbb{Z}\)
-\item
-  \textbf{Asymptotes} at
-    \(\theta = \frac{(2n + 1) \pi}{2} \> \vert \> n \in \mathbb{Z}\)
-\end{itemize}
 
-\subsubsection*{Cotangent}
+                  \section{Differential calculus}
 
-\begin{center}\includegraphics[width=0.7\columnwidth]{graphics/cot.png}\end{center}
+                  \subsection*{Limits}
+
+                  \[\lim_{x \rightarrow a}f(x)\]
+                  \(L^-,\quad L^+\) \qquad limit from below/above\\
+                  \(\lim_{x \to a} f(x)\) \quad limit of a point\\
+
+                  \noindent For solving \(x\rightarrow\infty\), put all \(x\) terms in denominators\\
+                  e.g. \[\lim_{x \rightarrow \infty}{{2x+3} \over {x-2}}={{2+{3 \over x}} \over {1-{2 \over x}}}={2 \over 1} = 2\]
+
+                  \subsubsection*{Limit theorems}
+
+                  \begin{enumerate}
+                    \item
+                      For constant function \(f(x)=k\), \(\lim_{x \rightarrow a} f(x) = k\)
+                    \item
+                      \(\lim_{x \rightarrow a} (f(x) \pm g(x)) = F \pm G\)
+                    \item
+                      \(\lim_{x \rightarrow a} (f(x) \times g(x)) = F \times G\)
+                    \item
+                      \(\therefore \lim_{x \rightarrow a} c \times f(x)=cF\) where \(c=\) constant
+                    \item
+                      \({\lim_{x \rightarrow a} {f(x) \over g(x)}} = {F \over G}, G \ne 0\)
+                    \item
+                      \(f(x)\) is continuous \(\iff L^-=L^+=f(x) \> \forall x\)
+                  \end{enumerate}
+
+                  \subsection*{Gradients of secants and tangents}
+
+                  \textbf{Secant (chord)} - line joining two points on curve\\
+                  \textbf{Tangent} - line that intersects curve at one point
+
+                  \subsection*{First principles derivative}
+
+                  \[f^\prime(x) = \lim_{\delta x \rightarrow 0}{\delta y \over \delta x}={\frac{dy}{dx}}\]
+
+                  \subsubsection*{Logarithmic identities}
+
+                  \(\log_b (xy)=\log_b x + \log_b y\)\\
+                  \(\log_b x^n = n \log_b x\)\\
+                  \(\log_b y^{x^n} = x^n \log_b y\)
+
+                  \subsubsection*{Index identities}
+
+                  \(b^{m+n}=b^m \cdot b^n\)\\
+                  \((b^m)^n=b^{m \cdot n}\)\\
+                  \((b \cdot c)^n = b^n \cdot c^n\)\\
+                  \({a^m \div a^n} = {a^{m-n}}\)
+
+                  \subsection*{Derivative rules}
+
+                  \renewcommand{\arraystretch}{1.4}
+                  \begin{tabularx}{\columnwidth}{rX}
+                    \hline
+                    \(f(x)\) & \(f^\prime(x)\)\\
+                    \hline
+                    \(\sin x\) & \(\cos x\)\\
+                    \(\sin ax\) & \(a\cos ax\)\\
+                    \(\cos x\) & \(-\sin x\)\\
+                    \(\cos ax\) & \(-a \sin ax\)\\
+                    \(\tan f(x)\) & \(f^2(x) \sec^2f(x)\)\\
+                    \(e^x\) & \(e^x\)\\
+                    \(e^{ax}\) & \(ae^{ax}\)\\
+                    \(ax^{nx}\) & \(an \cdot e^{nx}\)\\
+                    \(\log_e x\) & \(\dfrac{1}{x}\)\\
+                    \(\log_e {ax}\) & \(\dfrac{1}{x}\)\\
+                    \(\log_e f(x)\) & \(\dfrac{f^\prime (x)}{f(x)}\)\\
+                    \(\sin(f(x))\) & \(f^\prime(x) \cdot \cos(f(x))\)\\
+                    \(\sin^{-1} x\) & \(\dfrac{1}{\sqrt{1-x^2}}\)\\
+                    \(\cos^{-1} x\) & \(\dfrac{-1}{sqrt{1-x^2}}\)\\
+                    \(\tan^{-1} x\) & \(\dfrac{1}{1 + x^2}\)\\
+                    \(\frac{d}{dy}f(y)\) & \(\dfrac{1}{\frac{dx}{dy}}\) (reciprocal)\\
+                    \(uv\) & \(u \frac{dv}{dx}+v\frac{du}{dx} (product rule)\)\\
+                    \(\dfrac{u}{v}\) & \(\dfrac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}\) (quotient rule)\\
+                    \(f(g(x))\) & \(f^\prime(g(x))\cdot g^\prime(x)\)\\
+                    \hline
+                  \end{tabularx}
+
+                  \subsection*{Reciprocal derivatives}
+
+                  \[\frac{1}{\frac{dy}{dx}} = \frac{dx}{dy}\]
+
+                  \subsection*{Differentiating \(x=f(y)\)}
+                  \begin{align*}
+                    \text{Find }& \frac{dx}{dy}\\
+                    \text{Then, }\frac{dx}{dy} &= \frac{1}{\frac{dy}{dx}} \\
+                    \implies {\frac{dy}{dx}} &= \frac{1}{\frac{dx}{dy}}\\
+                    \therefore {\frac{dy}{dx}} &= \frac{1}{\frac{dx}{dy}}
+                  \end{align*}
+
+                  \subsubsection*{Second derivative}
+                  \begin{align*}f(x) \longrightarrow &f^\prime (x) \longrightarrow f^{\prime\prime}(x)\\
+                  \implies y \longrightarrow &\frac{dy}{dx} \longrightarrow \frac{d^2 y}{dx^2}\end{align*}
+
+                  \noindent Order of polynomial \(n\)th derivative decrements each time the derivative is taken
+
+                  \subsubsection*{Points of Inflection}
+
+                  \emph{Stationary point} - i.e.
+                  \(f^\prime(x)=0\)\\
+                  \emph{Point of inflection} - max \(|\)gradient\(|\) (i.e.
+                  \(f^{\prime\prime} = 0\))
+                  %\begin{table*}[ht]
+                  %\centering
+                  %  \begin{tabularx}{\textwidth}{XXXX}
+                  %\hline
+                  %    \rowcolor{shade2}
+                  %    & \(\dfrac{d^2 y}{dx^2} > 0\)  & \(\dfrac{d^2y}{dx^2}<0\) & \(\dfrac{d^2y}{dx^2}=0\) (inflection) \\
+                  %\hline
+                  %    \(\frac{dy}{dx}>0\) & \begin{tikzpicture} \draw[domain=1:2,smooth,variable=\x,blue] plot ({\x},{(1/10)*\x*\x*\x}) plot ({\x},{0.675*\x-0.677}); \end{tikzpicture} & cell 3\\
+                  %cell 1 & cell 2 & cell 3\\
+                  %\hline
+                  %\end{tabularx}
+                  %\end{table*}
 
-\[\operatorname{cot} \theta = {{\cos \theta} \over {\sin \theta}} \> \vert \> \sin \theta \ne 0\]
 
 \begin{itemize}
+  \item
+                      if \(f^\prime (a) = 0\) and \(f^{\prime\prime}(a) > 0\), then point
+                      \((a, f(a))\) is a local min (curve is concave up)
+                    \item
+                      if \(f^\prime (a) = 0\) and \(f^{\prime\prime} (a) < 0\), then point
+                      \((a, f(a))\) is local max (curve is concave down)
+                    \item
+                      if \(f^{\prime\prime}(a) = 0\), then point \((a, f(a))\) is a point of
+                      inflection
+                    \item
+                      if also \(f^\prime(a)=0\), then it is a stationary point of inflection
+                  \end{itemize}
 
-\item
-  \textbf{Domain} \(= \mathbb{R} \setminus \{n \pi: n \in \mathbb{Z}\}\)
-\item
-  \textbf{Range} \(= \mathbb{R}\)
-\item
-  \textbf{Asymptotes} at \(\theta = n\pi \> \vert \> n \in \mathbb{Z}\)
-\end{itemize}
-
-\subsubsection*{Symmetry properties}
-
-\[\begin{split}
-  \operatorname{sec} (\pi \pm x) & = -\operatorname{sec} x \\
-  \operatorname{sec} (-x) & = \operatorname{sec} x \\
-  \operatorname{cosec} (\pi \pm x) & = \mp \operatorname{cosec} x \\
-  \operatorname{cosec} (-x) & = - \operatorname{cosec} x \\
-  \operatorname{cot} (\pi \pm x) & = \pm \operatorname{cot} x \\
-  \operatorname{cot} (-x) & = - \operatorname{cot} x
-\end{split}\]
-
-\subsubsection*{Complementary properties}
-
-\[\begin{split}
-  \operatorname{sec} \left({\pi \over 2} - x\right) & = \operatorname{cosec} x \\
-  \operatorname{cosec} \left({\pi \over 2} - x\right) & = \operatorname{sec} x \\
-  \operatorname{cot} \left({\pi \over 2} - x\right) & = \tan x \\
-  \tan \left({\pi \over 2} - x\right) & = \operatorname{cot} x
-\end{split}\]
-
-\subsubsection*{Pythagorean identities}
+                  \subsection*{Implicit Differentiation}
 
-\[\begin{split}
-  1 + \operatorname{cot}^2 x & = \operatorname{cosec}^2 x, \quad \text{where } \sin x \ne 0 \\
-  1 + \tan^2 x & = \operatorname{sec}^2 x, \quad \text{where } \cos x \ne 0
-\end{split}\]
+                  \noindent Used for differentiating circles etc.
 
-\subsection*{Compound angle formulas}
+                  If \(p\) and \(q\) are expressions in \(x\) and \(y\) such that \(p=q\),
+                  for all \(x\) and \(y\), then:
 
-\[\cos(x \pm y) = \cos x + \cos y \mp \sin x \sin y\]
-\[\sin(x \pm y) = \sin x \cos y \pm \cos x \sin y\]
-\[\tan(x \pm y) = {{\tan x \pm \tan y} \over {1 \mp \tan x \tan y}}\]
+                  \[{\frac{dp}{dx}} = {\frac{dq}{dx}} \quad \text{and} \quad {\frac{dp}{dy}} = {\frac{dq}{dy}}\]
 
-\subsection*{Double angle formulas}
-
-\[\begin{split}
-  \cos 2x &= \cos^2 x - \sin^2 x \\
-  & = 1 - 2\sin^2 x \\
-  & = 2 \cos^2 x -1
-\end{split}\]
+                  \noindent \colorbox{cas}{\textbf{On CAS:}}\\
+                  Action \(\rightarrow\) Calculation \(\rightarrow\) \texttt{impDiff(y\^{}2+ax=5,\ x,\ y)}\\
+                  Returns \(y^\prime= \dots\).
 
-\[\sin 2x = 2 \sin x \cos x\]
+                  \subsection*{Integration}
 
-\[\tan 2x = {{2 \tan x} \over {1 - \tan^2 x}}\]
+                  \[\int f(x) \cdot dx = F(x) + c \quad \text{where } F^\prime(x) = f(x)\]
 
-\subsection*{Inverse circular functions}
-
-Inverse functions: \(f(f^{-1}(x)) = x, \quad f(f^{-1}(x)) = x\)\\
-Must be 1:1 to find inverse (reflection in \(y=x\)).\\
-Domain is restricted to make functions 1:1.
-
-\[\sin^{-1}: [-1, 1] \rightarrow \mathbb{R}, \quad \sin^{-1} x = y\]
-\hfill where \(\sin y = x, \> y \in [{-\pi \over 2}, {\pi \over 2}]\)
+                  \subsection*{Integral laws}
 
-\[\cos^{-1}: [-1,1] \rightarrow \mathbb{R}, \quad \cos^{-1} x = y\]
-\hfill where \(\cos y = x, \> y \in [0, \pi]\)
+                  \renewcommand{\arraystretch}{1.4}
+                  \begin{tabularx}{\columnwidth}{rX}
+                    \hline
+                    \(f(x)\) & \(\int f(x) \cdot dx\) \\
+                    \hline
+                    \(k\) (constant) & \(kx + c\)\\
+                    \(x^n\) & \(\dfrac{1}{n+1} x^{n+1}\) \\
+                    \(a x^{-n}\) &\(a \cdot \log_e |x| + c\)\\
+                    \(\dfrac{1}{ax+b}\) &\(\dfrac{1}{a} \log_e (ax+b) + c\)\\
+                    \((ax+b)^n\) & \(\dfrac{1}{a(n+1)}(ax+b)^{n-1} + c\>|\>n\ne 1\)\\
+                    \((ax+b)^{-1}\) & \(\dfrac{1}{a}\log_e |ax+b|+c\)\\
+                    \(e^{kx}\) & \(\dfrac{1}{k} e^{kx} + c\)\\
+                    \(e^k\) & \(e^kx + c\)\\
+                    \(\sin kx\) & \(\dfrac{-1}{k} \cos (kx) + c\)\\
+                    \(\cos kx\) & \(\dfrac{1}{k} \sin (kx) + c\)\\
+                    \(\sec^2 kx\) & \(\dfrac{1}{k} \tan(kx) + c\)\\
+                    \(\dfrac{1}{\sqrt{a^2-x^2}}\) & \(\sin^{-1} \dfrac{x}{a} + c \>\vert\> a>0\)\\
+                    \(\dfrac{-1}{\sqrt{a^2-x^2}}\) & \(\cos^{-1} \dfrac{x}{a} + c \>\vert\> a>0\)\\
+                    \(\frac{a}{a^2-x^2}\) & \(\tan^{-1} \frac{x}{a} + c\)\\
+                    \(\frac{f^\prime (x)}{f(x)}\) & \(\log_e f(x) + c\)\\
+                    \(\int f(u) \cdot \frac{du}{dx} \cdot dx\) & \(\int f(u) \cdot du\) (substitution)\\
+                    \(f(x) \cdot g(x)\) & \(\int [f^\prime(x) \cdot g(x)] dx + \int [g^\prime(x) f(x)] dx\)\\
+                    \hline
+                  \end{tabularx}
 
-\[\tan^{-1}: \mathbb{R} \rightarrow \mathbb{R}, \quad \tan^{-1} x = y\]
-\hfill where \(\tan y = x, \> y \in \left(-{\pi \over 2}, {\pi \over 2}\right)\)
+                  Note \(\sin^{-1} {x \over a} + \cos^{-1} {x \over a}\) is constant \(\forall x \in (-a, a)\)
 
+                  \subsection*{Definite integrals}
 
-\section{Differential calculus}
-
-\subsection*{Limits}
-
-\[\lim_{x \rightarrow a}f(x)\]
-\(L^-,\quad L^+\) \qquad limit from below/above\\
-\(\lim_{x \to a} f(x)\) \quad limit of a point\\
-
-\noindent For solving \(x\rightarrow\infty\), put all \(x\) terms in denominators\\
-    e.g. \[\lim_{x \rightarrow \infty}{{2x+3} \over {x-2}}={{2+{3 \over x}} \over {1-{2 \over x}}}={2 \over 1} = 2\]
-
-\subsubsection*{Limit theorems}
-
-\begin{enumerate}
-\item
-  For constant function \(f(x)=k\), \(\lim_{x \rightarrow a} f(x) = k\)
-\item
-  \(\lim_{x \rightarrow a} (f(x) \pm g(x)) = F \pm G\)
-\item
-  \(\lim_{x \rightarrow a} (f(x) \times g(x)) = F \times G\)
-    \item
-\(\therefore \lim_{x \rightarrow a} c \times f(x)=cF\) where \(c=\) constant
-\item
-  \({\lim_{x \rightarrow a} {f(x) \over g(x)}} = {F \over G}, G \ne 0\)
-\item
-  \(f(x)\) is continuous \(\iff L^-=L^+=f(x) \> \forall x\)
-\end{enumerate}
-
-\subsection*{Gradients of secants and tangents}
-
-\textbf{Secant (chord)} - line joining two points on curve\\
-\textbf{Tangent} - line that intersects curve at one point
-
-\subsection*{First principles derivative}
-
-\[f^\prime(x) = \lim_{\delta x \rightarrow 0}{\delta y \over \delta x}={\frac{dy}{dx}}\]
-
-\subsubsection*{Logarithmic identities}
-
-\(\log_b (xy)=\log_b x + \log_b y\)\\
-\(\log_b x^n = n \log_b x\)\\
-\(\log_b y^{x^n} = x^n \log_b y\)
-
-\subsubsection*{Index identities}
-
-\(b^{m+n}=b^m \cdot b^n\)\\
-\((b^m)^n=b^{m \cdot n}\)\\
-\((b \cdot c)^n = b^n \cdot c^n\)\\
-\({a^m \div a^n} = {a^{m-n}}\)
-
-\subsection*{Derivative rules}
-
-\renewcommand{\arraystretch}{1.4}
-\begin{tabularx}{\columnwidth}{rX}
-  \hline
-\(f(x)\) & \(f^\prime(x)\)\\
-\hline
-\(\sin x\) & \(\cos x\)\\
-\(\sin ax\) & \(a\cos ax\)\\
-\(\cos x\) & \(-\sin x\)\\
-\(\cos ax\) & \(-a \sin ax\)\\
-\(\tan f(x)\) & \(f^2(x) \sec^2f(x)\)\\
-\(e^x\) & \(e^x\)\\
-\(e^{ax}\) & \(ae^{ax}\)\\
-\(ax^{nx}\) & \(an \cdot e^{nx}\)\\
-  \(\log_e x\) & \(\dfrac{1}{x}\)\\
-  \(\log_e {ax}\) & \(\dfrac{1}{x}\)\\
-  \(\log_e f(x)\) & \(\dfrac{f^\prime (x)}{f(x)}\)\\
-\(\sin(f(x))\) & \(f^\prime(x) \cdot \cos(f(x))\)\\
-  \(\sin^{-1} x\) & \(\dfrac{1}{\sqrt{1-x^2}}\)\\
-  \(\cos^{-1} x\) & \(\dfrac{-1}{sqrt{1-x^2}}\)\\
-  \(\tan^{-1} x\) & \(\dfrac{1}{1 + x^2}\)\\
-  \hline
-\end{tabularx}
-
-\subsection*{Reciprocal derivatives}
-
-\[\frac{1}{\frac{dy}{dx}} = \frac{dx}{dy}\]
-
-\subsection*{Differentiating \(x=f(y)\)}
-\begin{align*}
-  \text{Find }& \frac{dx}{dy}\\
-  \text{Then, }\frac{dx}{dy} &= \frac{1}{\frac{dy}{dx}} \\
-  \implies {\frac{dy}{dx}} &= \frac{1}{\frac{dx}{dy}}\\
-  \therefore {\frac{dy}{dx}} &= \frac{1}{\frac{dx}{dy}}
-\end{align*}
-
-\subsection*{Second derivative}
-\begin{align*}f(x) \longrightarrow &f^\prime (x) \longrightarrow f^{\prime\prime}(x)\\
-\implies y \longrightarrow &\frac{dy}{dx} \longrightarrow \frac{d^2 y}{dx^2}\end{align*}
-
-\noindent Order of polynomial \(n\)th derivative decrements each time the derivative is taken
-
-\subsubsection*{Points of Inflection}
-
-\emph{Stationary point} - i.e.
-\(f^\prime(x)=0\)\\
-\emph{Point of inflection} - max \(|\)gradient\(|\) (i.e.
-\(f^{\prime\prime} = 0\))
-%\begin{table*}[ht]
-%\centering
-%  \begin{tabularx}{\textwidth}{XXXX}
-%\hline
-%    \rowcolor{shade2}
-%    & \(\dfrac{d^2 y}{dx^2} > 0\)  & \(\dfrac{d^2y}{dx^2}<0\) & \(\dfrac{d^2y}{dx^2}=0\) (inflection) \\
-%\hline
-%    \(\frac{dy}{dx}>0\) & \begin{tikzpicture} \draw[domain=1:2,smooth,variable=\x,blue] plot ({\x},{(1/10)*\x*\x*\x}) plot ({\x},{0.675*\x-0.677}); \end{tikzpicture} & cell 3\\
-%cell 1 & cell 2 & cell 3\\
-%\hline
-%\end{tabularx}
-%\end{table*}
-\begin{itemize}
+                  \[\int_a^b f(x) \cdot dx = [F(x)]_a^b=F(b)-F(a)\]
 
-\item
-  if \(f^\prime (a) = 0\) and \(f^{\prime\prime}(a) > 0\), then point
-  \((a, f(a))\) is a local min (curve is concave up)
-\item
-  if \(f^\prime (a) = 0\) and \(f^{\prime\prime} (a) < 0\), then point
-  \((a, f(a))\) is local max (curve is concave down)
-\item
-  if \(f^{\prime\prime}(a) = 0\), then point \((a, f(a))\) is a point of
-  inflection
-\item
-  if also \(f^\prime(a)=0\), then it is a stationary point of inflection
-\end{itemize}
-
-\begin{table*}[ht]
-  \centering
-  \includegraphics[width=0.7\textwidth]{graphics/second-derivatives.png}
-\end{table*}
-
-\subsection*{Implicit Differentiation}
-
-\noindent Used for differentiating circles etc.
-
-If \(p\) and \(q\) are expressions in \(x\) and \(y\) such that \(p=q\),
-for all \(x\) and \(y\), then:
-
-\[{\frac{dp}{dx}} = {\frac{dq}{dx}} \quad \text{and} \quad {\frac{dp}{dy}} = {\frac{dq}{dy}}\]
-
-\noindent \colorbox{cas}{\textbf{On CAS:}}\\
-Action \(\rightarrow\) Calculation \(\rightarrow\) \texttt{impDiff(y\^{}2+ax=5,\ x,\ y)}\\
-Returns \(y^\prime= \dots\).
-
-\subsection*{Integration}
-
-\[\int f(x) \cdot dx = F(x) + c \quad \text{where } F^\prime(x) = f(x)\]
-
-\subsection*{Integral laws}
-
-\renewcommand{\arraystretch}{1.4}
-\begin{tabularx}{\columnwidth}{rX}
-\hline
-  \(f(x)\) & \(\int f(x) \cdot dx\) \\
-  \hline
-  \(k\) (constant) & \(kx + c\)\\
-  \(x^n\) & \(\dfrac{1}{n+1} x^{n+1}\) \\
-  \(a x^{-n}\) &\(a \cdot \log_e x + c\)\\
-  \(\dfrac{1}{ax+b}\) &\(\dfrac{1}{a} \log_e (ax+b) + c\)\\
-  \((ax+b)^n\) & \(\dfrac{1}{a(n+1)}(ax+b)^{n-1} + c\)\\
-  \(e^{kx}\) & \(\dfrac{1}{k} e^{kx} + c\)\\
-  \(e^k\) & \(e^kx + c\)\\
-  \(\sin kx\) & \(\dfrac{-1}{k} \cos (kx) + c\)\\
-  \(\cos kx\) & \(\frac{1}{k} \sin (kx) + c\)\\
-  \(\sec^2 kx\) & \(\frac{1}{k} \tan(kx) + c\)\\
-  \(\dfrac{1}{\sqrt{a^2-x^2}}\) & \(\sin^{-1} \dfrac{x}{a} + c \>\vert\> a>0\)\\
-  \(\dfrac{-1}{\sqrt{a^2-x^2}}\) & \(\cos^{-1} \dfrac{x}{a} + c \>\vert\> a>0\)\\
-  \(\frac{a}{a^2-x^2}\) & \(\tan^{-1} \frac{x}{a} + c\)\\
-  \(\frac{f^\prime (x)}{f(x)}\) & \(\log_e f(x) + c\)\\
-  \(g^\prime(x)\cdot f^\prime(g(x)\) & \(f(g(x))\) (chain rule)\\
-  \(f(x) \cdot g(x)\) & \(\int [f^\prime(x) \cdot g(x)] dx + \int [g^\prime(x) f(x)] dx\)\\
-  \hline
-\end{tabularx}
-
-Note \(\sin^{-1} {x \over a} + \cos^{-1} {x \over a}\) is constant \(\forall x \in (-a, a)\)
-
-\subsection*{Definite integrals}
-
-\[\int_a^b f(x) \cdot dx = [F(x)]_a^b=F(b)-F(a)\]
+                  \begin{itemize}
 
-\begin{itemize}
+                    \item
+                      Signed area enclosed by\\
+                      \(\> y=f(x), \quad y=0, \quad x=a, \quad x=b\).
+                    \item
+                      \emph{Integrand} is \(f\).
+                  \end{itemize}
 
-\item
-  Signed area enclosed by\\
-  \(\> y=f(x), \quad y=0, \quad x=a, \quad x=b\).
-\item
-  \emph{Integrand} is \(f\).
-\end{itemize}
+                  \subsubsection*{Properties}
 
-\subsubsection*{Properties}
+                  \[\int^b_a f(x) \> dx = \int^c_a f(x) \> dx + \int^b_c f(x) \> dx\]
 
-\[\int^b_a f(x) \> dx = \int^c_a f(x) \> dx + \int^b_c f(x) \> dx\]
+                  \[\int^a_a f(x) \> dx = 0\]
 
-\[\int^a_a f(x) \> dx = 0\]
+                  \[\int^b_a k \cdot f(x) \> dx = k \int^b_a f(x) \> dx\]
 
-\[\int^b_a k \cdot f(x) \> dx = k \int^b_a f(x) \> dx\]
+                  \[\int^b_a f(x) \pm g(x) \> dx = \int^b_a f(x) \> dx \pm \int^b_a g(x) \> dx\]
 
-\[\int^b_a f(x) \pm g(x) \> dx = \int^b_a f(x) \> dx \pm \int^b_a g(x) \> dx\]
+                  \[\int^b_a f(x) \> dx = - \int^a_b f(x) \> dx\]
 
-\[\int^b_a f(x) \> dx = - \int^a_b f(x) \> dx\]
+                  \subsection*{Integration by substitution}
 
-\subsection*{Integration by substitution}
+                  \[\int f(u) {\frac{du}{dx}} \cdot dx = \int f(u) \cdot du\]
 
-\[\int f(u) {\frac{du}{dx}} \cdot dx = \int f(u) \cdot du\]
+                  \noindent Note \(f(u)\) must be 1:1 \(\implies\) one \(x\) for each \(y\)
+                  \begin{align*}\text{e.g. for } y&=\int(2x+1)\sqrt{x+4} \cdot dx\\
+                    \text{let } u&=x+4\\
+                    \implies& {\frac{du}{dx}} = 1\\
+                    \implies& x = u - 4\\
+                    \text{then } &y=\int (2(u-4)+1)u^{\frac{1}{2}} \cdot du\\
+                    &\text{(solve as  normal integral)}
+                  \end{align*}
 
-\noindent Note \(f(u)\) must be 1:1 \(\implies\) one \(x\) for each \(y\)
-\begin{align*}\text{e.g. for } y&=\int(2x+1)\sqrt{x+4} \cdot dx\\
-  \text{let } u&=x+4\\
-  \implies& {\frac{du}{dx}} = 1\\
-  \implies& x = u - 4\\
-  \text{then } &y=\int (2(u-4)+1)u^{\frac{1}{2}} \cdot du\\
-  &\text{(solve as  normal integral)}
-\end{align*}
+                  \subsubsection*{Definite integrals by substitution}
 
-\subsubsection*{Definite integrals by substitution}
+                  For \(\int^b_a f(x) {\frac{du}{dx}} \cdot dx\), evaluate new \(a\) and
+                  \(b\) for \(f(u) \cdot du\).
 
-For \(\int^b_a f(x) {\frac{du}{dx}} \cdot dx\), evaluate new \(a\) and
-\(b\) for \(f(u) \cdot du\).
+                  \subsubsection*{Trigonometric integration}
 
-\subsubsection*{Trigonometric integration}
+                  \[\sin^m x \cos^n x \cdot dx\]
 
-\[\sin^m x \cos^n x \cdot dx\]
+                  \paragraph{\textbf{\(m\) is odd:}}
+                  \(m=2k+1\) where \(k \in \mathbb{Z}\)\\
+                  \(\implies \sin^{2k+1} x = (\sin^2 z)^k \sin x = (1 - \cos^2 x)^k \sin x\)\\
+                  Substitute \(u=\cos x\)
 
-\paragraph{\textbf{\(m\) is odd:}}
-\(m=2k+1\) where \(k \in \mathbb{Z}\)\\
-\(\implies \sin^{2k+1} x = (\sin^2 z)^k \sin x = (1 - \cos^2 x)^k \sin x\)\\
-Substitute \(u=\cos x\)
+                  \paragraph{\textbf{\(n\) is odd:}}
+                  \(n=2k+1\) where \(k \in \mathbb{Z}\)\\
+                  \(\implies \cos^{2k+1} x = (\cos^2 x)^k \cos x = (1-\sin^2 x)^k \cos x\)\\
+                  Substitute \(u=\sin x\)
 
-\paragraph{\textbf{\(n\) is odd:}}
-\(n=2k+1\) where \(k \in \mathbb{Z}\)\\
-\(\implies \cos^{2k+1} x = (\cos^2 x)^k \cos x = (1-\sin^2 x)^k \cos x\)\\
-Substitute \(u=\sin x\)
+                  \paragraph{\textbf{\(m\) and \(n\) are even:}}
+                  use identities...
 
-\paragraph{\textbf{\(m\) and \(n\) are even:}}
-use identities...
+                  \begin{itemize}
 
-\begin{itemize}
+                    \item
+                      \(\sin^2x={1 \over 2}(1-\cos 2x)\)
+                    \item
+                      \(\cos^2x={1 \over 2}(1+\cos 2x)\)
+                    \item
+                      \(\sin 2x = 2 \sin x \cos x\)
+                  \end{itemize}
 
-\item
-  \(\sin^2x={1 \over 2}(1-\cos 2x)\)
-\item
-  \(\cos^2x={1 \over 2}(1+\cos 2x)\)
-\item
-  \(\sin 2x = 2 \sin x \cos x\)
-\end{itemize}
+                  \subsection*{Partial fractions}
 
-\subsection*{Partial fractions}
+                  \colorbox{cas}{On CAS:}\\
+                  \indent Action \(\rightarrow\) Transformation \(\rightarrow\)
+                  \texttt{expand/combine}\\
+                  \indent Interactive \(\rightarrow\) Transformation \(\rightarrow\)
+                  Expand \(\rightarrow\) Partial
 
-\colorbox{cas}{On CAS:}\\
-\indent Action \(\rightarrow\) Transformation \(\rightarrow\)
-\texttt{expand/combine}\\
-\indent Interactive \(\rightarrow\) Transformation \(\rightarrow\)
-Expand \(\rightarrow\) Partial
+                  \subsection*{Graphing integrals on CAS}
 
-\subsection*{Graphing integrals on CAS}
+                  \colorbox{cas}{In main:} Interactive \(\rightarrow\) Calculation \(\rightarrow\)
+                  \(\int\) (\(\rightarrow\) Definite)\\
+                  Restrictions: \texttt{Define\ f(x)=..} then \texttt{f(x)\textbar{}x\textgreater{}..}
 
-\colorbox{cas}{In main:} Interactive \(\rightarrow\) Calculation \(\rightarrow\)
-\(\int\) (\(\rightarrow\) Definite)\\
-Restrictions: \texttt{Define\ f(x)=..} then \texttt{f(x)\textbar{}x\textgreater{}..}
+                  \subsection*{Applications of antidifferentiation}
 
-\subsection*{Applications of antidifferentiation}
+                  \begin{itemize}
 
-\begin{itemize}
+                    \item
+                      \(x\)-intercepts of \(y=f(x)\) identify \(x\)-coordinates of
+                      stationary points on \(y=F(x)\)
+                    \item
+                      nature of stationary points is determined by sign of \(y=f(x)\) on
+                      either side of its \(x\)-intercepts
+                    \item
+                      if \(f(x)\) is a polynomial of degree \(n\), then \(F(x)\) has degree
+                      \(n+1\)
+                  \end{itemize}
 
-\item
-  \(x\)-intercepts of \(y=f(x)\) identify \(x\)-coordinates of
-  stationary points on \(y=F(x)\)
-\item
-  nature of stationary points is determined by sign of \(y=f(x)\) on
-  either side of its \(x\)-intercepts
-\item
-  if \(f(x)\) is a polynomial of degree \(n\), then \(F(x)\) has degree
-  \(n+1\)
-\end{itemize}
+                  To find stationary points of a function, substitute \(x\) value of given
+                  point into derivative. Solve for \({\frac{dy}{dx}}=0\). Integrate to find
+                  original function.
 
-To find stationary points of a function, substitute \(x\) value of given
-point into derivative. Solve for \({\frac{dy}{dx}}=0\). Integrate to find
-original function.
+                  \subsection*{Solids of revolution}
 
-\subsection*{Solids of revolution}
+                  Approximate as sum of infinitesimally-thick cylinders
 
-Approximate as sum of infinitesimally-thick cylinders
+                  \subsubsection*{Rotation about \(x\)-axis}
 
-\subsubsection*{Rotation about \(x\)-axis}
+                  \begin{align*}
+                    V &= \int^{x=b}_{x-a} \pi y^2 \> dx \\
+                    &= \pi \int^b_a (f(x))^2 \> dx
+                  \end{align*}
 
-\begin{align*}
-  V &= \int^{x=b}_{x-a} \pi y^2 \> dx \\
-    &= \pi \int^b_a (f(x))^2 \> dx
-\end{align*}
+                  \subsubsection*{Rotation about \(y\)-axis}
 
-\subsubsection*{Rotation about \(y\)-axis}
+                  \begin{align*}
+                    V &= \int^{y=b}_{y=a} \pi x^2 \> dy \\
+                    &= \pi \int^b_a (f(y))^2 \> dy
+                  \end{align*}
 
-\begin{align*}
-  V &= \int^{y=b}_{y=a} \pi x^2 \> dy \\
-    &= \pi \int^b_a (f(y))^2 \> dy
-\end{align*}
+                  \subsubsection*{Regions not bound by \(y=0\)}
 
-\subsubsection*{Regions not bound by \(y=0\)}
+                  \[V = \pi \int^b_a f(x)^2 - g(x)^2 \> dx\]
+                  \hfill where \(f(x) > g(x)\)
 
-\[V = \pi \int^b_a f(x)^2 - g(x)^2 \> dx\]
-\hfill where \(f(x) > g(x)\)
+                  \subsection*{Length of a curve}
 
-\subsection*{Length of a curve}
+                  \[L = \int^b_a \sqrt{1 + ({\frac{dy}{dx}})^2} \> dx \quad \text{(Cartesian)}\]
 
-\[L = \int^b_a \sqrt{1 + ({\frac{dy}{dx}})^2} \> dx \quad \text{(Cartesian)}\]
+                  \[L = \int^b_a \sqrt{{\frac{dx}{dt}} + ({\frac{dy}{dt}})^2} \> dt \quad \text{(parametric)}\]
 
-\[L = \int^b_a \sqrt{{\frac{dx}{dt}} + ({\frac{dy}{dt}})^2} \> dt \quad \text{(parametric)}\]
+                  \noindent \colorbox{cas}{On CAS:}\\
+                  \indent Evaluate formula,\\
+                  \indent or Interactive \(\rightarrow\) Calculation
+                  \(\rightarrow\) Line \(\rightarrow\) \texttt{arcLen}
 
-\noindent \colorbox{cas}{On CAS:}\\
-\indent Evaluate formula,\\
-\indent or Interactive \(\rightarrow\) Calculation
-\(\rightarrow\) Line \(\rightarrow\) \texttt{arcLen}
+                  \subsection*{Rates}
 
-\subsection*{Rates}
+                  \subsubsection*{Gradient at a point on parametric curve}
 
-\subsubsection*{Gradient at a point on parametric curve}
+                  \[{\frac{dy}{dx}} = {{\frac{dy}{dt}} \div {\frac{dx}{dt}}} \> \vert \> {\frac{dx}{dt}} \ne 0\]
 
-\[{\frac{dy}{dx}} = {{\frac{dy}{dt}} \div {\frac{dx}{dt}}} \> \vert \> {\frac{dx}{dt}} \ne 0\]
+                  \[\frac{d^2}{dx^2} = \frac{d(y^\prime)}{dx} = {\frac{dy^\prime}{dt} \div {\frac{dx}{dt}}} \> \vert \> y^\prime = {\frac{dy}{dx}}\]
 
-\[\frac{d^2}{dx^2} = \frac{d(y^\prime)}{dx} = {\frac{dy^\prime}{dt} \div {\frac{dx}{dt}}} \> \vert \> y^\prime = {\frac{dy}{dx}}\]
+                  \subsection*{Rational functions}
 
-\subsection*{Rational functions}
+                  \[f(x) = \frac{P(x)}{Q(x)} \quad \text{where } P, Q \text{ are polynomial functions}\]
 
-\[f(x) = \frac{P(x)}{Q(x)} \quad \text{where } P, Q \text{ are polynomial functions}\]
+                  \subsubsection*{Addition of ordinates}
 
-\subsubsection*{Addition of ordinates}
+                  \begin{itemize}
 
-\begin{itemize}
+                    \item
+                      when two graphs have the same ordinate, \(y\)-coordinate is double the
+                      ordinate
+                    \item
+                      when two graphs have opposite ordinates, \(y\)-coordinate is 0 i.e.
+                      (\(x\)-intercept)
+                    \item
+                      when one of the ordinates is 0, the resulting ordinate is equal to the
+                      other ordinate
+                  \end{itemize}
 
-\item
-  when two graphs have the same ordinate, \(y\)-coordinate is double the
-  ordinate
-\item
-  when two graphs have opposite ordinates, \(y\)-coordinate is 0 i.e.
-  (\(x\)-intercept)
-\item
-  when one of the ordinates is 0, the resulting ordinate is equal to the
-  other ordinate
-\end{itemize}
+                  \subsection*{Fundamental theorem of calculus}
 
-\subsection*{Fundamental theorem of calculus}
+                  If \(f\) is continuous on \([a, b]\), then
 
-If \(f\) is continuous on \([a, b]\), then
+                  \[\int^b_a f(x) \> dx = F(b) - F(a)\]
+                  \hfill where \(F = \int f \> dx\)
 
-\[\int^b_a f(x) \> dx = F(b) - F(a)\]
-\hfill where \(F = \int f \> dx\)
+                  \subsection*{Differential equations}
 
-\subsection*{Differential equations}
+                  \noindent\textbf{Order} - highest power inside derivative\\
+                  \textbf{Degree} - highest power of highest derivative\\
+                  e.g. \({\left(\dfrac{dy^2}{d^2} x\right)}^3\) \qquad order 2, degree 3
 
-\noindent\textbf{Order} - highest power inside derivative\\
-\textbf{Degree} - highest power of highest derivative\\
-e.g. \({\left(\dfrac{dy^2}{d^2} x\right)}^3\) \qquad order 2, degree 3
+                  \subsubsection*{Verifying solutions}
 
-\subsubsection*{Verifying solutions}
+                  Start with \(y=\dots\), and differentiate. Substitute into original
+                  equation.
 
-Start with \(y=\dots\), and differentiate. Substitute into original
-equation.
+                  \subsubsection*{Function of the dependent
+                  variable}
 
-\subsubsection*{Function of the dependent
-variable}
+                  If \({\frac{dy}{dx}}=g(y)\), then
+                  \(\frac{dx}{dy} = 1 \div \frac{dy}{dx} = \frac{1}{g(y)}\). Integrate both sides to solve equation. Only add \(c\) on one side. Express
+                  \(e^c\) as \(A\).
 
-If \({\frac{dy}{dx}}=g(y)\), then
-\(\frac{dx}{dy} = 1 \div \frac{dy}{dx} = \frac{1}{g(y)}\). Integrate both sides to solve equation. Only add \(c\) on one side. Express
-\(e^c\) as \(A\).
+                  \begin{table*}[ht]
+                    \centering
+                    \includegraphics[width=0.7\textwidth]{graphics/second-derivatives.png}
+                  \end{table*}
 
-\subsubsection*{Mixing problems}
+                  \subsubsection*{Mixing problems}
 
-\[\left(\frac{dm}{dt}\right)_\Sigma = \left(\frac{dm}{dt}\right)_{\text{in}} - \left(\frac{dm}{dt}_{\text{out}}\right)\]
+                  \[\left(\frac{dm}{dt}\right)_\Sigma = \left(\frac{dm}{dt}\right)_{\text{in}} - \left(\frac{dm}{dt}_{\text{out}}\right)\]
 
-\subsubsection*{Separation of variables}
+                  \subsubsection*{Separation of variables}
 
-If \({\frac{dy}{dx}}=f(x)g(y)\), then:
+                  If \({\frac{dy}{dx}}=f(x)g(y)\), then:
 
-\[\int f(x) \> dx = \int \frac{1}{g(y)} \> dy\]
+                  \[\int f(x) \> dx = \int \frac{1}{g(y)} \> dy\]
 
-\subsubsection*{Euler's method for solving DEs}
+                  \subsubsection*{Euler's method for solving DEs}
 
-\[\frac{f(x+h) - f(x)}{h} \approx f^\prime (x) \quad \text{for small } h\]
+                  \[\frac{f(x+h) - f(x)}{h} \approx f^\prime (x) \quad \text{for small } h\]
 
-\[\implies f(x+h) \approx f(x) + hf^\prime(x)\]
+                  \[\implies f(x+h) \approx f(x) + hf^\prime(x)\]
 
-  \end{multicols}
-\end{document}
+                \end{multicols}
+              \end{document}
-- 
2.43.2