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\fancyfoot[L]{\small May 2026}
\fancyfoot[R]{\small\itshape J.\ Murphy --- EOY Summary Sheet}
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\begin{document}

\begin{center}
{\LARGE\bfseries\color{sectionblue} EOY Cumulative --- Summary Sheet}\\[2pt]
{\small Year 11 Extended\;$\cdot$\;June 2026 EOY\;$\cdot$\;Probability $\cdot$ Functions $\cdot$ Exp/Log $\cdot$ Trigonometry}\\[2pt]
{\footnotesize\itshape\color{rulegray}v.\ \today}
\end{center}
{\color{tealmain}\rule{\linewidth}{0.4pt}}

% =============================================================
% PAGE 1: PROBABILITY + FUNCTIONS
% =============================================================
\begin{multicols}{2}

% =============================================================
% PROBABILITY
% =============================================================
\begin{topic}{Sets \& Set Notation}
$A \cup B$ = union (in $A$ or $B$); \;$A \cap B$ = intersection (in both); \;$A'$ = complement (not in $A$).\\
$n(A)$: number of elements in $A$. \;\; $\emptyset$: empty set.\\
\textbf{Subset:} $A \subseteq B$ if every element of $A$ is in $B$.
\end{topic}

\begin{topic}{Probability}
For equally likely outcomes:
\begin{formulabox}
\centering $P(A) = \dfrac{n(A)}{n(U)}$
\end{formulabox}

\textbf{Complement:} $P(A') = 1 - P(A)$.

\textbf{Union:}
\begin{formulabox}
\centering $P(A\cup B) = P(A) + P(B) - P(A\cap B)$
\end{formulabox}

\textbf{Mutually exclusive:} $P(A\cap B) = 0$, so $P(A\cup B) = P(A) + P(B)$.
\end{topic}

\begin{topic}{Conditional \& Independence}
\textbf{Conditional:}
\begin{formulabox}
\centering $P(A\mid B) = \dfrac{P(A\cap B)}{P(B)}$
\end{formulabox}

\textbf{Independent} $\iff P(A\cap B) = P(A)\cdot P(B)$ \\
$\iff P(A\mid B) = P(A)$.

\textbf{Trees:} multiply along branches; add across paths.\\
\textit{Without replacement:} second-stage probabilities depend on the first outcome.
\end{topic}

% =============================================================
% FUNCTIONS
% =============================================================
\begin{topic}{Functions: Definition, Domain, Range}
\textbf{Function:} each input $\to$ exactly one output (vertical line test).\\
\textbf{Natural domain:} all $x$ for which $f(x)$ is defined.\\
\textit{Common restrictions:} $\sqrt{\cdot}$ requires non-negative argument; $\frac{1}{\cdot}$ excludes zero denominator.
\end{topic}

\begin{topic}{Composite \& Inverse}
\textbf{Composite:} $(f\circ g)(x) = f(g(x))$. Apply $g$ first, then $f$.

\textbf{Inverse $f^{-1}$:} swap $x \leftrightarrow y$ then solve.\\
$\bullet$ Domain of $f^{-1}$ = range of $f$.\\
$\bullet$ Graph of $f^{-1}$ is reflection of $f$ in $y = x$.\\
$\bullet$ Verify: $f(f^{-1}(x)) = x$.
\end{topic}

\begin{topic}{Quadratic Functions}
\textbf{Vertex form:} $f(x) = a(x-h)^2 + k$, vertex $(h, k)$.

\textbf{Quadratic formula} (gives surds):
\begin{formulabox}
\centering $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
\end{formulabox}

\textbf{Discriminant} $\Delta = b^2 - 4ac$:\\
$\Delta > 0$: two real roots; $\Delta = 0$: one (repeated); $\Delta < 0$: none.

\textbf{Completing the square:} $x^2 + bx = (x + b/2)^2 - (b/2)^2$.
\end{topic}

\begin{topic}{Function Transformations}
Starting from the graph of $y = f(x)$:
\begin{tabbing}
\hspace{2.4cm}\=\kill
$y = f(x) + a$ \> translate \textbf{up} by $a$\\
$y = f(x) - a$ \> translate \textbf{down} by $a$\\
$y = f(x - a)$ \> translate \textbf{right} by $a$\\
$y = f(x + a)$ \> translate \textbf{left} by $a$\\
$y = -f(x)$ \> reflect in the \textbf{$x$-axis}\\
$y = f(-x)$ \> reflect in the \textbf{$y$-axis}\\
$y = a\,f(x)$ \> vertical stretch, factor $a$\\
$y = f(ax)$ \> horizontal stretch, factor $\dfrac{1}{a}$
\end{tabbing}
\textit{Tip:} for $f(x-a)$ think ``$x$ has to grow by $a$ to get the same output'', so the graph moves \emph{right} by $a$.

\textbf{Tracking a key point.}\; If $(p, q)$ lies on $y = f(x)$, then on
\begin{itemize}[nosep, leftmargin=1.2em]
  \item $y = f(x) + a$: the point becomes $(p,\; q+a)$.
  \item $y = f(x - a)$: the point becomes $(p+a,\; q)$.
  \item $y = -f(x)$: the point becomes $(p,\; -q)$.
  \item $y = f(ax)$: the point becomes $\left(\dfrac{p}{a},\; q\right)$.
\end{itemize}
\textit{Quadratics special case:} $y = (x-h)^2 + k$ \,\textit{is}\, $y = x^2$ translated right by $h$ and up by $k$.
\end{topic}

\begin{topic}{Lines (Y10 carryover)}
Through $(x_1, y_1)$ with gradient $m$:\;\; $y - y_1 = m(x - x_1)$.\\
Two points: $m = \dfrac{y_2 - y_1}{x_2 - x_1}$.\\
\textbf{Perpendicular:} gradients multiply to $-1$.\\
\textbf{Midpoint:} $\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$.\\
\textbf{Distance:} $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
\end{topic}

\end{multicols}

\newpage

% =============================================================
% PAGE 2: EXPONENTIALS/LOGS + TRIGONOMETRY
% =============================================================
\begin{multicols}{2}

% =============================================================
% EXPONENTIALS & LOGS
% =============================================================
\begin{topic}{Index Laws}
$a^m \cdot a^n = a^{m+n}$ \quad
$\dfrac{a^m}{a^n} = a^{m-n}$ \quad
$(a^m)^n = a^{mn}$\\[2pt]
$a^0 = 1$ \quad $a^{-n} = \dfrac{1}{a^n}$ \quad $a^{1/n} = \sqrt[n]{a}$ \quad $a^{p/q} = \sqrt[q]{a^p}$
\end{topic}

\begin{topic}{Logarithms: Definition}
\begin{formulabox}
\centering $\log_b a = c \iff b^c = a$
\end{formulabox}
$\log_b 1 = 0$ \quad $\log_b b = 1$ \quad $\log_b b^x = x$ \quad $b^{\log_b x} = x$
\end{topic}

\begin{topic}{Log Laws}
\begin{formulabox}
$\log(xy) = \log x + \log y$ \\[2pt]
$\log\!\left(\dfrac{x}{y}\right) = \log x - \log y$ \\[2pt]
$\log(x^n) = n\log x$
\end{formulabox}

\textbf{Change of base:} $\log_b a = \dfrac{\ln a}{\ln b} = \dfrac{\log_{10} a}{\log_{10} b}$.

\textit{Logs of surds:} write the surd as a power, then bring out the exponent. \\
$\log_2(\sqrt{32}) = \log_2(2^{5/2}) = 5/2$.
\end{topic}

\begin{topic}{Solving Equations}
$\bullet$ \textbf{Common base:} rewrite both sides with the same base, equate exponents.\\
$\bullet$ \textbf{Take logs:} for $a^x = k$, use $x = \dfrac{\log k}{\log a}$.\\
$\bullet$ \textbf{Combine logs:} $\log_b X + \log_b Y = \log_b(XY)$, then $X\cdot Y = b^{\text{RHS}}$.\\
$\bullet$ \textbf{Substitution:} for $a^{2x} - p\cdot a^x + q = 0$, let $u = a^x$ to get a quadratic.\\
$\bullet$ \textbf{Always check} that any log argument is positive.
\end{topic}

\begin{topic}{Growth \& Decay}
$A(t) = A_0\cdot b^{\,t}$. \;\;Increase rate $r\%$: $b = 1 + r/100$. \;\;Decrease rate $r\%$: $b = 1 - r/100$.\\
\textbf{Half-life:} solve $A_0 b^{\,t} = \tfrac{1}{2}A_0 \Rightarrow t = \dfrac{\ln(1/2)}{\ln b}$.
\end{topic}

% =============================================================
% TRIGONOMETRY
% =============================================================
\begin{topic}{Right-Angle Trig (Y10 review)}
\begin{formulabox}
$\sin\theta = \dfrac{\text{opp}}{\text{hyp}}$ \quad
$\cos\theta = \dfrac{\text{adj}}{\text{hyp}}$ \quad
$\tan\theta = \dfrac{\text{opp}}{\text{adj}}$
\end{formulabox}
Pythagoras: $a^2 + b^2 = c^2$.\\
Angles of elevation/depression measured from \textit{horizontal}.
\end{topic}

\begin{topic}{Bearings}
\textit{Always 3 digits, clockwise from north.} \;
N$=000\degree$, E$=090\degree$, S$=180\degree$, W$=270\degree$.\\
Back-bearing: $\pm 180\degree$.
\end{topic}

\begin{topic}{Sine \& Cosine Rule, Area}
\begin{formulabox}
$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$ \\[3pt]
$a^2 = b^2 + c^2 - 2bc\cos A$ \\[3pt]
$\text{Area} = \tfrac{1}{2}\,ab\sin C$
\end{formulabox}
Sine rule: angle $+$ opposite side. Cosine rule: SAS or SSS.\\
\textbf{Ambiguous case (SSA):} both $\theta$ and $180\degree - \theta$ may be valid.
\end{topic}

\begin{topic}{Radians, Arcs \& Sectors}
$\pi$ rad $= 180\degree$. \;\; deg $\to$ rad: $\times \pi/180$.

\begin{formulabox}
Arc: $s = r\theta$ \quad
Sector: $A = \tfrac{1}{2}r^2\theta$ \\[2pt]
Segment: $A_{\text{seg}} = \tfrac{1}{2}r^2(\theta - \sin\theta)$
\end{formulabox}
\textbf{Sector perimeter:} $2r + r\theta$. ($\theta$ in radians throughout.)
\end{topic}

\begin{topic}{Unit Circle \& Exact Values}
On the unit circle, $\sin\theta$ = height ($y$), $\cos\theta$ = width ($x$).

\textbf{Exact values:}
\renewcommand{\arraystretch}{1.2}
\begin{tabular}{c|cccc}
$\theta$ & $0$ & $30\degree$ & $45\degree$ & $60\degree$ \\\hline
$\sin$ & $0$ & $\tfrac{1}{2}$ & $\tfrac{\sqrt{2}}{2}$ & $\tfrac{\sqrt{3}}{2}$ \\
$\cos$ & $1$ & $\tfrac{\sqrt{3}}{2}$ & $\tfrac{\sqrt{2}}{2}$ & $\tfrac{1}{2}$
\end{tabular}

$\sin 90\degree = 1$, $\cos 90\degree = 0$, $\cos 180\degree = -1$.

\textbf{Identities:} $\sin^2 x + \cos^2 x = 1$; \;\;$\tan x = \dfrac{\sin x}{\cos x}$.
\end{topic}

\begin{topic}{Solving $\sin x = k$, $\cos x = k$}
\textbf{Symmetry:}
\begin{itemize}[leftmargin=1em, itemsep=0pt, topsep=1pt]
  \item Same sine: angles add to $180\degree$ (e.g.\ $\sin 30\degree = \sin 150\degree$).
  \item Same cosine: angles add to $360\degree$ (e.g.\ $\cos 60\degree = \cos 300\degree$).
\end{itemize}
\textit{Recipe:}
\begin{enumerate}[leftmargin=1.2em, itemsep=0pt, topsep=1pt]
  \item Calculator gives one solution.
  \item Sketch the unit circle; find the second solution by symmetry.
  \item Add multiples of $360\degree$ ($2\pi$) for solutions in extended ranges.
\end{enumerate}
\end{topic}

\end{multicols}

\end{document}