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Module 2 · L15 of 15 ~35 min ⚡ +100 XP available

Review and Connections

You have reached the final lesson of Module 2. This lesson brings together everything you have learned about trigonometric functions and graphs — from exact values and identities to sketching, solving, and modelling. Use this review to solidify your understanding before tackling the Module Quiz.

Today's hook — Every concept in Module 2 traces back to the unit circle. The Pythagorean identity is just the circle equation in disguise. Phase shifts are horizontal translations. Solving equations is finding graph intersections. Can you see all the connections?
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Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery — or build your own from any module’s questions.

01
Recall — your gut answer first
+5 XP warm-up

Without looking back at your notes, try to list as many connections as you can between the ideas in this module. For example: how are the Pythagorean identities connected to the unit circle? How are phase shifts connected to horizontal translations? How is solving trig equations connected to graph intersections?

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02
Two moves — the big picture at a glance
+5 XP to read

Three pillars hold up everything in Module 2. Master these connections and you will see trig as one unified system instead of isolated facts.

The unit circle is the foundation: exact values, identities, signs in quadrants, and the shapes of graphs all trace back to it. Transformations follow the general form $y = a\sin(b(x-c)) + d$ — amplitude, period, phase shift, vertical shift. Modelling reverses this: given real data, extract $a = \dfrac{\text{max}-\text{min}}{2}$, $d = \dfrac{\text{max}+\text{min}}{2}$, and $b = \dfrac{2\pi}{P}$.

Big Three
Unit circle
Exact values, identities, signs, graphs — everything traces back to the unit circle. If you are ever stuck, return to $x^2 + y^2 = 1$.
Transformations
$y = a\sin(b(x-c)) + d$ — amplitude $|a|$, period $\dfrac{2\pi}{|b|}$, phase shift $c$, vertical shift $d$. Always factor out $b$ first.
Modelling
$a = \dfrac{\text{max}-\text{min}}{2}$, $d = \dfrac{\text{max}+\text{min}}{2}$, $b = \dfrac{2\pi}{P}$. Reverse the transformation process to fit real-world data.
03
What you'll master
Know

Key facts

  • All six trig functions and their reciprocal relationships
  • The three Pythagorean identities
  • The domains, ranges, periods, and key features of all trig graphs
Understand

Concepts

  • How all trig ideas connect back to the unit circle
  • How transformations affect trig graphs
  • How trig functions model periodic real-world phenomena
Can do

Skills

  • Solve problems combining identities, exact values, and equations
  • Analyse and sketch transformed trig graphs
  • Build and use trig models for applications
04
Key terms
Periodic FunctionA function that repeats its values at regular intervals.
RadianThe natural unit for angle measure in calculus and physics.
AmplitudeThe maximum displacement from the centre line of a periodic function.
Phase ShiftA horizontal translation of a trigonometric graph.
IdentityAn equation that is true for all valid values of the variable.
Exact ValueA trig value expressed using surds and fractions, not decimals.
05
Module 2 at a Glance
core concept

Inquiry Question 1: Trigonometry

  • Exact values of sine, cosine, and tangent at special angles ($0, \dfrac{\pi}{6}, \dfrac{\pi}{4}, \dfrac{\pi}{3}, \dfrac{\pi}{2}$ and multiples)
  • Reciprocal trig functions: $\csc x$, $\sec x$, $\cot x$ and their relationships to sine, cosine, and tangent
  • Pythagorean identities: $\sin^2 x + \cos^2 x = 1$, $1 + \tan^2 x = \sec^2 x$, $1 + \cot^2 x = \csc^2 x$
  • Complementary angle identities (co-functions)

Inquiry Question 2: Trigonometric Functions and Graphs

  • Domains and ranges of all six trig functions
  • Graphs of sine and cosine: amplitude, period, intercepts, maxima, minima
  • Graphs of tangent and cotangent: asymptotes, period $\pi$, branches
  • Transformations: $y = a\sin(b(x - c)) + d$ and the meaning of each parameter
  • Phase shifts as horizontal translations
  • Solving equations graphically by finding intersections
  • Modelling periodic phenomena with trig functions
The unit circle: the heart of it all. Every concept in this module — exact values, identities, signs in different quadrants, the shapes of graphs, domain restrictions — can be traced back to the unit circle. If you ever feel lost, return to the unit circle. It is the single most powerful diagram in trigonometry.
Unit Circle x² + y² = 1 Exact Values sin, cos, tan at special angles Identities Pythagorean Co-function Graphs Domain, range Period, amplitude Equations Solve graphically Multiple solutions Modelling Real-world waves Tides, sound, light

Module 2 concept map: the unit circle is the foundation — everything connects back to it

Unit circle: $x^2 + y^2 = 1$ underpins all exact values, identities, and graph shapes; Three Pythagorean identities: $\sin^2 x + \cos^2 x = 1$; $1 + \tan^2 x = \sec^2 x$; $1 + \cot^2 x = \csc^2 x$

Pause — copy the Module 2 quick-reference: unit circle $x^2+y^2=1$ underpins everything, plus the three Pythagorean identities ($\sin^2 x + \cos^2 x = 1$; $1 + \tan^2 x = \sec^2 x$; $1 + \cot^2 x = \csc^2 x$) into your book.

Did you get this? True or false: the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ can be derived directly from the equation of the unit circle $x^2 + y^2 = 1$.

06
Making Connections

We just saw all the key facts — unit circle, Pythagorean identities, domain/range, period, phase shift — as a module reference. That raises a question: how do these topics connect to each other, so you can see the module as a coherent whole rather than isolated skills? This card answers it → identities explain graph bounds; graphical intersection = algebraic solution; ASTC governs signs throughout.

From Identities to Graphs

The identity $\sin^2 x + \cos^2 x = 1$ is why the graphs of $y = \sin x$ and $y = \cos x$ are both bounded between $-1$ and $1$. The Pythagorean identity is the algebraic expression of the geometric fact that points on the unit circle satisfy $x^2 + y^2 = 1$.

From Equations to Graphs

Solving $\sin x = k$ is equivalent to finding the $x$-coordinates where the horizontal line $y = k$ intersects the sine curve. This graphical viewpoint explains why there can be zero, one, two, or infinitely many solutions depending on the value of $k$ and the domain.

From Graphs to Models

Once you understand how amplitude, period, phase shift, and vertical shift transform the basic sine and cosine graphs, you can reverse the process: take real-world data, identify these features, and write a mathematical model that predicts future behaviour.

Identities ↔ Graphs: $\sin^2 x + \cos^2 x = 1$ means both graphs are bounded in $[-1, 1]$; Equations ↔ Graphs: Solving $\sin x = k$ = finding intersections of $y = \sin x$ and $y = k$

Pause — copy the two cross-topic links: identities ↔ graphs ($\sin^2+\cos^2=1$ explains the $[-1,1]$ bound) and equations ↔ graphs (solving $\sin x = k$ = finding intersections on the sketch) into your book.

Quick check: The graph of $y = \cos x$ is the same as $y = \sin x$ shifted how far?

07
Worked Example — Combined Identity and Exact Value
+5 XP for trying first

If $\sin \theta = \dfrac{1}{3}$ and $\dfrac{\pi}{2} < \theta < \pi$, find the exact value of $\tan \theta$.

Your turn first. Try it yourself before viewing the solution.
08
Worked Example — Sketching a Transformed Graph
+5 XP for trying first

Sketch $y = 2\sin\!\left(x + \dfrac{\pi}{3}\right) - 1$ for $-\dfrac{\pi}{3} \leq x \leq \dfrac{5\pi}{3}$ and state the range.

Your turn first. Try it yourself before viewing the solution.
09
Worked Example — Counting Graphical Solutions
+5 XP for trying first

We just saw how identities, graphs, and equations are all facets of the same unit circle geometry. That raises a question: what does a full exam-quality response look like when these ideas are combined in one question? This card answers it → three worked examples integrating identity proofs, transformed graph sketching, and graphical solution counting.

How many solutions does $2\cos x = 1$ have in $0 \leq x \leq 4\pi$?

Your turn first. Try it yourself before viewing the solution.

Identity + exact value (W.E. 1): Use $\sin^2\theta + \cos^2\theta = 1$, then check the quadrant sign for $\cos\theta$, then compute $\tan\theta = \sin\theta / \cos\theta$.; Transformed sketch (W.E. 2): Always factor out $b$ first to read the phase shift....

Pause — copy the three exam strategies: W.E.1 (use Pythagorean identity + ASTC sign check), W.E.2 (factor out $b$ before reading phase shift), W.E.3 (count intersections per period) into your book.

Think to learn: What is the phase shift of $y = \sin\!\left(2x - \dfrac{\pi}{3}\right)$? Explain why you need to factor before reading off the phase shift.

10
Common traps

We just saw three worked examples combining identities, transformed sketches, and graphical solution counts. That raises a question: what exam-day errors most often cost marks across Module 2 specifically? This card answers it → the reciprocal-pair mix-ups ($\sin\leftrightarrow\csc$ etc.) and ASTC sign errors when applying identities.

Trap 1 — Mixing up reciprocal identities

Students sometimes write $\sec x = \dfrac{1}{\sin x}$ or $\csc x = \dfrac{1}{\cos x}$. Remember the pairs: sine-cosecant, cosine-secant, tangent-cotangent.

Trap 2 — Forgetting quadrant signs when finding exact values

Even if you calculate the correct magnitude, you can lose marks if you give the wrong sign for the quadrant. ASTC — All positive in Q1, Sin in Q2, Tan in Q3, Cos in Q4.

Trap 3 — Not factoring out $b$ when finding phase shifts

In $y = \sin\!\left(2x - \dfrac{\pi}{3}\right)$, the phase shift is $\dfrac{\pi}{6}$, not $\dfrac{\pi}{3}$. Always factor: $\sin\!\left(2\!\left(x - \dfrac{\pi}{6}\right)\right)$.

Reciprocal pairs: $\sin \leftrightarrow \csc$, $\cos \leftrightarrow \sec$, $\tan \leftrightarrow \cot$; ASTC for signs — commit this to memory

Pause — copy the Module 2 trap list: reciprocal pairs ($\sin\leftrightarrow\csc$, $\cos\leftrightarrow\sec$, $\tan\leftrightarrow\cot$) and ASTC — commit the quadrant sign rule to memory into your book.

Odd one out: Three of these statements about reciprocal trig functions are correct. Which one is wrong?

Did you get this? True or false: the equation $2\cos x = 1$ has exactly 2 solutions in $[0, 2\pi]$.

Work mode · how are you completing this lesson?
1

If $\cos \theta = -\dfrac{3}{5}$ and $\pi < \theta < \dfrac{3\pi}{2}$, find $\sin \theta$ and $\tan \theta$.

Show answer
In QIII, $\sin \theta = -\dfrac{4}{5}$ and $\tan \theta = \dfrac{4}{3}$.
2

Simplify $\dfrac{1 - \cos^2 x}{\sin x \cos x}$.

Show answer
$\dfrac{\sin^2 x}{\sin x \cos x} = \dfrac{\sin x}{\cos x} = \tan x$
3

State the amplitude, period, and range of $y = 4\cos(3x) + 2$.

Show answer
Amplitude = $4$, Period = $\dfrac{2\pi}{3}$, Range = $[-2, 6]$.
4

How many solutions does $\sin x = -0.5$ have in $[0, 4\pi]$?

Show answer
$4$ solutions (2 per period, 2 periods).
5

Sketch $y = 3\sin\!\left(2x - \dfrac{\pi}{3}\right)$ for one complete period. State key features.

Show answer
Rewrite: $y = 3\sin\!\left(2\!\left(x - \dfrac{\pi}{6}\right)\right)$. Amplitude = $3$, Period = $\pi$, Phase shift = $\dfrac{\pi}{6}$ right. Starts at midline going up at $x = \dfrac{\pi}{6}$. Max at $x = \dfrac{5\pi}{12}$ (value 3). Min at $x = \dfrac{11\pi}{12}$ (value $-3$).
13
Revisit — module connections
+5 XP for checking

Earlier you were asked to list connections between the ideas in this module.

Here are some key connections:

  • The Pythagorean identity $\sin^2 x + \cos^2 x = 1$ comes directly from the unit circle equation $x^2 + y^2 = 1$.
  • A phase shift is just a horizontal translation of the basic graph.
  • Solving $\sin x = k$ is the same as finding where $y = \sin x$ and $y = k$ intersect.
  • Modelling uses the same transformations (amplitude, period, phase shift, vertical shift) that we use to sketch graphs.
  • Exact values, identities, and graphs all depend on the symmetries of the unit circle.

Return to your original answer from Section 01. What did you get right? What has changed in your thinking?

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01
Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

02
Short answer
Apply Band 4

Sketch a transformed sine graph

Sketch $y = 3\sin\!\left(2x - \dfrac{\pi}{3}\right)$ for one complete period, starting from the first positive $x$-intercept. Label the amplitude, period, phase shift, and key points. [5 marks]

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View comprehensive answer

Rewrite: $y = 3\sin\!\left(2\!\left(x - \dfrac{\pi}{6}\right)\right)$ [1].

Amplitude = $3$, Period = $\pi$, Phase shift = $\dfrac{\pi}{6}$ right [1].

First positive intercept at $x = \dfrac{\pi}{6}$ [1].

Max at $x = \dfrac{5\pi}{12}$ (value 3) [1].

Min at $x = \dfrac{11\pi}{12}$ (value $-3$) [1].

Analyse Band 5

Water wheel model

A water wheel rotates so that the height $h$ (in metres) of a bucket above the water surface is modelled by $h = 2\sin\!\left(\dfrac{\pi}{4}t\right) + 1$, where $t$ is in seconds. (a) Find the maximum and minimum heights. (b) Find the first time after $t = 0$ when the bucket is at a height of 2 metres. (c) State one limitation of using a simple sine model for this situation. [5 marks]

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View comprehensive answer

(a) Max = $2 + 1 = 3$ m, Min = $-2 + 1 = -1$ m [2].

(b) $2\sin\!\left(\dfrac{\pi}{4}t\right) + 1 = 2 \Rightarrow \sin\!\left(\dfrac{\pi}{4}t\right) = \dfrac{1}{2} \Rightarrow \dfrac{\pi}{4}t = \dfrac{\pi}{6} \Rightarrow t = \dfrac{2}{3}$ s [2].

(c) Limitation: the model doesn't account for friction, variable rotation speed, or the bucket dipping below the water surface [1].

Evaluate Band 6

Evaluate a claim about sine and cosine

Evaluate the claim: "The graphs of $y = \sin x$ and $y = \cos x$ are identical except for a horizontal translation." Is this claim fully correct? Explain any limitations or exceptions. [3 marks]

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View comprehensive answer

The claim is correct for the standard functions: $\cos x = \sin\!\left(x + \dfrac{\pi}{2}\right)$ [1].

The graphs have the same shape, amplitude, and period, differing only by a horizontal shift of $\dfrac{\pi}{2}$ [1].

No exceptions for the basic functions, though transformed versions would need matching transformations [1].

📖 Comprehensive answers (click to reveal)

Drill 1: In QIII, $\sin \theta = -\dfrac{4}{5}$, $\tan \theta = \dfrac{4}{3}$.

Drill 2: $\dfrac{\sin^2 x}{\sin x \cos x} = \dfrac{\sin x}{\cos x} = \tan x$.

Drill 3: Amplitude = 4, Period = $\dfrac{2\pi}{3}$, Range = $[-2, 6]$.

Drill 4: 4 solutions (2 per period, 2 periods).

Drill 5: Rewrite: $y = 3\sin\!\left(2\!\left(x - \dfrac{\pi}{6}\right)\right)$. Amplitude = 3, Period = $\pi$, Phase shift = $\dfrac{\pi}{6}$ right. First positive intercept at $x = \dfrac{\pi}{6}$. Max at $x = \dfrac{5\pi}{12}$ (value 3). Min at $x = \dfrac{11\pi}{12}$ (value $-3$).

SAQ1 (5 marks): Rewrite [1], amplitude = 3 [0.5], period = $\pi$ [0.5], phase shift = $\dfrac{\pi}{6}$ right [1]. First positive intercept at $x = \dfrac{\pi}{6}$ [1]. Max at $x = \dfrac{5\pi}{12}$ (value 3) [0.5]. Min at $x = \dfrac{11\pi}{12}$ (value $-3$) [0.5].

SAQ2 (5 marks): (a) Max = 3 m, Min = $-1$ m [2]. (b) $t = \dfrac{2}{3}$ s [2]. (c) One valid limitation [1].

SAQ3 (3 marks): Correct for standard functions [1], same shape/amplitude/period [1], note about transformed versions [1].

01
Boss battle · Trigonometry Final
earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena
02
Science Jump · platform challenge

Climb platforms by answering Module 2 questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

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