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 graphs, equations, and problems in circles and triangles. 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. Radian measure turns arc length into simple multiplication. Solving equations is finding graph intersections. Can you see all the connections?
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Orient and recall
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
Worksheets
Practise this lesson
Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.
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 is radian measure connected to arc length? How is solving trig equations connected to graph intersections?
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Two moves, the big picture at a glance
Work through the core explanation before applying it.
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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. Radian measure defines the angle as $\theta = \dfrac{\ell}{r}$, which turns arc length into $\ell = r\theta$ and sector area into $A = \tfrac{1}{2}r^2\theta$, 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 $x^2 + y^2 = 1$.
Radians, arcs and sectors
$\theta = \dfrac{\ell}{r}$, so $\ell = r\theta$ and $A = \tfrac{1}{2}r^2\theta$. Both formulas need $\theta$ in radians, never degrees.
Sine and cosine rules
$\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$ and $c^2 = a^2 + b^2 - 2ab\cos C$. Area $= \tfrac{1}{2}ab\sin C$. Check for the ambiguous case whenever you find an angle from the sine rule.
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What you'll master
Meet the destination, bring back what you already know, and gather the terms and formulas this lesson leans on.
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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 radian measure links angles to arc length and area
How the sine and cosine rules extend trigonometry beyond right-angled triangles
Can do
Skills
Solve problems combining identities, exact values, and equations
Sketch the basic trig graphs and read off period and amplitude
Solve arc, sector and segment problems in radians
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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.
RadianThe angle subtended at the centre of a circle by an arc equal in length to the radius.
IdentityAn equation that is true for all valid values of the variable.
Exact ValueA trig value expressed using surds and fractions, not decimals.
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Module 2 at a Glance
Work through the core explanation before applying it.
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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
Radian measure: $\ell = r\theta$ and $A = \tfrac{1}{2}r^2\theta$
Sine and cosine rules, including the ambiguous case
Solving equations graphically by finding intersections
Arc, sector and segment problems in radians
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.
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 $x^2 + y^2 = 1$.
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Making Connections
We just saw all the key facts, unit circle, Pythagorean identities, domain/range, period, reciprocal ratios, 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 Triangles
The same unit circle that defines $\sin$ and $\cos$ also gives the sine and cosine rules, which extend trigonometry to triangles with no right angle. Radian measure then ties the angle directly to arc length, so $\ell = r\theta$ and $A = \tfrac{1}{2}r^2\theta$ need no conversion factor at all.
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?
Worked examples · 3 in a row, reveal as you go
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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.
SOLUTION
3 marks · HSC band 4+
Step 1: Find $\cos \theta$ using the Pythagorean identity.
Answer: $\angle B \approx 52^\circ$ or $\angle B \approx 128^\circ$, both valid.
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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, the ambiguous case of the sine rule, 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.
SOLUTION
2 marks · HSC band 4+
Step 1: Rewrite the equation.
$$\cos x = \frac{1}{2}$$
Step 2: Recall solutions in one period.
$$\text{In } [0, 2\pi], \cos x = \frac{1}{2} \text{ has solutions at } x = \frac{\pi}{3} \text{ and } x = \frac{5\pi}{3}$$
Identity + exact value (W.E. 1): Use $\sin^2\theta + \cos^2\theta = 1$, then check the quadrant sign for $\cos\theta$, then compute $\tan\theta = \sin\theta / \cos\theta$.; Ambiguous case (W.E. 2): find both the acute and the obtuse angle, then test each against the $180^\circ$ angle sum....
Pause, copy the three exam strategies: W.E.1 (use Pythagorean identity + ASTC sign check), W.E.2 (test both angles in the ambiguous case), W.E.3 (count intersections per period) into your book.
Think to learn: A sector of radius $6$ cm has arc length $9$ cm. Find the angle at the centre in radians, and explain why no degree conversion is needed.
$\theta = \dfrac{\ell}{r} = \dfrac{9}{6} = 1.5$ radians. A radian is defined as arc length divided by radius, so the ratio already is the angle. You would only convert if the question asked for degrees.
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Common traps
Meet the mistakes that cost marks, then do it yourself.
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Common traps
We just saw three worked examples combining identities, the ambiguous case, 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, Using degrees in the arc and sector formulas
$\ell = r\theta$ and $A = \tfrac{1}{2}r^2\theta$ require $\theta$ in radians. With $r = 6$ and $\theta = 30^\circ$, the arc is $6 \times \dfrac{\pi}{6} = \pi$ cm, not $6 \times 30 = 180$.
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 $[0, 2\pi]$.
Quick-fire practice · 5 reps
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Drill it, then lock it in
Run the quick drill and copy the summary into your book.
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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 $\tan \theta = \dfrac{4}{3}$.
A sector has radius $8$ cm and subtends $0.75$ radians. Find the arc length and the sector area.
Show answer
Amplitude = $4$, Period = $\dfrac{2\pi}{3}$, Range = $[-2, 6]$.
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How many solutions does $\sin x = -0.5$ have in $[0, 4\pi]$?
Show answer
$4$ solutions (2 per period, 2 periods).
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Solve $2\sin x = \sqrt{3}$ for $0 \leq x \leq 2\pi$.
Show answer
$\sin x = \dfrac{\sqrt{3}}{2}$, so $x = \dfrac{\pi}{3}$ or $x = \dfrac{2\pi}{3}$ in $[0, 2\pi]$.
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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 radian is defined by arc length over radius, which is why $\ell = r\theta$ needs no conversion factor.
Solving $\sin x = k$ is the same as finding where $y = \sin x$ and $y = k$ intersect.
The sine and cosine rules come from dropping a perpendicular, which is why they reduce to right-angled trigonometry.
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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Multiple choice
Answer the drill bank and rate your confidence.
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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.
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Short answer
Write full responses, then check them against the model answers.
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Short answer
ApplyBand 4
Arc, sector and segment
A sector of a circle has radius $12$ cm and subtends an angle of $\dfrac{\pi}{3}$ radians at the centre. (a) Find the exact arc length. (b) Find the exact area of the sector. (c) Find the exact area of the segment cut off by the chord joining the two radii. [5 marks]
Solve $2\cos^2 x + \sin x = 1$ for $0 \leq x \leq 2\pi$, giving exact values. [5 marks]
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View comprehensive answer
Use $\cos^2 x = 1 - \sin^2 x$: $2(1 - \sin^2 x) + \sin x = 1$ [1].
$2 - 2\sin^2 x + \sin x = 1 \Rightarrow 2\sin^2 x - \sin x - 1 = 0$ [1].
Factorise: $(2\sin x + 1)(\sin x - 1) = 0$ [1].
$\sin x = 1 \Rightarrow x = \dfrac{\pi}{2}$ [1].
$\sin x = -\dfrac{1}{2} \Rightarrow x = \dfrac{7\pi}{6}$ or $x = \dfrac{11\pi}{6}$ [1].
EvaluateBand 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].
This follows from the complementary angle relationship $\cos x = \sin\!\left(\dfrac{\pi}{2} - x\right)$ [1].
SAQ2 (5 marks): substitute $\cos^2 x = 1 - \sin^2 x$ [1], form $2\sin^2 x - \sin x - 1 = 0$ [1], factorise [1], $x = \dfrac{\pi}{2}$ [1], $x = \dfrac{7\pi}{6}, \dfrac{11\pi}{6}$ [1].
SAQ3 (3 marks): Correct for the basic functions [1], same shape, amplitude and period [1], note the complementary angle relationship $\cos x = \sin\!\left(\dfrac{\pi}{2} - x\right)$ [1].
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Review and finish
Take the module quiz if you are ready, then mark the lesson complete.
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Take the full module quiz
quiz
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
