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Module 7 · L20 of 20 ~45 min ⚡ +100 XP available

Module 7 Synthesis & Exam Technique

This is the final lesson of Module 7 — everything comes together here. You'll build a complete reference map of the module, work through a multi-part exam question from scratch, and practise the time-management tactics that turn thorough preparation into actual HSC marks. By the end you'll know exactly which technique to reach for, and why.

Today's hook — Module 7 covers 6 major technique families. Without looking at notes, how many can you list? Jot them here. You'll build a full technique map in card 05 — then compare against your list.
0/5QUESTS
01
Recall — your gut answer first
+5 XP warm-up

Module 7 covers a substantial set of techniques. Without looking at notes — list as many distinct technique families as you can recall, and for each, write the key formula or the type of problem it solves.

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02
The big picture — what Module 7 is really about
+5 XP to read

Module 7 builds a single unified skill: transforming trigonometric expressions into forms that are easier to solve or analyse. Every technique is a different transformation strategy.

The six families:

  1. Auxiliary angle — combine $a\sin x + b\cos x$ into $R\sin(x \pm \alpha)$
  2. t-formulae — rational substitution $t = \tan\tfrac x2$
  3. Inverse trig — principal values, identities, evaluation
  4. Identities & proofs — double-angle, compound angle, sum-to-product
  5. Multiple-angle equations — reduce to a standard form via identities
  6. Combined / harder — chain two or more techniques in one question
Module 7 Trig skills Auxiliary angle t-formulae Inverse Identities Multi-angle Combined
$a\sin x + b\cos x = R\sin(x+\alpha)$
Time management
In a 3-hour HSC paper worth 100 marks, each mark $\approx$ 1.8 min. A 4-mark trig question should take at most 8 minutes. Move on if stuck — partial marks are available.
Read the question
The question usually signals the method: "express in the form $R\sin(x+\alpha)$" = auxiliary; "use $t = \tan\tfrac x2$" = t-formulae; "hence find" = use your previous answer.
Show all working
Markers award method marks. Even a wrong final answer can earn 2 of 3 marks if the approach is sound. Never skip steps in a proof question.
03
What you'll consolidate
Know

Key formulae

  • All six technique families and when each applies
  • The identity chain: compound $\to$ double-angle $\to$ half-angle
  • Inverse trig principal value ranges and key identities
Understand

Connections

  • How auxiliary angle, t-formulae and identities all stem from the same underlying algebra
  • Why "hence" parts must use the previous result, not an alternative method
  • How to check solutions and identify missing/extraneous roots
Can do

Exam skills

  • Identify the correct technique within 30 seconds of reading a question
  • Manage time effectively — allocate marks-to-minutes correctly
  • Complete a multi-part Module 7 question with full working
04
Key terms
SynthesisCombining knowledge from different topics to solve a problem that draws on more than one technique or concept. Common in HSC Band 5/6 questions.
Compound angle identity$\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B$; $\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B$. Foundation for all other identities in the module.
Double-angle identity$\sin 2A = 2\sin A\cos A$; $\cos 2A = \cos^2 A - \sin^2 A = 1-2\sin^2 A = 2\cos^2 A - 1$. Derived from compound angle with $B = A$.
"Hence" instructionYou must use the result from the previous part. Solving independently (even correctly) may not earn marks if the examiner intended a specific approach.
Domain of solutionThe set of $x$-values in which you must find solutions. Always state the domain at the start of a solve question and verify that all final answers lie within it.
ME12-3NESA outcome: applies advanced concepts and techniques in simplifying expressions involving compound angles and solving trigonometric equations.
05
Complete technique selection map
core concept

Use this map to decide which technique to apply. The diagnostic question is always: what structure does the equation have?

Question type Key signal Technique Core formula
Linear trig equation $a\sin x + b\cos x = c$ Auxiliary angle $R = \sqrt{a^2+b^2}$
Rational trig equation fraction in $\sin x$, $\cos x$ t-substitution $t = \tan\tfrac x2$
Inverse trig value/identity $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$ Principal values Range definitions
Double/compound expression $\sin 2x$, $\cos(A+B)$ etc. Identities $\sin(A\pm B)$ etc.
Equation in $\sin nx$ or $\cos nx$ multiple angle, $x \in [0, 2\pi]$ Expand, substitute Substitute $u = nx$, solve in $u$
Multi-part "hence" question part (i) gives a form, part (ii) uses it Chain techniques Must use prior result
The "hence" trap. If part (i) asks you to express something in a specific form, and part (ii) says "hence solve", you must substitute your part (i) result into part (ii). Using a completely different method (even if correct) may score zero because the marker is looking for evidence you can connect parts of a multi-step problem.

Keep a one-page "technique selector" in the front of your notes book: the 6-row table above; For every question: (1) identify structure, (2) select technique, (3) execute, (4) check domain

Pause — copy the 6-row technique selection table into your book, mapping each equation structure (factorisation / quadratic-sub / auxiliary angle / $t$-formula / identity / multiple angle) to its required method.

Quick check: A question says "use $t = \tan\tfrac{x}{2}$ to solve $\ldots$". This instruction tells you to use which technique family?

06
Multi-part questions — exam technique
core concept

We just saw the 6-technique selection table mapping every trig equation structure (factorisation, quadratic-sub, auxiliary angle, $t$-formula, identity, multiple angle) to its method. That raises a question: in an HSC exam with linked multi-part questions, how do you manage the information flow between parts without re-deriving results? This card answers it → read all parts first; the result proved in part (i) is the scaffold for part (ii) — cite it, don’t reprove it.

HSC Module 7 questions often have 3–4 linked parts worth 8–12 marks total. The structure is predictable:

  • Part (i) (1–2 marks): Express or show. Derive a useful form of the expression. This is the easy entry point — always attempt it.
  • Part (ii) (2–3 marks): Hence solve/find. Must use part (i). A domain is given. Expect 2–3 solutions.
  • Part (iii) (2–3 marks): Harder application — may involve: maximum/minimum, number of solutions, or a contextual problem.
  • Part (iv) (1–2 marks, sometimes): Proof, extension, or "show that".

Strategy: Read all parts before starting. Part (iii) often reveals what "nice form" part (i) should produce. Working backwards from the end gives you strategic information.

The "show that" trick. In "show that $\sin^2 x = \frac{1-\cos 2x}{2}$", the answer is given to you. Your job is to derive it rigorously — write every step. One-line answers almost never earn full marks in show-that questions.

Before writing: read all parts of the question; Part (i) is usually the scaffolding for part (ii) — do (i) carefully

Pause — copy the multi-part strategy into your book: read all parts before writing; the result from part (i) feeds directly into part (ii) — never re-derive in (ii) what you proved in (i).

Did you get this? True or false: in a "hence solve" question, you may use any valid method to solve the equation, not just the result from the previous part.

PROBLEM 1 · MULTI-PART (8 marks)

(i) Express $f(x) = \sqrt{3}\sin x + \cos x$ in the form $R\sin(x + \alpha)$ where $R > 0$ and $0 < \alpha < \frac{\pi}{2}$. (2 marks)

(ii) Hence, or otherwise, solve $f(x) = \sqrt{2}$ for $x \in [0, 2\pi]$. (3 marks)

(iii) Hence find the maximum value of $f(x)$ and state all $x \in [0, 2\pi]$ where it occurs. (2 marks)

(iv) For how many values of $x \in [0, 2\pi]$ does $f(x) = k$ have solutions when $k > R$? (1 mark)

(i)
$R = \sqrt{3+1} = 2$; $\tan\alpha = \dfrac{1}{\sqrt{3}} \Rightarrow \alpha = \dfrac{\pi}{6}$. So $f(x) = 2\sin\!\left(x + \dfrac{\pi}{6}\right)$.
Identify $a = \sqrt{3}$, $b = 1$. Both positive, so $\alpha \in (0, \pi/2)$. $\tan\alpha = b/a = 1/\sqrt{3}$ gives the exact value $\pi/6$.
PROBLEM 2 · COMBINED (3 marks)

Show that $\cos 2x = 1 - 2\sin^2 x$, and hence solve $\cos 2x + 3\sin x = 2$ for $x \in [0, 2\pi]$. (3 marks)

1
$\cos 2x = \cos^2 x - \sin^2 x = (1-\sin^2 x) - \sin^2 x = 1 - 2\sin^2 x$. $\checkmark$
Start from the compound angle identity $\cos(A+A)$ and substitute $\cos^2 x = 1 - \sin^2 x$.
PROBLEM 3 · INVERSE TRIG IDENTITY (3 marks)

Given that $\theta = \cos^{-1}\!\left(-\dfrac{\sqrt{3}}{2}\right)$, find the exact value of $\sin\theta$. (3 marks)

1
$\cos^{-1}$ has range $[0, \pi]$. $\cos\theta = -\dfrac{\sqrt{3}}{2}$ and $\theta \in [0,\pi]$, so $\theta = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$.
Identify the principal value. $\cos\tfrac\pi6 = \tfrac{\sqrt3}{2}$, so the angle with $\cos = -\tfrac{\sqrt3}{2}$ in $[0,\pi]$ is the second-quadrant angle.

Fill the gap: The maximum value of $a\sin x + b\cos x$ is $= \sqrt{a^2 + b^2}$, and it is achieved when $\sin(x+\alpha) = 1$.

Trap 01
Ignoring the "hence"
If the question says "hence solve", you must connect to the previous part. Solving independently using, say, the t-formula when the previous part set up the auxiliary angle form, will likely receive zero for method. Always use the form derived in part (i).
Trap 02
Incorrect number of solutions
$\sin u = k$ has two solutions per period (for $-1 < k < 1$, $k \neq \pm1$). A common error is finding only one. Sketch a rough graph of $y = \sin u$ over the extended domain to visualise all intersections before writing your answer.
Trap 03
Careless sign error in $\alpha$
When $b < 0$ in $a\sin x + b\cos x$, you must use $R\sin(x - \alpha)$ (not $R\sin(x + \alpha)$) and find $\alpha > 0$. Mixing up the sign of the shift puts every solution off by $2\alpha$ — a systematic error that loses all method marks in the solve part.

Did you get this? True or false: for $x \in [0, 2\pi]$, the equation $\sin x = \dfrac{1}{2}$ has exactly two solutions.

Work mode · how are you completing this lesson?
1

Express $f(x) = \sin x - \cos x$ in the form $R\sin(x - \alpha)$. Find $R$ and $\alpha$ exactly.

2

Hence solve $f(x) \geq 1$ for $x \in [0, 2\pi]$, expressing your answer as an interval.

3

Use $\cos 2x = 2\cos^2 x - 1$ to solve $2\cos^2 x + \cos x - 1 = 0$ for $x \in [0, 2\pi]$.

4

Find the exact value of $\tan\!\left(\cos^{-1}\!\left(\dfrac{3}{5}\right)\right)$, showing all reasoning.

5

A Module 7 exam question is worth 10 marks and you have 3 hours (180 min) for the paper (100 marks total). How many minutes should you budget for this question?

Odd one out: Three of these statements about Module 7 are correct. Which one is NOT?

11
Revisit your thinking — Module 7 complete

Earlier you listed as many Module 7 technique families as you could from memory. The full list is: auxiliary angle, t-formulae, inverse trig, identities/proofs, multiple-angle equations, combined/harder questions. How many did you get?

More importantly: can you now describe, in one sentence each, what triggers you to reach for each technique? Jot your descriptions below — these are the heuristics that translate revision into examination performance.

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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.

02
Short answer
ApplyBand 32 marks

Q1. Express $\sin x - \cos x$ in the form $R\sin(x - \alpha)$ where $R > 0$ and $0 < \alpha < \dfrac\pi2$. Give exact values for $R$ and $\alpha$. (2 marks)

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ApplyBand 43 marks

Q2. Hence solve $\sin x - \cos x = 1$ for $x \in [0, 2\pi]$. (3 marks)

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AnalyseBand 53 marks

Q3. Find all solutions of $\cos 2x + 3\sin x - 2 = 0$ for $x \in [0, 2\pi]$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. $R = \sqrt{1+1} = \sqrt{2}$; $\alpha = \pi/4$; $f(x) = \sqrt{2}\sin(x - \pi/4)$.

2. $\sqrt{2}\sin(x-\pi/4) \geq 1 \Rightarrow \sin(x-\pi/4) \geq \tfrac{1}{\sqrt{2}}$. Let $u = x - \pi/4 \in [-\pi/4, 7\pi/4]$. $\sin u \geq \tfrac{1}{\sqrt{2}}$ when $u \in [\pi/4, 3\pi/4]$. So $x \in [\pi/4+\pi/4, 3\pi/4+\pi/4] = [\pi/2, \pi]$.

3. $(2\cos x - 1)(\cos x + 1) = 0$; $\cos x = \tfrac12 \Rightarrow x = \pi/3, 5\pi/3$; $\cos x = -1 \Rightarrow x = \pi$. Solutions: $x = \pi/3, \pi, 5\pi/3$.

4. $\cos\theta = 3/5$, $\theta \in [0,\pi]$. By Pythagoras: $\sin\theta = 4/5$ (positive). $\tan\theta = \tfrac{4/5}{3/5} = \tfrac{4}{3}$.

5. Time per mark $= 180/100 = 1.8$ min. Budget $= 1.8 \times 10 = 18$ minutes.

Q1 (2 marks): $R = \sqrt{1+1} = \sqrt{2}$ [1]; $\tan\alpha = 1 \Rightarrow \alpha = \pi/4$ [1].

Q2 (3 marks): $\sqrt{2}\sin(x-\pi/4) = 1 \Rightarrow \sin(x-\pi/4) = \tfrac{1}{\sqrt{2}}$ [1]. $x-\pi/4 = \pi/4$ or $3\pi/4$ [1]. $x = \pi/2$ or $x = \pi$ [1].

Q3 (3 marks): $\cos 2x = 1-2\sin^2 x$: $1-2\sin^2 x + 3\sin x - 2 = 0 \Rightarrow 2\sin^2 x - 3\sin x + 1 = 0 \Rightarrow (2\sin x - 1)(\sin x - 1) = 0$ [1]. $\sin x = \tfrac12 \Rightarrow x = \pi/6, 5\pi/6$ [1]. $\sin x = 1 \Rightarrow x = \pi/2$ [1]. Three solutions.

01
Boss battle · The Module 7 Final
earn bronze · silver · gold

Five timed questions spanning all Module 7 techniques. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). This is your module sign-off.

⚔ Enter the arena
02
Science Jump · platform challenge

Climb platforms by answering Module 7 synthesis questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review. Congratulations on completing Module 7!

🎓
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