Review of Trigonometric Foundations
Compound angles, double angles, t-formulae — these three families of identities are the toolkit for everything in Module 7. In this lesson you'll rebuild them from scratch so every derivation in lessons 2–20 feels inevitable rather than surprising.
Without looking anything up, write everything you remember about compound angle and double angle identities from Year 11. Which ones can you recall exactly? Which feel fuzzy?
Module 7 rests on three families of identities you first met in Year 11. Master their forms and derivations and every advanced technique becomes accessible.
Family 1 — Compound angles expand $\sin$, $\cos$, $\tan$ of sums and differences of two angles.
Family 2 — Double angles are special cases with both angles equal; they give three useful forms for $\cos 2A$.
Family 3 — t-formulae (half-angle substitution $t = \tan\frac{\theta}{2}$) convert rational trigonometric expressions into algebraic ones — essential for solving equations.
Key facts
- $\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$
- $\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$
- Three forms of $\cos 2A$; the form $\sin 2A = 2\sin A \cos A$
- t-formulae: $\sin\theta = \dfrac{2t}{1+t^2}$, $\cos\theta = \dfrac{1-t^2}{1+t^2}$, $\tan\theta = \dfrac{2t}{1-t^2}$ where $t = \tan\dfrac{\theta}{2}$
Concepts
- How the double angle formulas derive from compound angle formulas with $A = B$
- Why three equivalent forms of $\cos 2A$ are each useful in different situations
- The substitution mechanism behind the t-formulae and its algebraic power
Skills
- Expand $\sin(A \pm B)$, $\cos(A \pm B)$, $\tan(A \pm B)$ from memory
- Choose the most convenient form of $\cos 2A$ for a given problem
- Apply the t-formulae substitution to simplify trigonometric expressions
The compound angle identities expand trigonometric functions of sums and differences. You must know these exactly — they are not given in the HSC reference sheet (the double angle forms are derived from them, so they flow naturally).
Memory tip for cosine: The signs are opposite — where there is a $+$ in the angle, there is a $-$ between the terms, and vice versa. "Cosine fights the sign."
(A B) = A B A B — signs match; (A B) = A B A B — signs flip ("cosine fights")
Pause — copy the compound angle formulas into your book: $\sin(A\pm B) = \sin A\cos B \pm \cos A\sin B$ (sign matches), $\cos(A\pm B) = \cos A\cos B \mp \sin A\sin B$ (sign flips).
Quick check: Which of the following correctly expands $\cos(A - B)$?
We just saw that $\sin(A+B) = \sin A\cos B + \cos A\sin B$ and $\cos(A+B) = \cos A\cos B - \sin A\sin B$. That raises a question: what if $A = B$ — can we write $\sin 2A$ and $\cos 2A$ using only a single angle? This card answers it → setting $B = A$ gives $\sin 2A = 2\sin A\cos A$ and three equivalent forms of $\cos 2A$.
Set $B = A$ in the compound angle formulas to get the double angle identities. For $\cos 2A$ there are three equivalent forms — learn all three because HSC questions often require a specific one.
When to use which form of $\cos 2A$:
- Use $\cos^2 A - \sin^2 A$ when the expression already contains both $\sin^2$ and $\cos^2$.
- Use $2\cos^2 A - 1$ (or rearranged: $\cos^2 A = \frac{1+\cos 2A}{2}$) to eliminate $\cos^2$ terms.
- Use $1 - 2\sin^2 A$ (or rearranged: $\sin^2 A = \frac{1-\cos 2A}{2}$) to eliminate $\sin^2$ terms.
2A = 2 A A — exactly one form; 2A = ^2 A - ^2 A = 2^2 A - 1 = 1 - 2^2 A — all three forms
Pause — copy $\sin 2A = 2\sin A\cos A$ and all three forms of $\cos 2A$ into your book, with a note on which form to prefer when only $\sin$ or only $\cos$ is present.
Did you get this? True or false: $\cos 2A = 1 - 2\sin^2 A$ is derived from the compound angle formula $\cos(A+B)$ by setting $B = A$ and then replacing $\cos^2 A$ with $1 - \sin^2 A$.
Worked examples · 3 in a row, reveal as you go
Find the exact value of $\sin 75°$.
If $\sin\theta = \frac{3}{5}$ and $\theta$ is acute, find $\cos 2\theta$ and $\sin 2\theta$.
Using $t = \tan\frac{\theta}{2}$, express $\dfrac{1 + \cos\theta}{1 - \cos\theta}$ purely in terms of $t$.
Fill the gap: Using $\sin\theta = \frac{4}{5}$ with $\theta$ acute, $\cos 2\theta = 1 - 2\sin^2\theta = 1 - 2 \times \frac{16}{25} = $ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: when solving a trigonometric equation using the t-substitution $t = \tan\frac{\theta}{2}$, the solution $\theta = 180°$ may be lost and must be checked separately.
Activities · practice with the ideas
Find the exact value of $\cos 15°$ using $\cos(45° - 30°)$.
If $\cos\theta = -\frac{5}{13}$ and $\theta$ is obtuse, find $\sin 2\theta$.
Verify that $\tan(A - B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B}$ by starting from $\tan(A + (-B))$ and the addition formula for tangent.
Using $t = \tan\frac{\theta}{2}$, express $\sin\theta + \cos\theta$ entirely in terms of $t$. Simplify fully.
Show that $\cos 3A = 4\cos^3 A - 3\cos A$ by writing $\cos 3A = \cos(2A + A)$ and applying the compound angle and double angle formulas.
Odd one out: Three of these are valid forms of $\cos 2A$. Which one is NOT?
Return to card 01 where you wrote your recall of compound angle identities. How close were you? The sign rule for $\cos(A \pm B)$ trips most students. The t-formulae are often the shakiest — that is normal; they are less frequently practised but crucial for Lessons 6–7.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Find the exact value of $\sin 105°$, leaving your answer in simplest surd form. (2 marks)
Q2. Given $\tan\theta = 2$ where $0 < \theta < \frac{\pi}{2}$, find the exact values of $\sin 2\theta$ and $\cos 2\theta$. (3 marks)
Q3. Using $t = \tan\frac{\theta}{2}$, show that $\dfrac{\sin\theta}{1 + \cos\theta} = t$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $\cos 15° = \cos(45°-30°) = \cos 45°\cos 30° + \sin 45°\sin 30° = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6}+\sqrt{2}}{4}$
2. $\cos\theta = -\frac{5}{13}$, obtuse $\Rightarrow \sin\theta = \frac{12}{13}$. $\sin 2\theta = 2 \cdot \frac{12}{13} \cdot (-\frac{5}{13}) = -\frac{120}{169}$
4. $\sin\theta + \cos\theta = \frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2} = \frac{1+2t-t^2}{1+t^2}$
5. $\cos(2A+A) = \cos 2A\cos A - \sin 2A\sin A = (2\cos^2A-1)\cos A - 2\sin^2A\cos A = 2\cos^3A - \cos A - 2(1-\cos^2A)\cos A = 4\cos^3A - 3\cos A$ ✓
Q1 (2 marks): $\sin 105° = \sin(60°+45°) = \sin 60°\cos 45° + \cos 60°\sin 45° = \frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2} + \frac{1}{2}\cdot\frac{\sqrt{2}}{2}$ [1] $= \frac{\sqrt{6}+\sqrt{2}}{4}$ [1].
Q2 (3 marks): $\tan\theta = 2$, acute $\Rightarrow$ hypotenuse $= \sqrt{5}$, $\sin\theta = \frac{2}{\sqrt{5}}$, $\cos\theta = \frac{1}{\sqrt{5}}$ [1]. $\sin 2\theta = 2\cdot\frac{2}{\sqrt{5}}\cdot\frac{1}{\sqrt{5}} = \frac{4}{5}$ [1]. $\cos 2\theta = \cos^2\theta - \sin^2\theta = \frac{1}{5} - \frac{4}{5} = -\frac{3}{5}$ [1].
Q3 (3 marks): $\frac{\sin\theta}{1+\cos\theta} = \frac{2t/(1+t^2)}{1+(1-t^2)/(1+t^2)}$ [1] $= \frac{2t/(1+t^2)}{(1+t^2+1-t^2)/(1+t^2)} = \frac{2t/(1+t^2)}{2/(1+t^2)}$ [1] $= \frac{2t}{2} = t$ ✓ [1].
Five timed questions on compound and double angle identities. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering trigonometric identity questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.