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Module 7 · L5 of 20 ~40 min ⚡ +100 XP available

Proving Trigonometric Identities

You already know the compound angle formulas. Now comes the craftwork: showing that two different-looking expressions are actually the same thing. Identity proofs are a core HSC skill — the method matters as much as the answer. In this lesson you'll build a reliable strategy and practise it on the identities that appear most often in exam questions.

Today's hook — Without expanding anything, can you tell whether $\cos^2 x - \sin^2 x$ and $\cos 2x$ are equal? Jot your gut answer. You'll confirm it after card 05.

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Recall — your gut answer first

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Without expanding, do you think $\cos^2 x - \sin^2 x$ equals $\cos 2x$? Write your reasoning before we check together.

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The golden rule of identity proofs

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An identity is an equation that is true for all values of the variable (within the domain). Proving one means showing the two sides are equivalent without assuming they are — so you never cross-multiply or move terms across the equals sign.

The four-step strategy:

Choose one side (usually the more complex one).
Apply known identities — compound angle, Pythagorean, double angle.
Simplify algebraically until you reach the other side.
Write "= RHS" (or "= LHS") and conclude.

Start LHS $\longrightarrow$ work $\longrightarrow$ = RHS

Start complex

The more complex side has more "room to move". Simplifying complexity toward simplicity is much easier than building complexity from nothing.

Convert to sin/cos

If you're stuck, express everything in terms of $\sin$ and $\cos$. Tan, sec, csc, cot can all be rewritten, and cancellation often becomes visible.

No shortcutting

Never write $\text{LHS} = \text{RHS}$ and then verify numerically. That is not a proof. The whole proof must follow a chain of valid equalities.

What you'll master

Know

Key facts

The compound angle formulas for $\sin$, $\cos$ and $\tan$
The three double-angle formulas for $\cos 2A$
The half-angle (squared) forms: $\cos^2 A = \frac{1+\cos 2A}{2}$, etc.

Understand

Concepts

Why an identity proof works on one side only
When to convert to $\sin/\cos$ versus when to factor
The role of Pythagorean identities as substitution tools

Can do

Skills

Choose the correct identity to apply at each step
Write a complete, logically valid proof
Prove identities involving compound, double and half angles

Key terms & formulas

IdentityAn equation true for all values of the variable in its domain, e.g. $\sin^2 x + \cos^2 x = 1$.

Compound angle$\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$; $\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$.

Double angle$\sin 2A = 2\sin A\cos A$; $\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$.

Half-angle squared$\cos^2 A = \dfrac{1+\cos 2A}{2}$; $\sin^2 A = \dfrac{1-\cos 2A}{2}$.

Pythagorean$\sin^2 x + \cos^2 x = 1$; $1 + \tan^2 x = \sec^2 x$; $1 + \cot^2 x = \csc^2 x$.

QED / hence shownHow you end a proof — write "= RHS" and state "as required" or use the QED symbol $\square$.

Core approach — one side at a time

core concept

Proving $\cos^2 x - \sin^2 x = \cos 2x$ illustrates the method perfectly. The right-hand side is already a named identity, so work on the left:

$$\text{LHS} = \cos^2 x - \sin^2 x$$

$$= \cos 2x \qquad [\text{double angle: } \cos 2A = \cos^2 A - \sin^2 A]$$

$$= \text{RHS} \quad \square$$

That was only one step because the double-angle formula is exactly the definition. Most proofs need several steps — but the logic is always the same: a chain of valid equalities, each justified by a known identity or algebraic rule.

What about starting from RHS? Sometimes it's easier. You can also prove LHS = RHS by showing LHS = $k$ and RHS = $k$ independently — but then you must make that clear. The default is always: pick one side, transform it, reach the other.

Exam tip. In the HSC, a proof question will say "Show that …". Every line of working earns partial credit. A clear layout — stating which identity you use at each step — is worth more marks than a fast but unexplained simplification.

Choose the more complex side. Work on only that side.; Apply compound / double / half-angle formulas, Pythagorean identities, algebraic factoring.

Pause — copy the identity-proof rules into your book: choose the more complex side; work on that side only; never move terms across the equals sign; state which identity you are using at each step.

Quick check: When proving a trigonometric identity, which of the following approaches is valid?

Using Pythagorean identities as substitution tools

core concept

We just saw the golden rules: work on the more complex side only, use trig identities and algebraic factoring, never cross the equals sign. That raises a question: when the complex side is a fraction with $1 \pm \sin x$ or $1 \pm \cos x$ in the denominator, which Pythagorean technique unlocks the simplification? This card answers it → $\cos^2 x = (1-\sin x)(1+\sin x)$ is the difference-of-squares form useful for rationalising such fractions.

The three Pythagorean identities let you swap between $\sin^2$, $\cos^2$ and $1$ at any step:

$$\sin^2 x + \cos^2 x = 1 \implies \cos^2 x = 1 - \sin^2 x \text{ or } \sin^2 x = 1 - \cos^2 x$$

Example: prove $\dfrac{\sin^2 x}{1 - \cos x} = 1 + \cos x$.

$$\text{LHS} = \frac{\sin^2 x}{1-\cos x} = \frac{1-\cos^2 x}{1-\cos x} \quad [\sin^2 x = 1-\cos^2 x]$$

$$= \frac{(1-\cos x)(1+\cos x)}{1-\cos x} \quad [\text{difference of two squares}]$$

$$= 1 + \cos x = \text{RHS} \quad \square$$

The key insight here is recognising $1 - \cos^2 x$ as a difference of squares after applying the Pythagorean swap.

Pattern to remember. Whenever you see a denominator of the form $1 \pm \cos x$ or $1 \pm \sin x$, the Pythagorean identity (possibly combined with a difference-of-squares factorisation) is almost certainly the tool to reach for.

^2 x = 1 - ^2 x = (1- x)(1+ x) — the difference-of-squares factor is very useful in fractions; Spot 1 x or 1 x in a denominator — think Pythagorean + difference of squares

Pause — copy the Pythagorean substitution patterns into your book: $\cos^2 x = (1-\sin x)(1+\sin x)$; spot $1\pm\sin x$ in a denominator and think difference-of-squares to rationalise.

Did you get this? True or false: $\dfrac{1 - \cos^2 x}{1 - \cos x} = 1 + \cos x$ (for $\cos x \neq 1$).

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · DOUBLE ANGLE

Prove that $\sin 2x = \dfrac{2\tan x}{1+\tan^2 x}$.

Start with RHS: $\dfrac{2\tan x}{1+\tan^2 x}$

The RHS is more complex (fraction with $\tan$). Rewrite $\tan x = \dfrac{\sin x}{\cos x}$ and $1+\tan^2 x = \sec^2 x = \dfrac{1}{\cos^2 x}$.

PROBLEM 2 · COMPOUND ANGLE

Prove that $\cos(A+B)\cos(A-B) = \cos^2 A - \sin^2 B$.

$\text{LHS} = \cos(A+B)\cos(A-B)$

Expand each factor using the compound angle formula: $\cos(A+B)=\cos A\cos B - \sin A\sin B$ and $\cos(A-B)=\cos A\cos B + \sin A\sin B$.

PROBLEM 3 · HALF ANGLE

Prove that $\dfrac{1 - \cos 2x}{\sin 2x} = \tan x$.

$\text{LHS} = \dfrac{1-\cos 2x}{\sin 2x}$

The LHS has double-angle expressions. Substitute: $1-\cos 2x = 1-(1-2\sin^2 x) = 2\sin^2 x$ and $\sin 2x = 2\sin x\cos x$.

Fill the gap: Using the double-angle formula, $1 - \cos 2x =$ . (Write your answer as 2sin²x.)

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Working on both sides at once

A common mistake is to write manipulations on both sides simultaneously (e.g. "LHS = … = RHS = …"). This is circular reasoning. Pick one side and transform it until you arrive at the other — only then write "= RHS".

Trap 02

Using the wrong double-angle form

There are three forms of $\cos 2A$. Choose the one that matches your current expression: use $\cos^2 A - \sin^2 A$ when both $\sin^2$ and $\cos^2$ are present; use $1-2\sin^2 A$ when only $\sin$ appears; use $2\cos^2 A - 1$ when only $\cos$ appears.

Trap 03

Forgetting domain restrictions

When you cancel a factor (e.g. $\sin x$) you must note the implied restriction ($\sin x \neq 0$). Not doing so loses a mark in formal proofs and is technically incorrect.

Did you get this? True or false: it is valid to begin an identity proof by writing "Assume LHS = RHS" and then deriving a true statement.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Prove that $\cos 2x = 1 - 2\sin^2 x$ by starting from $\cos 2x = \cos^2 x - \sin^2 x$ and applying $\cos^2 x = 1 - \sin^2 x$.

Prove that $\dfrac{\cos 2x}{\cos x - \sin x} = \cos x + \sin x$. (Hint: factorise the numerator as a difference of squares.)

Prove that $\sin^4 x - \cos^4 x = -\cos 2x$. (Hint: factor as difference of squares, then use a Pythagorean identity.)

Prove that $\dfrac{\sin 3x + \sin x}{2\sin 2x} = \cos x$. (Hint: use $\sin 3x = \sin(2x+x)$ and expand.)

Prove that $\tan 2x = \dfrac{2\tan x}{1-\tan^2 x}$ directly from the compound angle formula for $\tan(A+B)$ with $A = B = x$.

Odd one out: Three of these are valid forms of $\cos 2x$. Which one is NOT?

Revisit your thinking

Earlier you predicted whether $\cos^2 x - \sin^2 x = \cos 2x$.

They are equal — it is one of the three forms of the double-angle formula for cosine. The key insight: the double-angle formula is that equation. Recognising this stops you from spending time on an unnecessary proof. Did your intuition match the result?

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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.

Short answer

ApplyBand 32 marks

Q1. Prove that $\dfrac{\cos 2x}{\cos x - \sin x} = \cos x + \sin x$. (2 marks)

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ApplyBand 42 marks

Q2. Prove that $\dfrac{1-\cos 2x}{\sin 2x} = \tan x$. (2 marks)

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AnalyseBand 53 marks

Q3. Prove that $\sin^4 x - \cos^4 x = -\cos 2x$. (3 marks)

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Comprehensive answers (click to reveal)

Q1 (2 marks): $\text{LHS} = \dfrac{\cos^2 x - \sin^2 x}{\cos x - \sin x} = \dfrac{(\cos x-\sin x)(\cos x+\sin x)}{\cos x - \sin x} = \cos x + \sin x = \text{RHS}$ [2]. (Award 1 for factoring numerator, 1 for cancellation and conclusion.)

Q2 (2 marks): $\text{LHS} = \dfrac{1-(1-2\sin^2 x)}{2\sin x\cos x} = \dfrac{2\sin^2 x}{2\sin x\cos x} = \dfrac{\sin x}{\cos x} = \tan x = \text{RHS}$ [2].

Q3 (3 marks): $\text{LHS} = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x)$ [1] $= (\sin^2 x - \cos^2 x)(1)$ [1] $= -(\cos^2 x - \sin^2 x) = -\cos 2x = \text{RHS}$ [1].

Boss battle · The Identity Architect

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

Science Jump · platform challenge

Climb platforms by answering identity questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 4 · Solving a cos x + b sin x = c Lesson 6 · The t-Formulae Revisited →

Module overview · Extension 1 · Checkpoint 1