Harder Identity Proofs
Proving a trigonometric identity isn't about guessing — it's about choosing the right opening move and following the algebra until both sides agree. In this lesson you'll tackle three-star proofs that combine double angle, compound angle, t-formula, and Pythagorean identities. Each example reveals a new strategic toolkit so you can crack any identity the HSC throws at you.
Try to prove that $\dfrac{1 - \cos 2x}{\sin 2x} = \tan x$. Write your attempt — even a single first step is valuable. What identity would you use to rewrite $\cos 2x$ or $\sin 2x$?
Harder proofs rarely yield to a single identity. You need a strategy. The four most powerful moves are:
- Work from the more complex side. The side with fractions, products, or double angles usually has more to simplify.
- Express everything in $\sin$ and $\cos$. Convert $\tan$, $\cot$, $\sec$, $\csc$ to reveal hidden cancellations.
- Use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$ (and its variants) to trade between $\sin^2$ and $\cos^2$.
- Use double angle / compound angle / t-formula to change the "angle type" when it's blocking progress.
Key facts
- The double angle identities: $\sin 2x = 2\sin x\cos x$ and $\cos 2x = \cos^2 x - \sin^2 x = 1-2\sin^2 x = 2\cos^2 x-1$
- The t-substitution: $t = \tan\tfrac{x}{2}$ gives $\sin x = \tfrac{2t}{1+t^2}$, $\cos x = \tfrac{1-t^2}{1+t^2}$
- Factor forms of $\sin^2 x + \cos^2 x = 1$
Concepts
- Why working from one side only preserves logical validity
- How to spot which identity unlocks a stuck proof
- When the t-formula is the most efficient tool
Skills
- Complete three-star identity proofs in exam conditions
- Choose and apply the correct strategy without prompting
- Present proofs with clear, justified steps
Many harder proofs are unlocked by substituting $1-\cos 2x = 2\sin^2 x$ and $\sin 2x = 2\sin x\cos x$.
Proof that $\dfrac{1-\cos 2x}{\sin 2x} = \tan x$:
The key move was replacing $1-\cos 2x$ with $2\sin^2 x$ and $\sin 2x$ with $2\sin x\cos x$, then cancelling common factors.
1 - 2x = 2^2 x (use when 1- 2x appears in a fraction numerator); 1 + 2x = 2^2 x (use when 1+ 2x appears)
Pause — copy the two key double-angle substitutions into your book: $1 - \cos 2x = 2\sin^2 x$ and $1 + \cos 2x = 2\cos^2 x$ — the first is useful when $1-\cos 2x$ appears in a numerator.
Quick check: In the proof of $\dfrac{1-\cos 2x}{\sin 2x}=\tan x$, the substitution $1-\cos 2x = 2\sin^2 x$ uses which identity?
We just saw that $1 \pm \cos 2x$ can be replaced by $2\cos^2 x$ or $2\sin^2 x$ to clear complex numerators. That raises a question: for proofs involving $\sin x$ and $\cos x$ simultaneously in fractions, the $t = \tan\frac{x}{2}$ substitution unifies everything — but does using $t$ always work, and when does it fail? This card answers it → the $t$-substitution works unless $x = \pi$ is in the domain, where $\tan\frac{\pi}{2}$ is undefined.
The t-substitution $t = \tan\tfrac{x}{2}$ converts a trigonometric expression entirely into a rational function of $t$. It is especially powerful when $\sin x$ and $\cos x$ appear together in a fraction.
Identities to memorise:
Example: Prove that $\dfrac{2\sin x}{1 + \cos x} = 2\tan\dfrac{x}{2}$.
$\text{LHS} = \dfrac{2 \cdot \tfrac{2t}{1+t^2}}{1 + \tfrac{1-t^2}{1+t^2}} = \dfrac{\tfrac{4t}{1+t^2}}{\tfrac{(1+t^2)+(1-t^2)}{1+t^2}} = \dfrac{\tfrac{4t}{1+t^2}}{\tfrac{2}{1+t^2}} = \dfrac{4t}{2} = 2t = 2\tan\dfrac{x}{2} = \text{RHS}$ ✓
t = (x/2): x = 2t/(1+t^2), x = (1-t^2)/(1+t^2), x = 2t/(1-t^2); 1 + x = 2{1+t^2}, 1 - x = 2t^2{1+t^2} (useful derived forms)
Pause — copy the derived $t$-forms used in identity proofs: $1 + \cos x = \frac{2}{1+t^2}$, $1 - \cos x = \frac{2t^2}{1+t^2}$, where $t = \tan\frac{x}{2}$ (undefined at $x = \pi$).
Did you get this? True or false: when using the t-substitution $t = \tan(x/2)$, we have $1 + \cos x = \dfrac{2}{1+t^2}$.
Worked examples · 3 in a row, reveal as you go
Prove that $\dfrac{\sin 2x}{1 + \cos 2x} = \tan x$.
Prove that $\cos^4 x - \sin^4 x = \cos 2x$.
Using the t-substitution, prove that $\dfrac{1 - \cos x}{\sin x} = \tan\dfrac{x}{2}$.
Fill the gap: In the proof of $\cos^4 x - \sin^4 x = \cos 2x$, we first factorise as $(\cos^2 x + \sin^2 x)(\cos^2 x - \sin^2 x)$ and then use $\cos^2 x + \sin^2 x = $ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: in a trigonometric identity proof, it is valid to add the same expression to both sides if it simplifies the algebra.
Activities · practice with the ideas
Prove that $\dfrac{1-\cos 2\theta}{\sin 2\theta} = \tan\theta$.
Prove that $\sin^4 x - \cos^4 x = \sin^2 x - \cos^2 x$.
Using t-formula, prove that $\dfrac{2\tan(x/2)}{1+\tan^2(x/2)} = \sin x$.
Prove that $\cos^2\!\left(\dfrac{x}{2}\right) - \sin^2\!\left(\dfrac{x}{2}\right) = \cos x$.
Identify the first move you would make to prove $\dfrac{\tan x - \sin x}{\tan x + \sin x} = \dfrac{\tan x - \sin x}{\tan x + \sin x}$. No — instead prove: $\dfrac{\cos^2 x - 1}{\sin x} = -\sin x$.
Odd one out: Three of these are valid proof moves. Which one is NOT a valid proof technique?
At the start you attempted to prove $\dfrac{1-\cos 2x}{\sin 2x} = \tan x$. The key was replacing $1 - \cos 2x$ with $2\sin^2 x$ and $\sin 2x$ with $2\sin x\cos x$. Did you choose that substitution? If you tried a different path, what happened?
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. Prove that $\cos^4 x - \sin^4 x = \cos 2x$. (2 marks)
Q2. Prove that $\dfrac{\sin 2x}{1 - \cos 2x} = \cot x$. (3 marks)
Q3. Using the substitution $t = \tan\tfrac{x}{2}$, prove that $\dfrac{1-\cos x}{1+\cos x} = \tan^2\!\dfrac{x}{2}$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers: 1. LHS $= 2\sin^2\theta/(2\sin\theta\cos\theta)=\tan\theta$ ✓ · 2. $(\sin^2x-\cos^2x)(\sin^2x+\cos^2x)=\sin^2x-\cos^2x$ ✓ · 3. $2t/(1+t^2) = \sin x$ ✓ · 4. LHS $=\cos(2\cdot x/2)=\cos x$ ✓ · 5. $(\cos^2x-1)/\sin x = -\sin^2x/\sin x=-\sin x$ ✓.
Q1 (2 marks): LHS $=(\cos^2x+\sin^2x)(\cos^2x-\sin^2x)$ [1 — difference of squares]. $=1\cdot(\cos^2x-\sin^2x)=\cos 2x=$ RHS [1 — Pythagorean then double angle].
Q2 (3 marks): LHS $=\dfrac{2\sin x\cos x}{2\sin^2 x}$ [1 — double angle sub]. $=\dfrac{\cos x}{\sin x}$ [1 — cancel $2\sin x$]. $=\cot x=$ RHS [1 — definition of cot].
Q3 (3 marks): Let $t=\tan(x/2)$; $1-\cos x = \tfrac{2t^2}{1+t^2}$ and $1+\cos x = \tfrac{2}{1+t^2}$ [1]. LHS $=\dfrac{2t^2/(1+t^2)}{2/(1+t^2)}=t^2$ [1]. $=\tan^2(x/2)=$ RHS [1].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering identity proof questions. A lighter alternative to the boss battle.
Mark lesson as complete
Tick when you've finished the practice and review.