Introduction to Trigonometric Identities Revision Notes | GCSE AQA Higher Further Maths
The two key identities are tan θ = sin θ ÷ cos θ and sin² θ + cos² θ = 1, which comes from Pythagoras on the unit circle. Free notes and 11 questions answered.
Updated 14 Mar 2026 · Revise Right Team
Introduction to Trigonometric Identities is part of Know and use tan theta = sin theta / cos theta and sin^2 theta + cos^2 theta = 1 in GCSE AQA Higher Further Maths.
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The two key identities are tan θ = sin θ ÷ cos θ and sin² θ + cos² θ = 1, which comes from Pythagoras on the unit circle. Free notes and 11 questions answered.
State the two trigonometric identities $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and $\sin^2\theta + \cos^2\theta = 1$
Apply $\tan\theta = \frac{\sin\theta}{\cos\theta}$ to simplify expressions and solve problems
Use both identities together to prove or simplify trigonometric expressions
Key ideas to keep in view
Use this lesson to stay inside Know and use tan theta = sin theta / cos theta and sin^2 theta + cos^2 theta = 1 while connecting the detail back to the wider Geometry topic.
Geometry
Know and use tan theta = sin theta / cos theta and sin^2 theta + cos^2 theta = 1
GCSE AQA Higher Further Maths
Revision notes
Read the core explanations from Introduction to Trigonometric Identities before testing yourself from memory.
1. Why Trigonometric Identities Matter
You already know $\sin\theta$, $\cos\theta$, and $\tan\theta$ as ratios in right-angled triangles. Trigonometric identities take that further: they are equations that hold true for every angle, not just a specific one. That makes them powerful tools for simplifying expressions and solving problems without a calculator.
There are two identities you need at GCSE. The first connects $\tan\theta$ to $\sin\theta$ and $\cos\theta$. The second is a Pythagorean relationship between $\sin\theta$ and $\cos\theta$. Both appear regularly in exam questions — sometimes hidden inside a longer problem.
2. The First Identity: $\tan\theta = \frac{\sin\theta}{\cos\theta}$
Think back to a right-angled triangle with angle $\theta$. The three sides are opposite (O), adjacent (A), and hypotenuse (H). You know $\sin\theta = \frac{O}{H}$, $\cos\theta = \frac{A}{H}$, and $\tan\theta = \frac{O}{A}$.
Now watch what happens when you divide $\sin\theta$ by $\cos\theta$: $\frac{\sin\theta}{\cos\theta} = \frac{O/H}{A/H} = \frac{O}{H} \times \frac{H}{A} = \frac{O}{A} = \tan\theta$. The hypotenuse cancels, and you land exactly on $\tan\theta$. This is where the identity comes from — it is not a rule to memorise blindly, it follows directly from the triangle ratios.
The condition $\cos\theta
eq 0$ matters: if $\cos\theta = 0$ (which happens at $\theta = 90°$ and $\theta = 270°$), then $\tan\theta$ is undefined because you cannot divide by zero. Examiners occasionally test whether students recognise this restriction.
3. The Second Identity: $\sin^2\theta + \cos^2\theta = 1$
The second identity comes from Pythagoras' theorem applied to the unit circle — a circle of radius 1 centred at the origin. For any point on that circle at angle $\theta$, the coordinates are $(\cos\theta,\, \sin\theta)$. Pythagoras says $x^2 + y^2 = r^2$, so $\cos^2\theta + \sin^2\theta = 1^2 = 1$.
You do not need to derive this from the unit circle in the exam — you need to use it. The key skill is recognising when to rearrange it: $\sin^2\theta = 1 - \cos^2\theta$ and $\cos^2\theta = 1 - \sin^2\theta$ are both equally valid forms.
4. Using the Identities: Worked Examples
Knowing the identities is only half the job — the exam tests whether you can deploy them under pressure. The two most common tasks are: finding an unknown trig value given another, and simplifying or proving an expression. Here is how each one works.
Worked example: Finding $\cos\theta$ given $\sin\theta$
Given that $\sin\theta = \frac{3}{5}$ and $\theta$ is acute, find the exact value of $\cos\theta$.
$\sin^2\theta + \cos^2\theta = 1$ — Start with the identity — this is the tool for linking $\sin\theta$ and $\cos\theta$. [M1]
$\left(\frac{3}{5}\right)^2 + \cos^2\theta = 1$ — Replace $\sin\theta$ with $\frac{3}{5}$. Show the substitution explicitly for method marks.
$\frac{3}{5} \times \frac{3}{5} = \frac{9}{25}$, so $\frac{9}{25} + \cos^2\theta = 1$ — Multiply numerator by numerator and denominator by denominator. Do not skip this step on a non-calculator paper.
$\cos^2\theta = 1 - \frac{9}{25} = \frac{25}{25} - \frac{9}{25} = \frac{16}{25}$ — Subtract $\frac{9}{25}$ from both sides. Keep the fraction exact. [M1]
$\cos\theta = \sqrt{\frac{16}{25}} = \frac{4}{5}$ — Since $\theta$ is acute, $\cos\theta$ is positive. Take the positive root only. [A1]
Answer: $\cos\theta = \dfrac{4}{5}$
Worked example: Simplifying an expression using $\tan\theta = \frac{\sin\theta}{\cos\theta}$
Simplify $\frac{\sin\theta}{\tan\theta}$.
$\frac{\sin\theta}{\tan\theta} = \frac{\sin\theta}{\dfrac{\sin\theta}{\cos\theta}}$ — Substitute $\tan\theta = \frac{\sin\theta}{\cos\theta}$ into the denominator. [M1]
$= \sin\theta \times \frac{\cos\theta}{\sin\theta}$ — Dividing by $\frac{\sin\theta}{\cos\theta}$ is the same as multiplying by $\frac{\cos\theta}{\sin\theta}$.
$= \frac{\sin\theta \times \cos\theta}{\sin\theta} = \cos\theta$ — $\sin\theta$ appears in both numerator and denominator and cancels (provided $\sin\theta
eq 0$). [A1]
Answer: $\cos\theta$
5. Proving Expressions Using Both Identities
Proof questions ask you to show that one expression equals another. The approach is always the same: start with the more complicated side, substitute using the identities, and simplify until you reach the target expression. Never work on both sides simultaneously — examiners expect a one-directional argument.
Worked example: Proving a trigonometric identity
Show that $\frac{\sin^2\theta}{1 - \cos\theta} = 1 + \cos\theta$.
LHS $= \frac{\sin^2\theta}{1 - \cos\theta}$ — Always start with the more complex side. State this clearly in your working.
$= \frac{1 - \cos^2\theta}{1 - \cos\theta}$ — From $\sin^2\theta + \cos^2\theta = 1$, we get $\sin^2\theta = 1 - \cos^2\theta$. Substitute this in. [M1]
$= \frac{(1 - \cos\theta)(1 + \cos\theta)}{1 - \cos\theta}$ — $1 - \cos^2\theta$ is a difference of two squares: $(1 - \cos\theta)(1 + \cos\theta)$. [M1]
$= 1 + \cos\theta$ = RHS — $(1 - \cos\theta)$ cancels from numerator and denominator (valid provided $\cos\theta
eq 1$). This equals the right-hand side, so the identity is proved. [A1]
The worked examples above cover the main question types. Now try these problems yourself before checking the solutions — working through them independently is where the understanding sticks.
7. Exam Technique and Common Mistakes
These two identities appear in a range of question styles — from one-line simplifications worth one mark to multi-step proofs worth four or five marks. The method marks are generous, but only if your working is clear and flows logically from one line to the next.
For 'show that' and 'prove' questions, always work from one side only. Starting with both sides and meeting in the middle is not accepted as a valid proof at GCSE. Write 'LHS =' at the start of each line and end with '= RHS' to signal the proof is complete.
Both identities — $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and $\sin^2\theta + \cos^2\theta = 1$ — are the core tools for the GCSE trigonometric identities specification point. Mastering when and how to substitute them is what separates a confident answer from a blank page.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-03-14.
sin² θ + cos² θ = 1 comes from Pythagoras on the unit circle, where a point at angle θ has coordinates (cos θ, sin θ) and x² + y² = 1. It can be rearranged to sin² θ = 1 − cos² θ or cos² θ = 1 − sin² θ.
How do you prove a trigonometric identity?
Start with the more complicated side and work towards the other, never on both sides at once. For sin² θ ÷ (1 − cos θ), replace sin² θ with 1 − cos² θ, factorise as (1 − cos θ)(1 + cos θ), then cancel to get 1 + cos θ, the right-hand side.
When is tan θ undefined?
tan θ is undefined when cos θ = 0, which happens at θ = 90° and θ = 270°, because tan θ = sin θ ÷ cos θ would mean dividing by zero. The identity tan θ = sin θ ÷ cos θ only holds where cos θ ≠ 0.
Why does tan θ = sin θ ÷ cos θ?
In a right-angled triangle, sin θ = O/H and cos θ = A/H, so sin θ ÷ cos θ = (O/H) × (H/A) = O/A, which is tan θ: the hypotenuse cancels. The identity holds for any angle where cos θ is not 0; at 90° and 270°, tan θ is undefined.
How do you find cos θ given sin θ?
Substitute into sin² θ + cos² θ = 1 and solve. If sin θ = 3/5 and θ is acute, then 9/25 + cos² θ = 1, so cos² θ = 16/25 and cos θ = 4/5, taking the positive root because θ is acute.
How do you simplify sin θ ÷ tan θ?
Replace tan θ with sin θ ÷ cos θ, then flip and multiply: sin θ ÷ (sin θ ÷ cos θ) = sin θ × cos θ ÷ sin θ = cos θ, since sin θ cancels (provided it is not 0).
How do you rearrange sin²θ + cos²θ = 1?
sin²θ + cos²θ = 1 can be rearranged to sin²θ = 1 − cos²θ or cos²θ = 1 − sin²θ, and both forms are equally valid. The key skill is recognising when to rearrange it, such as finding cos θ when sin θ is known.
Where does sin²θ + cos²θ = 1 come from?
sin²θ + cos²θ = 1 comes from Pythagoras' theorem on the unit circle, a circle of radius 1 centred at the origin. A point at angle θ has coordinates (cos θ, sin θ), and x² + y² = r² gives cos²θ + sin²θ = 1² = 1.
How do you prove sin²θ ÷ (1 − cos θ) = 1 + cos θ?
Starting from the left-hand side, sin²θ is replaced by 1 − cos²θ, which factorises as a difference of two squares into (1 − cos θ)(1 + cos θ). The common factor (1 − cos θ) cancels, giving 1 + cos θ, which equals the right-hand side.
Why do you take the positive root when θ is acute?
The positive root is taken when θ is acute because sin θ and cos θ are both positive for acute angles. Given sin θ = 3/5, cos²θ = 1 − 9/25 = 16/25, so cos θ = 4/5, not −4/5.
Can you work on both sides when proving a trig identity?
A trig identity proof must work from one side only; starting with both sides and meeting in the middle is not accepted as a valid proof at GCSE. Writing 'LHS =' at the start of each line and ending with '= RHS' shows the proof is complete.