Proofs, Simplification, and Equations with Trigonometric Identities Revision Notes | GCSE AQA Higher Further Maths
To prove an identity, start with the more complicated side and work towards the other, using tan θ ≡ sin θ ÷ cos θ. Free notes and 11 questions answered.
Updated 14 Mar 2026 · Revise Right Team
Proofs, Simplification, and Equations with 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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What this lesson covers
To prove an identity, start with the more complicated side and work towards the other, using tan θ ≡ sin θ ÷ cos θ. Free notes and 11 questions answered.
Apply the identities $\tan\theta \equiv \frac{\sin\theta}{\cos\theta}$ and $\sin^2\theta + \cos^2\theta \equiv 1$ to simplify trigonometric expressions
Prove trigonometric identities by transforming one side into the other using known identities
Solve trigonometric equations by substituting identities to reduce them to a single trigonometric function
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 Proofs, Simplification, and Equations with Trigonometric Identities before testing yourself from memory.
1. Why These Two Identities Change Everything
Imagine an exam question hands you an expression tangled up with $\sin$, $\cos$, and $\tan$ all at once. Without a strategy, it looks unsolvable. With two identities, it collapses into something manageable.
Those two identities are $\tan\theta \equiv \frac{\sin\theta}{\cos\theta}$ and $\sin^2\theta + \cos^2\theta \equiv 1$. Every simplification, every proof, and every equation in this topic traces back to one or both of them.
Notice the Pythagorean identity has three useful forms. You pick the rearrangement that removes the function you don't want. That decision — which form to use — is the core skill this lesson builds.
2. Simplifying Expressions
Simplification means rewriting an expression in its most compact form. The target is usually a single trig function, a constant, or a neat fraction. The route is almost always: spot which identity fits, substitute, then cancel or combine.
A reliable starting move: if you see $\tan\theta$ alongside $\sin\theta$ or $\cos\theta$, replace $\tan\theta$ with $\frac{\sin\theta}{\cos\theta}$ immediately. If you see $\sin^2\theta$ and $\cos^2\theta$ together, look for the Pythagorean identity.
Worked example: Simplifying an Expression Using the Quotient Identity
Simplify $\dfrac{\sin\theta}{\tan\theta}$.
$\dfrac{\sin\theta}{\tan\theta} = \dfrac{\sin\theta}{\dfrac{\sin\theta}{\cos\theta}}$ — Substitute $\tan\theta \equiv \frac{\sin\theta}{\cos\theta}$. This converts the expression to a fraction divided by a fraction. [M1]
$= \sin\theta \times \dfrac{\cos\theta}{\sin\theta}$ — Dividing by $\frac{\sin\theta}{\cos\theta}$ is the same as multiplying by $\frac{\cos\theta}{\sin\theta}$.
$= \dfrac{\sin\theta \cdot \cos\theta}{\sin\theta} = \cos\theta$ — $\sin\theta$ appears in both numerator and denominator, so it cancels. [A1]
Answer: $\cos\theta$
Worked example: Simplifying Using the Pythagorean Identity
$\dfrac{1 - \cos^2\theta}{\sin\theta} = \dfrac{\sin^2\theta}{\sin\theta}$ — The numerator is now $\sin^2\theta$.
$= \dfrac{\sin\theta \cdot \sin\theta}{\sin\theta} = \sin\theta$ — One $\sin\theta$ cancels from numerator and denominator. [A1]
Answer: $\sin\theta$
3. Proving Trigonometric Identities
A proof question gives you two expressions and asks you to show they are identical. The golden rule: work on one side only. Pick the more complicated side, apply identities step by step, and arrive at the other side.
Never move terms across the equals sign during a proof — that assumes what you are trying to show. Each line must follow logically from the one before it, using only the two identities and standard algebra.
Worked example: Proving a Trigonometric Identity
Prove that $\tan\theta \cdot \cos\theta \equiv \sin\theta$.
Start with the left-hand side (LHS): $\tan\theta \cdot \cos\theta$ — The LHS contains $\tan\theta$, which can be expanded. The RHS is already a single function, so we work on the LHS.
Prove that $\dfrac{\sin^2\theta + \cos^2\theta}{\cos\theta} \equiv \dfrac{1}{\cos\theta}$.
$\text{LHS} = \dfrac{\sin^2\theta + \cos^2\theta}{\cos\theta}$ — The numerator contains $\sin^2\theta + \cos^2\theta$, which is the Pythagorean identity.
$= \dfrac{1}{\cos\theta}$ — Since $\sin^2\theta + \cos^2\theta \equiv 1$, the numerator becomes 1. [M1]
$\text{LHS} = \dfrac{1}{\cos\theta} = \text{RHS}$ $\therefore$ proved. — The expression matches the RHS exactly. [A1]
Equations become solvable when they contain only one trig function. If an equation mixes $\sin\theta$ and $\cos\theta$, or includes $\tan\theta$, the identities let you reduce everything to a single function — then solve as normal.
The most common scenario: an equation has both $\sin^2\theta$ and $\cos^2\theta$. Replace one using $\sin^2\theta \equiv 1 - \cos^2\theta$ (or vice versa), then factorise or use the quadratic formula on what remains.
Worked example: Solving an Equation Using the Pythagorean Identity
$\cos^2\theta \equiv 1 - \sin^2\theta$
$2\sin^2\theta - (1 - \sin^2\theta) = 1$ — Substitute to get an equation in $\sin\theta$ only. [M1]
$2\sin^2\theta - 1 + \sin^2\theta = 1$
$3\sin^2\theta - 1 = 1$
$3\sin^2\theta = 2$ — Expand the bracket, then gather all $\sin^2\theta$ terms on one side.
$\sin^2\theta = \dfrac{2}{3}$ — Divide both sides by 3. [M1]
$\sin\theta = \pm\sqrt{\dfrac{2}{3}} = \pm\dfrac{\sqrt{2}}{\sqrt{3}} = \pm\dfrac{\sqrt{6}}{3}$ — Both positive and negative roots are valid — do not discard the negative root. $\sqrt{\frac{2}{3}} = \frac{\sqrt{6}}{3} \approx 0.8165$.
$\sin\theta = +\dfrac{\sqrt{6}}{3}$: $\theta \approx 54.7^\circ$ (Q1) and $\theta \approx 125.3^\circ$ (Q2)
$\sin\theta = -\dfrac{\sqrt{6}}{3}$: $\theta \approx 234.7^\circ$ (Q3) and $\theta \approx 305.3^\circ$ (Q4) — Sine is positive in Q1 and Q2; negative in Q3 and Q4. Use the reference angle $\approx 54.7^\circ$ in each quadrant. [A1 for each correct pair]
Worked example: Solving an Equation Using the Quotient Identity
Solve $\sin\theta = 3\cos\theta$ for $0^\circ \leq \theta \leq 360^\circ$.
$\dfrac{\sin\theta}{\cos\theta} = 3$ — Dividing both sides by $\cos\theta$ (valid where $\cos\theta
eq 0$) gives a ratio that equals $\tan\theta$. [M1]
$\tan\theta = 3$ — Since $\tan\theta \equiv \frac{\sin\theta}{\cos\theta}$, the left side is $\tan\theta$.
$\theta = 71.6^\circ + 180^\circ = 251.6^\circ$ — Tangent is positive in Q1 and Q3. Add $180^\circ$ to find the Q3 solution. [A1]
Answer: $\theta \approx 71.6^\circ$ and $\theta \approx 251.6^\circ$
5. Exam Technique and Common Mistakes
Three question types appear in exams: simplify, prove, and solve. Each has a distinct method, and mixing them up is the most common source of lost marks. Simplification ends when you reach the simplest form. A proof ends when you write LHS = RHS. An equation ends when you list all solutions in the given interval.
For equations, always check whether the question asks for solutions in degrees or radians, and always find all solutions in the given interval — not just the principal value from your calculator.
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.
The symbol ≡ is read as 'is identically equal to'. Unlike =, which holds for specific values, ≡ means the relationship is true for every valid angle, so identities such as tan θ ≡ sin θ ÷ cos θ and sin² θ + cos² θ ≡ 1 should use ≡.
How do you simplify (1 − cos² θ) ÷ sin θ?
Rearrange the Pythagorean identity to 1 − cos² θ ≡ sin² θ, so the expression becomes sin² θ ÷ sin θ. One factor of sin θ cancels, so the answer is sin θ.
How do you prove tan θ cos θ ≡ sin θ?
Start with the left-hand side and replace tan θ with sin θ ÷ cos θ: LHS = (sin θ ÷ cos θ) × cos θ. The cos θ cancels, giving sin θ, so LHS = RHS. Always work on one side only and finish by stating LHS = RHS.
How do you solve 2 sin² θ − cos² θ = 1?
Replace cos² θ with 1 − sin² θ to get 3 sin² θ − 1 = 1, so sin² θ = 2/3 and sin θ = ±√6/3. Using the reference angle of about 54.7° in all four quadrants, θ ≈ 54.7°, 125.3°, 234.7° and 305.3° for 0° to 360°.
How do you solve sin θ = 3 cos θ?
Divide both sides by cos θ to get tan θ = 3, using tan θ ≡ sin θ ÷ cos θ. The principal value is θ ≈ 71.6°, and since tangent is also positive in Q3, the second solution is 71.6° + 180° = 251.6°.
What are the three forms of the Pythagorean trig identity?
The three forms are sin²θ + cos²θ ≡ 1, sin²θ ≡ 1 − cos²θ and cos²θ ≡ 1 − sin²θ. The form chosen is the one that removes the function that is not wanted.
What is the difference between simplify, prove and solve in trigonometry?
Simplify ends when the expression reaches its simplest form, prove ends with LHS = RHS, and solve ends when all solutions in the given interval are listed. Mixing up these three question types is the most common source of lost marks.
How do you prove (sin²θ + cos²θ) ÷ cos θ ≡ 1 ÷ cos θ?
Starting from the left-hand side, the numerator sin²θ + cos²θ is replaced by 1 using the Pythagorean identity, so the expression becomes 1 ÷ cos θ, which is the right-hand side. The proof ends by stating LHS = RHS.
Why can't you move terms across the equals sign in a trig proof?
Terms cannot be moved across the equals sign during a proof because that assumes what is being shown. Each line must follow logically from the one before, using only the two identities and standard algebra.
How do identities help solve trig equations?
Identities help solve trig equations by reducing them to a single trig function. Replacing cos²θ with 1 − sin²θ turns 2 sin²θ − cos²θ = 1 into 3 sin²θ = 2, and dividing sin θ = 3 cos θ by cos θ gives tan θ = 3.
Why must you keep the negative root in a trig equation?
The negative root must be kept because it gives more valid solutions. For sin²θ = 2/3, sin θ = ±√6/3, so the solutions in 0° to 360° are about 54.7° and 125.3° from the positive root and 234.7° and 305.3° from the negative root.