Trigonometry (Pure 3) — A-Level Mathematics Pure 3 Study Guide

For: A-Level Mathematics candidates sitting Paper 3 (Pure Mathematics 3).

Covers: reciprocal trigonometric functions and their identities, compound and double angle formulas, the $R\sin (\theta +\alpha )$ form for linear combinations of sine and cosine, solving trigonometric equations using identities, and proving trigonometric identities.

You should already know: A-Level Mathematics Pure 1 (functions, calculus, trig).

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Mathematics style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Trigonometry (Pure 3)?

Pure 3 trigonometry extends foundational trigonometric concepts from Pure 1 to complex identity manipulation, function transformation, and multi-step problem-solving that underpins calculus, vector, and differential equation topics later in the syllabus. Unlike Pure 1 trigonometry, which focuses on basic $\sin ,\cos ,\tan$ and simple linear equations, Pure 3 trigonometry introduces reciprocal functions, combined angle transformations, and structured identity proof frameworks that appear in 3–6 mark questions across almost every Paper 3 exam session.

2. Reciprocal trig: $\sec ,\csc ,\cot$ and their identities

The three reciprocal trigonometric functions are defined as follows, with restricted domains to avoid division by zero:

• $\sec \theta =\frac{1}{\cos \theta }$: Domain $\theta \ne \frac{\pi }{2}+n\pi$ (radians) or $90^{\circ }+180^{\circ }n$ (degrees), range $|\sec \theta |\ge 1$
• $\csc \theta =\frac{1}{\sin \theta }$: Domain $\theta \ne n\pi$ (radians) or $180^{\circ }n$ (degrees), range $|\csc \theta |\ge 1$
• $\cot \theta =\frac{\cos \theta }{\sin \theta }=\frac{1}{\tan \theta }$: Domain $\theta \ne n\pi$ (radians) or $180^{\circ }n$ (degrees), range all real numbers

Two core Pythagorean identities are derived directly from ${\sin }^2\theta +{\cos }^2\theta =1$:

1. Divide through by ${\cos }^2\theta$: ${\tan }^2\theta +1={\sec }^2\theta$
2. Divide through by ${\sin }^2\theta$: $1+{\cot }^2\theta ={\csc }^2\theta$

Worked Example

Find the exact value of $\cot \theta$ if $\sec \theta =\frac{5}{3}$ and $\theta$ lies in the fourth quadrant.

1. Substitute into ${\sec }^2\theta =1+{\tan }^2\theta$: ${\left(\frac{5}{3}\right)}^2=1+{\tan }^2\theta ⟹{\tan }^2\theta =\frac{25}{9}-1=\frac{16}{9}$
2. In the fourth quadrant, $\tan \theta$ is negative, so $\tan \theta =-\frac{4}{3}$
3. $\cot \theta =\frac{1}{\tan \theta }=-\frac{3}{4}$

Exam tip: Examiners regularly test these reciprocal Pythagorean identities in integration and differentiation questions, so memorize the link between $\tan$ and $\sec$, and $\cot$ and $\csc$ separately to avoid confusion.

3. Compound and double angle formulas

Compound angle formulas express trigonometric functions of sums or differences of two angles as combinations of functions of the individual angles. Critically, trigonometric functions are not linear, so $\sin (A+B)\ne \sin A+\sin B$, a common mistake for students rushing through problems.

The core compound angle formulas are: $$ \begin{align*} \sin(A \pm B) &= \sin A \cos B \pm \cos A \sin B \ \cos(A \pm B) &= \cos A \cos B \mp \sin A \sin B \ \tan(A \pm B) &= \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B} \end{align*} $$

Double angle formulas are derived by setting $A=B$ in the compound angle formulas: $$ \begin{align*} \sin 2A &= 2 \sin A \cos A \ \cos 2A &= \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A \ \tan 2A &= \frac{2 \tan A}{1 - \tan^2 A} \end{align*} $$

Worked Example

Find the exact value of $\cos 75^{\circ }$ using compound angle formulas.

1. Rewrite $75^{\circ }=45^{\circ }+30^{\circ }$
2. Substitute into the compound cosine formula: $$\cos (45^{\circ }+30^{\circ })=\cos 45^{\circ }\cos 30^{\circ }-\sin 45^{\circ }\sin 30^{\circ }$$
3. Substitute exact values: $$=\frac{\sqrt{2}}{2}\times \frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times \frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}$$

Exam tip: The rearranged double angle formulas ${\cos }^{2}A=\frac{1+\cos 2A}{2}$ and ${\sin }^{2}A=\frac{1-\cos 2A}{2}$ are the most frequently tested of all trig identities, as they are required to integrate squared trigonometric functions.

4. The $R\sin (\theta +\alpha )$ form for $a\sin \theta +b\cos \theta$

Any linear combination of $\sin \theta$ and $\cos \theta$ can be rewritten as a single trigonometric function with a phase shift, making it far easier to solve equations, find maximum/minimum values, and integrate these expressions.

The form is derived by equating coefficients: $$a\sin \theta +b\cos \theta =R\sin (\theta +\alpha )=R\sin \theta \cos \alpha +R\cos \theta \sin \alpha$$

1. Equate coefficients: $R\cos \alpha =a$, $R\sin \alpha =b$
2. Square and add to eliminate $\alpha$: $R^2({\cos }^2\alpha +{\sin }^2\alpha )=a^2+b^2⟹R=\sqrt{a^2+b^2}$ (always take $R>0$)
3. Calculate $\alpha$ using $\tan \alpha =\frac{b}{a}$, checking the signs of $\sin \alpha$ and $\cos \alpha$ to find the correct quadrant for $\alpha$.

You can also rewrite the expression as $R\cos (\theta -\alpha )$ or other forms using the same process, matching the expansion to the required form given in the question.

Worked Example

Express $3\sin \theta +4\cos \theta$ in the form $R\sin (\theta +\alpha )$, where $0<\alpha <90^{\circ }$, giving $\alpha$ to 1 decimal place.

1. Calculate $R=\sqrt{3^2+4^2}=5$
2. Equate coefficients: $5\cos \alpha =3⟹\cos \alpha =0.6$, $5\sin \alpha =4⟹\sin \alpha =0.8$
3. Both $\sin \alpha$ and $\cos \alpha$ are positive, so $\alpha$ is in the first quadrant: $\alpha =\arctan \left(\frac{4}{3}\right)=53.1^{\circ }$
4. Final form: $5\sin (\theta +53.1^{\circ })$

Exam tip: When asked for maximum/minimum values of these expressions, remember the maximum value of $R\sin (\theta +\alpha )$ is $R$, the minimum is $-R$, and the angle where this occurs is when the trigonometric term equals 1 or -1 respectively.

5. Solving trig equations using identities

Nearly all Paper 3 trigonometric equations require you to use identities to simplify the equation to a single trigonometric function of a single angle, before solving for the given interval. Follow these standard steps:

1. Identify which identity to use to eliminate compound angles, reciprocal functions, or squared terms.
2. Rearrange into a linear or quadratic form in one trigonometric function.
3. Solve for the angle, finding all solutions in the given interval, and check for extraneous solutions where the original function is undefined (e.g., $\sec \theta$ where $\cos \theta =0$).

Worked Example

Solve ${\sec }^2\theta -3\tan \theta =-1$ for $0^{\circ }\le \theta \le 360^{\circ }$.

1. Substitute ${\sec }^2\theta =1+{\tan }^2\theta$ into the equation: $$1+{\tan }^2\theta -3\tan \theta =-1⟹{\tan }^2\theta -3\tan \theta +2=0$$
2. Factor the quadratic: $(\tan \theta -1)(\tan \theta -2)=0⟹\tan \theta =1$ or $\tan \theta =2$
3. Find all solutions:

• For $\tan \theta =1$: solutions are $45^{\circ },225^{\circ }$
• For $\tan \theta =2$: principal value is $63.4^{\circ }$, so solutions are $63.4^{\circ },243.4^{\circ }$

4. Check no solutions have $\cos \theta =0$ (all are valid), so full solution set: $45^{\circ },63.4^{\circ },225^{\circ },243.4^{\circ }$

Exam tip: Always note if the question specifies radians or degrees, and use the CAST diagram or a quick sketch of the function to ensure you find all solutions in the given interval, not just the principal value from your calculator.

6. Proving trig identities

Proving identities requires you to manipulate one side of the identity (usually the more complex side) step-by-step until it equals the other side. You should never move terms across the equals sign as you would when solving an equation, as this assumes the identity is true before you prove it.

Common strategies for proofs:

1. Rewrite all reciprocal functions in terms of $\sin ,\cos ,\tan$ to simplify.
2. Use Pythagorean, compound, or double angle identities to replace complex terms.
3. Combine fractions using common denominators.
4. For fractions with trigonometric terms in the denominator, multiply numerator and denominator by the conjugate to use the difference of squares.

Worked Example

Prove that $\frac{1}{\sec \theta -\tan \theta }\equiv \sec \theta +\tan \theta$.

1. Start with the left-hand side (LHS), multiply numerator and denominator by the conjugate of the denominator $(\sec \theta +\tan \theta )$: $$\text{LHS}=\frac{1\times (\sec \theta +\tan \theta )}{(\sec \theta -\tan \theta )(\sec \theta +\tan \theta )}$$
2. Simplify the denominator using difference of squares and the Pythagorean identity ${\sec }^2\theta -{\tan }^2\theta =1$: $$\text{LHS}=\frac{\sec \theta +\tan \theta }{1}=\sec \theta +\tan \theta =\text{Right-hand side (RHS)}$$
3. The identity is proven.

Exam tip: Always show every step of your working for identity proofs, as examiners award marks for each correct manipulation even if you do not reach the final result.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Expanding $\sin (A+B)=\sin A+\sin B$ or assuming $\tan 2\theta =2\tan \theta$. Why you might do it: Confusing algebraic expansion with non-linear trigonometric functions. Correct move: Test with a sample value (e.g., $\sin (30^{\circ }+30^{\circ })=\sin 60^{\circ }=\frac{\sqrt{3}}{2}\ne 1=\sin 30^{\circ }+\sin 30^{\circ }$) if you are unsure, and memorize compound/double angle formulas explicitly.
• Wrong move: Using the wrong quadrant for $\alpha$ in the $R\sin (\theta +\alpha )$ form, only taking the principal value of $\arctan \left(\frac{b}{a}\right)$ without checking signs. Why you might do it: Rushing through the calculation to save time. Correct move: Cross-check the signs of $R\cos \alpha =a$ and $R\sin \alpha =b$ to confirm the quadrant of $\alpha$, adjusting your angle if needed.
• Wrong move: Missing solutions for trigonometric equations, only submitting the principal value from your calculator. Why you might do it: Forgetting the periodicity of trigonometric functions. Correct move: Count how many solutions you expect before solving, and use the CAST diagram or a quick function sketch to find all values in the given interval.
• Wrong move: Proving identities by moving terms across the equals sign. Why you might do it: Confusing identity proof with equation solving. Correct move: Manipulate only one side at a time, working towards the other side, or manipulate both sides to meet in the middle if you get stuck.
• Wrong move: Forgetting to check for extraneous solutions where the original function is undefined. Why you might do it: Focusing only on the simplified equation. Correct move: Cross-check all solutions against the original equation to ensure no denominators are zero or reciprocal functions are undefined.

8. Practice Questions (Paper 3 Style)

Question 1

(a) Express $2\sin x+5\cos x$ in the form $R\sin (x+\alpha )$, where $R>0$ and $0<\alpha <\pi$ radians, giving $\alpha$ to 3 decimal places. (3 marks) (b) Hence find the maximum value of $2\sin x+5\cos x$ and the smallest positive value of $x$ where this maximum occurs. (2 marks)

Solution 1

(a) Calculate $R=\sqrt{2^2+5^2}=\sqrt{29}\approx 5.385$. Equate coefficients: $R\cos \alpha =2⟹\cos \alpha =\frac{2}{\sqrt{29}}\approx 0.371$, $R\sin \alpha =5⟹\sin \alpha =\frac{5}{\sqrt{29}}\approx 0.928$. Both values are positive, so $\alpha$ is in the first quadrant: $\alpha =\arctan \left(\frac{5}{2}\right)\approx 1.190$ radians. Final form: $\sqrt{29}\sin (x+1.190)$. (b) The maximum value of $\sin (x+1.190)$ is 1, so the maximum value of the expression is $\sqrt{29}\approx 5.385$. This occurs when $x+1.190=\frac{\pi }{2}⟹x=\frac{\pi }{2}-1.190\approx 0.381$ radians.

Question 2

Solve the equation $\sin 2\theta -\tan \theta =0$ for $0\le \theta \le 2\pi$ radians. (5 marks)

Solution 2

Rewrite using identities: $\sin 2\theta =2\sin \theta \cos \theta$, $\tan \theta =\frac{\sin \theta }{\cos \theta }$. Substitute into the equation: $$2\sin \theta \cos \theta -\frac{\sin \theta }{\cos \theta }=0$$ Multiply through by $\cos \theta$ (note $\cos \theta \ne 0$, so $\theta \ne \frac{\pi }{2},\frac{3\pi }{2}$): $$2\sin \theta {\cos }^2\theta -\sin \theta =0$$ Factor out $\sin \theta$ and use the double angle identity for $\cos 2\theta$: $$\sin \theta (2{\cos }^2\theta -1)=\sin \theta \cos 2\theta =0$$ Solve for each factor:

• $\sin \theta =0$: solutions are $\theta =0,\pi ,2\pi$
• $\cos 2\theta =0$: $2\theta =\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},\frac{7\pi }{2}⟹\theta =\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$ All solutions are valid (no undefined values), so full solution set: $0,\frac{\pi }{4},\frac{3\pi }{4},\pi ,\frac{5\pi }{4},\frac{7\pi }{4},2\pi$

Question 3

Prove that $\frac{\cos 3A}{\sin A}+\frac{\sin 3A}{\cos A}\equiv 2\cot 2A$. (4 marks)

Solution 3

Start with the LHS, combine fractions with common denominator $\sin A\cos A$: $$\text{LHS}=\frac{\cos 3A\cos A+\sin 3A\sin A}{\sin A\cos A}$$ The numerator matches the compound angle identity for $\cos (3A-A)=\cos 2A$. The denominator can be rewritten using the double angle identity for $\sin 2A$: $\sin A\cos A=\frac{1}{2}\sin 2A$. Substitute these in: $$\text{LHS}=\frac{\cos 2A}{\frac{1}{2}\sin 2A}=2\times \frac{\cos 2A}{\sin 2A}=2\cot 2A=\text{RHS}$$ The identity is proven.

9. Quick Reference Cheatsheet

10. What's Next

The trigonometric techniques you learned here are foundational for almost all remaining topics in A-Level Mathematics Paper 3. You will use double angle identities to integrate squared and higher-power trigonometric functions, the R-form to solve differential equations for oscillating systems, and trigonometric identities to simplify scalar product calculations in vectors and complex number argument problems. Examiners frequently combine trigonometry with other Pure 3 topics in 8–10 mark extended questions, so mastering these identities now will save you significant time when studying later topics.

If you are stuck on any identity, practice problem, or want more exam-style questions tailored to your weak spots, you can ask Ollie, our AI tutor, for personalized support at any time on the homepage. You can also access our dedicated study guides for related Pure 3 topics including integration, differential equations, and vectors to build a complete understanding of the full Paper 3 syllabus.

Aligned with the Cambridge International AS & A Level Mathematics 9709 syllabus. OwlsAi is not affiliated with Cambridge Assessment International Education.