Welcome to A2 Pure Trigonometry
Welcome to one of the most powerful and rewarding chapters in CCEA A2 1 Pure Mathematics. In AS Mathematics, you worked with angles in degrees and solved fundamental trigonometric equations. In A2, we take trigonometry to a professional mathematical standard by introducing radians, new reciprocal functions, compound angle identities, and the harmonic form.
Don't worry if this seems like a big step up! Every concept here builds logically on what you already know. Master these building blocks step by step, and you will unlock essential tools for calculus, mechanics, and coordinate geometry.
---1. Radian Measure and Circular Measure
What is a Radian?
In everyday life, measuring angles in degrees (\(360^\circ\) in a full circle) is familiar, but mathematically arbitrary. A radian is a natural geometric unit based directly on the radius of a circle.
Formal Definition: One radian (\(1\text{ rad}\)) is the angle subtended at the centre of a circle by an arc whose length is exactly equal to the radius \(r\) of the circle.
Because the circumference of a circle is \(2\pi r\), there are \(2\pi\) radians in a complete turn (\(360^\circ\)):
\(2\pi\text{ rad} = 360^\circ \implies \pi\text{ rad} = 180^\circ\)
\(1\text{ rad} = \frac{180^\circ}{\pi} \approx 57.3^\circ\)
Key Conversion Reference
You should know these common angles instantly in radians:
• \(30^\circ = \frac{\pi}{6}\text{ rad}\)
• \(45^\circ = \frac{\pi}{4}\text{ rad}\)
• \(60^\circ = \frac{\pi}{3}\text{ rad}\)
• \(90^\circ = \frac{\pi}{2}\text{ rad}\)
• \(180^\circ = \pi\text{ rad}\)
• \(270^\circ = \frac{3\pi}{2}\text{ rad}\)
• \(360^\circ = 2\pi\text{ rad}\)
Arc Length, Sector Area, and Segment Area
When the angle \(\theta\) is measured strictly in radians, standard circle formulas simplify elegantly:
1. Arc Length (\(s\)):
\(s = r\theta\)
2. Area of a Sector (\(A\)):
\(A = \frac{1}{2}r^2\theta\)
3. Area of a Segment (\(A_{\text{seg}}\)):
A segment is formed by cutting off a sector with a triangle. Using Area of Triangle = \(\frac{1}{2}ab\sin C = \frac{1}{2}r^2\sin\theta\):
\(A_{\text{seg}} = \text{Area of Sector} - \text{Area of Triangle}\)
\(A_{\text{seg}} = \frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta = \frac{1}{2}r^2(\theta - \sin\theta)\)
Step-by-Step Example:
A circle has radius \(r = 6\text{ cm}\). A sector has an angle of \(\theta = \frac{2\pi}{3}\text{ radians}\). Find the exact perimeter and exact area of the sector.
• Arc Length: \(s = r\theta = 6 \times \frac{2\pi}{3} = 4\pi\text{ cm}\)
• Perimeter: \(P = s + 2r = 4\pi + 2(6) = (4\pi + 12)\text{ cm}\)
• Sector Area: \(A = \frac{1}{2}r^2\theta = \frac{1}{2}(6^2)\left(\frac{2\pi}{3}\right) = \frac{1}{2}(36)\left(\frac{2\pi}{3}\right) = 12\pi\text{ cm}^2\)
Key Takeaway: All circular measure formulas (\(s = r\theta\), \(A = \frac{1}{2}r^2\theta\)) work only when \(\theta\) is in radians. Ensure your calculator is set to RAD mode.
---2. Small-Angle Approximations
When an angle \(\theta\) is very small and measured in radians (close to \(0\)), the trigonometric functions can be approximated using simple algebraic terms:
• \(\sin\theta \approx \theta\)
• \(\tan\theta \approx \theta\)
• \(\cos\theta \approx 1 - \frac{\theta^2}{2}\)
Why Does This Work?
Did you know? If you type \(\sin(0.05)\) into your calculator in radian mode, you get \(0.049979...\), which is virtually identical to \(0.05\). These approximations are derived from Maclaurin series expansions and allow complex trigonometric limits and physics equations to be simplified easily.
Applying Small-Angle Approximations
Be careful when expanding composite terms such as \(\cos(3\theta)\) or \(\sin(4\theta)\). Substitute the entire input into the formula:
• \(\sin(4\theta) \approx 4\theta\)
• \(\tan(2\theta) \approx 2\theta\)
• \(\cos(3\theta) \approx 1 - \frac{(3\theta)^2}{2} = 1 - \frac{9\theta^2}{2}\)
Worked Example:
Find the value of \(\lim_{\theta \to 0}\frac{1 - \cos(4\theta)}{2\theta\sin(3\theta)}\).
• Using small-angle approximations:
\(\cos(4\theta) \approx 1 - \frac{(4\theta)^2}{2} = 1 - 8\theta^2\)
\(\sin(3\theta) \approx 3\theta\)
• Substitute into the expression:
\(\frac{1 - (1 - 8\theta^2)}{2\theta(3\theta)} = \frac{8\theta^2}{6\theta^2} = \frac{8}{6} = \frac{4}{3}\)
Key Takeaway: Always place brackets around the full angle when squaring for cosine: \(\cos(k\theta) \approx 1 - \frac{(k\theta)^2}{2} = 1 - \frac{k^2\theta^2}{2}\), not \(1 - \frac{k\theta^2}{2}\).
---3. Reciprocal and Inverse Trigonometric Functions
The Reciprocal Functions
The three reciprocal trigonometric functions are defined as:
• Secant: \(\sec\theta = \frac{1}{\cos\theta}\) (defined where \(\cos\theta \neq 0\))
• Cosecant: \(\text{cosec}\,\theta = \frac{1}{\sin\theta}\) (defined where \(\sin\theta \neq 0\))
• Cotangent: \(\cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}\) (defined where \(\sin\theta \neq 0\))
Memory Trick (The 3rd Letter Rule):
Look at the 3rd letter of each reciprocal function name:
• sec \(\theta \to \frac{1}{\mathbf{c}\text{os}\,\theta}\)
• cosec \(\theta \to \frac{1}{\mathbf{s}\text{in}\,\theta}\)
• cot \(\theta \to \frac{1}{\mathbf{t}\text{an}\,\theta}\)
Pythagorean Identities
From AS mathematics, you know that \(\sin^2\theta + \cos^2\theta \equiv 1\). By dividing this identity, we generate two new A2 identities:
1. Divide by \(\cos^2\theta\):
\(\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} \equiv \frac{1}{\cos^2\theta} \implies \mathbf{1 + \tan^2\theta \equiv \sec^2\theta}\)
2. Divide by \(\sin^2\theta\):
\(\frac{\sin^2\theta}{\sin^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} \equiv \frac{1}{\sin^2\theta} \implies \mathbf{1 + \cot^2\theta \equiv \text{cosec}^2\theta}\)
Inverse Trigonometric Functions
To make the inverse trigonometric relations true one-to-one functions, their domains must be restricted:
• \(y = \arcsin x\) (or \(\sin^{-1}x\)):
Domain: \(-1 \le x \le 1\)
Range: \(-\frac{\pi}{2} \le y \le \frac{\pi}{2}\)
Symmetry: Odd function, rotational symmetry of order 2 about the origin.
• \(y = \arccos x\) (or \(\cos^{-1}x\)):
Domain: \(-1 \le x \le 1\)
Range: \(0 \le y \le \pi\)
• \(y = \arctan x\) (or \(\tan^{-1}x\)):
Domain: \(x \in \mathbb{R}\) (all real numbers)
Range: \(-\frac{\pi}{2} < y < \frac{\pi}{2}\)
Horizontal Asymptotes: \(y = \frac{\pi}{2}\) and \(y = -\frac{\pi}{2}\)
Crucial Distinction:
\(\sec x = (\cos x)^{-1} = \frac{1}{\cos x}\), whereas \(\arccos x = \cos^{-1} x\) is the inverse angle function. Never confuse reciprocals with inverse functions!
Key Takeaway: Keep the two new Pythagorean identities (\(1 + \tan^2\theta \equiv \sec^2\theta\) and \(1 + \cot^2\theta \equiv \text{cosec}^2\theta\)) memorised; they are crucial for solving quadratic-form trigonometric equations.
---4. Compound and Double-Angle Identities
Compound (Addition and Subtraction) Formulae
These identities let you break down functions containing a sum or difference of angles:
• \(\sin(A + B) \equiv \sin A \cos B + \cos A \sin B\)
• \(\sin(A - B) \equiv \sin A \cos B - \cos A \sin B\)
• \(\cos(A + B) \equiv \cos A \cos B - \sin A \sin B\) (Note the sign change!)
• \(\cos(A - B) \equiv \cos A \cos B + \sin A \sin B\)
• \(\tan(A + B) \equiv \frac{\tan A + \tan B}{1 - \tan A \tan B}\)
• \(\tan(A - B) \equiv \frac{\tan A - \tan B}{1 + \tan A \tan B}\)
Double-Angle Formulae
By letting \(B = A\) in the addition formulae, we obtain the double-angle formulae:
1. Sine Double Angle:
\(\sin(2A) \equiv 2\sin A \cos A\)
2. Cosine Double Angle (Three Useful Forms):
• \(\cos(2A) \equiv \cos^2 A - \sin^2 A\)
• \(\cos(2A) \equiv 2\cos^2 A - 1\)
• \(\cos(2A) \equiv 1 - 2\sin^2 A\)
3. Tangent Double Angle:
\(\tan(2A) \equiv \frac{2\tan A}{1 - \tan^2 A}\)
Rearrangements for Integration / Squared Terms
Rearranging the cosine double-angle identities expresses \(\sin^2 A\) and \(\cos^2 A\) in terms of linear cosine functions (essential for A2 calculus):
• \(\sin^2 A \equiv \frac{1 - \cos(2A)}{2}\)
• \(\cos^2 A \equiv \frac{1 + \cos(2A)}{2}\)
Proving Trigonometric Identities
When asked to prove an identity:
1. Start strictly with one side (usually the more complex side, typically the LHS).
2. Apply standard identities step-by-step.
3. Show every algebraic transition until it matches the RHS exactly.
Exam Warning: Do not treat an identity proof as an equation by moving terms across the equals sign. Keep the sides separate!
Key Takeaway: Watch out for the sign change in cosine addition: \(\cos(A+B)\) has a minus sign between terms, while \(\cos(A-B)\) has a plus sign.
---5. Harmonic Form (\(R\)-Formulae)
An expression combining sine and cosine of the same angle, such as \(a\sin\theta + b\cos\theta\) or \(a\cos\theta + b\sin\theta\), can be rewritten as a single scaled wave function.
Standard Harmonic Forms
• \(a\sin\theta \pm b\cos\theta \equiv R\sin(\theta \pm \alpha)\)
• \(a\cos\theta \pm b\sin\theta \equiv R\cos(\theta \mp \alpha)\)
Where:
• \(R = \sqrt{a^2 + b^2}\) with \(R > 0\)
• \(\alpha\) is an acute angle (\(0 < \alpha < \frac{\pi}{2}\) or \(0^\circ < \alpha < 90^\circ\))
Step-by-Step Method to Express \(a\cos\theta + b\sin\theta\) as \(R\cos(\theta - \alpha)\)
1. Expand: Write \(R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha\).
2. Equate coefficients:
• \(\cos\theta\) coefficient: \(R\cos\alpha = a\)
• \(\sin\theta\) coefficient: \(R\sin\alpha = b\)
3. Find \(R\): Square and add the equations: \(R^2(\cos^2\alpha + \sin^2\alpha) = a^2 + b^2 \implies R = \sqrt{a^2 + b^2}\).
4. Find \(\alpha\): Divide the equations: \(\frac{R\sin\alpha}{R\cos\alpha} = \tan\alpha = \frac{b}{a} \implies \alpha = \arctan\left(\frac{b}{a}\right)\).
Maximum and Minimum Values
Since the maximum value of \(\cos(\text{anything})\) is \(1\) and the minimum is \(-1\):
• Maximum value of \(R\cos(\theta - \alpha) = R \times 1 = R\)
• Minimum value of \(R\cos(\theta - \alpha) = R \times (-1) = -R\)
Worked Example:
Express \(3\sin\theta + 4\cos\theta\) in the form \(R\sin(\theta + \alpha)\), where \(R > 0\) and \(0 < \alpha < \frac{\pi}{2}\). Hence solve \(3\sin\theta + 4\cos\theta = 2.5\) for \(0 \le \theta \le 2\pi\).
Step 1: Find \(R\) and \(\alpha\):
\(R\sin(\theta + \alpha) = R\sin\theta\cos\alpha + R\cos\theta\sin\alpha\)
\(R\cos\alpha = 3\) and \(R\sin\alpha = 4\)
\(R = \sqrt{3^2 + 4^2} = \sqrt{25} = 5\)
\(\tan\alpha = \frac{4}{3} \implies \alpha = \arctan\left(\frac{4}{3}\right) \approx 0.9273\text{ rad}\)
So, \(3\sin\theta + 4\cos\theta \equiv 5\sin(\theta + 0.9273)\)
Step 2: Solve the equation:
\(5\sin(\theta + 0.9273) = 2.5\)
\(\sin(\theta + 0.9273) = 0.5\)
Principal value: \(\theta + 0.9273 = \arcsin(0.5) = \frac{\pi}{6} \approx 0.5236\text{ rad}\)
Second value in \([0, 2\pi]\): \(\pi - 0.5236 = 2.6180\text{ rad}\)
Third value (adding \(2\pi\)): \(0.5236 + 2\pi = 6.8068\text{ rad}\)
Subtract \(0.9273\):
• \(\theta = 0.5236 - 0.9273 = -0.4037\) (outside \([0, 2\pi]\))
• \(\theta = 2.6180 - 0.9273 = 1.69\text{ rad}\) (to 3 s.f.)
• \(\theta = 6.8068 - 0.9273 = 5.88\text{ rad}\) (to 3 s.f.)
Solutions: \(\theta = 1.69\text{ rad}\) and \(\theta = 5.88\text{ rad}\).
Key Takeaway: Never round \(R\) or \(\alpha\) too early in your working. Store full precision in your calculator to avoid final rounding errors.
---6. Solving Equations & Avoiding Classic Exam Pitfalls
Adjusting the Angle Interval
When solving an equation with a transformed angle, such as \(\cos\left(2\theta - \frac{\pi}{4}\right) = \frac{1}{2}\) for \(0 \le \theta \le 2\pi\):
1. Transform the interval first: if \(0 \le \theta \le 2\pi\), then \(-\frac{\pi}{4} \le 2\theta - \frac{\pi}{4} \le 4\pi - \frac{\pi}{4} = \frac{15\pi}{4}\).
2. Find all solutions for the compound angle within this wider range.
3. Rearrange each solution to find \(\theta\) at the very end.
Top Examiner-Reported Traps
• Never Divide by a Trigonometric Term: If you have \(\sin(2\theta) = \sin\theta\), expand to \(2\sin\theta\cos\theta - \sin\theta = 0\) and factorise: \(\sin\theta(2\cos\theta - 1) = 0\). If you simply divide by \(\sin\theta\), you eliminate the solutions where \(\sin\theta = 0\) and lose marks.
• Radian Mode Vigilance: Check your calculator status before starting every question. If the question gives the domain as \([0, 2\pi]\), your calculator must be in radians.
• Accuracy Standards: Non-exact answers must be given to 3 significant figures. Exact answers must be given using surds or multiples of \(\pi\).
Quick Revision Checklist
Before sitting your CCEA A2 1 examination, make sure you can:
• Convert fluently between degrees and radians.
• Calculate arc length (\(s = r\theta\)), sector area (\(A = \frac{1}{2}r^2\theta\)), and segment area (\(A = \frac{1}{2}r^2(\theta - \sin\theta)\)).
• Apply small-angle approximations for \(\sin\theta\), \(\cos\theta\), and \(\tan\theta\).
• Use reciprocal functions (\(\sec\theta, \text{cosec}\,\theta, \cot\theta\)) and their Pythagorean identities.
• State the domains, ranges, and sketch graphs of \(\arcsin x\), \(\arccos x\), and \(\arctan x\).
• Expand and simplify expressions using compound angle and double-angle identities.
• Convert \(a\sin\theta + b\cos\theta\) into harmonic form \(R\sin(\theta \pm \alpha)\) or \(R\cos(\theta \mp \alpha)\) to find maxima/minima and solve equations.
• Prove identities rigorously working strictly from LHS to RHS.