CIE AS/A Level Maths-2.3 Trigonometry- Study Notes- New Syllabus - 2026-2027
CIE AS/A Level Maths-2.3 Trigonometry- Study Notes- New Syllabus
Ace AS/A Level Maths Exam with CIE AS/A Level Maths-2.3 Trigonometry- Study Notes
Key Concepts:
- Relationships Between Trigonometric Functions
- Trigonometric Identities and Their Applications
Relationships Between Trigonometric Functions
Relationships Between Trigonometric Functions
The trigonometric functions secant, cosecant, and cotangent are defined as reciprocals of cosine, sine, and tangent respectively:
- \(\sec \theta = \dfrac{1}{\cos \theta}\)
- \(\csc \theta = \dfrac{1}{\sin \theta}\)
- \(\cot \theta = \dfrac{1}{\tan \theta} = \dfrac{\cos \theta}{\sin \theta}\)
Key Properties of the Six Trigonometric Functions
| Function | Definition | Domain (restrictions) | Range | Period | Symmetry |
|---|---|---|---|---|---|
| \(\sin \theta\) | Opposite/Hypotenuse | All real \(\theta\) | \([-1, 1]\) | \(2\pi\) | Odd: \(\sin(-\theta) = -\sin(\theta)\) |
| \(\cos \theta\) | Adjacent/Hypotenuse | All real \(\theta\) | \([-1, 1]\) | \(2\pi\) | Even: \(\cos(-\theta) = \cos(\theta)\) |
| \(\tan \theta\) | \(\dfrac{\sin \theta}{\cos \theta}\) | \(\theta \neq \dfrac{\pi}{2} + n\pi\) | All real numbers | \(\pi\) | Odd |
| \(\csc \theta\) | \(\dfrac{1}{\sin \theta}\) | \(\theta \neq n\pi\) | \((-\infty, -1] \cup [1, \infty)\) | \(2\pi\) | Odd |
| \(\sec \theta\) | \(\dfrac{1}{\cos \theta}\) | \(\theta \neq \dfrac{\pi}{2} + n\pi\) | \((-\infty, -1] \cup [1, \infty)\) | \(2\pi\) | Even |
| \(\cot \theta\) | \(\dfrac{\cos \theta}{\sin \theta}\) | \(\theta \neq n\pi\) | All real numbers | \(\pi\) | Odd |
Graphs of the Six Trigonometric Functions
- \(\sin \theta\): Smooth wave oscillating between -1 and 1, period \(2\pi\).
- \(\cos \theta\): Same shape as sine, but phase-shifted by \(\dfrac{\pi}{2}\).
- \(\tan \theta\): Repeating curve with vertical asymptotes at \(\theta = \dfrac{\pi}{2} + n\pi\), period \(\pi\).
- \(\csc \theta\): Reciprocal of sine. Has vertical asymptotes at multiples of \(\pi\). Curves lie outside [-1,1].
- \(\sec \theta\): Reciprocal of cosine. Has vertical asymptotes at \(\dfrac{\pi}{2} + n\pi\). Curves outside [-1,1].
- \(\cot \theta\): Reciprocal of tangent. Vertical asymptotes at \(\theta = n\pi\), period \(\pi\).
Example:
Solve the equation \(\sec \theta = 2\) for \(0 \leq \theta \leq 2\pi\).
▶️Answer/Explanation
Solution:
\(\sec \theta = 2 \implies \cos \theta = \dfrac{1}{2}\).
From the unit circle, \(\cos \theta = \dfrac{1}{2}\) at \(\theta = \dfrac{\pi}{3}, \dfrac{5\pi}{3}\).
Final Answer: \(\theta = \dfrac{\pi}{3}, \dfrac{5\pi}{3}\).
Example:
Simplify the expression \(\csc^2 \theta – \cot^2 \theta\).
▶️Answer/Explanation
Solution:
Recall the Pythagorean identity: \(1 + \cot^2 \theta = \csc^2 \theta\).
So, \(\csc^2 \theta – \cot^2 \theta = (1 + \cot^2 \theta) – \cot^2 \theta = 1\).
Final Answer: \(\csc^2 \theta – \cot^2 \theta = 1\).
Example:
Solve the inequality \(\csc \theta > 2\) for \(0 < \theta < 2\pi\).
▶️Answer/Explanation
Solution:
\(\csc \theta > 2 \implies \dfrac{1}{\sin \theta} > 2 \implies \sin \theta < \dfrac{1}{2}\), since \(\sin \theta > 0\) in Quadrants I and II.
For \(0 < \theta < 2\pi\), \(\sin \theta < \dfrac{1}{2}\) in two ranges:
- Quadrant I: \(0 < \theta < \dfrac{\pi}{6}\)
- Quadrant II: \(\dfrac{5\pi}{6} < \theta < \pi\)
- Also in Quadrants III and IV, where \(\sin \theta < 0\), inequality is automatically true (since \(\csc \theta\) will be negative and less than 2 does not apply). So only Quadrant I and II solutions matter.
Final Answer: \(\theta \in (0, \dfrac{\pi}{6}) \cup (\dfrac{5\pi}{6}, \pi)\).
Trigonometric Identities and Their Applications
Trigonometric Identities and Their Applications
Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables for which both sides are defined. These are used for simplification, solving equations, and exact evaluation of expressions.
Pythagorean Identities for Secant and Cosecant
From the fundamental identity \(\sin^2 \theta + \cos^2 \theta = 1\), we obtain:
- \(\sec^2 \theta = 1 + \tan^2 \theta\)
- \(\csc^2 \theta = 1 + \cot^2 \theta\)
These are particularly useful for converting between tangent and secant, or cotangent and cosecant, when simplifying or solving equations.
Angle Addition and Subtraction Formulae
The expansions are:
- \(\sin (A \pm B) = \sin A \cos B \pm \cos A \sin B\)
- \(\cos (A \pm B) = \cos A \cos B \mp \sin A \sin B\)
- \(\tan (A \pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A \tan B}\)
These formulas allow us to break down trigonometric functions of compound angles into simpler functions, and are often used in exact trigonometric evaluations.
Double Angle Formulae
Special cases of the above identities when \(A = B\):
- \(\sin 2A = 2 \sin A \cos A\)
- \(\cos 2A = \cos^2 A – \sin^2 A = 2\cos^2 A – 1 = 1 – 2\sin^2 A\)
- \(\tan 2A = \dfrac{2 \tan A}{1 – \tan^2 A}\)
These are useful for reducing expressions to involve single angles.
Expressions of the Form \(a \sin \theta + b \cos \theta\)
Any expression of the form \(a \sin \theta + b \cos \theta\) can be written as:
- \(R \sin(\theta + \alpha)\), where \(R = \sqrt{a^2 + b^2}\) and \(\tan \alpha = \dfrac{b}{a}\).
- Alternatively, as \(R \cos(\theta – \alpha)\).
This is especially useful in solving trigonometric equations and inequalities, as it reduces a two-term expression into a single sine or cosine function with a phase shift.
Example:
Simplify \(\cos(x – 30^\circ) – 3 \sin(x – 60^\circ)\).
▶️Answer/Explanation
Solution:
\(\cos(x – 30^\circ) = \cos x \cos 30^\circ + \sin x \sin 30^\circ = \dfrac{\sqrt{3}}{2} \cos x + \dfrac{1}{2} \sin x\).
\(\sin(x – 60^\circ) = \sin x \cos 60^\circ – \cos x \sin 60^\circ = \dfrac{1}{2}\sin x – \dfrac{\sqrt{3}}{2}\cos x\).
Thus, \(\cos(x – 30^\circ) – 3 \sin(x – 60^\circ) = \left(\dfrac{\sqrt{3}}{2} \cos x + \dfrac{1}{2} \sin x\right) – 3\left(\dfrac{1}{2}\sin x – \dfrac{\sqrt{3}}{2}\cos x\right)\).
Simplify: \(= \dfrac{\sqrt{3}}{2}\cos x + \dfrac{1}{2}\sin x – \dfrac{3}{2}\sin x + \dfrac{3\sqrt{3}}{2}\cos x\).
\(= 2\sqrt{3}\cos x – \sin x\).
Final Answer: \(2\sqrt{3}\cos x – \sin x\).
Example:
Solve for \(\theta\): \(3\cos \theta + 2\sin \theta = 1\), where \(0 \leq \theta \leq 2\pi\).
▶️Answer/Explanation
Solution:
Write in the form \(R \sin(\theta + \alpha)\):
\(R = \sqrt{3^2 + 2^2} = \sqrt{13}, \quad \tan \alpha = \dfrac{3}{2}\).
So, \(3\cos \theta + 2\sin \theta = \sqrt{13}\sin(\theta + \alpha)\).
Equation becomes \(\sqrt{13}\sin(\theta + \alpha) = 1 \implies \sin(\theta + \alpha) = \dfrac{1}{\sqrt{13}}\).
Solutions: \(\theta + \alpha = \arcsin\left(\dfrac{1}{\sqrt{13}}\right)\) or \(\pi – \arcsin\left(\dfrac{1}{\sqrt{13}}\right)\).
Hence, \(\theta = -\alpha + \arcsin\left(\dfrac{1}{\sqrt{13}}\right)\) or \(\theta = -\alpha + \pi – \arcsin\left(\dfrac{1}{\sqrt{13}}\right)\).