Exam Marking Strategy
Trigonometry carries 8 to 12 marks in ECZ Grade 12 Mathematics Paper 2. Always state the domain range (e.g. $0^\circ \le \theta \le 360^\circ$) and double check whether your calculator is in DEGREE or RADIAN mode!
Table of Contents
1. Fundamental Trigonometric Identities
Every trigonometric proof and equation substitution in ECZ Mathematics relies on two key identities:
Primary Identities
$$\tan \theta = \frac{\sin \theta}{\cos \theta} \quad (\text{where } \cos \theta \neq 0)$$
$$\sin^2 \theta + \cos^2 \theta = 1$$
Rearrangements:
$\sin^2 \theta = 1 - \cos^2 \theta$
$\cos^2 \theta = 1 - \sin^2 \theta$
$\sin^2 \theta = 1 - \cos^2 \theta$
$\cos^2 \theta = 1 - \sin^2 \theta$
2. The CAST Diagram & Quadrant Rules
The sign (+ or -) of a trigonometric ratio depends on the quadrant in which the angle lies:
| Quadrant | Angle Range ($\theta$) | Positive Ratios | Reference Angle Formula ($\alpha$) |
|---|---|---|---|
| Quadrant I (A) | $0^\circ < \theta < 90^\circ$ | All ($\sin, \cos, \tan$) | $\theta = \alpha$ |
| Quadrant II (S) | $90^\circ < \theta < 180^\circ$ | Sine only | $\theta = 180^\circ - \alpha$ |
| Quadrant III (T) | $180^\circ < \theta < 270^\circ$ | Tangent only | $\theta = 180^\circ + \alpha$ |
| Quadrant IV (C) | $270^\circ < \theta < 360^\circ$ | Cosine only | $\theta = 360^\circ - \alpha$ |
3. Proving Trigonometric Identities
When asked to "Prove that LHS = RHS" in ECZ exams:
- Start with the more complex side (usually the Left-Hand Side).
- Convert all terms into terms of $\sin \theta$ and $\cos \theta$ using $\tan \theta = \frac{\sin \theta}{\cos \theta}$.
- Combine fractions over a common denominator.
- Use $\sin^2 \theta + \cos^2 \theta = 1$ to simplify expressions.
Worked Proof Example: Prove that $\frac{1}{\cos \theta} - \cos \theta = \sin \theta \tan \theta$.
Proof:
$\text{LHS} = \frac{1 - \cos^2 \theta}{\cos \theta} = \frac{\sin^2 \theta}{\cos \theta} = \sin \theta \times \frac{\sin \theta}{\cos \theta} = \sin \theta \tan \theta = \text{RHS}$. $\quad \blacksquare$
Proof:
$\text{LHS} = \frac{1 - \cos^2 \theta}{\cos \theta} = \frac{\sin^2 \theta}{\cos \theta} = \sin \theta \times \frac{\sin \theta}{\cos \theta} = \sin \theta \tan \theta = \text{RHS}$. $\quad \blacksquare$
4. Solving Quadratic Trig Equations
Example: Solve $2\sin^2 \theta - \cos \theta - 1 = 0$ for $0^\circ \le \theta \le 360^\circ$.
- Substitute $\sin^2 \theta = 1 - \cos^2 \theta$: $$2(1 - \cos^2 \theta) - \cos \theta - 1 = 0$$ $$2 - 2\cos^2 \theta - \cos \theta - 1 = 0 \implies 2\cos^2 \theta + \cos \theta - 1 = 0$$
- Let $x = \cos \theta$, getting quadratic $2x^2 + x - 1 = 0 \implies (2x - 1)(x + 1) = 0$.
- Roots: $\cos \theta = \frac{1}{2}$ or $\cos \theta = -1$.
- Case 1: $\cos \theta = \frac{1}{2}$ (Positive in Quadrants I & IV). Reference angle $\alpha = 60^\circ$.
$\theta_1 = 60^\circ$, $\theta_2 = 360^\circ - 60^\circ = 300^\circ$. - Case 2: $\cos \theta = -1 \implies \theta_3 = 180^\circ$.
- Case 1: $\cos \theta = \frac{1}{2}$ (Positive in Quadrants I & IV). Reference angle $\alpha = 60^\circ$.
- Final Solution Set: $\theta = \{60^\circ, 180^\circ, 300^\circ\}$.
Frequently Asked Questions
The two primary identities are: 1) $\tan \theta = \frac{\sin \theta}{\cos \theta}$, and 2) $\sin^2 \theta + \cos^2 \theta = 1$ (from which $1 - \sin^2 \theta = \cos^2 \theta$ and $1 - \cos^2 \theta = \sin^2 \theta$ are derived).