The Core Identity Framework
Trigonometric proofs and equations represent a significant proportion of marks in pure mathematics papers. The trick is to quickly convert mixtures of $\sin$, $\cos$, $\tan$, and their reciprocal functions into a single quadratic structure.
Pythagorean & Double-Angle Identities
Memorize these identities and master their derivations:
Pythagorean Trios
$$\sin^2\theta + \cos^2\theta = 1$$
Dividing by $\cos^2\theta$: $$1 + \tan^2\theta = \sec^2\theta$$
Dividing by $\sin^2\theta$: $$1 + \cot^2\theta = \csc^2\theta$$
Sine Double-Angle
$$\sin(2\theta) = 2\sin\theta\cos\theta$$
Cosine Double-Angle (Three Forms)
$$\cos(2\theta) = \cos^2\theta - \sin^2\theta$$
Using $\sin^2\theta = 1-\cos^2\theta$: $$\cos(2\theta) = 2\cos^2\theta - 1$$
Using $\cos^2\theta = 1-\sin^2\theta$: $$\cos(2\theta) = 1 - 2\sin^2\theta$$
Worked Exam Example
Worked A-Level Example
Question: Solve the equation $\cos(2\theta) - 3\cos\theta + 2 = 0$ in the interval $0^\circ \leq \theta \leq 360^\circ$.
Solution:
- Note that we have a mixture of double-angle ($\cos 2\theta$) and single-angle ($\cos \theta$). We must convert $\cos(2\theta)$ into terms of $\cos\theta$ using: $$\cos(2\theta) = 2\cos^2\theta - 1$$
- Substitute this identity back into the original equation: $$(2\cos^2\theta - 1) - 3\cos\theta + 2 = 0$$ $$2\cos^2\theta - 3\cos\theta + 1 = 0$$
- Let $y = \cos\theta$. This forms a quadratic equation: $$2y^2 - 3y + 1 = 0$$ $$(2y - 1)(y - 1) = 0$$ $$y = \frac{1}{2} \quad \text{or} \quad y = 1$$
- Solve for $\theta$:
Case 1: $\cos\theta = \frac{1}{2}$ $$\theta = \arccos\left(\frac{1}{2}\right) = 60^\circ$$ By symmetry (using $360^\circ - \theta$ for cosine): $$\theta = 360^\circ - 60^\circ = 300^\circ$$
Case 2: $\cos\theta = 1$ $$\theta = \arccos(1) = 0^\circ \text{ or } 360^\circ$$
Solutions: $\theta = 0^\circ, 60^\circ, 300^\circ, 360^\circ$.
Frequently Asked Questions
An identity is true for **every** possible value of the variable (e.g. $\sin^2\theta + \cos^2\theta \equiv 1$), which is why we use the equivalence symbol $\equiv$. An equation is only true for specific values of the variable (e.g., $\sin\theta = 0.5$).
Using the compound angle formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A\tan B}$, substituting $A = \theta$ and $B = \theta$ yields the identity: $$\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}$$
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