Trigonometric Identities and Equations

Radians and Degrees

At A-Level, angles are frequently measured in radians rather than degrees. The conversion is:

$\pi \text{ radians} = 180°$

Key equivalences: $\frac{\pi}{6} = 30°$, $\frac{\pi}{4} = 45°$, $\frac{\pi}{3} = 60°$, $\frac{\pi}{2} = 90°$, $\pi = 180°$, $2\pi = 360°$.

You must be comfortable working in radians, especially for calculus-based problems where radian measure is essential.

The Fundamental Trigonometric Functions

For an angle $\theta$, the six trigonometric functions are defined, but at A-Level we focus primarily on three:

• $\sin\theta$, $\cos\theta$, $\tan\theta = \frac{\sin\theta}{\cos\theta}$

The reciprocal functions (A2 content for some boards) are:

• $\sec\theta = \frac{1}{\cos\theta}$, $\cosec\theta = \frac{1}{\sin\theta}$, $\cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}$

Pythagorean Identities

The most fundamental trigonometric identity is:

$\sin^2\theta + \cos^2\theta = 1$

This follows directly from the unit circle definition: a point on the unit circle has coordinates $(\cos\theta, \sin\theta)$, and since it lies on $x^2 + y^2 = 1$, the identity holds.

Dividing through by $\cos^2\theta$ gives:

$\tan^2\theta + 1 = \sec^2\theta$

Dividing through by $\sin^2\theta$ gives:

$1 + \cot^2\theta = \cosec^2\theta$

These three identities are used extensively to simplify expressions and solve equations.

Double Angle Formulae

The double angle formulae express trigonometric functions of $2\theta$ in terms of $\theta$:

$\sin 2\theta = 2\sin\theta\cos\theta$

$\cos 2\theta = \cos^2\theta - \sin^2\theta$

The cosine double angle formula has two equivalent forms (obtained using $\sin^2\theta + \cos^2\theta = 1$):

$\cos 2\theta = 2\cos^2\theta - 1$

$\cos 2\theta = 1 - 2\sin^2\theta$

For tangent:

$\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$

These formulae are provided on the formula sheet for all exam boards, but you should understand how they are derived and be able to apply them fluently.

Addition Formulae

The double angle formulae are special cases of the addition formulae (also called compound angle formulae):

$\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$

$\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$

$\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$

Setting $A = B = \theta$ in these formulae yields the double angle formulae.

Solving Trigonometric Equations

The general approach to solving trigonometric equations is:

1. Rearrange the equation to isolate a trigonometric function (or a recognisable form).
2. Find the principal value using inverse trigonometric functions or exact values.
3. Find all solutions in the given range using the symmetry properties of the trigonometric functions.

For $\sin\theta = k$ (where $-1 \leq k \leq 1$):

• Principal value: $\theta_0 = \arcsin(k)$
• General solutions: $\theta = \theta_0 + 360°n$ or $\theta = (180° - \theta_0) + 360°n$

For $\cos\theta = k$:

• Principal value: $\theta_0 = \arccos(k)$
• General solutions: $\theta = \theta_0 + 360°n$ or $\theta = -\theta_0 + 360°n$ (i.e., $\theta = \pm\theta_0 + 360°n$)

For $\tan\theta = k$:

• Principal value: $\theta_0 = \arctan(k)$
• General solutions: $\theta = \theta_0 + 180°n$

Exact Trigonometric Values

You must know these exact values:

$\theta$	$\sin\theta$	$\cos\theta$	$\tan\theta$
$0°$	$0$	$1$	$0$
$30°$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{3}}$
$45°$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$	$1$
$60°$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$
$90°$	$1$	$0$	undefined

Tip 1: Use CAST Diagrams

The CAST diagram tells you which trigonometric functions are positive in each quadrant. Starting from the fourth quadrant and going anticlockwise: Cos (4th), All (1st), Sin (2nd), Tan (3rd). This is invaluable for finding all solutions in a given range.

Tip 2: Look for Identities to Simplify

Before solving a trigonometric equation, see if an identity can reduce it to a simpler form. Common substitutions include replacing $\sin^2\theta$ with $1 - \cos^2\theta$ (or vice versa) to create a quadratic in one trigonometric function.

Tip 3: Be Careful with the Range

Always check the range specified in the question (e.g., $0° \leq \theta < 360°$ or $0 \leq \theta < 2\pi$). List all solutions within this range — missing solutions is a very common error.

Tip 4: Don't Divide by Trigonometric Functions

Dividing both sides by $\sin\theta$ or $\cos\theta$ can lose solutions (where those functions equal zero). Instead, factorise. For example, $\sin\theta\cos\theta = \sin\theta$ becomes $\sin\theta(\cos\theta - 1) = 0$.

Tip 5: Recognise Disguised Quadratics

Equations like $2\cos^2\theta - \cos\theta - 1 = 0$ are quadratics in $\cos\theta$. Let $u = \cos\theta$, solve the quadratic in $u$, then find $\theta$ from each value of $u$.