Trigonometric identities

IB Mathematics: Analysis and Approaches HL· 20 min read

1. Fundamental Trigonometric Identities★★☆☆☆⏱ 15 min

The most basic identities are derived directly from the unit circle definition of sine and cosine. For any angle $θ$ , the point on the unit circle has coordinates $(cos θ, sin θ)$ , so by Pythagoras' theorem we get the core Pythagorean identity.

📘 Definition

Pythagorean Identity

For all $θ$ where the functions are defined

Three related core identities derived from the unit circle Pythagorean relationship

Example:

$sin^{2} θ + cos^{2} θ = 1$ , $1 + tan^{2} θ = sec^{2} θ$ , $1 + cot^{2} θ = csc^{2} θ$

Other fundamental identities include reciprocal and quotient identities, which follow directly from the definition of reciprocal trigonometric functions:

Reciprocal: $csc θ = \frac{1}{s i n θ}$ , $sec θ = \frac{1}{c o s θ}$ , $cot θ = \frac{1}{t a n θ}$
Quotient: $tan θ = \frac{s i n θ}{c o s θ}$ , $cot θ = \frac{c o s θ}{s i n θ}$

📐 Worked Example

Simplify the expression $\frac{( 1 - s i n θ ) ( 1 + s i n θ )}{c o s ^{2} θ}$ , stating any domain restrictions.

1
Expand the numerator using the difference of squares identity:
2
$(1 - \sin\theta)(1 + \sin\theta) = 1 - \sin^2\theta$
3
Use the core Pythagorean identity to rewrite the numerator:
4
$1 - \sin^2\theta = \cos^2\theta$
5
Substitute back and simplify, noting the domain restriction:
6
$\frac{\cos^2\theta}{\cos^2\theta} = 1, \quad \cos\theta \neq 0$

Exam tip:

Always state domain restrictions where the original expression is undefined; exam markers regularly award marks for this step.

2. Compound Angle Identities★★★☆☆⏱ 20 min

Compound angle identities relate the trigonometric function of a sum or difference of two angles to the functions of the individual angles. A common mistake is to assume linearity: $sin (A + B) \neq = sin A + sin B$ , so always use the formal identity.

📘 Definition

Compound-Angle Identity

An identity that expresses a trigonometric function of $A \pm B$ in terms of trigonometric functions of $A$ and $B$

Example:

$sin (A + B) = sin A cos B + cos A sin B$

$sin (A \pm B) = sin A cos B \pm cos A sin B$
$cos (A \pm B) = cos A cos B \mp sin A sin B$
$tan (A \pm B) = \frac{t a n A \pm t a n B}{1 \mp t a n A t a n B}$

📐 Worked Example

Find the exact value of $cos (1 5^{\circ})$ using a compound angle identity.

1
Write $1 5^{\circ}$ as a difference of two angles with known exact trig values:
2
$15^\circ = 45^\circ - 30^\circ$
3
Substitute into the cosine difference identity:
4
$\cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ$
5
Substitute known exact values and simplify:
6
$= \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{6} + \sqrt{2}}{4}$

3. Double Angle Identities★★★☆☆⏱ 20 min

Double angle identities are a special case of compound angle identities where $A = B = θ$ , so we get expressions for trigonometric functions of $2 θ$ in terms of functions of $θ$ . We can derive them directly from the compound angle formulas.

🔬 Derivation

Goal:

Derive the double angle identity for $sin 2 θ$

Starting from:

The sine compound addition identity $sin (A + B) = sin A cos B + cos A sin B$

1
Set $A = B = θ$ , so $sin 2 θ = sin (θ + θ)$
2
$\sin(\theta + \theta) = \sin\theta \cos\theta + \cos\theta \sin\theta$
3
Combine like terms on the right-hand side

Result:

$sin 2 θ = 2 sin θ cos θ$

For cosine double angle, there are three equivalent forms, derived by substituting the Pythagorean identity into the base form. These are extremely useful for solving equations and later integration.

Base form: $cos 2 θ = cos^{2} θ - sin^{2} θ$
In terms of cosine only: $cos 2 θ = 2 cos^{2} θ - 1$
In terms of sine only: $cos 2 θ = 1 - 2 sin^{2} θ$
Tangent double angle: $tan 2 θ = \frac{2 t a n θ}{1 - t a n ^{2} θ}$

📐 Worked Example

Rewrite $sin θ cos θ$ in terms of a double angle trigonometric function.

1
Start with the standard sine double angle identity:
2
$\sin 2\theta = 2 \sin\theta \cos\theta$
3
Rearrange to isolate $sin θ cos θ$ :
4
$\sin\theta \cos\theta = \frac{1}{2} \sin 2\theta$

4. Proving Trigonometric Identities★★★★☆⏱ 25 min

Proving trigonometric identities is a common exam question. The standard strategy is to start with the more complex side of the equation, simplify it step-by-step using known identities, and transform it into the simpler side.

📐 Worked Example

Prove that $\frac{1 - t a n ^{2} θ}{1 + t a n ^{2} θ} = cos 2 θ$ .

1
Start with the left-hand side (LHS), use the Pythagorean identity $1 + tan^{2} θ = sec^{2} θ$ :
2
$\text{LHS} = \frac{1 - \frac{\sin^2\theta}{\cos^2\theta}}{\sec^2\theta}$
3
Substitute $sec^{2} θ = \frac{1}{c o s ^{2} θ}$ and multiply numerator/denominator by $cos^{2} θ$ :
4
$= \frac{\cos^2\theta - \sin^2\theta}{\frac{1}{\cos^2\theta} \cdot \cos^2\theta} = \cos^2\theta - \sin^2\theta$
5
Use the double angle identity for cosine to get the right-hand side (RHS):
6
$\cos^2\theta - \sin^2\theta = \cos 2\theta = \text{RHS}$
7
The identity is proven.

5. Common Pitfalls

Wrong move:

Writing $sin (A + B) = sin A + sin B$ or $cos (A + B) = cos A + cos B$

Why:

Trigonometric functions are not linear, so this common assumption is incorrect

Correct move:

Always use the full compound angle identity for sums or differences of angles

Wrong move:

Forgetting to state domain restrictions when simplifying expressions

Why:

The simplified expression may be defined for more values than the original, so the identity is not fully correct without restrictions

Correct move:

Always note any values of $θ$ that make the original expression undefined (e.g. $cos θ \neq = 0$ where $cos θ$ was in a denominator)

Wrong move:

Mixing up the sign in the cosine compound angle formula

Why:

The sign of the second term is opposite the sign between $A$ and $B$ , which is easy to misremember

Correct move:

Recall the rule: for $cos (A \pm B)$ , the second term has the opposite sign: $cos (A - B) = cos A cos B + sin A sin B$

Wrong move:

Working on both sides of the identity when proving it

Why:

This implicitly assumes the identity is true before you prove it, which is circular reasoning

Correct move:

Start only with the more complex side, and manipulate it step-by-step to get the other side

Wrong move:

Writing $sin^{2} θ$ as $sin θ^{2}$

Why:

This notation is ambiguous: $sin (θ^{2})$ is not the same as $(sin θ)^{2}$

Correct move:

Always write $(sin θ)^{2}$ as $sin^{2} θ$ to avoid confusion

6. Quick Reference Cheatsheet

Identity Type	Core Formulas
Fundamental	$sin^{2} θ + cos^{2} θ = 1$ ; $1 + tan^{2} θ = sec^{2} θ$ ; $tan θ = \frac{s i n θ}{c o s θ}$
Compound Angle	$sin (A \pm B) = sin A cos B \pm cos A sin B$ ; $cos (A \pm B) = cos A cos B \mp sin A sin B$
Double Angle (Sine)	$sin 2 θ = 2 sin θ cos θ$
Double Angle (Cosine)	$cos 2 θ = cos^{2} θ - sin^{2} θ = 2 cos^{2} θ - 1 = 1 - 2 sin^{2} θ$
Double Angle (Tangent)	$tan 2 θ = \frac{2 t a n θ}{1 - t a n ^{2} θ}$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · 1
Prove trigonometric identity, 6 marks
2024 · 2
Use double-angle identity to solve equation
2023 · 1
Find exact value using compound angle

Going deeper

syllabusIB AA HL Topic 2 FunctionsOfficial syllabus reference

What's Next

Trigonometric identities are a foundational tool for almost all further trigonometric topics in IB AA HL, from solving trigonometric equations to integration of trigonometric functions and working with polar form of complex numbers. Mastery of these identities is critical for both Paper 1 and Paper 2, as they appear in questions across many different topic areas, even outside of pure trigonometry. Next, you will apply these identities to solve trigonometric equations and work with inverse trigonometric functions.