Compound and double angle identities

IB Mathematics: Analysis and Approaches HL· Topic 3.5 Trigonometric identities· 20 min read

1. Compound Angle Identities★★☆☆☆⏱ 5 min

📘 Definition

Compound Angle Identity

A trigonometric identity that expresses a function of a sum or difference of two angles in terms of functions of the individual angles

Example:

$cos (A - B)$ written in terms of $cos A, cos B, sin A, sin B$

The core compound angle identities for sine and cosine are the foundation for all other identities in this subtopic:

\cos(A + B) = \cos A \cos B - \sin A \sin B

\cos(A - B) = \cos A \cos B + \sin A \sin B

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

\sin(A - B) = \sin A \cos B - \cos A \sin B

For tangent, divide $sin (A \pm B)$ by $cos (A \pm B)$ to get:

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

📐 Worked Example

Find the exact value of $cos 1 5^{\circ}$ .

1
Write $1 5^{\circ}$ as the difference of two angles with known exact values: $1 5^{\circ} = 4 5^{\circ} - 3 0^{\circ}$ .
2
Apply the compound angle identity for cosine of a difference:
3
$\cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ$
4
Substitute known exact values:
5
$= \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right)$
6
Simplify to get the final result:
7
$= \frac{\sqrt{6} + \sqrt{2}}{4}$

Exam tip:

Remember the sign rule: cosine flips the sign, sine keeps the sign for the compound term.

2. Double Angle Identities★★☆☆☆⏱ 5 min

📘 Definition

Double Angle Identity

A special case of the compound angle identity where $A = B$ , giving a function of $2 A$ in terms of functions of $A$

Derive double angle identities by setting $B = A$ in the compound angle formulas. For cosine, we get three equivalent forms using the Pythagorean identity $sin^{2} A + cos^{2} A = 1$ :

\sin 2A = 2\sin A \cos A

\begin{aligned} \cos 2A &= \cos^2 A - \sin^2 A \\ &= 2\cos^2 A - 1 \\ &= 1 - 2\sin^2 A \end{aligned}

\tan 2A = \frac{2\tan A}{1 - \tan^2 A}

📐 Worked Example

Simplify $4 sin θ cos θ$ to a single multiple-angle trigonometric term.

1
Factor out 2 from the expression:
2
$4\sin \theta \cos \theta = 2 \left(2\sin \theta \cos \theta\right)$
3
Recognize $2 sin θ cos θ = sin 2 θ$ from the double angle identity for sine:
4
$2(\sin 2\theta) = 2\sin 2\theta$

3. Proving Trigonometric Identities★★★☆☆⏱ 5 min

Proving identities is a very common IB exam question. The standard approach is to start with the more complicated side of the identity and apply identities to simplify it until it matches the other side.

📐 Worked Example

Prove that $\frac{1 - c o s 2 θ}{s i n 2 θ} = tan θ$ .

1
Start with the left-hand side (LHS). Choose the double angle form of $cos 2 θ = 1 - 2 sin^{2} θ$ to match the $1 - cos 2 θ$ term:
2
$\text{LHS} = \frac{1 - (1 - 2\sin^2 \theta)}{\sin 2\theta}$
3
Simplify the numerator, then substitute $sin 2 θ = 2 sin θ cos θ$ :
4
$= \frac{2\sin^2 \theta}{2\sin \theta \cos \theta}$
5
Cancel common factors ( $2 sin θ$ , for $sin θ \neq = 0$ where the identity is defined):
6
$= \frac{\sin \theta}{\cos \theta} = \tan \theta = \text{Right-hand side (RHS)}$
7
The identity is proven.

✓ Quick check

Test your understanding of which $cos 2 A$ form to choose:

Which form of $cos 2 θ$ is optimal to prove $\frac{1 - c o s 2 θ}{s i n 2 θ} = tan θ$ ?
- A. $cos^{2} θ - sin^{2} θ$
- B. $2 cos^{2} θ - 1$
- C. $1 - 2 sin^{2} θ$
Reveal answer
C —
Correct! This form cancels the 1 in the numerator, leaving a simple $sin^{2} θ$ term that simplifies perfectly with the denominator.

4. Solving Trigonometric Equations★★★★☆⏱ 5 min

🚫 No Calculator

Most trigonometric equations in exams have multiple angles (e.g. $θ$ and $2 θ$ ). We use double angle identities to rewrite the entire equation in terms of a single angle, then solve as a polynomial or standard trigonometric equation.

📐 Worked Example

Solve $2 sin 2 θ = tan θ$ for $0 \leq θ \leq π$ .

1
Substitute $sin 2 θ = 2 sin θ cos θ$ and $tan θ = \frac{s i n θ}{c o s θ}$ :
2
$2(2\sin \theta \cos \theta) = \frac{\sin \theta}{\cos \theta}$
3
Multiply through by $cos θ$ (for $cos θ \neq = 0$ ) and rearrange:
4
$4\sin \theta \cos^2 \theta - \sin \theta = 0$
5
Factor out the common term $sin θ$ :
6
$\sin \theta (4\cos^2 \theta - 1) = 0$
7
Set each factor equal to zero and solve in the interval: $sin θ = 0$ gives $θ = 0, π$ . $4 cos^{2} θ - 1 = 0$ gives $cos θ = \pm \frac{1}{2}$ , giving $θ = \frac{π}{3}, \frac{2 π}{3}$ .
8
Check that $cos θ = 0$ gives $θ = \frac{π}{2}$ , which makes $tan θ$ undefined, so it is not a solution.
9
The full solution set is:
10
$\theta = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi$

5. Common Pitfalls

Wrong move:

Memorizing the wrong sign for $cos (A + B)$ , writing $cos A cos B + sin A sin B$ .

Why:

The sign for cosine compound angles is opposite to the angle operation.

Correct move:

Use the mnemonic: cos flips the sign, sine keeps the sign. So $A + B$ gives a minus for cosine, plus for sine.

Wrong move:

Canceling $sin θ$ from both sides of an equation, leading to lost solutions.

Why:

Canceling removes the case where $sin θ = 0$ , which is a valid solution.

Correct move:

Move all terms to one side and factor out common terms, then set each factor equal to zero to find all solutions.

Wrong move:

Incorrectly assuming $sin 2 A = 2 sin A$ .

Why:

The 2 cannot be factored out of the trigonometric function.

Correct move:

Remember the correct identity: $sin 2 A = 2 sin A cos A$ , which is not equal to $2 sin A$ for most values of $A$ .

Wrong move:

Using the wrong form of $cos 2 A$ , leading to unnecessary complicated algebra.

Why:

Not matching the form of $cos 2 A$ to the existing terms in the expression.

Correct move:

Choose the form of $cos 2 A$ that will cancel constants or simplify with existing terms, e.g. use $1 - 2 sin^{2} A$ when you have a $1 - cos 2 A$ term.

Wrong move:

Forgetting to exclude points that make denominators zero when proving identities.

Why:

Identities are only valid for values where both sides are defined.

Correct move:

Always note that the identity holds for all values where both sides are defined, and exclude any points that make denominators zero.

6. Quick Reference Cheatsheet

Identity Type	Core Formulas
Compound Angle	$cos (A \pm B) = cos A cos B \mp sin A sin B$
	$sin (A \pm B) = sin A cos B \pm cos A sin B$
	$tan (A \pm B) = \frac{t a n A \pm t a n B}{1 \mp t a n A t a n B}$
Double Angle	$sin 2 A = 2 sin A cos A$
	$cos 2 A = cos^{2} A - sin^{2} A = 2 cos^{2} A - 1 = 1 - 2 sin^{2} A$
	$tan 2 A = \frac{2 t a n A}{1 - t a n ^{2} A}$

7. Frequently Asked

Do I get these identities in the IB formula booklet?

No, compound and double angle identities are not provided in the IB AA HL formula booklet. You must memorize them or be able to derive them quickly in the exam.

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
Simplify trigonometric expression
2024 · 2
Solve trigonometric equation
2023 · 1
Prove trigonometric identity

Going deeper

syllabusOfficial IB AA HL Topic 3 GuideSyllabus reference for this subtopic

What's Next

Compound and double angle identities are the foundation for all advanced trigonometry in IB AA HL, appearing in topics from integration to differential equations. Mastery of these identities is critical to solving almost every complex trigonometry problem you will encounter in the exam. Next, you will extend these identities to other forms, and apply them to more complex problem types.