# Trigonometric identities

> IB Mathematics: Analysis and Approaches HL · Unit 2: Functions
> Source: https://www.owlsprep.com/study/ib-math-aa-hl-u2-trigonometric-identities/

This sub-topic introduces core trigonometric identities for IB AA HL, including Pythagorean, reciprocal, quotient, compound-angle and double-angle identities, used to simplify expressions, prove relationships and solve trigonometric problems.

**Prerequisites:** [Unit circle definitions of trigonometric functions](https://www.owlsprep.com/study/ib-math-aa-hl-u2-unit-circle-trig/); [Basic properties of sine, cosine and tangent](https://www.owlsprep.com/study/ib-math-aa-hl-u2-trigonometric-functions/)

## Learning objectives

- Recall and apply Pythagorean, reciprocal, quotient, compound-angle and double-angle identities
- Simplify trigonometric expressions and prove trigonometric identities
- Use identities to solve trigonometric equations and find exact values

## Fundamental Trigonometric Identities

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

**Pythagorean Identity** — Three related core identities derived from the unit circle Pythagorean relationship

*Notation:* For all $\theta$ where the functions are defined

*Example:* $\sin^2\theta + \cos^2\theta = 1$, $1 + \tan^2\theta = \sec^2\theta$, $1 + \cot^2\theta = \csc^2\theta$

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

- Reciprocal: $\csc\theta = \frac{1}{\sin\theta}$, $\sec\theta = \frac{1}{\cos\theta}$, $\cot\theta = \frac{1}{\tan\theta}$
- Quotient: $\tan\theta = \frac{\sin\theta}{\cos\theta}$, $\cot\theta = \frac{\cos\theta}{\sin\theta}$

**Worked example:** Simplify the expression $\frac{(1 - \sin\theta)(1 + \sin\theta)}{\cos^2\theta}$, 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.

## Compound Angle Identities

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.

**Compound-Angle Identity** — An identity that expresses a trigonometric function of $A \pm B$ in terms of trigonometric functions of $A$ and $B$

*Example:* $\sin(A+B) = \sin A \cos B + \cos A \sin B$

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

**Worked example:** Find the exact value of $\cos(15^\circ)$ using a compound angle identity.

1. Write $15^\circ$ 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. $$= \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{6} + \sqrt{2}}{4}$$

## Double Angle Identities

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

**Derivation:** Derive the double angle identity for $\sin 2\theta$

*Starting from:* The sine compound addition identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$

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

*Conclusion:* $\sin 2\theta = 2 \sin\theta \cos\theta$

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: $\cos 2\theta = \cos^2\theta - \sin^2\theta$
- In terms of cosine only: $\cos 2\theta = 2\cos^2\theta - 1$
- In terms of sine only: $\cos 2\theta = 1 - 2\sin^2\theta$
- Tangent double angle: $\tan 2\theta = \frac{2 \tan\theta}{1 - \tan^2\theta}$

**Worked example:** Rewrite $\sin\theta \cos\theta$ 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 $\sin\theta \cos\theta$:
4. $$\sin\theta \cos\theta = \frac{1}{2} \sin 2\theta$$

> **tip**
>
> The rearranged forms of the cosine double angle identity are used extensively to reduce powers of sine and cosine for integration later in the course.

## Proving Trigonometric Identities

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.

**Exam command terms**

The command term "Prove that" for trig identities has clear exam expectations:

- **Prove that** — Show full step-by-step algebraic manipulation using identities to confirm both sides are equal *(Full marks require all intermediate steps; skipping steps will lose marks)*

**Worked example:** Prove that $\frac{1 - \tan^2\theta}{1 + \tan^2\theta} = \cos 2\theta$.

1. Start with the left-hand side (LHS), use the Pythagorean identity $1 + \tan^2\theta = \sec^2\theta$:
2. $$\text{LHS} = \frac{1 - \frac{\sin^2\theta}{\cos^2\theta}}{\sec^2\theta}$$
3. Substitute $\sec^2\theta = \frac{1}{\cos^2\theta}$ and multiply numerator/denominator by $\cos^2\theta$:
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.

> **warning**
>
> Do not treat the identity like an equation and move terms between sides. This is circular reasoning, as you assume the identity is true before proving it, and will lose marks.

## Common pitfalls

- **Wrong:** Writing $\sin(A+B) = \sin A + \sin B$ or $\cos(A+B) = \cos A + \cos B$
  - Why it fails: Trigonometric functions are not linear, so this common assumption is incorrect
  - Correct: Always use the full compound angle identity for sums or differences of angles
- **Wrong:** Forgetting to state domain restrictions when simplifying expressions
  - Why it fails: The simplified expression may be defined for more values than the original, so the identity is not fully correct without restrictions
  - Correct: Always note any values of $\theta$ that make the original expression undefined (e.g. $\cos\theta \neq 0$ where $\cos\theta$ was in a denominator)
- **Wrong:** Mixing up the sign in the cosine compound angle formula
  - Why it fails: The sign of the second term is opposite the sign between $A$ and $B$, which is easy to misremember
  - Correct: 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:** Working on both sides of the identity when proving it
  - Why it fails: This implicitly assumes the identity is true before you prove it, which is circular reasoning
  - Correct: Start only with the more complex side, and manipulate it step-by-step to get the other side
- **Wrong:** Writing $\sin^2\theta$ as $\sin \theta^2$
  - Why it fails: This notation is ambiguous: $\sin(\theta^2)$ is not the same as $(\sin \theta)^2$
  - Correct: Always write $(\sin \theta)^2$ as $\sin^2\theta$ to avoid confusion

## Cheatsheet

| Identity Type | Core Formulas |
| --- | --- |
| Fundamental | $\sin^2\theta + \cos^2\theta = 1$; $1 + \tan^2\theta = \sec^2\theta$; $\tan\theta = \frac{\sin\theta}{\cos\theta}$ |
| 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) | $\sin 2\theta = 2\sin\theta \cos\theta$ |
| Double Angle (Cosine) | $\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$ |
| Double Angle (Tangent) | $\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$ |

## 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.

- [Function properties: parity and periodicity](https://www.owlsprep.com/study/ib-math-aa-hl-u2-function-properties-parity-and-periodicity/)
- [Solving function equations and inequalities](https://www.owlsprep.com/study/ib-math-aa-hl-u2-solving-function-equations-and-inequalities/)
- [Geometry & Trigonometry](https://www.owlsprep.com/study/ib-math-aa-hl-u3-overview/)

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