# Pythagorean and double angle trigonometric identities

> IB Mathematics: Analysis and Approaches SL · Geometry & Trigonometry
> Source: https://www.owlsprep.com/study/ib-math-aa-sl-u3-pythagorean-and-double-angle-trigonometric/

This subtopic covers core Pythagorean trigonometric identities and double angle identities, key tools for simplifying expressions, proving identities and solving trigonometric equations for IB AA SL exams.

**Prerequisites:** [Basic trigonometric ratios](https://www.owlsprep.com/study/ib-math-aa-sl-basic-trigonometric-ratios/); [Unit circle and radian measure](https://www.owlsprep.com/study/ib-math-aa-sl-unit-circle-radian-measure/)

## Learning objectives

- Recall and apply the three core Pythagorean trigonometric identities
- Derive and use double angle identities for sine, cosine and tangent
- Simplify trigonometric expressions and solve equations using these identities
- Recognize when to apply these identities in IB exam problems

## Pythagorean Trigonometric Identities

The fundamental Pythagorean identity comes directly from the unit circle, where any point $(
cos\theta, \sin\theta)$ lies on the circle $x^2 + y^2 = 1$. From this core identity, we can derive two additional identities by dividing through by $
cos^2\theta$ or $
sin^2\theta$ respectively.

**Pythagorean Trigonometric Identities** — Three core identities derived from Pythagoras' theorem on the unit circle that relate powers of basic trigonometric functions

*Notation:* For all defined $\theta$

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

**Worked example:** Given $\sin\theta = \frac{2}{3}$ and $\frac{\pi}{2} < \theta < \pi$, find the values of $\cos\theta$ and $\tan\theta$.

1. Use the fundamental Pythagorean identity to solve for $\cos^2\theta$:
2. $$\sin^2\theta + \cos^2\theta = 1 \implies \left(\frac{2}{3}\right)^2 + \cos^2\theta = 1$$
3. Simplify and solve for $\cos\theta$:
4. $$\frac{4}{9} + \cos^2\theta = 1 \implies \cos^2\theta = \frac{5}{9} \implies \cos\theta = \pm \frac{\sqrt{5}}{3}$$
5. We know $\theta$ is in the second quadrant, where cosine is negative, so we take the negative root: $\cos\theta = -\frac{\sqrt{5}}{3}$.
6. Finally, calculate $\tan\theta = \frac{\sin\theta}{\cos\theta}$:
7. $$\tan\theta = \frac{\frac{2}{3}}{-\frac{\sqrt{5}}{3}} = -\frac{2}{\sqrt{5}} = -\frac{2\sqrt{5}}{5}$$

> **Exam tip:** Always check the quadrant of the given angle to select the correct sign for your result.

## Double Angle Identities

Double angle identities relate trigonometric functions of $2\theta$ to functions of $\theta$, and are derived directly from the compound angle addition formulas. There are three equivalent forms of the cosine double angle identity, which can be simplified using Pythagorean identities for different use cases.

**Derivation:** Derive the double angle identity for sine

*Starting from:* Sine addition formula: $\sin(A+B) = \sin A \cos B + \cos A \sin B$

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

*Conclusion:* $\sin 2\theta = 2 \sin\theta \cos\theta$, the double angle identity for sine

**Double Angle Identities** — All core identities for double angles are listed below, valid for all $\theta$ where both sides are defined:

*Example:* $\sin 2\theta = 2\sin\theta \cos\theta$<br>$\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$<br>$\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$

**Worked example:** Express $4\sin\theta \cos\theta$ as a single simplified trigonometric function.

1. Recognize the expression matches the form of the sine double angle identity. Factor out the constant 2:
2. $$4\sin\theta \cos\theta = 2(2\sin\theta \cos\theta)$$
3. Substitute $\sin 2\theta = 2\sin\theta \cos\theta$ into the expression:
4. $$2(2\sin\theta \cos\theta) = 2\sin 2\theta$$

## Applications to Exam Problems

These identities are most commonly used for three exam tasks: simplifying trigonometric expressions, proving identities, and solving quadratic trigonometric equations. They are required in nearly every trigonometry problem on the IB AA SL exam.

**Worked example:** Solve $2\sin^2\theta = \cos\theta + 1$ for $0 \leq \theta < 2\pi$.

1. Use the Pythagorean identity to rewrite the equation in terms of $\cos\theta$ only:
2. $$2(1 - \cos^2\theta) = \cos\theta + 1$$
3. Expand and rearrange into standard quadratic form:
4. $$2 - 2\cos^2\theta = \cos\theta + 1 \implies 2\cos^2\theta + \cos\theta - 1 = 0$$
5. Factor the quadratic in $\cos\theta$:
6. $$(2\cos\theta - 1)(\cos\theta + 1) = 0$$
7. Solve each factor for $\theta$ in the given domain: <br>1. $2\cos\theta - 1 = 0 \implies \cos\theta = \frac{1}{2}$, solutions $\theta = \frac{\pi}{3}, \frac{5\pi}{3}$ <br>2. $\cos\theta + 1 = 0 \implies \cos\theta = -1$, solution $\theta = \pi$

**Check your understanding**

Test your knowledge of the double angle identity for cosine:

1. Which of the following is NOT a valid form of $\cos 2\theta$?

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

   *Answer:* $2\sin^2\theta - 1$

   *Why:* This is incorrect. The correct form when written in terms of sine is $\cos 2\theta = 1 - 2\sin^2\theta$, not $2\sin^2\theta - 1$.

## Common pitfalls

- **Wrong:** Forgetting to check the sign of a trigonometric value after using a Pythagorean identity.
  - Why it fails: Squaring removes the sign of the original value, so you need to use the quadrant of the angle to get the correct result.
  - Correct: Always check the given domain or quadrant before choosing between positive and negative roots.
- **Wrong:** Mixing up the sign in the double angle identity for cosine, writing $\cos 2\theta = 2\cos^2\theta + 1$.
  - Why it fails: This is a common sign error when rearranging the identity from the Pythagorean theorem.
  - Correct: Remember the constant term is negative: $\cos 2\theta = 2\cos^2\theta - 1$.
- **Wrong:** Dividing both sides of a trigonometric equation by $\sin\theta$ or $\cos\theta$ to simplify.
  - Why it fails: This removes the solution where $\sin\theta = 0$ or $\cos\theta = 0$ from your final result.
  - Correct: Factor the expression instead of dividing to retain all possible solutions in the domain.
- **Wrong:** Writing $(\sin\theta)^2$ as $\sin \theta^2$.
  - Why it fails: This notation error changes the meaning of the expression to sine of (theta squared).
  - Correct: Always write the power after the function name: $\sin^2\theta$ for $(\sin\theta)^2$.

## Cheatsheet

| Identity Type | Identity |
| --- | --- |
| Pythagorean | $\sin^2\theta + \cos^2\theta = 1$ |
| Pythagorean | $1 + \tan^2\theta = \sec^2\theta$ |
| Pythagorean | $1 + \cot^2\theta = \csc^2\theta$ |
| 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

Pythagorean and double angle identities form the foundation for all advanced trigonometric work in IB AA SL, including compound angle identities, solving complex trigonometric equations, and calculus applications of trigonometry. Mastery of these identities is critical, as they are required to simplify expressions before most integration, differentiation and problem-solving tasks on the exam. Next, you will extend these ideas to work with any sum or difference of angles, then apply them to increasingly complex problem types.

- [Solving trigonometric equations](https://www.owlsprep.com/study/ib-math-aa-sl-u3-solving-trigonometric-equations/)
- [Statistics & Probability](https://www.owlsprep.com/study/ib-math-aa-sl-u4-overview/)

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