Pythagorean and double angle trigonometric identities

IB Mathematics: Analysis and Approaches SL· 3.6 Trigonometry· 15 min read

1. Pythagorean Trigonometric Identities★★☆☆☆⏱ 5 min

The fundamental Pythagorean identity comes directly from the unit circle, where any point $( cos\theta, \sin\theta) $l i eso n t h ec i r c l e$ x^2 + y^2 = 1 $. F r o m t hi scor e i d e n t i t y, w ec an d er i v e tw o a dd i t i o na l i d e n t i t i es b y d i v i d in g t h r o ug hb y$ cos^2\theta $or$ sin^2\theta$ respectively.

📘 Definition

Pythagorean Trigonometric Identities

For all defined $θ$

Three core identities derived from Pythagoras' theorem on the unit circle that relate powers of basic trigonometric functions

Example:

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

📐 Worked Example

Given $sin θ = \frac{2}{3}$ and $\frac{π}{2} < θ < π$ , find the values of $cos θ$ and $tan θ$ .

1
Use the fundamental Pythagorean identity to solve for $cos^{2} θ$ :
2
$\sin^2\theta + \cos^2\theta = 1 \implies \left(\frac{2}{3}\right)^2 + \cos^2\theta = 1$
3
Simplify and solve for $cos θ$ :
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 $θ$ is in the second quadrant, where cosine is negative, so we take the negative root: $cos θ = - \frac{5}{3}$ .
6
Finally, calculate $tan θ = \frac{s i n θ}{c o s θ}$ :
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.

2. Double Angle Identities★★☆☆☆⏱ 6 min

Double angle identities relate trigonometric functions of $2 θ$ to functions of $θ$ , 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

Goal:

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

Result:

$sin 2 θ = 2 sin θ cos θ$ , the double angle identity for sine

📘 Definition

Double Angle Identities

All core identities for double angles are listed below, valid for all $θ$ where both sides are defined:

Example:

$sin 2 θ = 2 sin θ cos θ$
$cos 2 θ = cos^{2} θ - sin^{2} θ = 2 cos^{2} θ - 1 = 1 - 2 sin^{2} θ$
$tan 2 θ = \frac{2 t a n θ}{1 - t a n ^{2} θ}$

📐 Worked Example

Express $4 sin θ cos θ$ 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 $sin 2 θ = 2 sin θ cos θ$ into the expression:
4
$2(2\sin\theta \cos\theta) = 2\sin 2\theta$

3. Applications to Exam Problems★★★☆☆⏱ 7 min

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 $2 sin^{2} θ = cos θ + 1$ for $0 \leq θ < 2 π$ .

1
Use the Pythagorean identity to rewrite the equation in terms of $cos θ$ 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 $cos θ$ :
6
$(2\cos\theta - 1)(\cos\theta + 1) = 0$
7
Solve each factor for $θ$ in the given domain:
1. $2 cos θ - 1 = 0 ⟹ cos θ = \frac{1}{2}$ , solutions $θ = \frac{π}{3}, \frac{5 π}{3}$
2. $cos θ + 1 = 0 ⟹ cos θ = - 1$ , solution $θ = π$

✓ Quick check

Test your knowledge of the double angle identity for cosine:

Which of the following is NOT a valid form of $cos 2 θ$ ?
- $cos^{2} θ - sin^{2} θ$
- $1 - 2 sin^{2} θ$
- $2 cos^{2} θ - 1$
- $2 sin^{2} θ - 1$
Reveal answer
3 —
This is incorrect. The correct form when written in terms of sine is $cos 2 θ = 1 - 2 sin^{2} θ$ , not $2 sin^{2} θ - 1$ .

4. Common Pitfalls

Wrong move:

Forgetting to check the sign of a trigonometric value after using a Pythagorean identity.

Why:

Squaring removes the sign of the original value, so you need to use the quadrant of the angle to get the correct result.

Correct move:

Always check the given domain or quadrant before choosing between positive and negative roots.

Wrong move:

Mixing up the sign in the double angle identity for cosine, writing $cos 2 θ = 2 cos^{2} θ + 1$ .

Why:

This is a common sign error when rearranging the identity from the Pythagorean theorem.

Correct move:

Remember the constant term is negative: $cos 2 θ = 2 cos^{2} θ - 1$ .

Wrong move:

Dividing both sides of a trigonometric equation by $sin θ$ or $cos θ$ to simplify.

Why:

This removes the solution where $sin θ = 0$ or $cos θ = 0$ from your final result.

Correct move:

Factor the expression instead of dividing to retain all possible solutions in the domain.

Wrong move:

Writing $(sin θ)^{2}$ as $sin θ^{2}$ .

Why:

This notation error changes the meaning of the expression to sine of (theta squared).

Correct move:

Always write the power after the function name: $sin^{2} θ$ for $(sin θ)^{2}$ .

5. Quick Reference Cheatsheet

Identity Type	Identity
Pythagorean	$sin^{2} θ + cos^{2} θ = 1$
Pythagorean	$1 + tan^{2} θ = sec^{2} θ$
Pythagorean	$1 + cot^{2} θ = csc^{2} θ$
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.

2021 · Paper 1
Simplify a trigonometric expression
2022 · Paper 2
Solve a quadratic trigonometric equation
2023 · Paper 1
Prove a trigonometric identity

Going deeper

Official ResourceIB AA SL Formula BookletAll identities here are listed for reference in the exam

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.