Equivalent representations of trigonometric functions

AP Precalculus· AP Precalculus CED — Trigonometric and Polar Functions· 14 min read

1. Pythagorean Identities and Simplification★★☆☆☆⏱ 3 min

Pythagorean identities are the most commonly used tools for rewriting trigonometric expressions, derived directly from the unit circle equation. For any angle $θ$ corresponding to a point $(x, y)$ on the unit circle, $x^{2} + y^{2} = 1$ , so substituting $x = cos θ$ and $y = sin θ$ gives the core identity:

cos^2 \theta + \sin^2 \theta = 1

Dividing both sides by $cos^{2} θ$ (for $cos θ \neq = 0$ ) gives the tangent-secant form: $1 + tan^{2} θ = sec^{2} θ$ , and dividing by $sin^{2} θ$ (for $sin θ \neq = 0$ ) gives the cotangent-cosecant form: $1 + cot^{2} θ = csc^{2} θ$ . These identities are used to replace quadratic terms, cancel common factors, and simplify to a single basic trigonometric term.

📘 Definition

Equivalent Trigonometric Representations

Different algebraic expressions that produce identical output values for all inputs in their shared domain. Equivalence requires matching domains, not just matching output where both are defined.

Example:

$\frac{1 - s i n ^{2} θ}{c o s θ}$ and $cos θ$ are only equivalent when $cos θ \neq = 0$

📐 Worked Example

Simplify $\frac{1 - s i n ^{2} θ}{1 - c o s ^{2} θ}$ to an equivalent expression in terms of a single basic trigonometric function, for all $θ$ where the original expression is defined.

1
Apply the core Pythagorean identity to the numerator:
$1 - \sin^2 \theta = \cos^2 \theta$
2
Apply the core Pythagorean identity to the denominator:
$1 - \cos^2 \theta = \sin^2 \theta$
3
Rewrite the fraction and use the definition of cotangent:
$\frac{\cos^2 \theta}{\sin^2 \theta} = \left(\frac{\cos \theta}{\sin \theta}\right)^2 = \cot^2 \theta$
4
The simplified expression $cot^{2} θ$ has the same domain ( $sin θ \neq = 0$ ) as the original, so they are fully equivalent.

Exam tip:

Always confirm that your simplified expression has the same domain as the original. If the original excludes input values allowed in the simplified form, explicitly note the excluded values for full credit on FRQs.

2. Double-Angle Identities★★★☆☆⏱ 3 min

Double-angle identities relate trigonometric functions of $2 θ$ to functions of $θ$ . They are derived from the sum identities for sine and cosine. Setting $A = B = θ$ in the sine sum identity gives:

\sin 2\theta = 2\sin \theta \cos \theta

For cosine, setting $A = B = θ$ gives three equivalent forms:

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

Exam tip:

Know all three forms of the cosine double-angle identity; choosing the form that matches the rest of your expression (all in cosine or all in sine) is often the key to a clean simplification.

3. AP-Style Worked Practice★★★☆☆⏱ 5 min

✓ Quick check

Which of the following expressions is equivalent to $\frac{c o s 2 θ}{c o s θ - s i n θ}$ for all $θ$ where the expression is defined?

- A) $1$
- B) $cos θ + sin θ$
- C) $cos 2 θ$
- D) $\frac{1}{c o s θ + s i n θ}$
Reveal answer
B —
Rewrite $cos 2 θ = cos^{2} θ - sin^{2} θ$ , which factors to $(cos θ - sin θ) (cos θ + sin θ)$ . Cancel the common factor (valid since $cos θ - sin θ \neq = 0$ for the original expression) to get $cos θ + sin θ$ .

4. Common Pitfalls

Wrong move:

After simplifying $\frac{1 - s i n ^{2} θ}{c o s θ} = cos θ$ , claiming the two expressions are equivalent for all real $θ$

Why:

The original expression excludes all $θ$ where $cos θ = 0$ , while $cos θ$ is defined for all $θ$ , so they are not fully equivalent without noting exclusions

Correct move:

Always compare the domain of the original and simplified expression, and explicitly list any excluded input values when stating equivalence

Wrong move:

Forgetting the $cos θ$ term in the double-angle identity for sine, writing $sin 2 θ = 2 sin θ$

Why:

Students remember the factor of 2 but drop the cosine term when simplifying quickly

Correct move:

Always write the full identity $sin 2 θ = 2 sin θ cos θ$ before simplifying, never skip writing the cosine term

5. Quick Reference Cheatsheet

Category	Formula	Notes
Core Pythagorean Identity	$cos^{2} θ + sin^{2} θ = 1$	Holds for all real $θ$ , swap quadratic terms
Tangent-Secant Pythagorean	$1 + tan^{2} θ = sec^{2} θ$	Defined when $cos θ \neq = 0$
Cotangent-Cosecant Pythagorean	$1 + cot^{2} θ = csc^{2} θ$	Defined when $sin θ \neq = 0$
Double-Angle (Sine)	$sin 2 θ = 2 sin θ cos θ$	Holds for all real $θ$
Double-Angle (Cosine)	$cos 2 θ = cos^{2} θ - sin^{2} θ = 2 cos^{2} θ - 1 = 1 - 2 sin^{2} θ$	Three equivalent forms for different uses

What's Next

This topic is the foundational prerequisite for the next key topics in Unit 3: solving trigonometric equations and modeling periodic phenomena, which make up a much larger portion of the AP Precalculus exam. Without the ability to rewrite trigonometric expressions into equivalent simplified forms — using the Pythagorean and double-angle identities — you cannot factor complex trigonometric equations or reduce expressions to a solvable form, both commonly tested on the MCQ and FRQ sections. These manipulation skills also build a strong foundation for later courses.