Trigonometric Identities

12 min read

Quick Reference

A trigonometric identity is an equation that holds for every angle in the common domain of both sides. The fundamental families in Section 7.1 are the Pythagorean identities, the reciprocal and quotient identities, and the even-odd identities. They form a connected family that one rebuilds from the unit circle, not a list to memorize.

Try This First

Before any formula, work one small case by hand.

Pick the angle $\theta = \frac{\pi}{6}$ (that is, $30°$). Its point on the unit circle is $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$, so $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$ and $\sin\frac{\pi}{6} = \frac{1}{2}$.

1. Compute $\cos^2\frac{\pi}{6} + \sin^2\frac{\pi}{6}$. Square each coordinate and add.
2. Now try $\theta = \frac{\pi}{4}$ where $\cos\frac{\pi}{4} = \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$. Compute $\cos^2\frac{\pi}{4} + \sin^2\frac{\pi}{4}$.
3. Predict the answer for a third angle of your choice before you compute it.

Why Identities Exist: The Unit Circle

The unit circle is the set of points $(x, y)$ with $x^2 + y^2 = 1$. Every angle $\theta$, measured counterclockwise from the positive $x$-axis, labels exactly one point on this circle. That point is defined to be $(\cos\theta, \sin\theta)$.

Read this as a process: feed in an angle, and trigonometry returns a horizontal coordinate (cosine) and a vertical coordinate (sine). Because the point sits on the circle, those two outputs are tied together for every angle at once. An identity is the written form of one of those ties.

Two pictures generate the whole family.

Picture 1: the point is on the circle. Its coordinates satisfy $x^2 + y^2 = 1$, which is exactly $\cos^2\theta + \sin^2\theta = 1$. The other Pythagorean identities, the reciprocal and quotient relations, and the even-odd symmetry all come from this.

Picture 2: the mirror across the horizontal axis. The point for $-\theta$ is the reflection of the point for $\theta$ across the $x$-axis. The horizontal coordinate stays the same and the vertical coordinate flips sign. That single observation is the even-odd identities.

If you forget an identity mid-problem, you can rebuild it from these two pictures. The identities form a network, not a memorized list.

Prerequisite Hub

Builds on

Unlocks

The scope does not list a downstream node for this skill yet. In the wider course, the identities here support solving trigonometric equations (rewrite an equation in terms of one trig function, then solve) and, later, calculus techniques that depend on the Pythagorean and power-reduction relations.

Prerequisite self-check

Official Definitions

Pythagorean Identities. The fundamental Pythagorean identities are $$\sin^2\theta + \cos^2\theta = 1, \qquad 1 + \cot^2\theta = \csc^2\theta, \qquad 1 + \tan^2\theta = \sec^2\theta.$$

Source: OpenStax Precalculus 2e, Section 7.1 Solving Trigonometric Equations with Identities.

Even-Odd Identities. Cosine and secant are even: $$\cos(-\theta) = \cos\theta, \qquad \sec(-\theta) = \sec\theta.$$ Sine, cosecant, tangent, and cotangent are odd: $$\sin(-\theta) = -\sin\theta, \qquad \tan(-\theta) = -\tan\theta.$$

Source: OpenStax Precalculus 2e, Section 7.1 Solving Trigonometric Equations with Identities.

The reciprocal identities ($\csc\theta = \frac{1}{\sin\theta}$, $\sec\theta = \frac{1}{\cos\theta}$, $\cot\theta = \frac{1}{\tan\theta}$) and the quotient identities ($\tan\theta = \frac{\sin\theta}{\cos\theta}$, $\cot\theta = \frac{\cos\theta}{\sin\theta}$) are the definitions of the four secondary trig functions written as equations. They are part of the toolkit used throughout Section 7.1.

Quick Reference

Pythagorean identities: $$\sin^2\theta + \cos^2\theta = 1$$ $$1 + \tan^2\theta = \sec^2\theta \qquad 1 + \cot^2\theta = \csc^2\theta$$

Reciprocal identities: $$\csc\theta = \frac{1}{\sin\theta} \qquad \sec\theta = \frac{1}{\cos\theta} \qquad \cot\theta = \frac{1}{\tan\theta}$$

Quotient identities: $$\tan\theta = \frac{\sin\theta}{\cos\theta} \qquad \cot\theta = \frac{\cos\theta}{\sin\theta}$$

Even-odd identities: $$\cos(-\theta) = \cos\theta \quad (\text{even}) \qquad \sin(-\theta) = -\sin\theta \quad (\text{odd})$$ $$\sec(-\theta) = \sec\theta \qquad \tan(-\theta) = -\tan\theta$$

Key Concepts

1. The master identity and its two siblings

The point $(\cos\theta, \sin\theta)$ lies on the circle $x^2 + y^2 = 1$, so for every angle $\theta$: $$\cos^2\theta + \sin^2\theta = 1.$$

The other two Pythagorean identities are this same equation divided by a square.

Divide by $\cos^2\theta$ (where $\cos\theta \neq 0$): $$\frac{\cos^2\theta}{\cos^2\theta} + \frac{\sin^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta} \quad\Longrightarrow\quad 1 + \tan^2\theta = \sec^2\theta.$$

Divide by $\sin^2\theta$ (where $\sin\theta \neq 0$): $$\frac{\cos^2\theta}{\sin^2\theta} + \frac{\sin^2\theta}{\sin^2\theta} = \frac{1}{\sin^2\theta} \quad\Longrightarrow\quad \cot^2\theta + 1 = \csc^2\theta.$$

The three Pythagorean identities are not three separate facts. They are the same fact written three ways. This connects two representations of one idea: the geometric circle equation and the three algebraic relations.

2. Reciprocal and quotient identities

These follow directly from the definitions of the four secondary functions. Reading them as products is sometimes useful in a proof.

3. Even and odd symmetry

The point for $-\theta$ is the mirror image of $(\cos\theta, \sin\theta)$ across the $x$-axis. The $x$-coordinate (cosine) is unchanged; the $y$-coordinate (sine) is negated: $$\cos(-\theta) = \cos\theta \qquad \sin(-\theta) = -\sin\theta.$$

The behavior of the other four functions follows by definition. For example: $$\tan(-\theta) = \frac{\sin(-\theta)}{\cos(-\theta)} = \frac{-\sin\theta}{\cos\theta} = -\tan\theta.$$

So tangent is odd. Secant inherits cosine’s even behavior: $\sec(-\theta) = \frac{1}{\cos(-\theta)} = \frac{1}{\cos\theta} = \sec\theta$.

One way to see this is through the mirror picture above. Another way to see it is purely algebraic, by substituting $-\theta$ into the reciprocal and quotient definitions, as just shown for tangent. Which path feels more convincing to you, and why?

4. Verifying an identity

To verify that a proposed equation is an identity, transform one side until it becomes the other side, using only known identities and algebra. Do not move terms across the equal sign, because that would assume the equality you are trying to establish.

A working strategy:

1. Start with the more complicated side.
2. Look for structural cues. A square of a trig function suggests a Pythagorean substitution. A sum of fractions suggests a common denominator. A product suggests expanding or factoring.
3. Drive toward the simpler side. Do not manipulate both sides at once.

Worked Examples

Each example asks for a prediction first, then a computation, then a comparison.

Example 1: simplify using reciprocal and quotient identities

Simplify $\dfrac{\sin\theta}{\cos\theta} + \cos\theta\sec\theta$.

Predict. The first term is a quotient identity and the second is a reciprocal pairing. Before computing, guess: will the result still contain $\theta$, or will part of it collapse to a constant?

Compute. $$\frac{\sin\theta}{\cos\theta} + \cos\theta\sec\theta = \tan\theta + \cos\theta \cdot \frac{1}{\cos\theta} = \tan\theta + 1.$$

Compare. The second term collapsed to the constant $1$ because $\cos\theta$ and $\sec\theta$ are reciprocals, while the first term stayed as $\tan\theta$. The simplified form is $\tan\theta + 1$.

Example 2: find a missing function value with the Pythagorean identity

Given $\cos\theta = -\frac{5}{13}$ with $\theta$ in Quadrant II, find $\sin\theta$.

Predict. In Quadrant II the $y$-coordinate is positive, so $\sin\theta$ should come out positive. Predict the sign before the algebra.

Compute. From $\sin^2\theta + \cos^2\theta = 1$: $$\sin^2\theta = 1 - \left(-\frac{5}{13}\right)^2 = 1 - \frac{25}{169} = \frac{144}{169}.$$ So $\sin\theta = \pm\frac{12}{13}$. Quadrant II forces the positive root.

Compare. The result $\sin\theta = \frac{12}{13}$ is positive, matching the prediction. Choosing the wrong sign is the single most common error here, which is why the quadrant is checked before the root is reported.

Example 3: use even-odd identities to rewrite an expression

Rewrite $\sin(-\theta)\cos(-\theta)$ using only $\sin\theta$ and $\cos\theta$.

Predict. Cosine is even and sine is odd. Predict whether the overall expression keeps its sign or flips.

Compute. $$\sin(-\theta)\cos(-\theta) = \big(-\sin\theta\big)\big(\cos\theta\big) = -\sin\theta\cos\theta.$$

Compare. Exactly one factor (the sine) flipped sign, so the product flipped sign as predicted. The cosine factor was unchanged because cosine is even.

Example 4: verify an identity

Verify $\tan\theta + \cot\theta = \sec\theta\csc\theta$.

Predict. The left side is a sum of two fractions over different denominators. Predict that combining them over a common denominator will produce the numerator $\sin^2\theta + \cos^2\theta$, which is $1$.

Compute. Start with the left side and write each function with sine and cosine: $$\tan\theta + \cot\theta = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}.$$ Now the right side: $$\sec\theta\csc\theta = \frac{1}{\cos\theta}\cdot\frac{1}{\sin\theta} = \frac{1}{\sin\theta\cos\theta}.$$

Compare. Both sides reduce to $\frac{1}{\sin\theta\cos\theta}$, and the Pythagorean identity supplied the $1$ in the numerator exactly as predicted. The identity is verified. Notice that the proof never moved a term across the equal sign.

Example 5: verify an identity by multiplying by a conjugate

Verify $\dfrac{1 - \cos\theta}{\sin\theta} = \dfrac{\sin\theta}{1 + \cos\theta}$.

Predict. The right side has $1 + \cos\theta$ in its denominator while the left side has $1 - \cos\theta$ in its numerator. Predict that multiplying the left side by $\frac{1 + \cos\theta}{1 + \cos\theta}$ will create a difference of squares that the Pythagorean identity simplifies.

Compute. Start with the left side and multiply by a form of $1$: $$\frac{1 - \cos\theta}{\sin\theta}\cdot\frac{1 + \cos\theta}{1 + \cos\theta} = \frac{1 - \cos^2\theta}{\sin\theta(1 + \cos\theta)} = \frac{\sin^2\theta}{\sin\theta(1 + \cos\theta)} = \frac{\sin\theta}{1 + \cos\theta}.$$

Compare. The result is the right side, and the difference of squares $1 - \cos^2\theta$ became $\sin^2\theta$ through the Pythagorean identity, as predicted. The proof transformed one side only.

Common Misconceptions

Common errors summary

Leveled Practice

Level 1: read the identity off the unit circle

Problem 1. Using $\cos\frac{\pi}{3} = \frac{1}{2}$, find $\sin\frac{\pi}{3}$, given that $\frac{\pi}{3}$ is in Quadrant I.

Level 2: apply reciprocal, quotient, and even-odd identities

Problem 2. Given $\cos\theta = -\frac{5}{13}$ with $\theta$ in Quadrant II, find $\sin\theta$, $\tan\theta$, $\csc\theta$, and $\sec\theta$.

Problem 3. Simplify $\dfrac{\sin(-\theta)}{\cos(-\theta)}$ using even-odd identities.

Level 3: verify a one-step identity

Problem 4. Verify $\sec^2\theta - \tan^2\theta = 1$.

Problem 5. Verify $\dfrac{\sin^2\theta}{1 - \cos\theta} = 1 + \cos\theta$ (for $\cos\theta \neq 1$).

Level 4: a multi-step verification

Problem 6. Verify $\dfrac{\cos\theta}{1 - \sin\theta} = \dfrac{1 + \sin\theta}{\cos\theta}$ (for $\sin\theta \neq 1$).

Level 5: build a new identity from the family

Problem 7. Show that $\dfrac{1 - \tan^2\theta}{1 + \tan^2\theta} = \cos^2\theta - \sin^2\theta$.

Mastery Checklist

Foundational (Level 1-2):

• I can state $\sin^2\theta + \cos^2\theta = 1$ and explain why it is the equation of the unit circle
• I can find a missing trig value with the Pythagorean identity and pick the sign from the quadrant
• I can apply the reciprocal and quotient identities to rewrite an expression
• I can use the even-odd identities $\cos(-\theta) = \cos\theta$ and $\sin(-\theta) = -\sin\theta$

Competent (Level 3-4):

• I can derive $1 + \tan^2\theta = \sec^2\theta$ and $1 + \cot^2\theta = \csc^2\theta$ by dividing the master identity
• I can verify an identity by transforming one side into the other
• I can multiply by a conjugate to create a difference of squares in a verification

Proficient (Level 5):

• I can build a less familiar identity from the Pythagorean and reciprocal/quotient family
• I can explain why a verification may not move terms across the equal sign
• I can describe more than one valid route through a verification and say which I prefer

Mental Model

All of these identities live on the unit circle.

The unit circle is the set of points $(x, y)$ with $x^2 + y^2 = 1$. Every angle $\theta$ labels the point you reach after traveling arc length $\theta$ counterclockwise from $(1, 0)$, and that point is $(\cos\theta, \sin\theta)$.

The Pythagorean identity $\cos^2\theta + \sin^2\theta = 1$ is just $x^2 + y^2 = 1$ for that point. It is always true because it is the equation of the circle. Divide it by $\cos^2\theta$ or $\sin^2\theta$ and you get the other two Pythagorean identities. Reflect the point across the horizontal axis and you get the even-odd identities. The reciprocal and quotient identities are the definitions of the four secondary functions written as equations.

When a new identity appears, ask which part of this picture it comes from and which algebraic step takes you there. That question turns a list of memorizations into a network of derived facts.

Connections

Looking back:

• Algebraic simplification supplies the common-denominator and difference-of-squares moves that every verification uses.
• Graphing basics give the unit-circle point and the mirror symmetry that the even-odd identities describe.

Looking ahead:

• Solving trigonometric equations rewrites an equation in terms of a single trig function using these identities, then solves it. For example, $\sin 2\theta = \cos\theta$ becomes $2\sin\theta\cos\theta = \cos\theta$, which factors as $\cos\theta(2\sin\theta - 1) = 0$.
• Graphing trig functions relies on the Pythagorean identity to bound the range of $\sin$ and $\cos$ to $[-1, 1]$, and on even-odd symmetry to explain the shape of the graphs.

Audience notes:

For students who find math intimidating: The single most important identity is $\sin^2\theta + \cos^2\theta = 1$. If you know this one and know how to divide it by $\cos^2\theta$ and $\sin^2\theta$, you have all three Pythagorean identities. There is no need to rush the rest. Practice recognizing when a Pythagorean substitution helps, such as replacing $1 - \sin^2\theta$ with $\cos^2\theta$, because that one move appears in almost every problem.

For students interested in proof: The reflection argument for the even-odd identities is fully rigorous on its own. Reflecting the unit-circle point for $\theta$ across the $x$-axis lands exactly on the point for $-\theta$, which fixes the $x$-coordinate and negates the $y$-coordinate. Writing that out gives $\cos(-\theta) = \cos\theta$ and $\sin(-\theta) = -\sin\theta$ with nothing else assumed.

For students interested in careers: Trigonometric identities are the foundation of Fourier analysis, the technique of writing a signal as a sum of sines and cosines. In signal processing, audio compression, and medical imaging, this decomposition is the underlying mathematical structure, and the identities on this page are the algebra that makes it work.

Resources

The primary source for every definition and identity on this page:

Last updated: 2026-06-16