Fundamental Trigonometric Identities

13 min read

Tell me one trig value, say $\sin t = \tfrac{3}{5}$, and the quadrant of $t$. From that alone I can hand you the other five values. No table, no calculator. The six functions are not six separate numbers to look up. They are six views of the same point on the unit circle, and a few equations lock them together so tightly that one value forces the rest.

By the end of this page you can:

• State the three reciprocal identities, the three Pythagorean identities, and the even-odd identity for each function.
• Given one trig value at $t$ and the quadrant of $t$, find the remaining five values.
• Pick the correct Pythagorean form for the pair in front of you ($\sin/\cos$, $\tan/\sec$, or $\cot/\csc$).
• Resolve the plus-or-minus from a square root by reading the sign in the stated quadrant.

The floor. Memorize $\sin^2 t + \cos^2 t = 1$ and the three reciprocals ($\csc t = 1/\sin t$, $\sec t = 1/\cos t$, $\cot t = 1/\tan t$). With those plus the quadrant to fix each sign, one trig value recovers all six. That recovery is the graded skill.

Quick Reference

Before You Start

Check each box you can do from memory. A box you cannot check yet points to a quick refresher, not a grade.

• I can read the terminal point $(x, y)$ on the unit circle as $\cos t = x$ and $\sin t = y$. (Unit Circle: Sine and Cosine)
• I can name the sign of $\sin t$ and of $\cos t$ in each of the four quadrants. (Signs and Domains of the Trig Functions)
• I can take a square root of a fraction such as $\tfrac{16}{25}$ and simplify $\sqrt{\tfrac{16}{25}} = \tfrac{4}{5}$.

A 60-second self-check. Pick an answer, then reveal the reasoning.

The Core Idea

Start with the move that does the work. Suppose $\sin t = \tfrac{3}{5}$ and $t$ is in Quadrant I, where both coordinates are positive. Find $\cos t$ using only the picture.

The terminal point is $(\cos t, \sin t)$, and it sits on the unit circle, so $x^2 + y^2 = 1$. Here $y = \sin t = \tfrac{3}{5}$, so

\[ x^2 + \left(\tfrac{3}{5}\right)^2 = 1 \quad\Longrightarrow\quad x^2 = 1 - \tfrac{9}{25} = \tfrac{16}{25}. \]

The square root gives $x = \pm\tfrac{4}{5}$. Quadrant I has $x > 0$, so $\cos t = \tfrac{4}{5}$.

That is the whole skill in one example. The circle equation $x^2 + y^2 = 1$, written with $x = \cos t$ and $y = \sin t$, is the Pythagorean identity $\cos^2 t + \sin^2 t = 1$. The $\pm$ from the square root is real, and the quadrant removes it. Every recovery problem here is this move, sometimes one reciprocal further out.

An identity is an equation true for every value where both sides are defined. That sets it apart from an equation you solve, such as $\sin t = \tfrac{1}{2}$, which holds only at special values. The fundamental identities are always true, so they are tools you apply at any $t$. There are three families, each with one job.

The reciprocal identities turn top-row functions into bottom-row ones. Since $\csc t$, $\sec t$, and $\cot t$ are defined as $1/\sin t$, $1/\cos t$, and $\cos t/\sin t$, each bottom-row value is one division away. This family carries no sign ambiguity, because $1$ over a number keeps that number’s sign.

The Pythagorean identities come from $x^2 + y^2 = 1$. Dividing $\sin^2 t + \cos^2 t = 1$ by $\cos^2 t$ gives $\tan^2 t + 1 = \sec^2 t$; dividing by $\sin^2 t$ gives $1 + \cot^2 t = \csc^2 t$. Each form pairs two functions, so using it well means matching the form to the pair you already have. This is where the $\pm$ enters, since isolating a square and taking a root leaves two candidate signs.

The even-odd identities record what happens when $t$ flips to $-t$. Reflecting the terminal point across the horizontal axis keeps $x$ and negates $y$. So cosine and its reciprocal secant are unchanged (even), while sine, cosecant, tangent, and cotangent flip sign (odd). People overgeneralize this family because it does not treat all six functions alike.

Prerequisite Hub

graph LR
    subgraph Builds_On["Builds On"]
        A["Unit Circle:<br/>Sine and Cosine"]
        B["Signs and Domains<br/>of Trig Functions"]
    end

    subgraph ThisSkill["This Skill"]
        C["Fundamental<br/>Identities"]
    end

    subgraph Unlocks
        D["Inverse Trig<br/>Functions"]
    end

    A --> C
    B --> C
    C --> D

    style C fill:#d1fae5,stroke:#a565f0,stroke-width:3px

    click C "fundamental-trig-identities.html"
    click D "../week-2/inverse-trig-functions.html"

Builds on:

Unlocks (what this section feeds into next):

Cross-course prerequisites: none.

Official Definitions

Each statement below is the canonical form to put on a study sheet. The primary source is Stewart 7e, Section 5.2; OpenStax 2e Section 5.3 states the same identities for the reader who uses that text.

Identity

An identity is an equation that is true for all values of the variable for which both sides are defined.

Stewart 7e, Section 5.2 Trigonometric Functions of Real Numbers.

Reciprocal Identities

$\displaystyle \csc t = \frac{1}{\sin t}, \qquad \sec t = \frac{1}{\cos t}, \qquad \cot t = \frac{1}{\tan t}, \qquad \tan t = \frac{\sin t}{\cos t}, \qquad \cot t = \frac{\cos t}{\sin t}.$

Stewart 7e, Section 5.2 (Fundamental Identities). Each holds wherever the denominator is not zero.

Pythagorean Identities

$\displaystyle \sin^2 t + \cos^2 t = 1, \qquad \tan^2 t + 1 = \sec^2 t, \qquad 1 + \cot^2 t = \csc^2 t.$

Stewart 7e, Section 5.2 (Fundamental Identities).

The first form pairs $\sin$ with $\cos$. The second pairs $\tan$ with $\sec$. The third pairs $\cot$ with $\csc$. Matching the form to the pair in front of you is half of using it correctly.

Even-Odd Identities

Sine, cosecant, tangent, and cotangent are odd functions, and cosine and secant are even functions: $\displaystyle \sin(-t) = -\sin t, \quad \csc(-t) = -\csc t, \quad \tan(-t) = -\tan t, \quad \cot(-t) = -\cot t,$ $\displaystyle \cos(-t) = \cos t, \quad \sec(-t) = \sec t.$

Stewart 7e, Section 5.2 (Even-Odd Properties).

The Three Families at a Glance

Read across each row. The left column names what you already have, and the right column names the move that gets you the next value.

A first check on the three pieces most often misused: which Pythagorean form to pick, where the $\pm$ comes from, and which functions are even.

Worked Examples

Worked Example 1: One Value Plus a Quadrant Recovers All Six

Problem. Given $\sin t = \tfrac{3}{5}$ and $t$ in Quadrant II, find the remaining five trig values at $t$.

Predict first. Quadrant II has $x < 0$ and $y > 0$, so $\cos t$ comes out negative while $\sin t$ stays positive. Tangent, a positive over a negative, is negative. Hold those signs and check them at the end.

Step 1. Find $\cos t$ from the first Pythagorean identity.

\[ \sin^2 t + \cos^2 t = 1 \quad\Longrightarrow\quad \cos^2 t = 1 - \left(\tfrac{3}{5}\right)^2 = 1 - \tfrac{9}{25} = \tfrac{16}{25}. \]

So $\cos t = \pm\tfrac{4}{5}$. In Quadrant II cosine is negative, so $\cos t = -\tfrac{4}{5}$.

Step 2. Find $\tan t$ from the quotient identity.

\[ \tan t = \frac{\sin t}{\cos t} = \frac{3/5}{-4/5} = -\frac{3}{4}. \]

Step 3. Take reciprocals for the bottom row.

\[ \csc t = \frac{1}{\sin t} = \frac{5}{3}, \qquad \sec t = \frac{1}{\cos t} = -\frac{5}{4}, \qquad \cot t = \frac{1}{\tan t} = -\frac{4}{3}. \]

Check the prediction. Sine and cosecant are positive, cosine and secant are negative, tangent and cotangent are negative. That matches Quadrant II. A second check: $\sin^2 t + \cos^2 t = \tfrac{9}{25} + \tfrac{16}{25} = 1$, so the pair sits on the unit circle.

Worked Example 2: Choosing the Right Pythagorean Form

Problem. Given $\tan t = -\tfrac{5}{12}$ and $t$ in Quadrant IV, find $\sec t$ and then $\cos t$.

Predict first. Quadrant IV has $x > 0$ and $y < 0$, so $\cos t$ is positive and therefore $\sec t$, its reciprocal, is positive. Hold that sign.

Step 1. The value present is a tangent and the unknown is a secant, so the matching form is $\tan^2 t + 1 = \sec^2 t$, not the $\sin/\cos$ form.

\[ \sec^2 t = \tan^2 t + 1 = \left(-\tfrac{5}{12}\right)^2 + 1 = \tfrac{25}{144} + \tfrac{144}{144} = \tfrac{169}{144}. \]

Step 2. Take the square root.

\[ \sec t = \pm\sqrt{\tfrac{169}{144}} = \pm\tfrac{13}{12}. \]

In Quadrant IV cosine is positive, so secant is positive: $\sec t = \tfrac{13}{12}$.

Step 3. Take the reciprocal for cosine.

\[ \cos t = \frac{1}{\sec t} = \frac{12}{13}. \]

Check the prediction. $\cos t = \tfrac{12}{13} > 0$, as predicted for Quadrant IV. A cross-check with the other form: from $\tan t = -\tfrac{5}{12}$ and $\cos t = \tfrac{12}{13}$, the value $\sin t = \tan t \cdot \cos t = -\tfrac{5}{12}\cdot\tfrac{12}{13} = -\tfrac{5}{13}$, and $\sin^2 t + \cos^2 t = \tfrac{25}{169} + \tfrac{144}{169} = 1$. The point is on the unit circle, and $\sin t < 0$ fits Quadrant IV.

Worked Example 3: Using the Even-Odd Identities

Problem. Evaluate $\sin\!\left(-\tfrac{\pi}{6}\right)$, $\cos\!\left(-\tfrac{\pi}{6}\right)$, and $\tan\!\left(-\tfrac{\pi}{6}\right)$ using the even-odd identities and the known values at $\tfrac{\pi}{6}$.

Predict first. Sine and tangent are odd, so the answers at $-\tfrac{\pi}{6}$ are the negatives of the values at $\tfrac{\pi}{6}$. Cosine is even, so its value is unchanged. Hold that.

Step 1. Recall the values at $\tfrac{\pi}{6}$: $\sin\tfrac{\pi}{6} = \tfrac{1}{2}$, $\cos\tfrac{\pi}{6} = \tfrac{\sqrt{3}}{2}$, $\tan\tfrac{\pi}{6} = \tfrac{1/2}{\sqrt{3}/2} = \tfrac{1}{\sqrt{3}}$.

Step 2. Apply the even-odd identities.

\[ \sin\!\left(-\tfrac{\pi}{6}\right) = -\sin\tfrac{\pi}{6} = -\tfrac{1}{2}, \qquad \cos\!\left(-\tfrac{\pi}{6}\right) = \cos\tfrac{\pi}{6} = \tfrac{\sqrt{3}}{2}, \] \[ \tan\!\left(-\tfrac{\pi}{6}\right) = -\tan\tfrac{\pi}{6} = -\tfrac{1}{\sqrt{3}}. \]

Check the prediction. Sine and tangent flipped sign, cosine did not. That matches. A geometric check: the terminal point of $-\tfrac{\pi}{6}$ is the reflection of the terminal point of $\tfrac{\pi}{6}$ across the horizontal axis, which keeps $x = \cos$ the same and negates $y = \sin$, exactly the signs computed.

Worked Example 4: A Recovery That Starts From a Reciprocal Value

Problem. Given $\sec t = -\tfrac{5}{4}$ and $\sin t > 0$, find $\sin t$ and $\tan t$.

Predict first. A negative secant means a negative cosine ($x < 0$), and $\sin t > 0$ means $y > 0$. Both together put $t$ in Quadrant II, where tangent is negative. Hold that.

Step 1. Take the reciprocal to get cosine.

\[ \cos t = \frac{1}{\sec t} = -\frac{4}{5}. \]

Step 2. Use the Pythagorean identity for sine.

\[ \sin^2 t = 1 - \cos^2 t = 1 - \tfrac{16}{25} = \tfrac{9}{25} \quad\Longrightarrow\quad \sin t = \pm\tfrac{3}{5}. \]

The condition $\sin t > 0$ selects the positive root, so $\sin t = \tfrac{3}{5}$.

Step 3. Quotient identity for tangent.

\[ \tan t = \frac{\sin t}{\cos t} = \frac{3/5}{-4/5} = -\frac{3}{4}. \]

Check the prediction. $\sin t > 0$, $\cos t < 0$, $\tan t < 0$: the Quadrant II signature, as predicted. These are the same numbers as Worked Example 1, reached from a different starting clue, which shows that any one value plus a sign condition lands on the same point.

Common Misconceptions

Practice Problems

Check Yourself

Close the notes and answer each from memory, then reveal it. Pulling an idea back from memory is one of the strongest ways to make it stick.

Go Deeper (optional)

None of this is needed to recover the six values or to pass the section. It is here for readers who want to see where the identities lead.

The Pythagorean identity is the Pythagorean theorem on a circle of radius one. The right triangle with legs $|x|$ and $|y|$ and hypotenuse $1$ gives $x^2 + y^2 = 1$, and writing $x = \cos t$, $y = \sin t$ turns the theorem into $\cos^2 t + \sin^2 t = 1$. This is why a single trigonometric equation encodes a geometric fact, and why the recovery move feels like solving a right triangle: it is one.

Why the recovery matters in applied work. Surveying, navigation, signal processing, and physics constantly read one component of a rotation or wave and need the rest. A sextant fixes a position by measuring one angle to the horizon, and the remaining geometry follows from identities of exactly this kind. In computer graphics, a renderer often stores a cosine and reconstructs the sine through $\sin t = \pm\sqrt{1 - \cos^2 t}$ to rotate a point, choosing the sign from known orientation.

A note on the quotient identities and where the signs go. Because $\tan t = \sin t / \cos t$, the sign of tangent is the product of the signs of sine and cosine. That is why tangent and cotangent are positive in Quadrants I and III (where sine and cosine share a sign) and negative in Quadrants II and IV. Reading the signs this way, rather than memorizing a chart, is a check you can run at the end of every recovery problem.

Mastery Checklist

Novice (Level 1-2):

• I can state the three reciprocal identities and write each bottom-row value as one over its partner.
• I can write $\sin^2 t + \cos^2 t = 1$ from memory and isolate either square.
• I can take a square root, write the $\pm$, and use a stated quadrant to pick the sign.

Competent (Level 3-4):

• I can recover all six trig values from one value plus the quadrant.
• I can choose the matching Pythagorean form for a $\tan/\sec$ or $\cot/\csc$ pair, not only $\sin/\cos$.
• I can deduce the quadrant from two sign conditions (for example, $\cos t < 0$ and $\sin t > 0$).

Proficient (Level 5):

• I can apply the even-odd identities to rewrite each of the six functions at $-t$.
• I can simplify a compound expression by combining even-odd, reciprocal, and quotient identities.
• I can verify a recovered point by checking $\sin^2 t + \cos^2 t = 1$ and the quadrant signs.

Connections

Looking back:

• Unit Circle: Sine and Cosine defines $\cos t = x$ and $\sin t = y$, the coordinates the identities relate.
• Signs and Domains of the Trig Functions gives the per-quadrant signs that resolve the plus-or-minus.

Looking ahead:

• Inverse Trig Functions: reading a missing side off a reference triangle is the same recovery move applied to an inverse-function output.
• Verifying general trigonometric identities (later in the course) builds proofs on top of these fundamental relations rather than recovering single values.

Real-world connections:

• Navigation and surveying read one angle or one component and reconstruct the rest by relations of this kind.
• Computer graphics stores a cosine and rebuilds the sine through $\sin t = \pm\sqrt{1 - \cos^2 t}$, choosing the sign from known orientation.

Resources

Last updated: 2026-06-24