Trigonometric Functions

Define sine, cosine, and their companions

You have probably heard of sine and cosine before, perhaps in a geometry class or while playing with a graphing calculator. Maybe someone once told you these functions have something to do with triangles, or you have seen them pop up in physics problems involving waves or circles. But what actually are these functions, and why do mathematicians treat them like old friends?

Here is the good news: if you have read the previous chapter on the unit circle, you already know what sine and cosine are. The x-coordinate of a point on the unit circle is cosine, and the y-coordinate is sine. That is it. In this chapter, we are going to build on that foundation by introducing four more trigonometric functions, exploring their properties, and discovering the elegant relationships that connect them all.

By the end, you will see that these six functions are not arbitrary inventions; they are natural ways to describe the geometry of circles and triangles, and they unlock a powerful toolkit for solving problems throughout mathematics, physics, and engineering.

Core Concepts

The Six Trigonometric Functions

In the previous chapter, we defined sine and cosine using the unit circle: for any angle $\theta$, if you draw the angle in standard position and find the point where the terminal side intersects the unit circle, that point has coordinates $(\cos\theta, \sin\theta)$.

From these two fundamental functions, we can define four more. Here is the complete family:

The Primary Pair:

  • Sine: $\sin\theta = y$ (the y-coordinate on the unit circle)
  • Cosine: $\cos\theta = x$ (the x-coordinate on the unit circle)

The Tangent Family:

  • Tangent: $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{y}{x}$
  • Cotangent: $\cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{x}{y}$

The Reciprocal Functions:

  • Secant: $\sec\theta = \frac{1}{\cos\theta} = \frac{1}{x}$
  • Cosecant: $\csc\theta = \frac{1}{\sin\theta} = \frac{1}{y}$

Why do we need six functions when two seem to capture everything? The answer is convenience. Certain problems become much simpler when expressed using tangent or secant rather than sine and cosine. Over centuries of mathematical practice, these six functions emerged as the most useful combination for solving a wide variety of problems.

Understanding Tangent

The tangent function deserves special attention because it appears so frequently in applications. Since $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{y}{x}$, tangent represents the slope of the line from the origin to the point on the unit circle.

Think about what this means: as you rotate around the unit circle, the tangent tells you how steep that line is. At $\theta = 0$, the point is at $(1, 0)$, so the “slope” is $\frac{0}{1} = 0$. At $\theta = 45°$ (or $\frac{\pi}{4}$), the point is at $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$, and the slope is $\frac{\sqrt{2}/2}{\sqrt{2}/2} = 1$.

What happens at $\theta = 90°$? The point is at $(0, 1)$, and we would need to compute $\frac{1}{0}$, which is undefined. This makes geometric sense: a vertical line has undefined slope. So $\tan 90°$ does not exist; we say it is undefined.

Exact Values for Special Angles

Certain angles appear so often in mathematics that knowing their exact trigonometric values is essential. Here is the complete table for the first quadrant:

Angle (degrees) Angle (radians) $\sin\theta$ $\cos\theta$ $\tan\theta$
0 0 1 0
30° $\frac{\pi}{6}$ $\frac{1}{2}$ $\frac{\sqrt{3}}{2}$ $\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$
45° $\frac{\pi}{4}$ $\frac{\sqrt{2}}{2}$ $\frac{\sqrt{2}}{2}$ 1
60° $\frac{\pi}{3}$ $\frac{\sqrt{3}}{2}$ $\frac{1}{2}$ $\sqrt{3}$
90° $\frac{\pi}{2}$ 1 0 undefined

Notice the beautiful pattern: the sine values at 0°, 30°, 45°, 60°, and 90° are $\frac{\sqrt{0}}{2}$, $\frac{\sqrt{1}}{2}$, $\frac{\sqrt{2}}{2}$, $\frac{\sqrt{3}}{2}$, and $\frac{\sqrt{4}}{2}$. The cosine values follow the same pattern but in reverse order.

For the reciprocal functions at these angles:

Angle $\csc\theta$ $\sec\theta$ $\cot\theta$
undefined 1 undefined
30° 2 $\frac{2\sqrt{3}}{3}$ $\sqrt{3}$
45° $\sqrt{2}$ $\sqrt{2}$ 1
60° $\frac{2\sqrt{3}}{3}$ 2 $\frac{\sqrt{3}}{3}$
90° 1 undefined 0

Signs in Each Quadrant (ASTC)

As you move around the unit circle into different quadrants, the signs of your trigonometric functions change. We saw in the previous chapter that sine and cosine take their signs from the y and x coordinates, respectively. The other four functions inherit their signs from these.

Here is the complete picture:

Quadrant Sine Cosine Tangent Cosecant Secant Cotangent
I (0° to 90°) + + + + + +
II (90° to 180°) + - - + - -
III (180° to 270°) - - + - - +
IV (270° to 360°) - + - - + -

The classic mnemonic is “All Students Take Calculus” (ASTC), which tells you what is positive in each quadrant:

  • All functions positive in Quadrant I
  • Sine (and its reciprocal, cosecant) positive in Quadrant II
  • Tangent (and its reciprocal, cotangent) positive in Quadrant III
  • Cosine (and its reciprocal, secant) positive in Quadrant IV

The Pythagorean Identities

From the definition of the unit circle, every point $(x, y) = (\cos\theta, \sin\theta)$ satisfies $x^2 + y^2 = 1$. This gives us the most fundamental trigonometric identity:

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

This is called the Pythagorean identity because it comes directly from the Pythagorean theorem applied to the unit circle.

By dividing this identity by different quantities, we get two more Pythagorean identities:

Dividing by $\cos^2\theta$: $$\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\cos^2\theta} = \frac{1}{\cos^2\theta}$$ $$\tan^2\theta + 1 = \sec^2\theta$$

Dividing by $\sin^2\theta$: $$\frac{\sin^2\theta}{\sin^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \frac{1}{\sin^2\theta}$$ $$1 + \cot^2\theta = \csc^2\theta$$

These three Pythagorean identities are workhorses in trigonometry. Memorize the first one; the other two follow naturally.

Reciprocal and Quotient Identities

The relationships between the six trigonometric functions can be summarized in two categories:

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

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

These are not really new information; they are the definitions we started with. But organizing them this way helps you see how the six functions form a coherent family.

Even and Odd Properties

Functions can be classified as even, odd, or neither, based on their symmetry. This classification turns out to be very useful for simplifying expressions.

A function $f$ is even if $f(-x) = f(x)$ (symmetric about the y-axis). A function $f$ is odd if $f(-x) = -f(x)$ (symmetric about the origin).

For the trigonometric functions:

Even functions: $\cos(-\theta) = \cos\theta$ and $\sec(-\theta) = \sec\theta$

Odd functions: $\sin(-\theta) = -\sin\theta$, $\tan(-\theta) = -\tan\theta$, $\csc(-\theta) = -\csc\theta$, $\cot(-\theta) = -\cot\theta$

Why does this make sense geometrically? On the unit circle, if you go to angle $\theta$, you reach the point $(\cos\theta, \sin\theta)$. If you go to angle $-\theta$ (the same angle but clockwise), you reach the point $(\cos\theta, -\sin\theta)$. The x-coordinate stays the same (cosine is even), but the y-coordinate flips sign (sine is odd).

Evaluating Trig Functions for Any Angle

To find the exact value of a trigonometric function at any angle, follow these steps:

  1. Find a coterminal angle between 0° and 360° (or between 0 and $2\pi$) by adding or subtracting multiples of 360° (or $2\pi$).

  2. Identify the quadrant where the terminal side lands.

  3. Find the reference angle (the acute angle to the x-axis).

  4. Look up the trig values for the reference angle (one of the special angles 0°, 30°, 45°, 60°, or 90°).

  5. Adjust the signs based on the quadrant (using ASTC).

This method works because the trigonometric functions are periodic and the reference angle captures the “core” angle that determines the magnitude of each function value.

Notation and Terminology

Term Symbol Meaning
Sine $\sin\theta$ The y-coordinate on the unit circle at angle $\theta$
Cosine $\cos\theta$ The x-coordinate on the unit circle at angle $\theta$
Tangent $\tan\theta$ $\frac{\sin\theta}{\cos\theta}$; the slope of the terminal side
Cosecant $\csc\theta$ $\frac{1}{\sin\theta}$; the reciprocal of sine
Secant $\sec\theta$ $\frac{1}{\cos\theta}$; the reciprocal of cosine
Cotangent $\cot\theta$ $\frac{\cos\theta}{\sin\theta}$; the reciprocal of tangent
Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$ Fundamental relationship from the unit circle
Even function $f(-x) = f(x)$ Symmetric about the y-axis
Odd function $f(-x) = -f(x)$ Symmetric about the origin
ASTC “All Students Take Calculus” Mnemonic for signs in quadrants

Examples

Example 1: Evaluating Trig Functions at Special Angles

Find all six trigonometric function values for $\theta = 30°$.

Step 1: From the special angles table, we know: $$\sin 30° = \frac{1}{2} \qquad \cos 30° = \frac{\sqrt{3}}{2}$$

Step 2: Calculate tangent: $$\tan 30° = \frac{\sin 30°}{\cos 30°} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$$

Step 3: Calculate the reciprocals: $$\csc 30° = \frac{1}{\sin 30°} = \frac{1}{1/2} = 2$$ $$\sec 30° = \frac{1}{\cos 30°} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$$ $$\cot 30° = \frac{1}{\tan 30°} = \sqrt{3}$$

Answer: $\sin 30° = \frac{1}{2}$, $\cos 30° = \frac{\sqrt{3}}{2}$, $\tan 30° = \frac{\sqrt{3}}{3}$, $\csc 30° = 2$, $\sec 30° = \frac{2\sqrt{3}}{3}$, $\cot 30° = \sqrt{3}$

Example 2: Using Even/Odd Properties

Given that $\sin 40° \approx 0.643$ and $\cos 40° \approx 0.766$, find $\sin(-40°)$ and $\cos(-40°)$.

Solution: Since sine is an odd function: $$\sin(-40°) = -\sin 40° \approx -0.643$$

Since cosine is an even function: $$\cos(-40°) = \cos 40° \approx 0.766$$

Answer: $\sin(-40°) \approx -0.643$ and $\cos(-40°) \approx 0.766$

Example 3: Evaluating Trig Functions in Other Quadrants

Find the exact value of $\tan 150°$.

Step 1: Identify the quadrant. Since $90° < 150° < 180°$, this is in Quadrant II.

Step 2: Find the reference angle: $$\theta_{\text{ref}} = 180° - 150° = 30°$$

Step 3: Find $\tan 30°$: $$\tan 30° = \frac{\sqrt{3}}{3}$$

Step 4: Determine the sign. In Quadrant II, tangent is negative (using ASTC, only Sine is positive in Quadrant II).

Answer: $\tan 150° = -\frac{\sqrt{3}}{3}$

Example 4: Using the Pythagorean Identity

If $\sin\theta = \frac{3}{5}$ and $\theta$ is in Quadrant II, find $\cos\theta$ and $\tan\theta$.

Step 1: Use the Pythagorean identity to find $\cos\theta$: $$\sin^2\theta + \cos^2\theta = 1$$ $$\left(\frac{3}{5}\right)^2 + \cos^2\theta = 1$$ $$\frac{9}{25} + \cos^2\theta = 1$$ $$\cos^2\theta = 1 - \frac{9}{25} = \frac{16}{25}$$ $$\cos\theta = \pm\frac{4}{5}$$

Step 2: Determine the sign. In Quadrant II, cosine is negative: $$\cos\theta = -\frac{4}{5}$$

Step 3: Find tangent using the quotient identity: $$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3/5}{-4/5} = -\frac{3}{4}$$

Answer: $\cos\theta = -\frac{4}{5}$ and $\tan\theta = -\frac{3}{4}$

Example 5: Finding All Six Functions from One Value

If $\tan\theta = -2$ and $\cos\theta > 0$, find all six trigonometric function values.

Step 1: Determine the quadrant. Tangent is negative and cosine is positive, so $\theta$ is in Quadrant IV (where x is positive but y is negative).

Step 2: Use the identity $\tan^2\theta + 1 = \sec^2\theta$: $$(-2)^2 + 1 = \sec^2\theta$$ $$4 + 1 = \sec^2\theta$$ $$\sec^2\theta = 5$$ $$\sec\theta = \pm\sqrt{5}$$

Since $\theta$ is in Quadrant IV and secant has the same sign as cosine (positive): $$\sec\theta = \sqrt{5}$$

Step 3: Find cosine: $$\cos\theta = \frac{1}{\sec\theta} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5}$$

Step 4: Find sine using $\tan\theta = \frac{\sin\theta}{\cos\theta}$: $$\sin\theta = \tan\theta \cdot \cos\theta = (-2) \cdot \frac{\sqrt{5}}{5} = -\frac{2\sqrt{5}}{5}$$

Step 5: Find the remaining functions: $$\csc\theta = \frac{1}{\sin\theta} = \frac{1}{-2\sqrt{5}/5} = -\frac{5}{2\sqrt{5}} = -\frac{\sqrt{5}}{2}$$ $$\cot\theta = \frac{1}{\tan\theta} = -\frac{1}{2}$$

Answer: $\sin\theta = -\frac{2\sqrt{5}}{5}$, $\cos\theta = \frac{\sqrt{5}}{5}$, $\tan\theta = -2$, $\csc\theta = -\frac{\sqrt{5}}{2}$, $\sec\theta = \sqrt{5}$, $\cot\theta = -\frac{1}{2}$

Example 6: Evaluating Trig Functions at Non-Standard Angles

Find the exact value of $\sec\left(\frac{7\pi}{6}\right)$.

Step 1: Convert to understand the angle better. $\frac{7\pi}{6} = \frac{7 \cdot 180°}{6} = 210°$.

Step 2: Identify the quadrant. Since $180° < 210° < 270°$, this is in Quadrant III.

Step 3: Find the reference angle: $$\theta_{\text{ref}} = 210° - 180° = 30° = \frac{\pi}{6}$$

Step 4: Find $\sec\frac{\pi}{6}$. First, $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$, so: $$\sec\frac{\pi}{6} = \frac{1}{\cos(\pi/6)} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$$

Step 5: Determine the sign. In Quadrant III, cosine is negative, so secant is also negative.

Answer: $\sec\left(\frac{7\pi}{6}\right) = -\frac{2\sqrt{3}}{3}$

Example 7: Simplifying Using Identities

Simplify the expression: $\frac{\sin^2\theta}{\cos\theta} + \cos\theta$

Step 1: Get a common denominator: $$\frac{\sin^2\theta}{\cos\theta} + \frac{\cos^2\theta}{\cos\theta} = \frac{\sin^2\theta + \cos^2\theta}{\cos\theta}$$

Step 2: Apply the Pythagorean identity to the numerator: $$\frac{1}{\cos\theta}$$

Step 3: Recognize this as a reciprocal function: $$\sec\theta$$

Answer: $\frac{\sin^2\theta}{\cos\theta} + \cos\theta = \sec\theta$

Example 8: Verifying an Identity

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

Step 1: Convert the left side to sines and cosines: $$\tan\theta + \cot\theta = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta}$$

Step 2: Get a common denominator: $$= \frac{\sin^2\theta + \cos^2\theta}{\cos\theta \cdot \sin\theta}$$

Step 3: Apply the Pythagorean identity: $$= \frac{1}{\cos\theta \cdot \sin\theta}$$

Step 4: Split into two reciprocals: $$= \frac{1}{\cos\theta} \cdot \frac{1}{\sin\theta} = \sec\theta \cdot \csc\theta$$

Conclusion: The left side equals the right side, so the identity is verified.

Key Properties and Rules

The Three Pythagorean Identities

$$\sin^2\theta + \cos^2\theta = 1$$ $$\tan^2\theta + 1 = \sec^2\theta$$ $$1 + \cot^2\theta = \csc^2\theta$$

Reciprocal Relationships

$$\sin\theta \cdot \csc\theta = 1$$ $$\cos\theta \cdot \sec\theta = 1$$ $$\tan\theta \cdot \cot\theta = 1$$

Quotient Relationships

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

Even/Odd Properties

Even Functions Odd Functions
$\cos(-\theta) = \cos\theta$ $\sin(-\theta) = -\sin\theta$
$\sec(-\theta) = \sec\theta$ $\tan(-\theta) = -\tan\theta$
$\csc(-\theta) = -\csc\theta$
$\cot(-\theta) = -\cot\theta$

Domain Restrictions

Not all trigonometric functions are defined for all angles:

Function Undefined When Because
$\tan\theta$ $\theta = 90° + 180°n$ (i.e., $\frac{\pi}{2} + n\pi$) $\cos\theta = 0$
$\cot\theta$ $\theta = 180°n$ (i.e., $n\pi$) $\sin\theta = 0$
$\sec\theta$ $\theta = 90° + 180°n$ (i.e., $\frac{\pi}{2} + n\pi$) $\cos\theta = 0$
$\csc\theta$ $\theta = 180°n$ (i.e., $n\pi$) $\sin\theta = 0$

Complete Special Angles Reference Table

Angle $\sin$ $\cos$ $\tan$ $\csc$ $\sec$ $\cot$
0° (0) 0 1 0 undef 1 undef
30° ($\frac{\pi}{6}$) $\frac{1}{2}$ $\frac{\sqrt{3}}{2}$ $\frac{\sqrt{3}}{3}$ 2 $\frac{2\sqrt{3}}{3}$ $\sqrt{3}$
45° ($\frac{\pi}{4}$) $\frac{\sqrt{2}}{2}$ $\frac{\sqrt{2}}{2}$ 1 $\sqrt{2}$ $\sqrt{2}$ 1
60° ($\frac{\pi}{3}$) $\frac{\sqrt{3}}{2}$ $\frac{1}{2}$ $\sqrt{3}$ $\frac{2\sqrt{3}}{3}$ 2 $\frac{\sqrt{3}}{3}$
90° ($\frac{\pi}{2}$) 1 0 undef 1 undef 0

Real-World Applications

Surveyors use trigonometric functions to calculate distances and heights that cannot be measured directly. By measuring angles and using tangent, they can determine the height of a building or the distance across a river without actually crossing it. GPS systems rely heavily on trigonometric calculations to determine your position from satellite signals.

Physics: Waves and Oscillations

The motion of a pendulum, the vibration of a guitar string, and the oscillation of electrical current in a circuit are all described using sine and cosine functions. When you see a sound wave displayed on a screen, that wave shape is the graph of a trigonometric function. The “frequency” of a note is directly related to how rapidly the sine function oscillates.

Engineering: Force Analysis

Engineers use trigonometric functions to break forces into components. When a cable supports a bridge at an angle, the tension in the cable can be decomposed into horizontal and vertical components using sine and cosine. This allows engineers to ensure structures can handle the loads they will experience.

Computer Graphics and Game Development

Every rotation in a video game or 3D animation involves trigonometric functions. When a character turns or a camera pans, the computer calculates new positions using sine and cosine. The smooth circular motion of objects, from planets in a space simulator to spinning wheels in a racing game, relies on the unit circle and its associated functions.

Medicine: Imaging and Diagnostics

CT scans and MRI machines use mathematical techniques that rely on trigonometric functions to reconstruct images of the inside of your body. The Fourier transform, which is built on sine and cosine functions, allows doctors to see detailed internal structures from raw scan data.

Music and Audio Processing

Digital audio relies on representing sound waves as combinations of sine waves (a technique called Fourier analysis). When you use an equalizer to boost the bass or reduce treble, you are adjusting the amplitudes of different sine-wave components. Noise-canceling headphones use trigonometric principles to generate waves that cancel out unwanted sound.

Self-Test Problems

Problem 1: Find the exact value of $\sin 60°$, $\cos 60°$, and $\tan 60°$.

Show Answer

From the special angles table:

  • $\sin 60° = \frac{\sqrt{3}}{2}$
  • $\cos 60° = \frac{1}{2}$
  • $\tan 60° = \frac{\sin 60°}{\cos 60°} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$

Problem 2: If $\cos\theta = -\frac{5}{13}$ and $\theta$ is in Quadrant III, find $\sin\theta$.

Show Answer

Using the Pythagorean identity: $$\sin^2\theta + \cos^2\theta = 1$$ $$\sin^2\theta + \left(-\frac{5}{13}\right)^2 = 1$$ $$\sin^2\theta + \frac{25}{169} = 1$$ $$\sin^2\theta = \frac{144}{169}$$ $$\sin\theta = \pm\frac{12}{13}$$

Since $\theta$ is in Quadrant III, sine is negative: $$\sin\theta = -\frac{12}{13}$$

Problem 3: Find the exact value of $\csc\left(\frac{5\pi}{4}\right)$.

Show Answer

$\frac{5\pi}{4}$ is in Quadrant III (between $\pi$ and $\frac{3\pi}{2}$).

Reference angle: $\frac{5\pi}{4} - \pi = \frac{\pi}{4}$

$\sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$, so $\csc\frac{\pi}{4} = \sqrt{2}$

In Quadrant III, sine is negative, so cosecant is also negative: $$\csc\left(\frac{5\pi}{4}\right) = -\sqrt{2}$$

Problem 4: Simplify: $\sec^2\theta - \tan^2\theta$

Show Answer

Using the Pythagorean identity $\tan^2\theta + 1 = \sec^2\theta$, we can rearrange: $$\sec^2\theta - \tan^2\theta = 1$$

This is always equal to 1 (for all $\theta$ where these functions are defined).

Problem 5: Given that $\tan 70° \approx 2.747$, find $\tan(-70°)$ and $\cot 70°$.

Show Answer

Since tangent is an odd function: $$\tan(-70°) = -\tan 70° \approx -2.747$$

Since cotangent is the reciprocal of tangent: $$\cot 70° = \frac{1}{\tan 70°} \approx \frac{1}{2.747} \approx 0.364$$

Problem 6: In which quadrant(s) is $\sin\theta < 0$ and $\tan\theta > 0$?

Show Answer

$\sin\theta < 0$ in Quadrants III and IV (where y is negative).

$\tan\theta > 0$ in Quadrants I and III (where sine and cosine have the same sign).

The intersection is Quadrant III.

Problem 7: Find all six trigonometric function values for $\theta = 315°$.

Show Answer

$315°$ is in Quadrant IV. Reference angle: $360° - 315° = 45°$.

At $45°$: $\sin 45° = \cos 45° = \frac{\sqrt{2}}{2}$

In Quadrant IV, cosine is positive and sine is negative:

  • $\sin 315° = -\frac{\sqrt{2}}{2}$
  • $\cos 315° = \frac{\sqrt{2}}{2}$
  • $\tan 315° = \frac{-\sqrt{2}/2}{\sqrt{2}/2} = -1$
  • $\csc 315° = -\sqrt{2}$
  • $\sec 315° = \sqrt{2}$
  • $\cot 315° = -1$

Summary

  • The six trigonometric functions are sine, cosine, tangent, cosecant, secant, and cotangent. Sine and cosine are defined as coordinates on the unit circle; the others are built from these two.

  • Tangent equals $\frac{\sin\theta}{\cos\theta}$ and represents the slope of the terminal side. Cotangent is its reciprocal.

  • Secant and cosecant are the reciprocals of cosine and sine, respectively.

  • Special angles (0°, 30°, 45°, 60°, 90°) have exact trigonometric values that should be memorized. The values involve 0, 1, $\frac{1}{2}$, $\frac{\sqrt{2}}{2}$, and $\frac{\sqrt{3}}{2}$.

  • ASTC (“All Students Take Calculus”) tells you which functions are positive in each quadrant: All in I, Sine in II, Tangent in III, Cosine in IV.

  • The Pythagorean identities ($\sin^2\theta + \cos^2\theta = 1$ and its variations) are essential for simplifying expressions and solving equations.

  • Cosine and secant are even functions (symmetric about the y-axis); the other four are odd functions (symmetric about the origin).

  • To evaluate trig functions at any angle: (1) find a coterminal angle, (2) identify the quadrant, (3) find the reference angle, (4) use special angle values, (5) adjust signs.

  • Trigonometric functions appear throughout science and engineering, from analyzing waves and forces to powering computer graphics and medical imaging.