Trigonometric Functions and the Unit Circle: Key Concepts and Identities

Trigonometric Functions and the Unit Circle

Evaluating Trigonometric Functions at Special Angles

Trigonometric functions can be evaluated at special angles using the unit circle and known values. Understanding these values is essential for solving trigonometric equations and analyzing periodic phenomena.

• Key Point 1: The unit circle allows us to evaluate trigonometric functions at common angles such as $\frac{\pi}{6}$, $\frac{\pi}{3}$, and $\frac{\pi}{2}$.

• Key Point 2: Trigonometric values for these angles are often memorized or derived from special right triangles (30-60-90 and 45-45-90 triangles).

• Key Point 3: Reciprocal and quotient identities help relate different trigonometric functions.

• Example: $\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}$, $\cot\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{3}$, $\csc\left(\frac{5\pi}{3}\right) = -\frac{2}{\sqrt{3}}$

The Unit Circle and Coordinates

The unit circle is a circle of radius 1 centered at the origin. Each point $(x, y)$ on the unit circle corresponds to $x = \cos \theta$ and $y = \sin \theta$ for some angle $\theta$.

• Key Point 1: The equation of the unit circle is $x^2 + y^2 = 1$.

• Key Point 2: To determine if a point lies on the unit circle, substitute its coordinates into the equation and check if the equality holds.

• Example: $(\frac{3}{5}, \frac{4}{5})$ lies on the unit circle because $\left(\frac{3}{5}\right)^2 + \left(\frac{4}{5}\right)^2 = \frac{9}{25} + \frac{16}{25} = 1$.

• Example: $(2, 1)$ does not lie on the unit circle because $2^2 + 1^2 = 5 \neq 1$.

Finding Trigonometric Values from a Point

Given a point $(x, y)$ on the terminal side of an angle $\theta$, trigonometric functions can be found using the definitions and the Pythagorean Theorem.

• Key Point 1: $\sin \theta = \frac{y}{r}$, $\cos \theta = \frac{x}{r}$, $\tan \theta = \frac{y}{x}$, where $r = \sqrt{x^2 + y^2}$.

• Key Point 2: The Pythagorean Theorem is used to find $r$ if only $x$ and $y$ are given.

• Example: For $(2, 5)$, $r = \sqrt{2^2 + 5^2} = \sqrt{29}$, so $\sin \theta = \frac{5}{\sqrt{29}}$, $\cos \theta = \frac{2}{\sqrt{29}}$.

Solving for Trigonometric Functions Given One Value

If one trigonometric value is known, others can be found using identities and the Pythagorean Theorem.

• Key Point 1: Use $\sin^2 \theta + \cos^2 \theta = 1$ to find missing values.

• Key Point 2: The sign of the function depends on the quadrant in which the angle lies.

• Example: If $\sin \theta = \frac{3}{5}$ and $\theta$ is in Quadrant II, $\cos \theta = -\frac{4}{5}$ (since cosine is negative in Quadrant II).

Trigonometric Function Properties: Even and Odd Functions

Trigonometric functions have symmetry properties that are useful for simplifying expressions and solving equations.

• Key Point 1: Even functions: $f(-x) = f(x)$. Cosine and secant are even functions.

• Key Point 2: Odd functions: $f(-x) = -f(x)$. Sine, tangent, cosecant, and cotangent are odd functions.

• Example Table:

Function	Even/Odd
$\sin \theta$	Odd
$\cos \theta$	Even
$\tan \theta$	Odd
$\csc \theta$	Odd
$\sec \theta$	Even
$\cot \theta$	Odd

Pythagorean Identities

Pythagorean identities are fundamental relationships among trigonometric functions, derived from the equation of the unit circle.

• Key Point 1: $\cos^2 \theta + \sin^2 \theta = 1$

• Key Point 2: $\sec^2 \theta = 1 + \tan^2 \theta$

• Key Point 3: $\csc^2 \theta = 1 + \cot^2 \theta$

• Example: If $\tan \theta = \frac{3}{4}$, then $\sec^2 \theta = 1 + \left(\frac{3}{4}\right)^2 = 1 + \frac{9}{16} = \frac{25}{16}$, so $\sec \theta = \pm \frac{5}{4}$ (sign depends on quadrant).

Summary of Trigonometric Function Symmetry

• Odd functions: $\tan \theta$, $\cot \theta$, $\sin \theta$, $\csc \theta$

• Even functions: $\sec \theta$, $\cos \theta$

Helpful Trigonometric Values and Relationships

• Key Point 1: $\cos^2 \theta + \sin^2 \theta = 1$

• Key Point 2: $\sec^2 \theta = 1 + \tan^2 \theta$

• Key Point 3: $\csc^2 \theta = 1 + \cot^2 \theta$

• Key Point 4: $\cos(\pi - x) = -\cos(x)$, $\sin(\pi - x) = \sin(x)$

Additional info:

• These notes cover core concepts from Precalculus Chapter 4 (Trigonometric Functions) and Chapter 5 (Analytic Trigonometry), including the unit circle, trigonometric identities, and function properties.