Comprehensive Study Notes on Functions, Implied Domains, Inequalities, and Trigonometry

Logarithmic and Radical Functions: Implied Domains

• Definition of Implied Domain:

  • The implied domain (or natural domain) of a function is the largest set of real numbers for which the mathematical expression defining the function yields a valid real number.

• Domain Conditions for Logarithmic Functions:

  • For a logarithmic function $f(x) = \log(g(x))$, the argument inside the logarithm must be strictly positive: $g(x) > 0$.
  • Taking the logarithm of zero or a negative number is impossible because the logarithmic function is the inverse of the exponential function, and any real exponential expression $e^y$ is strictly positive ($e^y > 0$).

• Domain Conditions for Radical Functions:

  • For a root function $f(x) = \sqrt[n]{h(x)}$:
  • If $n$ is an even positive integer ($n = 2, 4, 6, \dots$), the argument must be non-negative: $h(x) \ge 0$.
  • If $n$ is an odd positive integer ($n = 3, 5, 7, \dots$), there is no restriction on $h(x)$, so $h(x)$ can take any real value ($h(x) \in \mathbb{R}$).

• Implied Domain Calculations and Methodologies:

  • Example 1: Find the implied domain of $f(x) = \log(x^2 - 3x - 10)$.

  • Requirement: The quadratic argument must be strictly positive: $x^2 - 3x - 10 > 0$.

  • Decompose into factors: $(x + 2)(x - 5) > 0$.

  • Method 1 (Algebraic Case Analysis):

    • For the product of two real factors to be positive, either both factors are positive or both factors are negative.
    • Case 1 (Both Positive): $x + 2 > 0 \text{ and } x - 5 > 0 \implies x > -2 \text{ and } x > 5 \implies x > 5$.
    • Case 2 (Both Negative): x + 2 < 0 \text{ and } x - 5 < 0 \implies x < -2 \text{ and } x < 5 \implies x < -2$.\n - Combining both cases gives the implied domain: x < -2 \text{ or } x > 5$, represented in interval notation as$(-\infty, -2) \cup (5, \infty).\n - *Method 2 (Sign Table / Interval Partitioning):*\n - Critical points where factors change sign occur at x = -2$and$x = 5.\n - Partition the real line into three open intervals: x < -2$,$-2 < x < 5$, and$x > 5.\n - Test interval signs:\n - For x < -2$(e.g.,$x = -3$):$(x+2)$is negative,$(x-5)$is negative. Product:$\text{negative} \times \text{negative} = \text{positive}.\n - For -2 < x < 5$(e.g.,$x = 0$):$(x+2)$is positive,$(x-5)$is negative. Product:$\text{positive} \times \text{negative} = \text{negative}.\n - For x > 5$(e.g.,$x = 6$):$(x+2)$is positive,$(x-5)$is positive. Product:$\text{positive} \times \text{positive} = \text{positive}.\n - Conclusion: (-\infty, -2) \cup (5, \infty).\n - Note on application: Sign tables are especially recommended for polynomials of degree 3 or higher, or rational functions (such as a quadratic over a linear term).\n\n - **Example 2:** Find the implied domain of f(x) = \frac{1}{\sqrt[4]{3 - x}}.\n - Since the index n = 4$is even, the argument requires$3 - x \ge 0.\n - Since the expression is in the denominator of a fraction, the denominator cannot equal zero: \sqrt[4]{3 - x} e 0 \implies 3 - x e 0.\n - Combining these two restrictions yields a single strict inequality: 3 - x > 0 \implies x < 3$.

  • Implied domain in interval notation: $(-\infty, 3)$.

  • Example 3: Find the implied domain of $f(x) = \sqrt[3]{x + 2}$.

  • Since the index $n = 3$ is odd, taking the third root imposes no algebraic restrictions.

  • Implied domain: $(-\infty, \infty)$.

• Fundamental Rule on Simplification Prior to Domain Determination:

  • Always apply first principles directly to the original, unsimplified formulation of a function.
  • Never simplify or algebraically rewrite an expression before finding its implied domain, as algebraic manipulations can alter the domain of definition.

Operations on Functions and Composition

• Impact of Algebraic Formulation on Implied Domain:

  • Consider $f_1(x) = \log(x + 1) + \log(x + 2)$ versus $f_2(x) = \log((x + 1)(x + 2))$.
  • For $f_1(x)$, both individual logarithms must be defined:
  • $x + 1 > 0 \implies x > -1$
  • $x + 2 > 0 \implies x > -2$
  • The intersection of these conditions is $x > -1$, giving domain $(-1, \infty)$.
  • For $f_2(x)$, only the combined product must be positive: $(x + 1)(x + 2) > 0$, which allows $x < -2$ in addition to $x > -1$, giving domain $(-\infty, -2) \cup (-1, \infty)$.
  • Expressing a function differently changes its implied domain. To determine the correct implied domain, stick to the initial definition.

• Standard Arithmetic Operations on Functions:

  • Given two functions $f$ and $g$ with implied domains $D_f$ and $D_g$:
  • Sum: $(f + g)(x) = f(x) + g(x)$, Domain: $D_f \cap D_g$
  • Difference: $(f - g)(x) = f(x) - g(x)$, Domain: $D_f \cap D_g$
  • Product: $(f \cdot g)(x) = f(x)g(x)$, Domain: $D_f \cap D_g$
  • Quotient: $\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}$, Domain: $\{ x \in D_f \cap D_g \mid g(x) \ne 0 \}$

• Function Composition:

  • Definition: The composition $(g \circ f)(x) = g(f(x))$ is formed by applying $f$ to $x$ first, and then applying $g$ to the intermediate result $f(x)$.
  • Sequential Evaluation: Computing $\cos(2x)$ involves two discrete steps: first multiply $x$ by $2$ ($y = 2x$), then apply the cosine function ($\cos(y)$). Identifying these steps is critical for calculus rules such as the chain rule.
  • Domain of Composition: D_{g \circ f} = \{ x \in D_f \mid f(x) \n\in D_g \}.
  • First, $x$ must be in the domain of $f$.
  • Second, the output $f(x)$ must fall within the domain of $g$.

• Composition Worked Example:

  • Given $f(x) = x^2 - x$.

  • Specific point evaluations:

  • $f(3) = 3^2 - 3 = 6$

  • $f(1.6) = (1.6)^2 - 1.6 = 2.56 - 1.6 = 0.96$

  • Problem: Find the implied domain of $(g \circ f)(x) = g(f(x))$ where $g(u)$ is defined if and only if its argument $u > 2$.

  • Solution Steps:

  • Require f(x) > 2 \implies x^2 - x > 2 \implies x^2 - x - 2 > 0$.\n - Factor the quadratic: (x - 2)(x + 1) > 0$.

  • The product is positive when $x > 2$ or x < -1$.\n - Implied domain in interval notation: (-\infty, -1) \cup (2, \infty).\n\n# Classification and Properties of Functions\n\n- Increasing and Decreasing Functions:\n - **Non-strict Increasing:** A function f$is increasing on an interval if for all$a < b$,$f(a) \le f(b)$. Horizontal segments (where$f(a) = f(b)) are permitted.\n - **Strictly Increasing:** A function f$is strictly increasing if for all$a < b$,$f(a) < f(b). Equal values are strictly disallowed; the function must constantly move upward.\n - **Non-strict Decreasing:** A function f$is decreasing on an interval if for all$a < b$,$f(a) \ge f(b).\n - **Strictly Decreasing:** A function f$is strictly decreasing if for all$a < b$,$f(a) > f(b).\n - Geometric Distinction: A function that rises but contains horizontal flat regions is increasing, but not strictly increasing.\n\n- Even and Odd Functions:\n - **Even Functions:**\n - Geometric Definition: Symmetric with respect to the y-axis (reflectional symmetry across the y-axis).\n - Algebraic Definition: f(-x) = f(x)$for all$x in the domain.\n - Primary Examples: Even powers of x$(e.g.,$x^2$,$x^4$,$x^4 + x^2$) and the cosine function$\cos(x).\n - **Odd Functions:**\n - Geometric Definition: Symmetric with respect to the origin (point symmetry; invariant under a 180^\circ$rotation about$(0,0)).\n - Algebraic Definition: f(-x) = -f(x)$for all$x in the domain.\n - Primary Examples: Odd powers of x$(e.g.,$x^1$,$x^3$) and the sine function$\sin(x).\n\n- Algebraic Combinations and Basic Proofs of Symmetry:\n - *Sum of two even functions:* If f$and$g$are even,$(f + g)(-x) = f(-x) + g(-x) = f(x) + g(x) = (f + g)(x)$. Thus,$f + g is **even**.\n - *Sum of two odd functions:* If f$and$g$are odd,$(f + g)(-x) = f(-x) + g(-x) = -f(x) - g(x) = -(f(x) + g(x)) = -(f + g)(x)$. Thus,$f + g is **odd**.\n - *Product of an odd function and an even function:* Let f$be odd and$g$be even. Evaluate$(f \cdot g)(-x) = f(-x) \cdot g(-x) = (-f(x)) \cdot (g(x)) = -(f(x)g(x)) = -(f \cdot g)(x)$. Thus, the product$f \cdot g is **odd**.\n - *Product of two odd functions:* Let f$and$g$both be odd. Evaluate$(f \cdot g)(-x) = f(-x) \cdot g(-x) = (-f(x)) \cdot (-g(x)) = f(x)g(x) = (f \cdot g)(x)$. The two negative signs cancel out, making the product$f \cdot g an **even** function.\n\n# Power Functions and Piecewise Defined Functions\n\n- Power Functions f(x) = x^a:\n - **Positive Integer Exponents (a \in \mathbb{Z}^+):**\n - *Odd Exponents (a = 1, 3, 5, \dots$):* Domain is$(-\infty, \infty)$. The functions are odd ($(-x)^a = -x^a) and strictly increasing across their entire domain.\n - *Even Exponents (a = 2, 4, 6, \dots$):* Domain is$(-\infty, \infty)$. The functions are even ($(-x)^a = x^a$). They are strictly decreasing on$(-\infty, 0]$and strictly increasing on$[0, \infty).\n - **Negative Integer Exponents (a = -b$, where$b > 0 \implies x^a = \frac{1}{x^b}):**\n - The domain excludes x = 0$, i.e.,$(-\infty, 0) \cup (0, \infty).\n - *Odd Negative Exponents (e.g., a = -1$):* Odd function. Decreasing on$(-\infty, 0)$and decreasing on$(0, \infty).\n - *Even Negative Exponents (e.g., a = -2$):* Even function. Increasing on$(-\infty, 0)$and decreasing on$(0, \infty).\n - **Fractional Exponents (e.g., a = \frac{1}{2} \implies f(x) = \sqrt{x}):**\n - Requires x \ge 0$, domain$[0, \infty). Strictly increasing function.\n - **Irrational Exponents (e.g., a = \pi, \sqrt{2}):**\n - Rigorously defined using a limit of rational exponents r_n$that approach$a$($r_n \rightarrow a).\n\n- Piecewise Defined Functions:\n - Defined by applying different explicit formulas to distinct, non-overlapping subdomains.\n - Non-overlapping Domain Rule: Subdomains must be disjoint. If subdomains overlap, all formulas assigned to the overlapping region must yield identical values.\n - Example 1:\n    f(x) = \begin{cases} 1, & x < 0 \ 2x, & x > 0 \end{cases}\n - Consists of a constant horizontal ray y = 1$for$x < 0$and a ray$y = 2x$with slope 2 passing through the origin for$x > 0.\n - Example 2 (Signum-style function):\n    f(x) = \begin{cases} x + 1, & x > 0 \ 0, & x = 0 \ -1, & x < 0 \end{cases}\n - Graphical representation uses a solid circle at (0, 0) to show inclusion, and open circles at boundary endpoints where values are excluded.\n - Example 3 (Linear and Quadratic Combination):\n    f(x) = \begin{cases} 2x + 3, & x < 1 \ -2x^2 + 11x - 12, & x \ge 1 \end{cases}\n - Linear segment y = 2x + 3$applies for$x < 1$, ending with an open circle at$(1, 5).\n - Quadratic segment y = -2x^2 + 11x - 12$applies for$x \ge 1.\n - Quadratic analysis: Opens downward (a = -2 < 0$), x-intercepts at roots of$-2x^2 + 11x - 12 = 0$, axis of symmetry and vertex at$x = -\frac{b}{2a} = -\frac{11}{2(-2)} = \frac{11}{4}$. Point at boundary$x = 1$is included as a solid circle at$(1, -3).\n\n# Absolute Value Functions and Inequalities\n\n- Formal Definition of Absolute Value:\n  |x| = \begin{cases} x, & x \ge 0 \ -x, & x < 0 \end{cases}\n - Geometrically represents the distance between x$and$0 on the real number line, stripping away negative signs.\n\n- Fundamental Properties:\n 1. Non-negativity: |x| \ge 0$for all$x \in \mathbb{R}.\n 2. Even Symmetry: |-x| = |x|.\n 3. Relation to Squares: |x|^2 = x^2$and$\sqrt{x^2} = |x|.\n 4. Multiplicativity and Divisibility: |xy| = |x||y|$and$\left|\frac{x}{y}\right| = \frac{|x|}{|y|}$(for$y e 0).\n 5. Triangle Inequality: |x + y| \le |x| + |y|.\n 6. Distance Equivalence Rules (for c > 0):\n - |x| \le c \iff -c \le x \le c\n - |x| \ge c \iff x \ge c \text{ or } x \le -c\n\n- Absolute Value Worked Examples:\n - **Problem 1:** Solve |2x - 3| \le 7$.

  • Double inequality: $-7 \le 2x - 3 \le 7$

  • Add 3 throughout: $-4 \le 2x \le 10$

  • Divide by 2: $-2 \le x \le 5$

  • Solution Set: $[-2, 5]$.

  • Problem 2: Solve |x + 2| > 4$.\n - Split into two independent inequalities:\n - x + 2 > 4 \implies x > 2\n - x + 2 < -4 \implies x < -6\n - Solution Set: (-\infty, -6) \cup (2, \infty).\n\n- Graphing Absolute Value Expressions via Piecewise Partitioning:\n - **Problem:** Graph f(x) = |x - 2| + |x + 2|\n - Step 1: Identify boundary points where expressions inside the absolute values equal zero: x = 2$and$x = -2.\n - Step 2: Break domain into three intervals:\n - Interval 1 (x > 2$): Both$(x - 2)$and$(x + 2) are positive.\n      f(x) = (x - 2) + (x + 2) = 2x\n - Interval 2 (-2 \le x \le 2$):$(x + 2) \ge 0$(positive) and$(x - 2) \le 0 (negative).\n      f(x) = -(x - 2) + (x + 2) = -x + 2 + x + 2 = 4\n - Interval 3 (x < -2$): Both$(x - 2)$and$(x + 2) are negative.\n      f(x) = -(x - 2) - (x + 2) = -x + 2 - x - 2 = -2x\n - Resulting Shape: A flat horizontal line y = 4$on$[-2, 2]$, with ray$y = 2x$rising for$x > 2$and ray$y = -2x$rising for$x < -2$.

• Absolute Value Transformations:

  • Graph of $y = |f(x)|$: Take the graph of $y = f(x)$ and reflect any portion that lies below the x-axis upward across the x-axis into the upper half-plane.
  • Graph of $y = f(|x|)$: Erase the graph for $x < 0$, keep the graph for $x \ge 0$, and reflect the $x \ge 0$ portion symmetrically across the y-axis.

Quadratic and Rational Inequalities

• Quadratic Inequalities:
  • Standard form: $ax^2 + bx + c > 0$ or $ax^2 + bx + c < 0$.
  • Warning: Multiplying or dividing an inequality by a negative number reverses the direction of the inequality sign.
  • Factored Form Solutions (assuming roots $\alpha < \beta$):
  • $(x - \alpha)(x - \beta) > 0 \iff x < \alpha \text{ or } x > \beta$
  • $(x - \alpha)(x - \beta) < 0 \iff \alpha < x < \beta$
  • Example: Solve (2x - 1)(2x - 5) \ge 0$.\n - Roots are \alpha = \frac{1}{2}$and$\beta = \frac{5}{2}.\n - Solution: x \le \frac{1}{2} \text{ or } x \ge \frac{5}{2}.\n\n- Rational Inequalities:\n - Critical Rule: You **cannot** simply multiply both sides of an inequality by an expression containing a variable (e.g., multiplying by x + 1), because the sign of the variable expression is unknown.\n - **Method 1 (Multiplying by Denominator Squared):**\n - Solve \frac{2x - 1}{x + 1} < 1\n - Multiply both sides by (x + 1)^2$, which is guaranteed to be strictly positive for$x e -1:\n      (x + 1)(2x - 1) < (x + 1)^2\n - Move all terms to the left side:\n      (x + 1)(2x - 1) - (x + 1)^2 < 0\n - Factor out common term (x + 1):\n      (x + 1)[(2x - 1) - (x + 1)] < 0 \implies (x + 1)(x - 2) < 0\n - Solution: -1 < x < 2.\n\n - **Method 2 (Sign Table for Rational Expressions):**\n - Move all terms to one side to set the expression against 0: \frac{P(x)}{Q(x)} \ge 0\n - Find all zeros of the numerator P(x)$and denominator$Q(x). These form the critical boundary points.\n - Set up a sign table across open intervals bounded by these critical points.\n\n- Detailed Rational Function Domain Example:\n - **Problem:** Find the implied domain of f(x) = \sqrt[4]{\frac{(x - 3)(x + 1)}{x(x + 2)}}.\n - Condition: Rational argument must be non-negative: g(x) = \frac{(x - 3)(x + 1)}{x(x + 2)} \ge 0\n - Critical Points (Zeros of numerator and denominator): x = -2, -1, 0, 3\n - Partition into 9 intervals and boundary points:\n 1. x < -2$: All 4 factors$(x-3), (x+1), x, (x+2)$are negative. Four negatives give positive ($g(x) > 0).\n 2. x = -2$: Denominator is zero$\implies **Undefined**.\n 3. -2 < x < -1$: 3 factors negative, 1 positive$\implies$Negative ($g(x) < 0).\n 4. x = -1$: Numerator is zero, denominator non-zero$\implies g(-1) = 0**.
  1. $-1 < x < 0$: 2 factors negative, 2 positive $\implies$ Positive ($g(x) > 0$).
  2. $x = 0$: Denominator is zero $\implies$ Undefined.
  3. $0 < x < 3$: 1 factor negative, 3 positive $\implies$ Negative ($g(x) < 0$).
  4. $x = 3$: Numerator is zero \implies g(3) = 0**.\n 9. x > 3$: All 4 factors positive$\implies$Positive ($g(x) > 0).\n - Selecting regions where g(x) \ge 0:\n - (-\infty, -2)$(excluding$-2)\n - [-1, 0)$(including$-1$, excluding$0)\n - [3, \infty)$(including$3)\n - Combined Implied Domain: (-\infty, -2) \cup [-1, 0) \cup [3, \infty).\n\n- Interval Test-Point Method:\n - Because factor signs remain constant within open intervals between critical points, an alternative to testing every factor is picking a single test value x_0$inside each interval (e.g.,$x = -3, -1.5, -0.5, 1, 4$), evaluating$g(x_0), and taking that sign for the entire interval.\n\n# Radian Measure, Arc Length, and Sector Area\n\n- Units of Angle Measurement:\n - **Degree Measure (360^\circ):** An artificial, human-chosen convention originating in antiquity because 360 has many integer divisors (2, 3, 4, 5, 6, 8, 9, 10, 12, etc.).\n - **Radian Measure:** A natural, intrinsic geometric measure based on the unit circle.\n - **Unit Circle Definition:** A circle centered at the origin with radius r = 1.\n - Circumference of unit circle: C = 2\pi r = 2\pi(1) = 2\pi\n - Therefore, a full revolution (360^\circ$) corresponds to$2\pi radians.\n - Fundamental Relation: 180^\circ = \pi \text{ radians}\n\n- Angle Conversions:\n - Degree to Radian conversion factor: Multiply degrees by \frac{\pi}{180^\circ}\n - 1^\circ = \frac{\pi}{180} \text{ radians}\n - Example: 60^\circ = 60 \times \frac{\pi}{180} = \frac{\pi}{3} \text{ radians}\n - Radian to Degree conversion factor: Replace \pi$with$180^\circ$or multiply by$\frac{180^\circ}{\pi}\n\n- Calculator Mode Setting:\n - Always verify whether a scientific/graphing calculator is set to Degree mode (indicated by `D`) or Radian mode (indicated by `R` or `RAD`). Higher mathematics almost exclusively uses radians.\n\n- Arc Length and Sector Area Formulas:\n - **Arc Length (s):**\n - For an angle \theta$measured in radians on a circle of radius$r:\n      s = r\theta\n - Derivation: Proportion of full circumference 2\pi r \times \frac{\theta}{2\pi} = r\theta$.
  • Sector Area ($A$):
  • For an angle $\theta$ measured in radians on a circle of radius $r$:       $A = \frac{1}{2}r^2\theta$
  • Derivation: Proportion of total circle area \pi r^2 \times \frac{\theta}{2\pi} = \frac{1}{2}r^2\theta$.\n\n- Right Triangle Trigonometry Definitions (for acute angle \theta):\n - Hypotenuse: The longest side, opposite the right angle.\n - Opposite: The side across from angle \theta.\n - Adjacent: The side adjacent to angle \theta (other than the hypotenuse).\n - \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}\n - \cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}\n - \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\sin(\theta)}{\cos(\theta)}\n\n- Behavior of Functions as \theta$increases from$0$to$\frac{\pi}{2}$($0^\circ$to$90^\circ):\n - \sin(\theta)$is strictly increasing from$0$to$1.\n - \cos(\theta)$is strictly decreasing from$1$to$0.\n - \tan(\theta)$is strictly increasing from$0$to$\infty.\n\n- Special Right Triangles and Exact Trigonometric Values:\n - **45^\circ-45^\circ-90^\circ$Triangle ($\frac{\pi}{4} Radians):**\n - Isosceles right triangle with leg lengths 1, 1.\n - Hypotenuse by Pythagorean theorem: \sqrt{1^2 + 1^2} = \sqrt{2}.\n - \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\n - \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}\n - \tan\left(\frac{\pi}{4}\right) = 1\n\n - **30^\circ-60^\circ-90^\circ$Triangle ($\frac{\pi}{6}$and$\frac{\pi}{3} Radians):**\n - Formed by bisecting an equilateral triangle of side length 1.\n - Short leg = \frac{1}{2}$, Hypotenuse =$1$, Long leg =$\sqrt{1^2 - (1/2)^2} = \frac{\sqrt{3}}{2}.\n - For \theta = 60^\circ = \frac{\pi}{3}:\n - \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\n - \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\n - \tan\left(\frac{\pi}{3}\right) = \sqrt{3}\n - For \theta = 30^\circ = \frac{\pi}{6}:\n - \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\n - \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\n - \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}\n\n- Exact Values Summary Table:\n - 0\text{ rad } (0^\circ)$:$\sin(0) = 0$,$\cos(0) = 1$,$\tan(0) = 0\n - \frac{\pi}{6}\text{ rad } (30^\circ)$:$\sin = \frac{1}{2}$,$\cos = \frac{\sqrt{3}}{2}$,$\tan = \frac{1}{\sqrt{3}}\n - \frac{\pi}{4}\text{ rad } (45^\circ)$:$\sin = \frac{1}{\sqrt{2}}$,$\cos = \frac{1}{\sqrt{2}}$,$\tan = 1\n - \frac{\pi}{3}\text{ rad } (60^\circ)$:$\sin = \frac{\sqrt{3}}{2}$,$\cos = \frac{1}{2}$,$\tan = \sqrt{3}\n - \frac{\pi}{2}\text{ rad } (90^\circ)$:$\sin = 1$,$\cos = 0$,$\tan is Undefined\n\n# Trigonometric Functions on the Unit Circle and Implied Domains\n\n- Unit Circle Generalization for Arbitrary Angles:\n - Right triangle definitions fail for angles \theta > \frac{\pi}{2}$or$\theta < 0 because right triangles cannot accommodate obtuse or negative interior angles.\n - **Unit Circle Extension:**\n - Place angle \theta in standard position (vertex at origin, initial side along positive x-axis).\n - Let the terminal side intersect the unit circle x^2 + y^2 = 1$at point$(x, y).\n - Definitions for arbitrary \theta \in \mathbb{R}:\n - \cos(\theta) = x\n - \sin(\theta) = y\n - \tan(\theta) = \frac{y}{x} = \frac{\sin(\theta)}{\cos(\theta)}$(provided$x e 0)\n\n- Unit Circle Evaluation Examples:\n - **Example 1:** Evaluate \theta = \frac{4\pi}{3}$($240^\circ).\n - Position: Quadrant III. Reference angle is \frac{4\pi}{3} - \pi = \frac{\pi}{3}.\n - Point on unit circle: \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right).\n - \cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2}$,$\sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}.\n\n - **Example 2:** Evaluate \theta = -\frac{\pi}{4}$($-45^\circ).\n - Position: Quadrant IV. Reference angle is \frac{\pi}{4}.\n - Point on unit circle: \left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right).\n - \cos\left(-\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$,$\sin\left(-\frac{\pi}{4}\right) = -\frac{1}{\sqrt{2}}.\n\n - **Example 3:** Evaluate \theta = \frac{5\pi}{2}.\n - Coterminal angle: \frac{5\pi}{2} - 2\pi = \frac{\pi}{2}.\n - Point on unit circle: (0, 1).\n - \cos\left(\frac{5\pi}{2}\right) = 0$,$\sin\left(\frac{5\pi}{2}\right) = 1.\n\n- Implied Domains of Trigonometric Functions:\n - \sin(\theta)$and$\cos(\theta)$: Defined for all real numbers, domain$(-\infty, \infty).\n - Zeros of Cosine: \cos(\theta) = 0$occurs where the x-coordinate on the unit circle is zero (top and bottom points$(0, 1)$and$(0, -1)).\n - Angle values: \theta = \dots, -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots\n - General formula: \theta = \frac{\pi}{2} + k\pi$for$k \in \mathbb{Z}.\n - Domain of \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$and$\sec(\theta) = \frac{1}{\cos(\theta)}:\n    \mathbb{R} \setminus \left{ \frac{\pi}{2} + k\pi \;\middle|\; k \in \mathbb{Z} \right}\n - Zeros of Sine: \sin(\theta) = 0$occurs where the y-coordinate on the unit circle is zero (left and right points$(1, 0)$and$(-1, 0)).\n - General formula: \theta = k\pi$for$k \in \mathbb{Z}.\n - Domain of \cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$and$\csc(\theta) = \frac{1}{\sin(\theta)}:\n    \mathbb{R} \setminus { k\pi \mid k \in \mathbb{Z} }\n\n- Quadrant Signs and CAST Rule:\n - **Quadrant I (0 < \theta < \frac{\pi}{2}$):**$x > 0, y > 0 \implies **A**ll trigonometric functions are positive.\n - **Quadrant II (\frac{\pi}{2} < \theta < \pi$):**$x < 0, y > 0 \implies **S**ine is positive (cosine and tangent are negative).\n - **Quadrant III (\pi < \theta < \frac{3\pi}{2}$):**$x < 0, y < 0 \implies **T**angent is positive (sine and cosine are negative).\n - **Quadrant IV (\frac{3\pi}{2} < \theta < 2\pi$):**$x > 0, y < 0 \implies **C**osine is positive (sine and tangent are negative).\n - **CAST Rule Mnemonic:** Read starting from Quadrant IV counterclockwise through Quadrants I, II, and III:\n - Quadrant IV: **C** (Cosine positive)\n - Quadrant I: **A** (All positive)\n - Quadrant II: **S** (Sine positive)\n - Quadrant III: **T** (Tangent positive)\n - **Application Example:** Determine if \tan(155^\circ) is positive or negative.\n - Angle 155^\circ$falls in Quadrant II ($90^\circ < 155^\circ < 180^\circ).\n - In Quadrant II, only sine is positive; therefore, \tan(155^\circ) < 0$$ (negative).

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