Trigonometric Functions, Inverse Trigonometry, and Introduction to Limits

Course Logistics and Exam Guidelines

  • Class Schedule Details:

    • Friday class periods are short (single period duration).

    • The upcoming exam takes place on Wednesday during the full class period.

  • Homework and Notes Policy:

    • There is no daily graded homework; assessment preparation relies on quizzes.

    • Online versions of lecture notes are not provided. The course textbook serves as the primary reference resource.

    • Note-taking strategy recommendation: Model note-taking directly after the board work standard, as structured written mathematics reflects refined communication practices.

  • Exam Mechanics and Calculator Policy:

    • Calculators are strictly prohibited on exams to avoid security issues and administrative complexities.

    • Exam questions are explicitly designed to be completed without numerical computational tools.

    • Simplification standards:

      • Answers do not need to be simplified or converted into decimal or colon forms.

      • Exact trigonometric or functional values must be left as exact expressions (e.g., an expression like sin⁡(4)\sin(4) is a valid final exact number).

  • Exam Format and Preparation:

    • The exam consists of exactly 44 questions.

    • Students are given the entire class period to complete the exam.

    • Question structure: Exam problems are structured identically to quiz problems, which in turn reflect in-class examples.

    • Mastery of all quiz problems is sufficient preparation for the exam.

    • Required materials: Writing utensil and critical reasoning.

Right Triangle Definitions of Trigonometric Functions

  • Right Triangle Construction:

    • A right triangle contains one 90deg90^\text{deg} angle and two acute angles.

    • Let θ\theta represent one of the acute angles.

    • Let xx represent the horizontal leg length.

    • Let yy represent the vertical leg length.

    • Let rr represent the hypotenuse length.

  • Pythagorean Relationship:

    • The relationship between the side lengths is governed by the Pythagorean Theorem:         x2+y2=r2x^2 + y^2 = r^2

  • Primary Trigonometric Functions:

    • Sine Function:         sin⁡(θ)=yr=oppositehypotenuse\sin(\theta) = \frac{y}{r} = \frac{\text{opposite}}{\text{hypotenuse}}

    • Cosine Function:         cos⁡(θ)=xr=adjacenthypotenuse\cos(\theta) = \frac{x}{r} = \frac{\text{adjacent}}{\text{hypotenuse}}

    • Tangent Function:         tan⁡(θ)=sin⁡(θ)cos⁡(θ)=yx=oppositeadjacent\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{y}{x} = \frac{\text{opposite}}{\text{adjacent}}

Circle-Based Definitions and the Right-Hand Rule

  • Geometric Setup on a Circle:

    • Consider a circle centered at the origin (0,0)(0, 0) with radius rr.

    • Select a point (x,y)(x, y) on the circle at distance rr from the origin.

    • Define the angle θ\theta between the positive x-axis and the radial line segment connecting the origin to (x,y)(x, y).

  • Rules for Standard Angle Measurement:

    • Initial side: Must always start on the positive x-axis (never the negative x-axis).

    • Direction: Positive angle measurements move in the counterclockwise (CCW) direction.

    • Right-Hand Rule:

      • Using the right hand, curl the fingers in the direction of counterclockwise rotation (positive angular displacement).

      • The thumb points outwards, perpendicular to the plane (towards the observer).

      • Using the left hand gives the opposite direction (pointing away), illustrating why the right-hand convention is standard.

  • Parametric Representation of Circle Coordinates:

    • The coordinates of the point on the circle are:         (x,y)=(r×cos⁡(θ),r×sin⁡(θ))(x, y) = (r \times \cos(\theta), r \times \sin(\theta))

    • For a unit circle where r=1r = 1, the coordinates simplify to:         (x,y)=(cos⁡(θ),sin⁡(θ))(x, y) = (\cos(\theta), \sin(\theta))

  • Equivalence to Triangle Definitions:

    • Constructing a vertical segment from (x,y)(x, y) down to the x-axis forms a right triangle with base x=r×cos⁡(θ)x = r \times \cos(\theta), height y=r×sin⁡(θ)y = r \times \sin(\theta), and hypotenuse rr

    • Rearranging these coordinate equations yields the classical ratios:         cos⁡(θ)=xr\cos(\theta) = \frac{x}{r}         sin⁡(θ)=yr\sin(\theta) = \frac{y}{r}

  • Generality of the Circle Definition:

    • The circle definition is superior because it applies to arbitrary angles beyond acute right-triangle constraints (θ>90∘\theta > 90^\circ).

    • It defines trigonometric functions for:

      • θ=180∘\theta = 180^\circ

      • θ=270∘\theta = 270^\circ

      • Negative rotations (e.g., −45∘-45^\circ

      • Rotations exceeding 360∘360^\circ

Inverse Trigonometric Functions and Restricted Domains

  • Domain Restrictions for Invertibility:

    • Full trigonometric functions repeat outputs around the circle, failing the horizontal line test.

    • To construct single-valued inverse functions, domains are restricted to continuous intervals yielding unique outputs:

      • Cosine domain restriction: θ∈[0,π]\theta \in [0, \pi]

      • Sine domain restriction: θ∈[−π2,π2]\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]

  • Conceptual Meaning of Inverse Functions:

    • Notation: The superscript −1-1 in cos⁡−1(x)\cos^{-1}(x) or sin⁡−1(x)\sin^{-1}(x) designates a functional inverse, not an exponent or reciprocal.         cos⁡−1(x)≠1cos⁡(x)\cos^{-1}(x) \neq \frac{1}{\cos(x)}

    • Alternative notation: arccos⁡(x)\arccos(x) or arccos represents inverse cosine.

    • Conceptual core: An inverse trigonometric expression represents an angle within the restricted domain whose trigonometric output equals the input ratio.

  • Key Composition Properties:

    • For a general function ff and its inverse f−1f^{-1}:         f(f−1(x))=xf(f^{-1}(x)) = x         f−1(f(x))=xf^{-1}(f(x)) = x

    • For the cosine function on its domain:         cos⁡(cos⁡−1(x))=x\cos(\cos^{-1}(x)) = x

Geometric Visualization and Composite Inverse Trig Evaluations

  • Methodology for Composite Expressions:

    • To evaluate an expression like tan⁡(cos⁡−1(a))\tan(\cos^{-1}(a)), treat the inner inverse function as an angle: θ=cos⁡−1(a)\theta = \cos^{-1}(a).

    • Translate the algebraic relation into a geometric right triangle.

    • Use the Pythagorean theorem to derive the missing side length.

    • Evaluate the outer function directly from the triangle ratios.

    • Pedagogical note on visualization: Drawing physical geometric representations helps unpack abstract mathematical definitions from the inside out.

  • Example 1: Evaluating tan⁡(cos⁡−1(37))\tan\left(\cos^{-1}\left(\frac{3}{7}\right)\right)

    • Define the angle: Let θ=cos⁡−1(37)\theta = \cos^{-1}\left(\frac{3}{7}\right), which implies cos⁡(θ)=37\cos(\theta) = \frac{3}{7}.

    • Construct the right triangle:

      • Adjacent leg x=3x = 3

      • Hypotenuse r=7r = 7

    • Determine vertical leg yy via Pythagorean Theorem:         32+y2=723^2 + y^2 = 7^2         y2=72−32=49−9=40y^2 = 7^2 - 3^2 = 49 - 9 = 40         y=40y = \sqrt{40}         (Take the positive principal square root because geometric leg lengths are positive distances).

    • Evaluate functions of θ\theta:         sin⁡(θ)=407\sin(\theta) = \frac{\sqrt{40}}{7}         tan⁡(θ)=403\tan(\theta) = \frac{\sqrt{40}}{3}

    • Simplification of radical expressions:         40=4×10=210\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}         tan⁡(cos⁡−1(37))=2103\tan\left(\cos^{-1}\left(\frac{3}{7}\right)\right) = \frac{2\sqrt{10}}{3}

    • Both 403\frac{\sqrt{40}}{3} and 2103\frac{2\sqrt{10}}{3} represent valid equal quantities.

  • Example 2: Evaluating cos⁡(sin⁡−1(29))\cos\left(\sin^{-1}\left(\frac{2}{9}\right)\right)

    • Define the angle: Let θ=sin⁡−1(29)\theta = \sin^{-1}\left(\frac{2}{9}\right), which implies sin⁡(θ)=29\sin(\theta) = \frac{2}{9}.

    • Construct the right triangle:

      • Opposite leg y=2y = 2

      • Hypotenuse r=9r = 9

    • Determine adjacent leg xx:         x2+22=92x^2 + 2^2 = 9^2         x2=92−22=81−4=77x^2 = 9^2 - 2^2 = 81 - 4 = 77         x=77x = \sqrt{77}

    • Since 77=11×777 = 11 \times 7, there are no perfect square factors to pull out of the radical.

    • Evaluate cosine of the angle:         cos⁡(sin⁡−1(29))=adjacenthypotenuse=779\cos\left(\sin^{-1}\left(\frac{2}{9}\right)\right) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{77}}{9}

Introduction to Limits and Algebraic Resolution

  • Conceptual Definition of a Limit:

    • A limit formalizes the behavior of a function f(x)f(x) as the input xx approaches a target value aa.

    • Standard notation:         lim⁡x→af(x)\lim_{x \to a} f(x)

    • Reads: "The limit as xx approaches aa of f(x)f(x)."

    • A limit can exist and equal a concrete number LL even if the function itself is undefined at x=ax = a (f(a)f(a) does not exist).

  • Analyzing Discontinuities and Rational Functions:

    • Consider the rational function:         f(x)=x2−6x+8x2−4f(x) = \frac{x^2 - 6x + 8}{x^2 - 4}

    • Domain determination:

      • Setting denominator to zero: x2−4=0  ⟹  x=±2x^2 - 4 = 0 \implies x = \pm 2

      • The function f(x)f(x) is undefined at x=2x = 2 and x=−2x = -2 due to division by zero.

    • Evaluating Function Behavior near Undefined Points:

      • Standard point plotting ("plug and chug") works everywhere on the domain except at domain boundaries/holes (x=±2x = \pm 2).

      • Limits determine whether an undefined point corresponds to a vertical asymptote or a removable hole.

  • Evaluating lim⁡x→2x2−6x+8x2−4\lim_{x \to 2} \frac{x^2 - 6x + 8}{x^2 - 4} via Factoring:

    • Step 1: Factor the numerator:         x2−6x+8=(x−4)(x−2)x^2 - 6x + 8 = (x - 4)(x - 2)

    • Step 2: Factor the denominator:         x2−4=(x+2)(x−2)x^2 - 4 = (x + 2)(x - 2)

    • Step 3: Simplify the expression for all x≠2x \neq 2:         f(x)=(x−4)(x−2)(x+2)(x−2)=x−4x+2for x≠2f(x) = \frac{(x - 4)(x - 2)}{(x + 2)(x - 2)} = \frac{x - 4}{x + 2} \quad \text{for } x \neq 2

    • Step 4: Compute the limit by direct substitution into the reduced expression:         lim⁡x→2f(x)=lim⁡x→2x−4x+2=2−42+2=−24=−12\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{x - 4}{x + 2} = \frac{2 - 4}{2 + 2} = \frac{-2}{4} = -\frac{1}{2}

    • Conclusion:

      • f(2)f(2) is undefined.

      • The limit as x→2x \to 2 is −12-\frac{1}{2}.

      • Geometrically, the graph of f(x)f(x) possesses a removable hole at the point (2,−12)\left(2, -\frac{1}{2}\right).