MGU Mathematics Series — Algebra II
Part V: Geometry Connections, Trigonometry, Probability, and Modeling
Right Triangles · Unit Circle · Sine · Cosine · Tangent · Periodic Models
Chapter 13 of 14
MGU Mathematics Series  ·  Algebra II  ·  Library Textbook Edition

Chapter 13:
Trigonometric
Foundations

Right-triangle ratios, the unit circle, radian measure, graphs of sine, cosine, and tangent, and trigonometric models — the gateway to periodic functions and precalculus

■ Math Level: Algebra II ■

■ 14 Sections  •  Part V of 5 ■

Follows Chapter 12: Conic Sections  ·  Prepares students for Precalculus, Trigonometry, Physics, and Calculus

Chapter 13 — Front Matter

Purpose and Overview

Chapter 13 introduces trigonometry — the study of triangle ratios and periodic relationships — as a natural extension of the algebraic thinking developed throughout this course. Trigonometry is one of the oldest and most useful branches of mathematics, connecting geometry, algebra, and the behavior of repeating patterns in nature and engineering.

Students begin with the most concrete entry point: right triangles. A right triangle has predictable, measurable relationships among its sides and angles, and trigonometry gives names and structure to those relationships. From there, the chapter extends those ideas outward — first to all angles, then to the coordinate plane, and finally to the graphs of functions that repeat themselves in predictable cycles.

By the end of this chapter, students should be able to define and evaluate the three primary trigonometric ratios, solve problems involving right triangles, convert between degrees and radians, locate values on the unit circle, and sketch and interpret the graphs of sine, cosine, and tangent. These skills form the foundation for Precalculus and Trigonometry, where the study of periodic functions becomes a central theme. This chapter prepares students for that work without becoming a full trigonometry course. The emphasis remains on building understanding from right triangles first, then extending to functions and graphs.

Opening Question

A 50-foot cable is anchored to the ground and attached to the top of a vertical support pole. If the cable makes a 62° angle with the ground, how tall is the pole? Without a way to directly measure a height that may be unreachable or dangerous to climb, how can you calculate it? Trigonometry provides an exact method — and by the end of this chapter, you will know how to solve it.

Why This Chapter Matters. Trigonometry appears in nearly every quantitative field. Engineering and construction use triangle ratios to design structures, calculate forces, and measure distances. Physics uses sine and cosine functions to model waves, oscillations, pendulums, and alternating electrical current. Navigation and surveying use angle measurements to determine positions and distances across terrain. Computer graphics and animation rely on trigonometric functions to simulate motion and create realistic visual effects. Music and acoustics describe sound waves using periodic functions. Even finance and economics sometimes use cyclical models to represent repeating patterns in data. Understanding how periodic functions behave provides a powerful lens for interpreting patterns in the world.

14Sections
13Objectives
22Key Terms
15+Worked Examples

Learning Objectives

  1. 1Identify and label the sides of a right triangle relative to a given angle.
  2. 2Define sine, cosine, and tangent as ratios of sides in a right triangle.
  3. 3Use trigonometric ratios to solve for unknown sides and angles.
  4. 4Apply trigonometric reasoning to problems involving angles of elevation and depression.
  5. 5Use special right triangle relationships for the 30°–60°–90° and 45°–45°–90° triangles.
  6. 6Convert between degree and radian measures.
  7. 7Identify coordinates of key angles on the unit circle.
  8. 8Use the unit circle to find sine and cosine values beyond the first quadrant.
  9. 9Sketch the graphs of \(y = \sin(x)\) and \(y = \cos(x)\).
  10. 10Identify and interpret amplitude, period, and midline.
  11. 11Describe and graph the tangent function.
  12. 12Apply trigonometric functions to model periodic situations.
  13. 13Recognize and avoid common trigonometry mistakes.
Chapter 13 — Contents

Sections at a Glance

Sections 13.1 – 13.6
Right-Triangle Trigonometry: Ratios, Solving, and Special Triangles
What Trig Studies · Right Triangles · SOH-CAH-TOA · Solving · Elevation/Depression · Special Angles
Section 13.1

What Trigonometry Studies

Trigonometry begins with a simple observation: in any right triangle, the ratios between pairs of sides are determined entirely by the angles of the triangle, not by the size of the triangle. To see why this is true, imagine two right triangles with the same angle measures but different sizes. Even though one triangle might be twice as large as the other, the ratio of the opposite side to the hypotenuse will be identical in both. This consistency is what makes trigonometry so powerful. Once you know an angle, you know the ratios of the sides. Once you know the ratios, you can find missing side lengths.

This idea expands significantly beyond triangles. Trigonometric ratios can be extended to any angle — not just angles inside a right triangle — and the resulting functions repeat their values in predictable cycles. This is why trigonometry is also the mathematics of waves, vibrations, orbits, and oscillations.

Algebra II introduces the foundational layer of this subject. The full study of trigonometric identities, laws of sines and cosines, inverse functions, and parametric equations belongs to Precalculus and Trigonometry. Here, the goal is to build firm roots: understand the ratios, learn the key angle values, grasp the unit circle, and read the graphs of sine, cosine, and tangent.

Section 13.2

Right Triangles

A right triangle is any triangle that contains exactly one 90° angle. Because the angles of any triangle must sum to 180°, the other two angles in a right triangle must each be less than 90° and must together add to exactly 90°. Those two non-right angles are called acute angles.

When working with trigonometry, three sides must be correctly identified relative to a chosen angle. The hypotenuse is always the side directly across from the 90° angle — it is the longest side. The opposite side is directly across from the angle you are working with. The adjacent side forms the angle you are working with and is not the hypotenuse. These labels shift depending on which angle you are using.

Diagram Instruction

Draw a right triangle with the right angle at the lower right corner. Label the vertex at the lower left as angle θ. The hypotenuse runs from the lower left to the upper right. Label the vertical side (across from θ) as "opposite." Label the horizontal bottom side (next to θ) as "adjacent." Label the longest side (from lower left to upper right) as "hypotenuse." Place a small square at the right angle corner.

The Pythagorean Theorem

If \(a\) and \(b\) are the two legs and \(c\) is the hypotenuse of a right triangle:

\[ a^2 + b^2 = c^2 \]

If two sides are known, the third can always be found using this relationship.

Example 13.2.1
Identifying sides of a right triangle
  • ProblemIn a right triangle, angle \(\theta\) is at the bottom left. The horizontal leg has length 5, the vertical leg has length 12, and the hypotenuse has length 13. Identify the opposite, adjacent, and hypotenuse relative to \(\theta\).
  • ThinkThe hypotenuse is always across from the right angle. The adjacent side forms the angle \(\theta\) without being the hypotenuse. The opposite side is across from \(\theta\).
  • AnswerHypotenuse = 13  ·  Adjacent = 5 (horizontal leg)  ·  Opposite = 12 (vertical leg)
Example 13.2.2
Using the Pythagorean Theorem
  • ProblemA right triangle has legs of length 8 and 15. Find the hypotenuse.
  • Solve\[ 8^2 + 15^2 = c^2 \implies 64 + 225 = 289 \implies c = \sqrt{289} = 17 \]
  • Check\(8^2 + 15^2 = 289 = 17^2\). ✓
Answer: The hypotenuse is 17.
Section 13.3

Sine, Cosine, and Tangent

The three primary trigonometric ratios each compare two sides of a right triangle relative to a specific angle \(\theta\). These ratios hold true for any right triangle containing the same angle, regardless of size.

SOH-CAH-TOA

\[ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \qquad \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \qquad \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \]

SOH: Sine = Opposite / Hypotenuse    CAH: Cosine = Adjacent / Hypotenuse    TOA: Tangent = Opposite / Adjacent

Example 13.3.1
Writing the three trigonometric ratios
  • ProblemA right triangle has angle \(\theta\) with opposite = 7, adjacent = 24, hypotenuse = 25. Write all three ratios.
  • CheckVerify: \(7^2 + 24^2 = 49 + 576 = 625 = 25^2\). ✓
  • Solve\[ \sin(\theta) = \frac{7}{25} \qquad \cos(\theta) = \frac{24}{25} \qquad \tan(\theta) = \frac{7}{24} \]
Answer: sin(θ) = 7/25, cos(θ) = 24/25, tan(θ) = 7/24
Example 13.3.2
Evaluating trigonometric ratios with a calculator
  • ProblemFind \(\sin(35°)\), \(\cos(35°)\), and \(\tan(35°)\) to four decimal places.
  • ThinkUse a scientific calculator. Confirm the calculator is in degree mode before computing.
  • Solve\[ \sin(35°) \approx 0.5736 \qquad \cos(35°) \approx 0.8192 \qquad \tan(35°) \approx 0.7002 \]
Answer: sin(35°) ≈ 0.5736, cos(35°) ≈ 0.8192, tan(35°) ≈ 0.7002
Student Warning — Calculator Mode

One of the most common early errors is leaving a calculator in radian mode when working with degree problems. Before computing any trigonometric value, check the mode setting. A quick check: \(\sin(90°)\) should equal exactly 1 in degree mode.

Practice 13.3Writing Ratios and Using a Calculator
  1. opposite = 3, hyp = 5, adj = 4. Write sin(θ), cos(θ), tan(θ).
  2. opposite = 8, hyp = 17, adj = 15. Write all three ratios.
  3. Find sin(50°) to 4 decimal places.
  4. Find cos(72°) to 4 decimal places.
  5. Find tan(20°) to 4 decimal places.
  6. Find tan(80°) to 4 decimal places. What do you notice?
Section 13.4

Solving Right Triangles

When at least one side and one acute angle of a right triangle are known, trigonometric ratios can be used to find the remaining sides. When two sides are known, inverse trigonometric functions can find missing angles. The procedure: identify the known angle, the known side, and the unknown side; choose the ratio that connects them; solve.

Note on Inverse Functions

The notation \(\sin^{-1}(x)\) does not mean \(\frac{1}{\sin(x)}\). It means the inverse sine function — the function that returns an angle whose sine is \(x\). On most calculators this is labeled \(\sin^{-1}\) or arcsin. The same applies to \(\cos^{-1}\) and \(\tan^{-1}\).

Example 13.4.1
Finding a missing side
  • ProblemA right triangle has a 40° angle and a hypotenuse of 20. Find the side opposite the 40° angle.
  • ThinkKnown: hypotenuse = 20. Unknown: opposite. The ratio connecting opposite to hypotenuse is sine.
  • Solve\[ \sin(40°) = \frac{\text{opposite}}{20} \implies \text{opposite} = 20 \cdot \sin(40°) \approx 20 \cdot 0.6428 \approx 12.86 \]
  • CheckThe adjacent side: \(20 \cdot \cos(40°) \approx 15.32\). Pythagorean check: \(12.86^2 + 15.32^2 \approx 399.7 \approx 20^2\). ✓
Answer: The opposite side is approximately 12.86.
Example 13.4.2
Finding a missing angle
  • ProblemIn a right triangle, the opposite side is 9 and the hypotenuse is 15. Find angle \(\theta\).
  • ThinkThe ratio of opposite to hypotenuse is sine. To find the angle from a ratio, apply the inverse sine function.
  • Solve\[ \sin(\theta) = \frac{9}{15} = 0.6 \implies \theta = \sin^{-1}(0.6) \approx 36.87° \]
  • Check\(\sin(36.87°) \approx 0.6\). ✓
Answer: θ ≈ 36.87°
Example 13.4.3
Fully solving a right triangle
  • ProblemA right triangle has a 55° angle and an adjacent side of 10. Find the opposite leg and the hypotenuse.
  • ThinkUse tangent for the opposite leg (opposite/adjacent). Use cosine for the hypotenuse (adjacent/hypotenuse).
  • Solve\[ \tan(55°) = \frac{\text{opp}}{10} \implies \text{opp} = 10 \cdot \tan(55°) \approx 14.28 \] \[ \cos(55°) = \frac{10}{\text{hyp}} \implies \text{hyp} = \frac{10}{\cos(55°)} \approx 17.43 \]
  • Check\(10^2 + 14.28^2 \approx 303.9 \approx 17.43^2\). ✓
Answer: Opposite leg ≈ 14.28, hypotenuse ≈ 17.43.
Practice 13.4Solving Right Triangles
  1. Angle = 30°, hyp = 24. Find opposite.
  2. Angle = 60°, adjacent = 8. Find opposite and hyp.
  3. Angle = 45°, hyp = 10. Find both legs.
  4. opposite = 5, hyp = 13. Find the angle.
  5. adjacent = 6, opposite = 6. Find the angle.
  6. adjacent = 4, hyp = 9. Find the angle.
Section 13.5

Angles of Elevation and Depression

Two common real-world applications involve angles measured from a horizontal reference line. The angle of elevation is measured upward from horizontal toward an object higher than the observer. The angle of depression is measured downward from horizontal toward an object lower than the observer.

A key geometric fact: when two horizontal lines are parallel, the angle of elevation from the lower level and the angle of depression from the upper level to the same two points are equal — they are alternate interior angles formed by the same transversal.

Diagram Instruction

Draw two horizontal lines at different heights. At the lower level, draw a person looking upward at an object above; label the angle between horizontal and the line of sight as the angle of elevation. At the upper level, draw a person looking downward at an object below; label that angle as the angle of depression. Note the two angles are equal when the horizontal lines are parallel.

Example 13.5.1
Angle of elevation — building height
  • ProblemA person stands 40 meters from the base of a building. The angle of elevation to the top is 65°. How tall is the building?
  • ThinkThe 40 m distance is the adjacent side. The building height is the opposite side. Use tangent.
  • Solve\[ \tan(65°) = \frac{\text{height}}{40} \implies \text{height} = 40 \cdot \tan(65°) \approx 40 \cdot 2.1445 \approx 85.78 \text{ m} \]
Answer: The building is approximately 85.78 meters tall.
Example 13.5.2
Angle of depression — lifeguard tower
  • ProblemA lifeguard in a tower 12 feet above the ground spots a swimmer at an angle of depression of 20°. How far is the swimmer from the base of the tower?
  • ThinkThe tower height (12 ft) is the opposite side. The horizontal distance is the adjacent side. Use tangent.
  • Solve\[ \tan(20°) = \frac{12}{\text{distance}} \implies \text{distance} = \frac{12}{\tan(20°)} \approx \frac{12}{0.3640} \approx 32.97 \text{ ft} \]
Answer: The swimmer is approximately 32.97 feet from the base.
Student Tip — Draw First

Draw a diagram for every angle-of-elevation or angle-of-depression problem. Establish clearly what side is opposite, what side is adjacent, and where the right angle is before choosing any ratio. Rushing without a diagram is the single most common source of error in these problems.

Section 13.6

Special Right Triangles

Two types of right triangles have angle measures that produce exact, simple side ratios. They appear so frequently in mathematics, physics, and engineering that their relationships should be derived and internalized.

The 45°–45°–90° Triangle

An isosceles right triangle where the two legs are equal. If each leg is \(s\), the Pythagorean theorem gives the hypotenuse \(s\sqrt{2}\).

45°–45°–90° Ratios

\[ \text{leg} : \text{leg} : \text{hypotenuse} = 1 : 1 : \sqrt{2} \]

If each leg = \(s\), then hypotenuse = \(s\sqrt{2}\).

The 30°–60°–90° Triangle

Derived by cutting an equilateral triangle in half along its altitude. The short leg (opposite 30°) is half the hypotenuse.

30°–60°–90° Ratios

\[ \text{short leg} : \text{long leg} : \text{hyp} = 1 : \sqrt{3} : 2 \]

If short leg = \(s\), then long leg = \(s\sqrt{3}\) and hypotenuse = \(2s\).

Example 13.6.1
Using the 45°–45°–90° Triangle
  • ProblemA square garden has a diagonal path of length 10 feet. Find the length of each side.
  • ThinkThe diagonal of a square forms two 45°–45°–90° triangles. The diagonal is the hypotenuse: \(s\sqrt{2} = 10\).
  • Solve\[ s = \frac{10}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2} \approx 7.07 \text{ ft} \]
Answer: Each side is \(5\sqrt{2} \approx 7.07\) feet.
Example 13.6.2
Using the 30°–60°–90° Triangle
  • ProblemA right triangle has a 30° angle and a hypotenuse of 16. Find both legs.
  • ThinkShort leg = \(\frac{1}{2}\) × hypotenuse. Long leg = short leg × \(\sqrt{3}\).
  • Solve\[ \text{short leg} = \frac{1}{2}(16) = 8 \qquad \text{long leg} = 8\sqrt{3} \approx 13.86 \]
  • Check\(8^2 + (8\sqrt{3})^2 = 64 + 192 = 256 = 16^2\). ✓
Answer: Short leg = 8, long leg = \(8\sqrt{3} \approx 13.86\).
Student Warning

Do not confuse which leg is opposite 30° and which is opposite 60°. The shorter leg is always opposite the smaller angle (30°). The longer leg is always opposite the larger acute angle (60°).

Exact Values at Special Angles

Angle\(\sin\)\(\cos\)\(\tan\)
30°\(\dfrac{1}{2}\)\(\dfrac{\sqrt{3}}{2}\)\(\dfrac{\sqrt{3}}{3}\)
45°\(\dfrac{\sqrt{2}}{2}\)\(\dfrac{\sqrt{2}}{2}\)\(1\)
60°\(\dfrac{\sqrt{3}}{2}\)\(\dfrac{1}{2}\)\(\sqrt{3}\)
Practice 13.6Special Triangles
  1. 45°–45°–90°, hyp = 8. Find each leg.
  2. 30°–60°–90°, hyp = 10. Find both legs.
  3. 45°–45°–90°, leg = \(3\sqrt{2}\). Find hyp.
  4. 30°–60°–90°, short leg = 7. Find hyp and long leg.
  5. Find exact value: \(\sin(30°)\).
  6. Find exact value: \(\tan(60°)\).
Sections 13.7 – 13.9
Degrees, Radians, and the Unit Circle
Radian Measure · Conversion · Unit Circle Coordinates · ASTC · Reference Angles
Section 13.7

Degrees and Radians

So far, angles have been measured in degrees, where a full rotation equals 360°. In higher mathematics, a different unit called the radian is standard. A radian is the angle formed when an arc of length equal to the radius of a circle is cut off on the circle's circumference.

Because the entire circumference of a circle with radius \(r\) is \(2\pi r\), the total angle in a full rotation equals \(2\pi\) radians. This gives the fundamental conversion:

Degree–Radian Conversion

\[ 180° = \pi \text{ radians} \]

Degrees → Radians: multiply by \(\dfrac{\pi}{180}\)     Radians → Degrees: multiply by \(\dfrac{180}{\pi}\)

Example 13.7.1
Converting degrees to radians
  • ProblemConvert 120° to radians.
  • Solve\[ 120° \times \frac{\pi}{180} = \frac{120\pi}{180} = \frac{2\pi}{3} \]
Answer: \(120° = \dfrac{2\pi}{3}\) radians.
Example 13.7.2
Converting radians to degrees
  • ProblemConvert \(\dfrac{5\pi}{6}\) radians to degrees.
  • Solve\[ \frac{5\pi}{6} \times \frac{180}{\pi} = \frac{5 \times 180}{6} = 150° \]
Answer: \(\dfrac{5\pi}{6}\) radians = 150°.

Common Angle Conversions

DegreesRadiansDegreesRadians
0°\(0\)120°\(\dfrac{2\pi}{3}\)
30°\(\dfrac{\pi}{6}\)135°\(\dfrac{3\pi}{4}\)
45°\(\dfrac{\pi}{4}\)150°\(\dfrac{5\pi}{6}\)
60°\(\dfrac{\pi}{3}\)180°\(\pi\)
90°\(\dfrac{\pi}{2}\)270°\(\dfrac{3\pi}{2}\)
——360°\(2\pi\)
Looking Ahead

Radians may feel unfamiliar at first. In Precalculus and Calculus, they become the standard unit for angle measure. Becoming comfortable with radian measure now will significantly reduce confusion in future courses.

Practice 13.7Degrees and Radians
  1. Convert 30° to radians.
  2. Convert 45° to radians.
  3. Convert 270° to radians.
  4. Convert 210° to radians.
  5. Convert \(\dfrac{\pi}{4}\) to degrees.
  6. Convert \(\dfrac{7\pi}{4}\) to degrees.
Section 13.8

Introduction to the Unit Circle

The unit circle is a circle with radius 1 centered at the origin. It is the central organizing tool for extending trigonometric values beyond right triangles. When an angle \(\theta\) is measured from the positive x-axis and the terminal ray intersects the unit circle, the coordinates of that intersection point are exactly \((\cos\theta,\, \sin\theta)\).

Unit Circle Definition

At any point \(P\) on the unit circle at angle \(\theta\):

\[ P = (\cos\theta,\, \sin\theta) \]

The x-coordinate always equals \(\cos\theta\). The y-coordinate always equals \(\sin\theta\).

Diagram Instruction

Draw a unit circle (radius 1) centered at the origin. Draw a ray from the origin at angle θ from the positive x-axis. Mark the intersection point as P. Draw a vertical dashed line from P to the x-axis. Label the horizontal distance as cos(θ) and the vertical distance as sin(θ). Label P as (cos θ, sin θ). Also mark the four axis-intercept points: (1,0) at 0°, (0,1) at 90°, (−1,0) at 180°, (0,−1) at 270°.

For angles in the first quadrant, this definition matches the right-triangle definition exactly — the hypotenuse is the radius (= 1), the horizontal leg is the cosine, and the vertical leg is the sine. The unit circle's advantage is that it extends trigonometry to angles greater than 90°, negative angles, and multiple rotations.

Section 13.9

Sine and Cosine on the Unit Circle

First Quadrant Key Points

Angle (degrees)Angle (radians)Coordinates \((\cos\theta,\, \sin\theta)\)
0°\(0\)\((1,\, 0)\)
30°\(\dfrac{\pi}{6}\)\(\left(\dfrac{\sqrt{3}}{2},\, \dfrac{1}{2}\right)\)
45°\(\dfrac{\pi}{4}\)\(\left(\dfrac{\sqrt{2}}{2},\, \dfrac{\sqrt{2}}{2}\right)\)
60°\(\dfrac{\pi}{3}\)\(\left(\dfrac{1}{2},\, \dfrac{\sqrt{3}}{2}\right)\)
90°\(\dfrac{\pi}{2}\)\((0,\, 1)\)

Signs by Quadrant — ASTC

When angles extend beyond 90°, the signs of sine and cosine change based on the quadrant the terminal ray falls in. The memory phrase "All Students Take Calculus" encodes which functions are positive:

QuadrantASTC letterPositive functionsExample angle
IAllsin, cos, tan all positive45°
IIStudentssin positive; cos, tan negative150°
IIITaketan positive; sin, cos negative210°
IVCalculuscos positive; sin, tan negative315°
Reference Angles

The reference angle is the positive acute angle between the terminal side and the nearest x-axis. It is always between 0° and 90°. You find the trigonometric value from the reference angle, then apply the correct sign based on the quadrant.

Example 13.9.1
Reading unit circle values — Quadrant II
  • ProblemFind \(\sin(150°)\) and \(\cos(150°)\).
  • Think150° is in Quadrant II. Reference angle = \(180° - 150° = 30°\). In QII: sin positive, cos negative.
  • Solve\[ \sin(150°) = +\sin(30°) = \frac{1}{2} \qquad \cos(150°) = -\cos(30°) = -\frac{\sqrt{3}}{2} \]
Answer: \(\sin(150°) = \dfrac{1}{2}\), \(\cos(150°) = -\dfrac{\sqrt{3}}{2}\)
Example 13.9.2
Quadrant III — 225°
  • ProblemFind \(\sin(225°)\) and \(\cos(225°)\).
  • Think225° is in Quadrant III. Reference angle = \(225° - 180° = 45°\). In QIII: both sin and cos are negative.
  • Solve\[ \sin(225°) = -\frac{\sqrt{2}}{2} \qquad \cos(225°) = -\frac{\sqrt{2}}{2} \]
Answer: \(\sin(225°) = -\dfrac{\sqrt{2}}{2}\), \(\cos(225°) = -\dfrac{\sqrt{2}}{2}\)
Practice 13.9Unit Circle — Exact Values

For each angle: identify the quadrant, state the reference angle, and give exact values.

  1. \(\sin(120°)\) and \(\cos(120°)\)
  2. \(\sin(135°)\) and \(\cos(135°)\)
  3. \(\sin(210°)\) and \(\cos(210°)\)
  4. \(\sin(240°)\) and \(\cos(240°)\)
  5. \(\sin(300°)\) and \(\cos(300°)\)
  6. \(\sin(315°)\) and \(\cos(315°)\)
Sections 13.10 – 13.13
Graphs, Transformations, Tangent, and Trigonometric Models
Sine and Cosine Graphs · Amplitude · Period · Midline · Tangent Function · Periodic Models
Section 13.10

Graphs of Sine and Cosine

The sine and cosine functions can be graphed by placing angle measure on the horizontal axis and function value on the vertical axis. These graphs reveal the periodic, cyclical behavior of the functions.

Key Values — Sine

\(x\) (radians)\(\sin(x)\)\(x\) (radians)\(\sin(x)\)
\(0\)\(0\)\(\pi\)\(0\)
\(\dfrac{\pi}{6}\)\(\dfrac{1}{2}\)\(\dfrac{3\pi}{2}\)\(-1\)
\(\dfrac{\pi}{2}\)\(1\)\(2\pi\)\(0\)
Diagram Instruction — Graph of y = sin(x)

Draw a coordinate plane. Label the horizontal axis from 0 to 2π, marked at π/2, π, 3π/2, and 2π. The vertical axis runs from −1 to 1. Draw the smooth S-shaped curve: starting at (0, 0), rising to a maximum of 1 at π/2, returning to 0 at π, falling to a minimum of −1 at 3π/2, and returning to 0 at 2π. Label the maximum, minimum, and all x-intercepts.

Properties of y = sin(x) and y = cos(x)

Sine: starts at origin (0, 0); max at \(\frac{\pi}{2}\); returns to 0 at \(\pi\); min at \(\frac{3\pi}{2}\); x-intercepts at multiples of \(\pi\).

Cosine: starts at maximum (0, 1); crosses 0 at \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\); min at \(\pi\); x-intercepts at odd multiples of \(\frac{\pi}{2}\).

Both functions share: Domain = all reals  ·  Range = \([-1,\, 1]\)  ·  Period = \(2\pi\)  ·  Amplitude = 1  ·  Midline: \(y = 0\).

The cosine graph is the sine graph shifted left by \(\dfrac{\pi}{2}\) — identical shape, different starting position.

Section 13.11

Amplitude, Period, and Midline

Real-world periodic phenomena do not always have amplitude 1, period \(2\pi\), or midline \(y = 0\). The general forms of sine and cosine allow these features to be adjusted.

General Form

\[ y = a\sin(bx) + k \qquad \text{and} \qquad y = a\cos(bx) + k \]

Amplitude = \(|a|\) — maximum distance from the midline to a peak or valley.

Period = \(\dfrac{2\pi}{|b|}\) — horizontal length of one complete cycle.

Midline = \(y = k\) — the horizontal center line of oscillation.

If \(a < 0\), the graph is reflected vertically (flipped upside down). Amplitude remains \(|a|\).

Example 13.11.1
Identifying features from an equation
  • ProblemIdentify the amplitude, period, and midline of \(y = 3\sin(2x) - 1\).
  • ThinkMatch to \(y = a\sin(bx) + k\): here \(a = 3\), \(b = 2\), \(k = -1\).
  • Solve\[ \text{Amplitude} = |3| = 3 \qquad \text{Period} = \frac{2\pi}{2} = \pi \qquad \text{Midline}: y = -1 \]
  • AnswerThe graph oscillates between \(y = -4\) (min) and \(y = 2\) (max), centered on \(y = -1\), completing one cycle every \(\pi\) units.
Answer: Amplitude = 3, Period = π, Midline y = −1.
Example 13.11.2
Sketching a transformed cosine function
  • ProblemIdentify features and sketch \(y = 2\cos\!\left(\dfrac{x}{2}\right) + 1\).
  • Think\(a = 2\), \(b = \frac{1}{2}\), \(k = 1\).
  • Solve\[ \text{Amplitude} = 2 \qquad \text{Period} = \frac{2\pi}{1/2} = 4\pi \qquad \text{Midline}: y = 1 \] Max = \(1 + 2 = 3\), Min = \(1 - 2 = -1\). One cycle spans \(4\pi\) horizontal units, starting at the maximum.
Answer: Amplitude = 2, Period = 4π, Midline y = 1.
Diagram Instruction — Example 13.11.2

Sketch a cosine curve starting at maximum (0, 3), descending through the midline (π, 1), reaching minimum (2π, −1), returning through the midline (3π, 1), and completing the cycle at maximum (4π, 3). Draw the midline as a dashed horizontal line at y = 1. Label the maximum, minimum, and midline values.

Example 13.11.3
Writing an equation from graph features
  • ProblemA sine function has amplitude 5, period \(3\pi\), and midline \(y = 2\). Write its equation.
  • ThinkUse \(y = a\sin(bx) + k\). Find \(b\) from the period: \(\dfrac{2\pi}{b} = 3\pi\).
  • Solve\[ a = 5,\quad k = 2,\quad b = \frac{2\pi}{3\pi} = \frac{2}{3} \] \[ y = 5\sin\!\left(\frac{2}{3}x\right) + 2 \]
Answer: \(y = 5\sin\!\left(\dfrac{2}{3}x\right) + 2\)
Practice 13.11Amplitude, Period, Midline
  1. Identify features: \(y = 4\sin(x)\)
  2. Identify features: \(y = \sin(3x)\)
  3. Identify features: \(y = -2\cos(x) + 3\)
  4. Identify features: \(y = \frac{1}{2}\sin(\pi x) - 1\)
  5. Write equation: amplitude 6, period \(4\pi\), midline \(y = 0\), sine.
  6. Write equation: amplitude 2, period \(\pi\), midline \(y = 5\), cosine.
Section 13.12

Tangent as a Function

The tangent function is defined as the ratio of sine to cosine:

Tangent Definition and Key Properties

\[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \]

Wherever \(\cos(\theta) = 0\), tangent is undefined — vertical asymptotes occur at all odd multiples of \(\dfrac{\pi}{2}\).

Domain: all reals except \(\dfrac{\pi}{2} + n\pi\)  ·  Range: all real numbers  ·  Period: \(\pi\) (repeats every half-rotation)  ·  No amplitude (no maximum or minimum).

Diagram Instruction — Graph of y = tan(x)

Draw a coordinate plane with horizontal axis from −π/2 to 5π/2. Draw vertical dashed lines at x = π/2 and x = 3π/2 as vertical asymptotes. In each interval between asymptotes, sketch the tangent curve: it rises from negative infinity near the left asymptote, passes through (0, 0), crosses through (π/4, 1), and rises steeply toward the right asymptote. The pattern repeats every π units. Each branch has the same rising S-like shape.

Example 13.12.1
Evaluating tangent using sine and cosine
  • ProblemEvaluate \(\tan(150°)\).
  • ThinkUse \(\tan = \sin/\cos\). From Section 13.9: \(\sin(150°) = \frac{1}{2}\), \(\cos(150°) = -\frac{\sqrt{3}}{2}\).
  • Solve\[ \tan(150°) = \frac{1/2}{-\sqrt{3}/2} = \frac{1}{2} \cdot \frac{-2}{\sqrt{3}} = \frac{-1}{\sqrt{3}} = -\frac{\sqrt{3}}{3} \]
Answer: \(\tan(150°) = -\dfrac{\sqrt{3}}{3}\)
Section 13.13

Trigonometric Models

Periodic phenomena occur throughout science, nature, technology, and daily life. Any situation that repeats in regular intervals can potentially be modeled by a sine or cosine function. Given the maximum value, the minimum value, and the period of a periodic situation, the following procedure identifies the model.

Setting Up a Trigonometric Model

Step 1 — Amplitude: \(a = \dfrac{\text{max} - \text{min}}{2}\)

Step 2 — Midline: \(k = \dfrac{\text{max} + \text{min}}{2}\)

Step 3 — Find b from period: \(b = \dfrac{2\pi}{\text{period}}\)

Step 4 — Choose sine or cosine based on the starting value. If the function starts at the midline moving upward, use sine. If it starts at the maximum, use cosine.

Example 13.13.1
Daily temperature model
  • ProblemA city's average monthly temperature reaches a maximum of 95°F in July (month 7) and a minimum of 35°F in January (month 1). Write a cosine model for temperature \(T\) as a function of month \(m\), then estimate April (month 4).
  • ThinkPeriod = 12 months. Max = 95°F, min = 35°F. Maximum occurs at month 7, so shift the cosine peak there.
  • Solve\[ a = \frac{95-35}{2} = 30 \qquad k = \frac{95+35}{2} = 65 \qquad b = \frac{2\pi}{12} = \frac{\pi}{6} \] \[ T(m) = 30\cos\!\left(\frac{\pi}{6}(m - 7)\right) + 65 \] April: \(T(4) = 30\cos\!\left(\frac{\pi}{6}(4-7)\right) + 65 = 30\cos\!\left(-\frac{\pi}{2}\right) + 65 = 30(0) + 65 = 65°F\)
  • CheckApril is a transitional month midway between winter and summer, so the midline value (65°F) is a reasonable answer. ✓
Answer: \(T(m) = 30\cos\!\left(\dfrac{\pi}{6}(m-7)\right) + 65\). April temperature ≈ 65°F.
Example 13.13.2
Tidal height model
  • ProblemA harbor's tidal depth ranges from a low of 6 feet to a high of 14 feet. The period from high tide to high tide is 12.5 hours. Write a model for depth \(H(t)\) assuming the cycle begins at high tide.
  • Solve\[ a = \frac{14-6}{2} = 4 \qquad k = \frac{14+6}{2} = 10 \qquad b = \frac{2\pi}{12.5} = \frac{4\pi}{25} \] Starting at high tide (maximum), use cosine: \[ H(t) = 4\cos\!\left(\frac{4\pi}{25}t\right) + 10 \]
Answer: \(H(t) = 4\cos\!\left(\dfrac{4\pi}{25}t\right) + 10\), where \(t\) is hours after high tide.
Always Define Your Variable

In modeling problems, always state what your variable represents and what units you are using. Writing "where \(t\) is measured in hours" is part of a complete mathematical answer, not an optional detail.

Practice 13.13Trigonometric Models
  1. A city experiences 15 hours of daylight in June and 9 hours in December. Write a cosine model for daylight hours \(D(m)\) where \(m = 6\) is June.
  2. A Ferris wheel starts at its lowest point (2 feet above ground) and reaches a maximum of 50 feet above ground. One revolution takes 40 seconds. Write a sine model for height \(h(t)\).
  3. Room temperature varies between 68°F and 74°F on a 6-hour cycle, starting at the maximum. Write a cosine model for temperature \(T(t)\).
  4. A sound wave has amplitude 0.004 (pressure units) and frequency 440 Hz. Write a sine model for pressure \(P(t)\) where period = \(\frac{1}{440}\) seconds.
Section 13.14

Common Trigonometry Mistakes

Mistake 1 — Calculator in the Wrong Mode
sin(30) in radian mode ≈ −0.988

Correct: sin(30°) = 0.5. Always verify the mode setting first. A quick check: sin(90°) should equal exactly 1 in degree mode.

Mistake 2 — Confusing Opposite, Adjacent, and Hypotenuse
Using sine when opposite and hypotenuse are known, then writing opposite/adjacent

Correct: Always label all three sides explicitly before choosing a ratio. The hypotenuse is always across from the right angle. The opposite and adjacent depend on which angle you are using.

Mistake 3 — Treating sin⁻¹ as a Reciprocal
sin⁻¹(0.5) = 1/sin(0.5) ≈ 1.919

Correct: sin⁻¹(0.5) = 30°. The inverse sine function returns an angle whose sine is 0.5. It is not the same as 1/sin(0.5).

Mistake 4 — Wrong Sign in Non-First-Quadrant Angles
cos(210°) = √3/2 (positive first-quadrant value used without sign adjustment)

Correct: 210° is in Quadrant III where cosine is negative. cos(210°) = −√3/2. Always apply the ASTC sign rule after finding the reference-angle value.

Mistake 5 — Swapping Amplitude and Period Formula
In y = a·sin(bx), saying "the period is a" or "the amplitude is 2π/b"

Correct: Amplitude = |a| (the coefficient in front). Period = 2π/|b| (depends on b, the coefficient inside). These are not interchangeable.

Mistake 6 — Claiming Tangent Has an Amplitude
"The amplitude of y = tan(x) is 1."

Correct: Tangent has no amplitude. Its range is all real numbers — it grows without bound in both directions and has no maximum or minimum.

Mistake 7 — Backwards Degree–Radian Conversion
90° to radians: 90 × 180/π (multiplying by the reciprocal of the correct factor)

Correct: Degrees to radians uses π/180. So 90 × π/180 = π/2 radians. Remember: to go to radians, multiply by π/180.

Mistake 8 — Evaluating tan(90°) as a Large Number
tan(90°) = 1,000,000 (just a very large number on the calculator)

Correct: tan(90°) is undefined because cos(90°) = 0, and division by zero is undefined. The calculator error or overflow message is correct — this value does not exist.

Chapter 13 — Back Matter

Chapter Summary

Key Terms Review

trigonometry
The branch of mathematics that studies relationships between angles and sides in triangles, and the behavior of periodic functions.
hypotenuse
The side of a right triangle that is across from the 90° angle; always the longest side.
opposite side
The side of a right triangle directly across from the angle being considered (not the 90° angle).
adjacent side
The side of a right triangle that forms the angle being considered and is not the hypotenuse.
sine
The ratio opposite/hypotenuse for a given angle in a right triangle. Written sin(θ).
cosine
The ratio adjacent/hypotenuse for a given angle in a right triangle. Written cos(θ).
tangent
The ratio opposite/adjacent, also equal to sin/cos. Written tan(θ).
SOH-CAH-TOA
A memory device: Sine = Opp/Hyp, Cosine = Adj/Hyp, Tangent = Opp/Adj.
inverse trig functions
sin⁻¹, cos⁻¹, tan⁻¹: functions that return an angle when given a ratio. Also called arcsin, arccos, arctan.
angle of elevation
The angle measured upward from horizontal to a line of sight toward a higher object.
angle of depression
The angle measured downward from horizontal to a line of sight toward a lower object.
special right triangle
A 45°–45°–90° or 30°–60°–90° triangle with side ratios that produce exact trig values.
radian
A unit of angle measure; the angle formed when arc length equals the radius. 2π radians = 360°.
unit circle
A circle of radius 1 centered at the origin. Points on it at angle θ have coordinates (cos θ, sin θ).
reference angle
The acute angle between the terminal side of an angle and the nearest x-axis. Always between 0° and 90°.
ASTC
"All Students Take Calculus": QI all positive, QII sine positive, QIII tangent positive, QIV cosine positive.
periodic function
A function that repeats its output values at regular horizontal intervals called periods.
amplitude
The maximum distance a periodic function reaches above or below its midline. Equal to |a| in y = a·sin(bx) + k.
period
The horizontal length of one complete cycle of a periodic function. For y = a·sin(bx) + k, period = 2π/|b|.
midline
The horizontal line y = k centered between the maximum and minimum of a periodic function.
vertical asymptote
A vertical line where a function's value grows without bound; tangent has asymptotes at odd multiples of π/2.
trigonometric model
A sine or cosine function used to represent a real-world periodic situation, with amplitude, period, and midline derived from the data.

Concept Review Questions

  1. What does a trigonometric ratio measure? Why do two right triangles with the same angles but different sizes have identical trigonometric ratios?
  2. Explain how the labels "opposite" and "adjacent" depend on which angle you are using. Does the hypotenuse also shift?
  3. Describe the difference between an angle of elevation and an angle of depression. When are they equal to each other, and why?
  4. What are the side ratios in a 45°–45°–90° triangle? In a 30°–60°–90° triangle? How can both be derived rather than simply memorized?
  5. Explain the relationship between degrees and radians. Why is 180° = π, and not some other value?
  6. What is the unit circle? How does a point on the unit circle relate to the cosine and sine of the angle at that point?
  7. What is a reference angle? How is it used to evaluate trig functions in the second, third, and fourth quadrants?
  8. Describe the shape of y = sin(x). What are its key features: domain, range, period, amplitude, midline, x-intercepts?
  9. In y = a·sin(bx) + k, what does each constant control? What happens to the graph if a is negative?
  10. Why does the tangent function have vertical asymptotes? Where do they occur, and why does the period equal π rather than 2π?
  11. Give two real-world examples of periodic phenomena that could be modeled by sine or cosine. What would the period represent in each case?
  12. What distinguishes the graph of tangent from the graphs of sine and cosine?

Skill Practice

Section ARight Triangles — Writing Ratios
  1. opp = 5, adj = 12, hyp = 13. Write sin(θ), cos(θ), tan(θ).
  2. opp = 8, adj = 15, hyp = 17. Write all three ratios.
  3. sin(θ) = 3/5. Find cos(θ) and tan(θ).
  4. cos(θ) = 5/13. Find sin(θ) and tan(θ).
Section BRight Triangles — Solving for Missing Parts
  1. Angle = 28°, hyp = 30. Find opp and adj.
  2. Angle = 52°, adj = 11. Find opp and hyp.
  3. Angle = 70°, opp = 18. Find adj and hyp.
  4. opp = 6, hyp = 10. Find the angle.
  5. adj = 7, opp = 7. Find the angle.
  6. adj = 4, hyp = 9. Find the angle.
Section CAngles of Elevation and Depression
  1. Person stands 80 m from a cliff; elevation angle = 42°. How tall is the cliff?
  2. Observer tower 30 ft high; depression angle = 15°. Distance to object?
  3. 20-foot ladder at 65° angle with ground. How high on the wall?
  4. Angle of elevation to a kite: 55°; string length: 45 ft. Height of kite?
Section DSpecial Triangles and Exact Values
  1. Exact: sin(30°), cos(30°), tan(30°)
  2. Exact: sin(45°), cos(45°), tan(45°)
  3. Exact: sin(60°), cos(60°), tan(60°)
  4. 30°–60°–90°, hyp = 20. Find both legs.
  5. 45°–45°–90°, leg = \(6\sqrt{2}\). Find hyp.
  6. Convert 210° to radians.
Section EUnit Circle — Exact Values
  1. sin(120°) and cos(120°)
  2. sin(150°) and cos(150°)
  3. sin(240°) and cos(240°)
  4. sin(300°) and cos(300°)
  5. sin(315°) and cos(315°)
  6. tan(135°)
  7. tan(210°)
  8. All angles in [0°, 360°] where cos(θ) = √2/2
Section FGraphs — Features and Equations
  1. Amplitude, period, midline: \(y = 4\sin(x)\)
  2. Amplitude, period, midline: \(y = \sin(3x)\)
  3. Amplitude, period, midline: \(y = 2\sin(\pi x) + 3\)
  4. Amplitude, period, midline: \(y = -3\cos(2x)\)
  5. Write equation: sine, amp = 6, period = \(4\pi\), midline \(y=0\).
  6. Write equation: cosine, amp = 2, period = \(\pi\), midline \(y=5\).
  7. Write equation: max = 10, min = 2, period = 8, sine.
  8. Write equation: max = −1, min = −7, period = \(6\pi\), cosine.

Word Problems

For each problem: define your variables, draw a diagram if applicable, set up and solve, and interpret the answer in context.

  1. Seasonal Daylight. A city has a maximum of 15 hours of daylight in June and a minimum of 9 hours in December. The cycle is annual (period = 12 months). Write a cosine model for daylight hours \(D(m)\) where month 6 is June. Estimate daylight hours in September (month 9).
  2. Ferris Wheel. A passenger on a Ferris wheel starts at the lowest point, 2 feet above the ground. The highest point is 50 feet above the ground. The wheel completes one revolution every 40 seconds. Write a sine model for height \(h(t)\) in feet at time \(t\) seconds. Find the height at \(t = 10\) seconds.
  3. Tidal Cycle. The depth of water in a harbor ranges from 5 feet (low tide) to 17 feet (high tide). Tides cycle every 12.5 hours, beginning at low tide. Write a cosine model for depth \(D(t)\) and find the depth at \(t = 3\) hours.
  4. Guy Wire. A 50-foot cable is anchored to the ground and attached to the top of a vertical pole. The cable makes a 62° angle with the ground. How tall is the pole? (This was the opening question of the chapter — now you can answer it.)
  5. Building Height. From a point 120 feet from a building's base, the angle of elevation to the roof is 35° and the angle of elevation to the top of a rooftop antenna is 39°. Find the height of the building and the height of the antenna.
  6. Navigation. A ship leaves port and travels 40 miles due north, then turns and travels 30 miles due east. What angle does the ship's return path make with due south? How far is the ship from port?

Error Analysis

Each problem contains a student error. Identify the error, explain what went wrong, and provide the correct solution.

Problem 1
Find the side opposite a 50° angle in a right triangle with hyp = 14. Student: cos(50°) = opp/14, so opp = 14·cos(50°) ≈ 9.00.

What ratio connects the opposite side to the hypotenuse? What is the correct answer?

Problem 2
"Find the angle whose tangent is 2." Student wrote: θ = 1/tan(2) ≈ −0.457°.

What function is needed to find an angle from a tangent value? What is the correct answer?

Problem 3
"Find sin(240°)." Student: "240° is in QIII. Reference angle = 240° − 180° = 60°. So sin(240°) = sin(60°) = √3/2."

The reference angle is correct, but what critical step was omitted? What is the correct answer?

Problem 4
"Identify the period of y = 3sin(4x)." Student: "The period is b = 4."

What is the correct formula for period? What is the correct period of this function?

Problem 5
"Convert 150° to radians." Student: 150 × 180/π = 27000/π.

Which direction does this conversion go? What factor should be used, and what is the correct answer?

Chapter 13 Checkpoint

Chapter 13 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 13
Part A — Right Triangles
  1. A right triangle has a 38° angle and an adjacent side of 15. Find the opposite side and the hypotenuse. Round to two decimal places.
  2. In a right triangle, the two legs are 9 and 40. Find the hypotenuse and both acute angles.
  3. From the top of a 45-meter lighthouse, the angle of depression to a boat is 22°. How far is the boat from the base of the lighthouse?
Part B — Special Angles and Radians
  1. Without a calculator, find the exact value of \(\tan(60°)\).
  2. Convert \(\dfrac{7\pi}{6}\) to degrees.
  3. Convert 210° to radians. Leave the answer as an exact fraction involving \(\pi\).
Part C — Unit Circle and Quadrants
  1. Find the exact values of \(\sin(315°)\) and \(\cos(315°)\).
  2. In which quadrant does the terminal side of a 200° angle lie? What is its reference angle?
  3. Find all angles from 0° to 360° where \(\tan(\theta) = 1\).
Part D — Graphs
  1. Identify the amplitude, period, and midline of \(y = -5\cos(3x) + 4\).
  2. A sine function has amplitude 3, period \(2\pi\), and midline \(y = -1\). Write its equation.
  3. Sketch one complete cycle of \(y = 2\sin\!\left(\dfrac{\pi}{3}x\right)\). Label the maximum, minimum, midline crossings, and start and end of the cycle.
Part E — Modeling
  1. A buoy bobs in the water. Its maximum height above the seafloor is 20 feet and its minimum is 14 feet. It completes one full cycle every 8 seconds. Write a sine or cosine model for height \(H(t)\) and use it to find the height at \(t = 3\) seconds.
Looking Ahead
Chapter 14: Probability, Statistics, and Algebraic Decision-Making

Chapter 13 introduced the mathematical language of periodic change — trigonometric functions that describe situations repeating in regular cycles. Chapter 14 turns in a different direction. Rather than studying repeating patterns, it studies uncertainty and data: how to count outcomes, measure likelihood, and make mathematical decisions when the future is not fully predictable.

The larger project of Algebra II has been learning to choose and use mathematical models for the world. Exponential functions model growth and decay. Logarithmic functions reverse exponential behavior. Trigonometric functions model cycles. Chapter 14 adds the tools of probability and statistical reasoning — models for situations where the answer is not certain, but where mathematics can still tell us what is likely, what is reasonable, and what conclusions the evidence can and cannot support. Students should understand that every chapter of this course has added a new class of model; Chapter 14 will be the last addition before Precalculus.

MGU Library Connection
  • Trigonometry Reference Sheet — Appendix L of this textbook
  • Precalculus: Law of Sines, Law of Cosines, all six trig functions, identities, parametric equations
  • Calculus I: derivatives and integrals of trigonometric functions
  • Physics: wave behavior, oscillatory motion, forces as vector components
  • Statistics: connections to normal distribution and periodic data analysis (Chapter 14)
  • MGU Dictionary entries: sine, cosine, tangent, radian, unit circle, periodic function, amplitude, period