Chapter 5: Trigonometric Functions
Trigonometric functions are among the most important tools in all of mathematics. They arise naturally from the study of triangles and circles, yet their applications extend far beyond geometry into physics, engineering, music, and data science. In this chapter, we build trigonometry from the ground up: starting with angles and the unit circle, defining all six trigonometric functions, learning to graph them and their transformations, and finally introducing inverse trigonometric functions. If you master this chapter, you will have the foundation needed for calculus and beyond.
5.1 Angles and the Unit Circle
An angle is formed by rotating a ray from an initial side to a terminal side. When we place an angle in standard position, its vertex sits at the origin and its initial side lies along the positive $x$-axis. Angles measured counterclockwise are positive; angles measured clockwise are negative.
Degree and Radian Measure
There are two standard units for measuring angles. Degrees divide one full rotation into 360 equal parts, so a full rotation is $360°$. Radians measure the angle by the length of the arc it subtends on a unit circle. Since the circumference of the unit circle is $2\pi$, a full rotation is $2\pi$ radians.
Definition: Radian Measure
One radian is the measure of a central angle that intercepts an arc equal in length to the radius of the circle. In general, if an arc of length $s$ is subtended by a central angle $\theta$ in a circle of radius $r$, then:
$$\theta = \frac{s}{r}$$
The conversion between degrees and radians follows from the fact that $360° = 2\pi$ radians:
$$\text{Degrees to radians: } \theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180}$$
$$\text{Radians to degrees: } \theta_{\text{deg}} = \theta_{\text{rad}} \times \frac{180}{\pi}$$
The Unit Circle
The unit circle is the circle of radius 1 centered at the origin: $x^2 + y^2 = 1$. It is the single most important diagram in trigonometry. Given any angle $\theta$ in standard position, the terminal side intersects the unit circle at a unique point $P$. We define:
$$\cos\theta = x\text{-coordinate of }P, \quad \sin\theta = y\text{-coordinate of }P$$
This definition works for all angles, not just acute angles in right triangles. It unifies trigonometry across all four quadrants and for angles greater than $360°$ or less than $0°$.
Reference Angles
A reference angle is the acute angle $\alpha$ formed between the terminal side of an angle and the $x$-axis. The reference angle always lies between $0°$ and $90°$ (or between $0$ and $\frac{\pi}{2}$ radians). Here is how to find it in each quadrant:
- Quadrant I: $\alpha = \theta$
- Quadrant II: $\alpha = 180° - \theta$ (or $\pi - \theta$)
- Quadrant III: $\alpha = \theta - 180°$ (or $\theta - \pi$)
- Quadrant IV: $\alpha = 360° - \theta$ (or $2\pi - \theta$)
The value of any trigonometric function at angle $\theta$ equals $\pm$ the value of that function at the reference angle $\alpha$. The sign depends on which quadrant $\theta$ is in (we cover the sign rules in Section 5.2).
Special Angles
There are three special angles whose trigonometric values you must memorize. These come from the $30$-$60$-$90$ triangle (half of an equilateral triangle) and the $45$-$45$-$90$ triangle (half of a square cut along the diagonal).
| Degrees | Radians | $\sin\theta$ | $\cos\theta$ | $\tan\theta$ |
|---|---|---|---|---|
| $0°$ | $0$ | $0$ | $1$ | $0$ |
| $30°$ | $\dfrac{\pi}{6}$ | $\dfrac{1}{2}$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{\sqrt{3}}{3}$ |
| $45°$ | $\dfrac{\pi}{4}$ | $\dfrac{\sqrt{2}}{2}$ | $\dfrac{\sqrt{2}}{2}$ | $1$ |
| $60°$ | $\dfrac{\pi}{3}$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{1}{2}$ | $\sqrt{3}$ |
| $90°$ | $\dfrac{\pi}{2}$ | $1$ | $0$ | undefined |
Memory tip: For sine at $0°, 30°, 45°, 60°, 90°$, the values follow the pattern $\frac{\sqrt{0}}{2}, \frac{\sqrt{1}}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{4}}{2}$. For cosine, the same sequence runs in reverse.
Example 5.1.1 — Converting Between Degrees and Radians
Problem: Convert $150°$ to radians. Convert $\dfrac{5\pi}{4}$ radians to degrees.
Step 1 Degrees to radians: multiply by $\frac{\pi}{180}$.
$$150° \times \frac{\pi}{180} = \frac{150\pi}{180} = \frac{5\pi}{6}$$
Step 2 Radians to degrees: multiply by $\frac{180}{\pi}$.
$$\frac{5\pi}{4} \times \frac{180}{\pi} = \frac{5 \times 180}{4} = \frac{900}{4} = 225°$$
Example 5.1.2 — Finding Exact Trig Values Using the Unit Circle
Problem: Find $\sin\left(\dfrac{5\pi}{3}\right)$ and $\cos\left(\dfrac{5\pi}{3}\right)$.
Step 1 Convert to degrees to identify the quadrant: $\frac{5\pi}{3} = 300°$. This is in Quadrant IV.
Step 2 Find the reference angle: $\alpha = 360° - 300° = 60°$.
Step 3 Use the special angle values for $60°$ and apply quadrant signs. In Quadrant IV, cosine is positive and sine is negative.
$$\cos\left(\frac{5\pi}{3}\right) = +\cos 60° = \frac{1}{2}$$
$$\sin\left(\frac{5\pi}{3}\right) = -\sin 60° = -\frac{\sqrt{3}}{2}$$
5.2 The Six Trigonometric Functions
Using the unit circle, we have already defined sine and cosine. The remaining four trigonometric functions are built from these two.
Definitions of the Six Trigonometric Functions
For a point $P(x, y)$ on the unit circle corresponding to angle $\theta$:
$$\sin\theta = y \qquad \cos\theta = x \qquad \tan\theta = \frac{y}{x} = \frac{\sin\theta}{\cos\theta} \; (x \neq 0)$$
$$\csc\theta = \frac{1}{y} = \frac{1}{\sin\theta} \; (y \neq 0) \qquad \sec\theta = \frac{1}{x} = \frac{1}{\cos\theta} \; (x \neq 0) \qquad \cot\theta = \frac{x}{y} = \frac{\cos\theta}{\sin\theta} \; (y \neq 0)$$
Notice that tangent and secant are undefined when $\cos\theta = 0$ (at $\theta = \frac{\pi}{2}, \frac{3\pi}{2}, \ldots$), while cosecant and cotangent are undefined when $\sin\theta = 0$ (at $\theta = 0, \pi, 2\pi, \ldots$).
Signs in Each Quadrant: The ASTC Rule
Since $\cos\theta$ gives the $x$-coordinate and $\sin\theta$ gives the $y$-coordinate on the unit circle, the signs of the trigonometric functions depend on the quadrant:
| Quadrant | $\sin\theta$ | $\cos\theta$ | $\tan\theta$ | Positive functions |
|---|---|---|---|---|
| I ($0° < \theta < 90°$) | $+$ | $+$ | $+$ | All |
| II ($90° < \theta < 180°$) | $+$ | $-$ | $-$ | Sine (and csc) |
| III ($180° < \theta < 270°$) | $-$ | $-$ | $+$ | Tangent (and cot) |
| IV ($270° < \theta < 360°$) | $-$ | $+$ | $-$ | Cosine (and sec) |
A common mnemonic is "All Students Take Calculus" (ASTC), reading counterclockwise from Quadrant I. In Quadrant I, All functions are positive. In II, only Sine. In III, only Tangent. In IV, only Cosine.
Pythagorean Identities
Since any point $(x, y)$ on the unit circle satisfies $x^2 + y^2 = 1$, we immediately get the most fundamental identity in trigonometry:
Pythagorean Identities
$$\sin^2\theta + \cos^2\theta = 1$$
Dividing both sides by $\cos^2\theta$: $\quad \tan^2\theta + 1 = \sec^2\theta$
Dividing both sides by $\sin^2\theta$: $\quad 1 + \cot^2\theta = \csc^2\theta$
Example 5.2.1 — Finding All Six Trig Values from One
Problem: Given $\sin\theta = \dfrac{3}{5}$ and $\theta$ is in Quadrant II, find the remaining five trigonometric functions.
Step 1 Use the Pythagorean identity to find $\cos\theta$:
$$\cos^2\theta = 1 - \sin^2\theta = 1 - \frac{9}{25} = \frac{16}{25}$$
Since $\theta$ is in Quadrant II, cosine is negative: $\cos\theta = -\frac{4}{5}$.
Step 2 Compute the remaining functions:
$$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3/5}{-4/5} = -\frac{3}{4}$$
$$\csc\theta = \frac{1}{\sin\theta} = \frac{5}{3}, \quad \sec\theta = \frac{1}{\cos\theta} = -\frac{5}{4}, \quad \cot\theta = \frac{1}{\tan\theta} = -\frac{4}{3}$$
Example 5.2.2 — Evaluating Trig Functions at Non-Standard Angles
Problem: Find the exact value of $\tan\left(\dfrac{7\pi}{6}\right)$.
Step 1 Convert: $\frac{7\pi}{6}$ radians $= 210°$. This is in Quadrant III.
Step 2 Reference angle: $\alpha = 210° - 180° = 30°$.
Step 3 In Quadrant III, tangent is positive (both sine and cosine are negative, so their ratio is positive).
$$\tan\left(\frac{7\pi}{6}\right) = +\tan 30° = \frac{\sqrt{3}}{3}$$
5.3 Graphs of Sine and Cosine
Graphing trigonometric functions is one of the most important skills in precalculus. The graphs of sine and cosine are smooth, wave-like curves that repeat every $2\pi$ units. Understanding how to read and transform these graphs is essential for modeling periodic phenomena such as sound waves, tides, and seasonal temperatures.
The Basic Graphs
The graph of $y = \sin x$ starts at the origin, rises to a maximum of $1$ at $x = \frac{\pi}{2}$, returns to $0$ at $x = \pi$, falls to a minimum of $-1$ at $x = \frac{3\pi}{2}$, and returns to $0$ at $x = 2\pi$. The graph of $y = \cos x$ has the same shape but is shifted $\frac{\pi}{2}$ units to the left: it starts at $1$, falls to $0$ at $\frac{\pi}{2}$, reaches $-1$ at $\pi$, returns to $0$ at $\frac{3\pi}{2}$, and returns to $1$ at $2\pi$.
Key properties of the basic graphs:
- Domain: $(-\infty, \infty)$ for both
- Range: $[-1, 1]$ for both
- Period: $2\pi$ (the length of one complete cycle)
- Amplitude: $1$ (the distance from the midline to a peak or trough)
The General Sinusoidal Form
The general form of a sinusoidal function is:
$$y = A\sin\bigl(B(x - C)\bigr) + D$$
Each parameter controls a specific transformation:
| Parameter | Effect | Formula |
|---|---|---|
| $A$ | Amplitude (vertical stretch/compression). If $A < 0$, the graph is reflected vertically. | Amplitude $= |A|$ |
| $B$ | Horizontal stretch/compression, which changes the period. | Period $= \dfrac{2\pi}{|B|}$ |
| $C$ | Phase shift (horizontal translation). Positive $C$ shifts right; negative $C$ shifts left. | Phase shift $= C$ |
| $D$ | Vertical shift (moves the midline up or down). | Midline: $y = D$ |
Common mistake: Make sure the function is in the form $B(x - C)$, not $Bx - C$. For example, if you see $y = \sin(2x - \pi)$, you must factor: $y = \sin\bigl(2(x - \frac{\pi}{2})\bigr)$. The phase shift is $\frac{\pi}{2}$, not $\pi$.
Interactive Graph: Exploring Sinusoidal Transformations
Use the sliders below to change the values of $A$, $B$, $C$, and $D$ and watch how the graph of $y = A\sin(B(x - C)) + D$ (blue) transforms compared to the base graph $y = \sin x$ (dashed gray).
Drag the sliders to adjust amplitude, period, phase shift, and vertical shift.
Example 5.3.1 — Identifying Parameters from an Equation
Problem: For $y = -3\cos\left(2x + \pi\right) + 1$, find the amplitude, period, phase shift, and vertical shift.
Step 1 Rewrite in standard form by factoring the argument:
$$y = -3\cos\left(2\left(x + \frac{\pi}{2}\right)\right) + 1$$
Step 2 Identify parameters: $A = -3$, $B = 2$, $C = -\frac{\pi}{2}$, $D = 1$.
Step 3 Compute:
- Amplitude: $|A| = |-3| = 3$
- Period: $\frac{2\pi}{|B|} = \frac{2\pi}{2} = \pi$
- Phase shift: $C = -\frac{\pi}{2}$ (shifted $\frac{\pi}{2}$ units to the left)
- Vertical shift: $D = 1$ (midline is $y = 1$)
- The negative $A$ means the graph is reflected vertically (starts going down instead of up).
Example 5.3.2 — Writing an Equation from a Description
Problem: A sinusoidal function has amplitude $4$, period $\pi$, phase shift $\frac{\pi}{3}$ to the right, and midline $y = -2$. Write an equation using sine.
Step 1 From the given information: $|A| = 4$, so $A = 4$ (assuming no reflection). The midline is $D = -2$. The phase shift is $C = \frac{\pi}{3}$.
Step 2 Find $B$ from the period:
$$\text{Period} = \frac{2\pi}{|B|} = \pi \implies |B| = \frac{2\pi}{\pi} = 2$$
Step 3 Assemble the equation:
$$y = 4\sin\left(2\left(x - \frac{\pi}{3}\right)\right) - 2$$
Example 5.3.3 — Graphing a Transformed Sine Function by Hand
Problem: Sketch one full period of $y = 2\sin\left(\frac{1}{2}x - \frac{\pi}{4}\right)$.
Step 1 Rewrite: $y = 2\sin\left(\frac{1}{2}\left(x - \frac{\pi}{2}\right)\right)$.
So $A = 2$, $B = \frac{1}{2}$, $C = \frac{\pi}{2}$, $D = 0$.
Step 2 Period $= \frac{2\pi}{1/2} = 4\pi$. The cycle starts at $x = C = \frac{\pi}{2}$ and ends at $x = \frac{\pi}{2} + 4\pi = \frac{9\pi}{2}$.
Step 3 Find key points by dividing the period into four equal intervals of length $\pi$:
- Start: $x = \frac{\pi}{2}$, $y = 0$ (midline)
- Quarter: $x = \frac{\pi}{2} + \pi = \frac{3\pi}{2}$, $y = 2$ (maximum)
- Half: $x = \frac{\pi}{2} + 2\pi = \frac{5\pi}{2}$, $y = 0$ (midline)
- Three-quarter: $x = \frac{\pi}{2} + 3\pi = \frac{7\pi}{2}$, $y = -2$ (minimum)
- End: $x = \frac{9\pi}{2}$, $y = 0$ (midline)
Plot these five points and connect them with a smooth wave. The graph oscillates between $y = 2$ and $y = -2$ with midline $y = 0$.
5.4 Graphs of Other Trigonometric Functions
The four remaining trigonometric functions — tangent, cotangent, secant, and cosecant — have graphs that look quite different from the smooth waves of sine and cosine. Their most distinctive feature is the presence of vertical asymptotes where the function is undefined.
The Tangent Function
Since $\tan x = \frac{\sin x}{\cos x}$, the tangent function is undefined wherever $\cos x = 0$, which occurs at $x = \frac{\pi}{2} + n\pi$ for any integer $n$. These produce vertical asymptotes.
Key properties of $y = \tan x$:
- Domain: All real numbers except $x = \frac{\pi}{2} + n\pi$
- Range: $(-\infty, \infty)$
- Period: $\pi$ (not $2\pi$)
- No amplitude (the function is unbounded)
- The graph passes through the origin and is increasing on each branch
- It has an inflection point at each $x$-intercept
Interactive Graph: Tangent Function
The tangent function with its vertical asymptotes shown as dashed lines.
The Cotangent Function
Since $\cot x = \frac{\cos x}{\sin x}$, it is undefined wherever $\sin x = 0$, producing vertical asymptotes at $x = n\pi$. The cotangent also has period $\pi$ but decreases on each branch, which is the opposite behavior of tangent.
Secant and Cosecant
The secant function $y = \sec x = \frac{1}{\cos x}$ has vertical asymptotes wherever $\cos x = 0$ and its graph consists of U-shaped and inverted-U-shaped branches opening away from the $x$-axis. Similarly, $y = \csc x = \frac{1}{\sin x}$ has vertical asymptotes wherever $\sin x = 0$. Both secant and cosecant have period $2\pi$ and range $(-\infty, -1] \cup [1, \infty)$.
| Function | Period | Vertical Asymptotes | Range |
|---|---|---|---|
| $\tan x$ | $\pi$ | $x = \frac{\pi}{2} + n\pi$ | $(-\infty, \infty)$ |
| $\cot x$ | $\pi$ | $x = n\pi$ | $(-\infty, \infty)$ |
| $\sec x$ | $2\pi$ | $x = \frac{\pi}{2} + n\pi$ | $(-\infty, -1] \cup [1, \infty)$ |
| $\csc x$ | $2\pi$ | $x = n\pi$ | $(-\infty, -1] \cup [1, \infty)$ |
Example 5.4.1 — Graphing a Transformed Tangent
Problem: Find the period and asymptotes of $y = \tan(2x)$, and sketch one period.
Step 1 The period of $\tan(Bx)$ is $\frac{\pi}{|B|}$. Here $B = 2$, so the period is $\frac{\pi}{2}$.
Step 2 Asymptotes occur where $2x = \frac{\pi}{2} + n\pi$, i.e., $x = \frac{\pi}{4} + \frac{n\pi}{2}$.
Step 3 One period lies between consecutive asymptotes: from $x = -\frac{\pi}{4}$ to $x = \frac{\pi}{4}$. The graph passes through the origin $(0, 0)$ and increases from $-\infty$ to $+\infty$ within this interval.
Key points within this period:
- $x = -\frac{\pi}{8}$: $y = \tan\left(-\frac{\pi}{4}\right) = -1$
- $x = 0$: $y = \tan(0) = 0$
- $x = \frac{\pi}{8}$: $y = \tan\left(\frac{\pi}{4}\right) = 1$
Example 5.4.2 — Identifying a Cosecant Graph
Problem: Find the period and vertical asymptotes of $y = \csc\left(\frac{x}{2}\right)$.
Step 1 Since $\csc$ has the same period formula as $\sin$, the period is $\frac{2\pi}{|B|} = \frac{2\pi}{1/2} = 4\pi$.
Step 2 Vertical asymptotes occur where $\sin\left(\frac{x}{2}\right) = 0$, i.e., $\frac{x}{2} = n\pi$, so $x = 2n\pi$ (at $x = 0, \pm 2\pi, \pm 4\pi, \ldots$).
Step 3 The range is still $(-\infty, -1] \cup [1, \infty)$ because there is no vertical stretch. The graph has U-shaped branches between consecutive asymptotes, alternating between opening upward and downward.
5.5 Inverse Trigonometric Functions
Since trigonometric functions are periodic, they are not one-to-one on their entire domains and therefore do not have inverses unless we restrict their domains. By choosing a specific interval on which the function is one-to-one, we can define inverse trigonometric functions that "undo" the original function.
Inverse Trigonometric Functions
Arcsine: $y = \arcsin x = \sin^{-1} x$ means $\sin y = x$ where $-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}$.
Domain: $[-1, 1]$. Range: $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
Arccosine: $y = \arccos x = \cos^{-1} x$ means $\cos y = x$ where $0 \leq y \leq \pi$.
Domain: $[-1, 1]$. Range: $[0, \pi]$.
Arctangent: $y = \arctan x = \tan^{-1} x$ means $\tan y = x$ where $-\frac{\pi}{2} < y < \frac{\pi}{2}$.
Domain: $(-\infty, \infty)$. Range: $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.
The restricted domains are chosen carefully so that each inverse function gives a unique output for every input. For arcsine and arctangent, the range is centered around $0$. For arccosine, the range starts at $0$ and goes to $\pi$, which covers the top half of the unit circle.
Notation warning: $\sin^{-1}x$ means arcsine, NOT $\frac{1}{\sin x}$. To write the reciprocal, use $(\sin x)^{-1}$ or $\csc x$. This is an unfortunate but standard notational ambiguity.
Evaluating Inverse Trig Functions (Exact Values)
To evaluate an expression like $\arcsin\left(\frac{\sqrt{3}}{2}\right)$, ask: "What angle $\theta$ in the range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ has $\sin\theta = \frac{\sqrt{3}}{2}$?" The answer is $\frac{\pi}{3}$, since $\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$ and $\frac{\pi}{3}$ is in the required range.
For $\arccos\left(-\frac{1}{2}\right)$, we need an angle in $[0, \pi]$ with cosine equal to $-\frac{1}{2}$. The reference angle is $\frac{\pi}{3}$ (since $\cos\frac{\pi}{3} = \frac{1}{2}$), and to get a negative cosine we need Quadrant II: $\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.
Example 5.5.1 — Evaluating Inverse Trig Expressions
Problem: Find the exact values of: (a) $\arctan(-1)$, (b) $\arcsin\left(-\frac{\sqrt{2}}{2}\right)$, (c) $\cos\left(\arcsin\frac{3}{5}\right)$.
(a) We need $\theta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ with $\tan\theta = -1$. The reference angle is $\frac{\pi}{4}$ (since $\tan\frac{\pi}{4} = 1$). The negative value requires $\theta$ in the fourth quadrant portion of the range: $\theta = -\frac{\pi}{4}$.
$$\arctan(-1) = -\frac{\pi}{4}$$
(b) We need $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ with $\sin\theta = -\frac{\sqrt{2}}{2}$. The reference angle is $\frac{\pi}{4}$, and the negative value requires $\theta = -\frac{\pi}{4}$.
$$\arcsin\left(-\frac{\sqrt{2}}{2}\right) = -\frac{\pi}{4}$$
(c) Let $\theta = \arcsin\frac{3}{5}$, so $\sin\theta = \frac{3}{5}$ with $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Since $\frac{3}{5} > 0$, $\theta$ is in Quadrant I where cosine is positive.
$$\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$$
Example 5.5.2 — Composition of Trig and Inverse Trig
Problem: Find the exact value of $\sin\left(\arccos\left(-\frac{3}{4}\right)\right)$.
Step 1 Let $\theta = \arccos\left(-\frac{3}{4}\right)$. Then $\cos\theta = -\frac{3}{4}$ with $\theta \in [0, \pi]$.
Step 2 Since $\cos\theta < 0$, $\theta$ is in Quadrant II where sine is positive.
Step 3 Use the Pythagorean identity:
$$\sin\theta = \sqrt{1 - \cos^2\theta} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$$
We take the positive root because $\theta$ is in Quadrant II (sine is positive there).
5.6 Practice Problems
Test your understanding of Chapter 5 with these problems. Each has a full solution that you can reveal by clicking "Show Solution."
Problem 1
Convert $\dfrac{7\pi}{12}$ radians to degrees.
Show Solution
Multiply by $\frac{180}{\pi}$:
$$\frac{7\pi}{12} \times \frac{180}{\pi} = \frac{7 \times 180}{12} = \frac{1260}{12} = 105°$$
Problem 2
Find the exact values of $\sin\left(\dfrac{3\pi}{4}\right)$ and $\cos\left(\dfrac{3\pi}{4}\right)$.
Show Solution
$\frac{3\pi}{4} = 135°$, which is in Quadrant II. The reference angle is $180° - 135° = 45°$.
In Quadrant II, sine is positive and cosine is negative:
$$\sin\left(\frac{3\pi}{4}\right) = +\sin 45° = \frac{\sqrt{2}}{2}$$
$$\cos\left(\frac{3\pi}{4}\right) = -\cos 45° = -\frac{\sqrt{2}}{2}$$
Problem 3
If $\cos\theta = -\dfrac{5}{13}$ and $\theta$ is in Quadrant III, find $\sin\theta$ and $\tan\theta$.
Show Solution
Use $\sin^2\theta = 1 - \cos^2\theta = 1 - \frac{25}{169} = \frac{144}{169}$.
In Quadrant III, sine is negative: $\sin\theta = -\frac{12}{13}$.
$$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-12/13}{-5/13} = \frac{12}{5}$$
Note that tangent is positive in Quadrant III, which checks out.
Problem 4
Find the amplitude, period, phase shift, and midline of $y = -2\sin\left(3x - \pi\right) + 5$.
Show Solution
Rewrite in standard form: $y = -2\sin\left(3\left(x - \frac{\pi}{3}\right)\right) + 5$.
Parameters: $A = -2$, $B = 3$, $C = \frac{\pi}{3}$, $D = 5$.
- Amplitude: $|A| = 2$
- Period: $\frac{2\pi}{3}$
- Phase shift: $\frac{\pi}{3}$ to the right
- Midline: $y = 5$
Problem 5
Write a cosine equation with amplitude $3$, period $4\pi$, no phase shift, and midline $y = -1$.
Show Solution
We need $|A| = 3$, so $A = 3$. The midline gives $D = -1$. No phase shift means $C = 0$.
Find $B$: Period $= \frac{2\pi}{|B|} = 4\pi \implies |B| = \frac{2\pi}{4\pi} = \frac{1}{2}$.
$$y = 3\cos\left(\frac{1}{2}x\right) - 1$$
Problem 6
Find the period and two consecutive vertical asymptotes of $y = \tan\left(\dfrac{\pi x}{4}\right)$.
Show Solution
Period $= \frac{\pi}{|B|} = \frac{\pi}{\pi/4} = 4$.
Vertical asymptotes occur where $\frac{\pi x}{4} = \frac{\pi}{2} + n\pi$, i.e., $x = 2 + 4n$.
Two consecutive asymptotes: $x = -2$ (when $n = -1$) and $x = 2$ (when $n = 0$).
Problem 7
Evaluate: $\arccos\left(\dfrac{\sqrt{3}}{2}\right)$.
Show Solution
We need $\theta \in [0, \pi]$ such that $\cos\theta = \frac{\sqrt{3}}{2}$.
From the special angles table, $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$, and $\frac{\pi}{6} \in [0, \pi]$.
$$\arccos\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$$
Problem 8
Find the exact value of $\tan\left(\arcsin\dfrac{5}{13}\right)$.
Show Solution
Let $\theta = \arcsin\frac{5}{13}$, so $\sin\theta = \frac{5}{13}$ with $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
Since $\sin\theta > 0$, $\theta$ is in Quadrant I. Find $\cos\theta$:
$$\cos\theta = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$$
$$\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{5/13}{12/13} = \frac{5}{12}$$
Problem 9
A Ferris wheel has a diameter of 40 meters with its center 25 meters above the ground. It completes one revolution every 3 minutes. Write a sinusoidal equation for the height $h(t)$ (in meters) of a rider as a function of time $t$ (in minutes), assuming the rider starts at the bottom.
Show Solution
The amplitude equals the radius: $A = 20$. The midline is the center height: $D = 25$. The period is $3$ minutes, so $B = \frac{2\pi}{3}$.
Since the rider starts at the bottom (minimum height), we can use a negative cosine:
$$h(t) = -20\cos\left(\frac{2\pi}{3}\,t\right) + 25$$
Verification: At $t = 0$: $h(0) = -20(1) + 25 = 5$ m (bottom, since the center is at 25 m and radius is 20 m). At $t = 1.5$: $h(1.5) = -20(-1) + 25 = 45$ m (top). This is correct.
Problem 10
Find all six trigonometric function values for $\theta = \dfrac{11\pi}{6}$.
Show Solution
$\frac{11\pi}{6} = 330°$, which is in Quadrant IV. Reference angle: $360° - 330° = 30°$.
In Quadrant IV: cosine and secant are positive; sine, cosecant, tangent, and cotangent are negative.
$$\sin\frac{11\pi}{6} = -\sin 30° = -\frac{1}{2}$$
$$\cos\frac{11\pi}{6} = +\cos 30° = \frac{\sqrt{3}}{2}$$
$$\tan\frac{11\pi}{6} = -\tan 30° = -\frac{\sqrt{3}}{3}$$
$$\csc\frac{11\pi}{6} = -2, \quad \sec\frac{11\pi}{6} = \frac{2\sqrt{3}}{3}, \quad \cot\frac{11\pi}{6} = -\sqrt{3}$$