Precalculus ยท Chapter 5 โ Trigonometric Functions: Unit Circle Approach
Precalculus ยท Chapter 5
5Trigonometric Functions: Unit Circle Approach
This chapter introduces the trigonometric functions by viewing them as functions of real numbers. The coordinates of points on the unit circle become the values of the sine and cosine functions, leading naturally to graphs, inverse functions, and the modeling of periodic motion.
Applications are diverse โ signal processing, digital encoding of music and video, finding distances to stars, and medical imaging. A central application studied here is periodic motion: the repetitive patterns found in tides, sound waves, vibrating springs, and the daily rising and setting of the sun.
5.1
The Unit Circle
โ The Unit Circleโ Terminal Points on the Unit Circleโ The Reference Number
This section explores the properties of the circle of radius 1 centered at the origin. These properties are used to define the trigonometric functions.
The Unit Circle
The set of points at a distance 1 from the origin forms a circle of radius 1. The equation of this circle is \( x^2 + y^2 = 1 \).
The Unit Circle
The unit circle is the circle of radius 1 centered at the origin in the \(xy\)-plane. Its equation is \( x^2 + y^2 = 1 \).
Figure 1. The unit circle, radius 1, centered at the origin.
EXAMPLE 1A Point on the Unit Circle
Show that the point \( P\!\left(\dfrac{\sqrt{3}}{3},\dfrac{\sqrt{6}}{3}\right) \) is on the unit circle.
Solution
Check whether the point satisfies \( x^2 + y^2 = 1 \).
Since the sum equals 1, the point \(P\) is on the unit circle.
EXAMPLE 2Locating a Point on the Unit Circle
The point \( P\!\left(\dfrac{\sqrt{3}}{2},\,y\right) \) lies on the unit circle in Quadrant IV. Find its \(y\)-coordinate.
Solution
Substitute into the equation: \( \left(\dfrac{\sqrt{3}}{2}\right)^2 + y^2 = 1 \).
\( y^2 = 1 - \dfrac{3}{4} = \dfrac{1}{4} \), so \( y = \pm\dfrac{1}{2} \).
Because \(P\) is in Quadrant IV, the \(y\)-coordinate must be negative: \( y = -\dfrac{1}{2} \).
Terminal Points on the Unit Circle
For a real number \(t\), mark off a distance \(|t|\) along the unit circle starting at \((1,0)\) โ counterclockwise if \(t \geq 0\), clockwise if \(t < 0\). The point reached is the terminal point determined by \(t\). Since the circumference is \(C = 2\pi\), travelling a distance of \(2\pi\) returns to the start.
Terminal Point
The terminal point \(P(x,y)\) determined by \(t\) is the point reached by travelling a distance \(|t|\) along the unit circle from \((1,0)\). Different values of \(t\) can determine the same terminal point.
EXAMPLE 3Finding Terminal Points
Find the terminal point determined by each: (a) \( t = 3\pi \), (b) \( t = -\pi \), (c) \( t = -\dfrac{\pi}{2} \).
Solution
(a) \( t = 3\pi \) gives the terminal point \( (-1, 0) \).
(b) \( t = -\pi \) gives the terminal point \( (-1, 0) \).
(c) \( t = -\dfrac{\pi}{2} \) gives the terminal point \( (0, -1) \).
By the symmetry of the circle about the line \(y = x\), the terminal point for \( t = \dfrac{\pi}{4} \) is found to be \( \left(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}\right) \). Similar methods give the special terminal points below.
(c) Quadrant III: both negative โ \( \left(-\dfrac{\sqrt{3}}{2}, -\dfrac{1}{2}\right) \).
The Reference Number
Reference Number
Let \(t\) be a real number. The reference number \(\bar{t}\) associated with \(t\) is the shortest distance along the unit circle between the terminal point determined by \(t\) and the \(x\)-axis.
Using Reference Numbers to Find Terminal Points
(1) Find the reference number \(\bar{t}\). (2) Find the terminal point \(Q(a,b)\) determined by \(\bar{t}\). (3) The terminal point determined by \(t\) is \(P(\pm a, \pm b)\), with signs chosen by the quadrant of \(t\).
EXAMPLE 5Finding Reference Numbers
Find the reference number for (a) \( t=\dfrac{5\pi}{6} \), (b) \( t=\dfrac{7\pi}{4} \), (c) \( t=-\dfrac{2\pi}{3} \), (d) \( t=5.80 \).
So the terminal point is the same as for \(\dfrac{5\pi}{6}\): \( \left(-\dfrac{\sqrt{3}}{2}, \dfrac{1}{2}\right) \).
The unit circle is the single most important picture in this chapter โ burn it into memory. In mechanical engineering you'll meet it again as the phasor diagram for rotating machinery and AC signals. When you can instantly place \(\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}\) and their reflections in all four quadrants, every later topic (graphs, harmonic motion, phase angle) becomes much easier. Practice sketching it from memory until you no longer need the table.
5.2
Trigonometric Functions of Real Numbers
โ The Trigonometric Functionsโ Values of the Trigonometric Functionsโ Fundamental Identities
Using the coordinates of the terminal point \(P(x,y)\) determined by \(t\), we define six functions. Because they are defined from the unit circle, they are sometimes called the circular functions.
Definition of the Trigonometric Functions
\[ \sin t = y \qquad \cos t = x \qquad \tan t = \frac{y}{x}\ (x \neq 0) \]
\[ \csc t = \frac{1}{y}\ (y \neq 0) \qquad \sec t = \frac{1}{x}\ (x \neq 0) \qquad \cot t = \frac{x}{y}\ (y \neq 0) \]
EXAMPLE 1Evaluating Trigonometric Functions
Find the six trigonometric functions of (a) \( t = \dfrac{\pi}{3} \) and (b) \( t = \dfrac{\pi}{2} \).
(b) Terminal point \((0,1)\): \( \sin\dfrac{\pi}{2}=1,\ \cos\dfrac{\pi}{2}=0,\ \csc\dfrac{\pi}{2}=1,\ \cot\dfrac{\pi}{2}=0 \). Here \( \tan\dfrac{\pi}{2} \) and \( \sec\dfrac{\pi}{2} \) are undefined (\(x=0\)).
TABLE 1 โ Special Values of the Trigonometric Functions
\(t\)
\(\sin t\)
\(\cos t\)
\(\tan t\)
\(\csc t\)
\(\sec t\)
\(\cot t\)
\(0\)
0
1
0
โ
1
โ
\(\frac{\pi}{6}\)
\(\frac{1}{2}\)
\(\frac{\sqrt3}{2}\)
\(\frac{\sqrt3}{3}\)
2
\(\frac{2\sqrt3}{3}\)
\(\sqrt3\)
\(\frac{\pi}{4}\)
\(\frac{\sqrt2}{2}\)
\(\frac{\sqrt2}{2}\)
1
\(\sqrt2\)
\(\sqrt2\)
1
\(\frac{\pi}{3}\)
\(\frac{\sqrt3}{2}\)
\(\frac{1}{2}\)
\(\sqrt3\)
\(\frac{2\sqrt3}{3}\)
2
\(\frac{\sqrt3}{3}\)
\(\frac{\pi}{2}\)
1
0
โ
1
โ
0
Domains of the Trigonometric Functions
\(\sin\) and \(\cos\): all real numbers. \(\tan\) and \(\sec\): all reals except \(\dfrac{\pi}{2}+n\pi\). \(\cot\) and \(\csc\): all reals except \(n\pi\), for any integer \(n\).
Values and Signs
The signs of the trigonometric functions depend on the quadrant of the terminal point. A useful mnemonic is "All Students Take Calculus": All positive in QI, Sine in QII, Tangent in QIII, Cosine in QIV.
Signs of the Trigonometric Functions
QI: all positive. QII: \(\sin,\csc\) positive. QIII: \(\tan,\cot\) positive. QIV: \(\cos,\sec\) positive. All others negative in the respective quadrant.
Evaluating Trigonometric Functions for Any Real Number \(t\)
(1) Find the reference number \(\bar{t}\). (2) Determine the sign from the quadrant. (3) The value at \(t\) equals the value at \(\bar{t}\), up to sign.
The Pythagorean identity \( \sin^2 t + \cos^2 t = 1 \) is just the unit circle equation \( x^2 + y^2 = 1 \) in disguise โ that's why it's always true. The other two come from dividing through by \( \cos^2 t \) or \( \sin^2 t \). In practice (and on exams), if you're given one function value plus a quadrant, you can recover all six. Engineers lean on this constantly when resolving force or velocity vectors into components.
5.3
Trigonometric Graphs
โ Graphs of Sine and Cosineโ Transformationsโ Graphing Devices
Since the terminal point determined by \(t\) is the same as the one determined by \(t + 2\pi\), the sine and cosine functions repeat every \(2\pi\). A function \(f\) is periodic if \( f(t+p) = f(t) \) for some positive \(p\); the least such \(p\) is the period.
Periodic Properties of Sine and Cosine
\[ \sin(t+2\pi)=\sin t \qquad \cos(t+2\pi)=\cos t \]
Figures 2โ3. One full view of \(y=\sin t\) (solid, red) and \(y=\cos t\) (dashed, steel), period \(2\pi\), amplitude 1.
EXAMPLE 1Cosine Curves (Shifts & Reflections)
Sketch (a) \( f(x) = 2 + \cos x \) and (b) \( g(x) = -\cos x \).
Solution
(a) The graph of \( y = \cos x \) shifted up 2 units.
(b) The graph of \( y = \cos x \) reflected in the \(x\)-axis.
Sine and Cosine Curves
\[ y = a\sin kx \quad\text{and}\quad y = a\cos kx \quad (k>0) \]
have amplitude \( |a| \) and period \( \dfrac{2\pi}{k} \). One period graphs on \( \left[0,\dfrac{2\pi}{k}\right] \).
EXAMPLE 2Stretching a Cosine Curve
Find the amplitude of \( y = -3\cos x \) and describe its graph.
Solution
Amplitude \( = |-3| = 3 \); largest value 3, smallest \(-3\).
Start from \( y=\cos x \), stretch vertically by 3, reflect in the \(x\)-axis.
EXAMPLE 3Amplitude and Period
Find the amplitude and period of (a) \( y=4\cos 3x \), (b) \( y = -2\sin\dfrac{1}{2}x \).
Solution
(a) Amplitude 4, period \( \dfrac{2\pi}{3} \).
(b) Amplitude 2, period \( \dfrac{2\pi}{1/2} = 4\pi \).
Shifted Sine and Cosine Curves
\[ y = a\sin k(x-b) \quad\text{and}\quad y = a\cos k(x-b) \quad (k>0) \]
have amplitude \(|a|\), period \( \dfrac{2\pi}{k} \), and horizontal shift \(b\). One period graphs on \( \left[b,\, b+\dfrac{2\pi}{k}\right] \).
EXAMPLE 4A Horizontally Shifted Sine Curve
Find the amplitude, period, and horizontal shift of \( y = 3\sin 2\!\left(x - \dfrac{\pi}{4}\right) \).
One complete period on \( \left[\dfrac{\pi}{4}, \dfrac{5\pi}{4}\right] \).
EXAMPLE 5A Horizontally Shifted Cosine Curve
Find amplitude, period, and shift of \( y = \dfrac{3}{4}\cos\!\left(2x + \dfrac{2\pi}{3}\right) \).
Solution
Factor: \( y = \dfrac{3}{4}\cos 2\!\left(x + \dfrac{\pi}{3}\right) \).
Amplitude \( \dfrac{3}{4} \); period \( \pi \); shift \( -\dfrac{\pi}{3} \) (left). One period on \( \left[-\dfrac{\pi}{3}, \dfrac{2\pi}{3}\right] \).
EXAMPLE 6Choosing the Viewing Rectangle
Graph \( f(x) = \sin 50x \) in an appropriate viewing rectangle.
Solution
Period \( = \dfrac{2\pi}{50} = \dfrac{\pi}{25} \approx 0.126 \). The oscillations are rapid.
Choose a narrow rectangle such as \( [-0.25, 0.25] \) by \( [-1.5, 1.5] \) to show a few accurate oscillations; wide windows produce misleading aliased graphs.
EXAMPLE 7A Sum of Sine and Cosine Curves
Graph \( h(x) = 2\cos x + \sin 2x \) by graphical addition.
Solution
Since \( h = f + g \) with \( f=2\cos x \), \( g=\sin 2x \), add the corresponding \(y\)-coordinates.
The result is periodic with period \(2\pi\).
EXAMPLE 8A Cosine Curve with Variable Amplitude
Graph \( y = x^2 \), \( y = -x^2 \), and \( y = x^2\cos 6\pi x \) together.
Solution
Since \( -1 \le \cos 6\pi x \le 1 \), multiplying by \( x^2 \ge 0 \) gives \( -x^2 \le x^2\cos 6\pi x \le x^2 \).
So \( y = \pm x^2 \) form an envelope (boundary) for the oscillating graph.
EXAMPLE 10A Sine Curve with Decaying Amplitude
Graph \( f(x) = \dfrac{\sin x}{x} \) and comment on behavior near \( x = 0 \).
Solution
Although \( f \) is undefined at \( x = 0 \), the values approach 1 as \( x \to 0 \) โ a key fact in calculus.
View as \( f(x) = \dfrac{1}{x}\sin x \): a sine function with amplitude controlled by \( a(x)=\dfrac{1}{x} \).
Fourier's Insight
Nearly every periodic function arising in practice can be written as a (usually infinite) sum of sines and cosines โ discovered by J. B. J. Fourier. This means any periodic variation can be described using sine and cosine. A modern application is the digital encoding of sound on compact discs.
Master the "A-P-S" reading of \( y = a\sin k(x-b) \): Amplitude \(=|a|\), Period \(= 2\pi/k\), Shift \(= b\). The single most common student error is forgetting to factor out \(k\) before reading the shift โ \( \sin(2x + \frac{2\pi}{3}) \) is shifted by \( \frac{\pi}{3} \), not \( \frac{2\pi}{3} \). Example 6 (aliasing) is exactly why your oscilloscope or data-logger must sample fast enough โ undersampling a vibration signal gives a beautiful but completely wrong waveform.
This section graphs the tangent, cotangent, secant, and cosecant functions and their transformations.
Periodic Properties
\[ \tan(x+\pi)=\tan x \qquad \cot(x+\pi)=\cot x \]
\[ \csc(x+2\pi)=\csc x \qquad \sec(x+2\pi)=\sec x \]
Tangent and cotangent have period \(\pi\); cosecant and secant have period \(2\pi\). As \(x \to \dfrac{\pi}{2}^-\), \( \tan x \to \infty \); the lines \( x = \pm\dfrac{\pi}{2} \) are vertical asymptotes of \( y=\tan x \).
Figure 1. One period of \(y=\tan x\) with vertical asymptotes at \(x=\pm\frac{\pi}{2}\).
Tangent and Cotangent Curves
\[ y = a\tan kx \quad\text{and}\quad y = a\cot kx \quad (k>0) \quad\text{have period } \frac{\pi}{k}. \]
For \( y=a\tan kx \) graph on \( \left(-\dfrac{\pi}{2k}, \dfrac{\pi}{2k}\right) \); for \( y=a\cot kx \) graph on \( \left(0, \dfrac{\pi}{k}\right) \).
EXAMPLE 1Graphing Tangent Curves
Graph (a) \( y = 2\tan x \) and (b) \( y = -\tan x \).
Solution
(a) Multiply each \(y\)-coordinate of \( y=\tan x \) by 2 (vertical stretch).
(b) Reflect \( y=\tan x \) in the \(x\)-axis.
EXAMPLE 2Graphing Tangent with a Shift
Graph (a) \( y = \tan 2x \) and (b) \( y = \tan 2\!\left(x - \dfrac{\pi}{4}\right) \).
Solution
(a) Period \( \dfrac{\pi}{2} \); asymptotes at \( x=\pm\dfrac{\pi}{4} \); graph one period on \( \left(-\dfrac{\pi}{4}, \dfrac{\pi}{4}\right) \).
(b) Same shape as (a), shifted right \( \dfrac{\pi}{4} \).
EXAMPLE 3A Horizontally Shifted Cotangent Curve
Graph \( y = 2\cot\!\left(3x - \dfrac{\pi}{4}\right) \).
Solution
Factor: \( y = 2\cot 3\!\left(x - \dfrac{\pi}{12}\right) \).
Period \( \dfrac{\pi}{3} \), shifted right \( \dfrac{\pi}{12} \); one period on \( \left(\dfrac{\pi}{12}, \dfrac{5\pi}{12}\right) \).
Cosecant and Secant Curves
\[ y = a\csc kx \quad\text{and}\quad y = a\sec kx \quad (k>0) \quad\text{have period } \frac{2\pi}{k}. \]
One period graphs on \( \left(0, \dfrac{2\pi}{k}\right) \).
EXAMPLE 4Graphing Cosecant Curves
Graph (a) \( y = \dfrac{1}{2}\csc 2x \) and (b) \( y = \dfrac{1}{2}\csc\!\left(2x + \dfrac{\pi}{2}\right) \).
Solution
(a) Period \( \pi \); asymptotes where \( \sin 2x = 0 \): \( x = 0, \dfrac{\pi}{2}, \pi \).
(b) Factor: \( \dfrac{1}{2}\csc 2\!\left(x + \dfrac{\pi}{4}\right) \) โ same as (a), shifted left \( \dfrac{\pi}{4} \).
EXAMPLE 5Graphing a Secant Curve
Graph \( y = 3\sec\dfrac{1}{2}x \).
Solution
Period \( \dfrac{2\pi}{1/2} = 4\pi \); asymptotes where \( \cos\dfrac{1}{2}x = 0 \): \( x = \pi, 3\pi \).
Sketch one period on \( [0, 4\pi] \) with the secant shape, then repeat.
Symmetry of the Reciprocal Graphs
\( y=\tan x \), \( y=\cot x \), and \( y=\csc x \) are symmetric about the origin (odd functions); \( y=\sec x \) is symmetric about the \(y\)-axis (even function).
Tangent and cotangent break the "everything has period \(2\pi\)" habit โ they repeat every \(\pi\). The asymptotes are not decoration: they mark where the function is genuinely undefined (division by zero on the unit circle). When you sketch a tan/cot graph, draw the asymptotes first as dashed lines, then fill in the curve between them. For sec/csc, sketch the underlying cos/sin lightly and "flip" it โ the peaks of sine become the troughs of cosecant.
Because the trigonometric functions are not one-to-one, they have no inverse on their full domains. By restricting each domain to an interval where the function is one-to-one, we obtain inverse functions.
The Inverse Sine Function
Restrict the domain of sine to \( \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] \), where it is one-to-one and attains every value in its range.
Definition of the Inverse Sine Function
\[ \sin^{-1}x = y \iff \sin y = x, \qquad \text{domain } [-1,1],\ \text{range } \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right] \]
Cancellation Properties (Sine)
\[ \sin(\sin^{-1}x)=x \ \text{ for } -1\le x\le 1 \qquad \sin^{-1}(\sin x)=x \ \text{ for } -\tfrac{\pi}{2}\le x\le \tfrac{\pi}{2} \]
(c) By calculator (radian mode), \( \tan^{-1}(-20) \approx -1.52084 \).
Inverse Secant, Cosecant, and Cotangent
These are defined by restricting each function to an interval where it is one-to-one and attains all its values. The chosen restrictions simplify sign choices in computations and are the ones appropriate for calculus.
The subtle point students miss: \( \sin^{-1}(\sin x) \) does not always give back \(x\). It only does so when \(x\) is already inside the restricted range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \). Example 3(b) shows the trap. Whenever you compute \( \sin^{-1} \), \( \cos^{-1} \), or \( \tan^{-1} \), always ask "which interval must the answer land in?" first. On a calculator, \( \tan^{-1} \) is the everyday tool for finding the angle of a resultant vector from its components โ but remember it only returns angles between \( \pm 90^\circ \), so you must add \(180^\circ\) manually for QII/QIII directions.
Periodic behavior is common in nature: tides, vibrating springs, sound waves, alternating current, and pulsating stars. The trigonometric functions are ideally suited to modeling it.
Simple Harmonic Motion
Simple Harmonic Motion
If \( y = a\sin\omega t \) or \( y = a\cos\omega t \), the object is in simple harmonic motion, with
\[ \text{amplitude}=|a| \qquad \text{period}=\frac{2\pi}{\omega} \qquad \text{frequency}=\frac{\omega}{2\pi} \]
The difference between the sine and cosine forms is the starting point: at \(t=0\), \( a\sin(\omega\cdot 0)=0 \) (starts at equilibrium) while \( a\cos(\omega\cdot 0)=a \) (starts at maximum displacement). Writing \( y=a\sin 2\pi\nu t \) lets the frequency \(\nu\) be read directly.
EXAMPLE 1A Vibrating Spring
A mass has displacement \( y = 10\sin 4\pi t \) (inches, seconds). Find amplitude, period, frequency.
Solution
Amplitude \( = 10 \) in.
Period \( = \dfrac{2\pi}{4\pi} = \dfrac{1}{2} \) s.
Frequency \( = \dfrac{4\pi}{2\pi} = 2 \) cycles per second (Hz).
EXAMPLE 2Vibrations of a Musical Note
For a pure E, \( V(t) = 0.2\sin 80\pi t \) (psi, seconds). Find amplitude, period, frequency, and describe louder/flatter changes.
Solution
Amplitude \(0.2\); period \( \dfrac{2\pi}{80\pi} = \dfrac{1}{40} \); frequency \( 40 \) Hz.
Louder โ larger amplitude (replace 0.2 with a bigger number).
Flat note โ lower frequency (coefficient of \(t\) becomes less than \(80\pi\)).
EXAMPLE 3Modeling a Vibrating Spring
A spring is compressed 4 cm and released; it returns to the compressed position after \( \dfrac{1}{3} \) s. Model the displacement.
Solution
Maximum at \(t=0\) โ use \( y = a\cos\omega t \) with \( a=4 \).
Period \( \dfrac{1}{3} = \dfrac{2\pi}{\omega} \Rightarrow \omega = 6\pi \).
\( y = 4\cos 6\pi t \).
Shifted Harmonic Motion
\[ y = a\sin(\omega(t-c)) + b \qquad y = a\cos(\omega(t-c)) + b \]
where \(b\) is the average value (vertical shift) and \(c\) gives the position at \(t=0\).
EXAMPLE 4Modeling the Brightness of a Variable Star
Delta Cephei: period 5.4 days, average brightness 4.0, variation 0.35. Model brightness vs. time.
Taking \(t=0\) at \(155\) V: \( E(t) = 155\cos 120\pi t \).
Damped Harmonic Motion
Damped Harmonic Motion
\[ y = ke^{-ct}\sin\omega t \qquad y = ke^{-ct}\cos\omega t \qquad (c>0) \]
\(c\) is the damping constant, \(k\) the initial amplitude, and \( \dfrac{2\pi}{\omega} \) the period. The amplitude is governed by \( a(t)=ke^{-ct} \).
Figure 12. Damped harmonic motion: oscillation with a decaying amplitude envelope (gold dashed).
EXAMPLE 7Modeling Damped Harmonic Motion
Two systems, both 0.5 Hz and initial displacement 10 cm. Damping constants 0.5 and 0.1. Model each \( g(t)=ke^{-ct}\cos\omega t \).
Solution
\( k=10 \); \( \omega = 2\pi(0.5) = \pi \).
\( g_1(t) = 10e^{-0.5t}\cos\pi t \) and \( g_2(t) = 10e^{-0.1t}\cos\pi t \).
Larger \(c\) (first system) dies down faster; smaller \(c\) sustains motion longer.
EXAMPLE 8A Vibrating Violin String
G-string pulled 0.5 cm, damping \(c=1.4\), pure G (200 Hz), max at \(t=0\). Find the motion.
Solution
\( k=0.5 \); \( \omega = 2\pi(200) = 400\pi \).
\( f(t) = 0.5e^{-1.4t}\cos 400\pi t \).
EXAMPLE 9Ripples on a Pond
Amplitude drops to \( \dfrac{1}{10} \) of its value after 20 s. Find the damping constant \(c\).
For \( y_1 = A\sin(kt+b) \) and \( y_2 = A\sin(kt+c) \), the phase difference is \( b-c \). If it is a multiple of \(2\pi\), the waves are in phase; otherwise they are out of phase. The lag time (\(b>0\)) or lead time (\(b<0\)) is \( b/k \).
EXAMPLE 10Finding Phase and Phase Difference
Given \( y_1 = 10\sin\!\left(3t - \dfrac{\pi}{6}\right) \), \( y_2 = 10\sin\!\left(3t - \dfrac{\pi}{2}\right) \), \( y_3 = 10\sin\!\left(3t + \dfrac{23\pi}{6}\right) \): find amplitude/period/phase/shift of \(y_1\), and the phase differences.
\(y_1\) vs \(y_2\): difference \( -\dfrac{\pi}{6} - \left(-\dfrac{\pi}{2}\right) = \dfrac{\pi}{3} \) โ out of phase.
\(y_1\) vs \(y_3\): difference \( -\dfrac{\pi}{6} - \dfrac{23\pi}{6} = -4\pi \) โ in phase (same graph).
EXAMPLE 11Using Phase (Ferris Wheel)
A Ferris wheel (radius 5 m) turns at 2 rev/min. Riders start at phases \( -\dfrac{\pi}{2}, -\dfrac{3\pi}{4}, \dfrac{\pi}{4} \). Model heights and find phase differences.
Solution
Amplitude 5; period \( \dfrac{1}{2} \) min โ \( k = 4\pi \).
\( y_A = 5\sin\!\left(4\pi t + \dfrac{\pi}{2}\right) \), \( y_B = 5\sin\!\left(4\pi t + \dfrac{3\pi}{4}\right) \), \( y_C = 5\sin\!\left(4\pi t - \dfrac{\pi}{4}\right) \).
Ali's shift: \( y_A = 5\sin 4\pi\!\left(t + \dfrac{1}{8}\right) \) โ lead time \( \dfrac{1}{8} \) min.
This is the payoff section for mechanical engineers. Simple harmonic motion is the math of every spring-mass system, pendulum, and undamped vibration. Damped motion with \( ke^{-ct} \) is what every real system does once friction enters โ and it's the exact basis for designing car shock absorbers (large \(c\) to kill motion fast) versus a tuned musical instrument (small \(c\) to sustain). Phase tells you whether two oscillating things move together; phase difference governs resonance, beat frequencies, and the timing of AC generators. If you take Vibrations or Dynamics later, this section is the foundation โ know the meaning of \(a\), \(\omega\), \(c\), and \(b\) cold.
FM
Focus on Modeling: Fitting Sinusoidal Curves to Data
When a scatter plot of data suggests simple harmonic behavior, we can fit a sine or cosine model of the form \( y = a\cos(\omega(t-c)) + b \) by reading four quantities directly from the data.
Four-Step Fitting Procedure
Vertical shift \( b = \dfrac{1}{2}(\text{max} + \text{min}) \). Amplitude \( a = \dfrac{1}{2}(\text{max} - \text{min}) \). Period: \( \dfrac{2\pi}{\omega} = 2\,(\text{time of max} - \text{time of min}) \). Horizontal shift \( c \) = time of the maximum value.
EXAMPLE 1Modeling the Height of a Tide
Water depth varies over 12 h with max 11.6 ft (near \(t=2\)) and min 5.4 ft (near \(t=8\)). Fit a model and find when depth exceeds 11 ft.
Solution
\( b = \dfrac{1}{2}(11.6+5.4) = 8.5 \); \( a = \dfrac{1}{2}(11.6-5.4) = 3.1 \).
Period \( = 2(8-2) = 12 \Rightarrow \omega = \dfrac{2\pi}{12} \approx 0.52 \); \( c = 2.0 \).
Model: \( y = 3.1\cos(0.52(t-2.0)) + 8.5 \).
Solving \( y \geq 11 \) graphically: depth exceeds 11 ft from about \( t \approx 0.8 \) to \( t \approx 3.2 \) (12:48 a.m. to 3:12 a.m.).
EXAMPLE 2Fitting a Sine Curve to Data (Regression)
Use a calculator's SinReg command to fit \( y = a\sin(bt+c)+d \) to the same tide data, then compare.
Solution
SinReg gives \( a \approx 3.1,\ b \approx 0.53,\ c \approx 0.55,\ d \approx 8.42 \): \( y = 3.1\sin(0.53t + 0.55) + 8.42 \).
Converting via \( \sin u = \cos(u - \tfrac{\pi}{2}) \) gives \( y \approx 3.1\cos(0.53(t-1.92)) + 8.42 \) โ close to the by-hand model; small differences come from the two fitting methods.
This is real engineering workflow: you collect noisy sensor data (tides, temperature, vibration, sunspot counts) and fit a clean sinusoid to it. The "by-hand" four-step method gives you a quick, defensible estimate; regression (SinReg, or least-squares in Python/Excel) refines it. Always sanity-check the regression output against your hand estimate โ if amplitude or period disagree wildly, you likely have a units or data-entry error. The reduction formula \( \sin u = \cos(u - \tfrac{\pi}{2}) \) is the bridge for converting between sine and cosine forms.