The Central Idea
A transformed sinusoidal function can be written in the form
\[ \boxed{ y=A\sin(B(x-C))+D } \]or
\[ \boxed{ y=A\cos(B(x-C))+D }. \]The parameters control different features:
- \(|A|\) is the amplitude;
- \(B\) controls the period and angular frequency;
- \(C\) is the horizontal phase shift;
- \(D\) is the vertical shift and determines the midline.
These four parameters describe a large class of periodic phenomena.
The Parent Functions
The basic sine and cosine functions are
\[ y=\sin x \]and
\[ y=\cos x. \]Both have amplitude
\[ 1, \]period
\[ 2\pi, \]and midline
\[ y=0. \]Their difference is primarily where each cycle begins.
Amplitude
The amplitude of a sine or cosine function is the distance from its midline to a maximum or minimum.
For
\[ y=A\sin(B(x-C))+D \]or
\[ y=A\cos(B(x-C))+D, \]the amplitude is
\[ \boxed{|A|}. \]Why Amplitude Uses Absolute Value
Amplitude is a distance, so it cannot be negative.
For example,
\[ y=-3\sin x \]has amplitude
\[ \boxed{3}, \]not \(-3\).
The negative sign reflects the graph vertically, but the size of the oscillation remains \(3\).
Vertical Stretch and Compression
The coefficient \(A\) scales every output of the parent function.
If
\[ |A|>1, \]the graph is vertically stretched.
If
\[ 0<|A|<1, \]the graph is vertically compressed.
If
\[ A<0, \]the graph is also reflected across its midline.
The Midline
The midline is the horizontal line halfway between the maximum and minimum values of a sinusoidal function.
For
\[ y=A\sin(B(x-C))+D \]or
\[ y=A\cos(B(x-C))+D, \]the midline is
\[ \boxed{y=D}. \]Maximum and Minimum from Amplitude and Midline
If the amplitude is \(|A|\) and the midline is \(y=D\), then
\[ \boxed{\text{maximum}=D+|A|} \]and
\[ \boxed{\text{minimum}=D-|A|}. \]Therefore the range is
\[ \boxed{[D-|A|,\ D+|A|]}. \]Finding Amplitude from a Maximum and Minimum
If the maximum value is \(M\) and the minimum value is \(m\), then the total vertical distance is
\[ M-m. \]The amplitude is half that distance:
\[ \boxed{ \text{amplitude}=\frac{M-m}{2} }. \]Finding the Midline from a Maximum and Minimum
The midline lies halfway between the maximum and minimum:
\[ \boxed{ D=\frac{M+m}{2} }. \]This is simply the average of the two extreme values.
Example: Amplitude and Midline
Suppose a periodic quantity has maximum value
\[ 11 \]and minimum value
\[ 3. \]Its amplitude is
\[ \frac{11-3}{2}=4. \]Its midline is
\[ \frac{11+3}{2}=7. \]Therefore,
\[ \boxed{\text{amplitude}=4,\qquad \text{midline }y=7}. \]Period
The period is the smallest positive horizontal interval over which a periodic function completes one full cycle and begins to repeat.
For the parent sine and cosine functions,
\[ \boxed{T=2\pi}. \]Here \(T\) is commonly used to represent period.
Period of Transformed Sine and Cosine
For
\[ y=A\sin(B(x-C))+D \]or
\[ y=A\cos(B(x-C))+D, \]the period is
\[ \boxed{ T=\frac{2\pi}{|B|} }. \]The absolute value appears because period is a positive length.
Why \(B\) Changes the Period
The parent sine function repeats when its input changes by \(2\pi\).
For
\[ \sin(Bx), \]one full cycle occurs when
\[ B\Delta x=2\pi. \]Solving for the horizontal change gives
\[ \Delta x=\frac{2\pi}{|B|}. \]Therefore,
\[ \boxed{T=\frac{2\pi}{|B|}}. \]Horizontal Compression
If
\[ |B|>1, \]then
\[ \frac{2\pi}{|B|}<2\pi. \]The graph completes a cycle in less horizontal distance.
The graph is therefore horizontally compressed.
Horizontal Stretch
If
\[ 0<|B|<1, \]then
\[ \frac{2\pi}{|B|}>2\pi. \]One cycle requires more horizontal distance.
The graph is horizontally stretched.
Example: Period of \(y=\sin(2x)\)
Here
\[ B=2. \]Therefore,
\[ T=\frac{2\pi}{2} = \boxed{\pi}. \]The function completes two full cycles over an interval of length \(2\pi\).
Example: Period of \(y=\cos(x/3)\)
Write
\[ \frac{x}{3}=\frac13x, \]so
\[ B=\frac13. \]Then
\[ T=\frac{2\pi}{1/3} = \boxed{6\pi}. \]The graph is horizontally stretched by a factor of \(3\).
Frequency
Frequency measures how many cycles occur per unit of the independent variable.
If the period is \(T\), then ordinary frequency is
\[ \boxed{ f=\frac1T }. \]A shorter period means a higher frequency, while a longer period means a lower frequency.
Period and Frequency Are Reciprocals
The relationships
\[ \boxed{f=\frac1T} \]and
\[ \boxed{T=\frac1f} \]show that period and frequency contain the same timing information in reciprocal forms.
Frequency in a Time Model
If the horizontal variable \(t\) measures time in seconds and a cycle lasts \(4\) seconds, then
\[ T=4\text{ s}. \]The frequency is
\[ f=\frac14\text{ cycle/s}. \]Thus the motion completes one quarter of a cycle per second, or one full cycle every four seconds.
Hertz
When frequency measures cycles per second, the SI unit is the hertz, abbreviated Hz.
Thus,
\[ 1\text{ Hz}=1\text{ cycle per second}. \]For example, a process with period
\[ 0.02\text{ s} \]has frequency
\[ f=\frac1{0.02}=50\text{ Hz}. \]Angular Frequency
In a sinusoidal time model written
\[ y=A\sin(\omega t+\phi)+D, \]the coefficient
\[ \omega \]is called the angular frequency.
It measures radians of phase change per unit time.
The period is
\[ \boxed{ T=\frac{2\pi}{|\omega|} }. \]Ordinary Frequency versus Angular Frequency
Ordinary frequency \(f\) counts cycles per unit time.
Angular frequency \(\omega\) counts radians of phase per unit time.
Because one full cycle contains \(2\pi\) radians,
\[ \boxed{\omega=2\pi f} \]and
\[ \boxed{f=\frac{\omega}{2\pi}}. \]These quantities are related but should not be confused.
Example: Frequency from Angular Frequency
Suppose
\[ y=\sin(8\pi t). \]Then
\[ \omega=8\pi. \]Therefore,
\[ f=\frac{8\pi}{2\pi}=4. \]If \(t\) is measured in seconds,
\[ \boxed{f=4\text{ Hz}}. \]Phase
Phase describes a position within a repeating cycle.
Two sinusoidal motions can have the same amplitude and frequency but reach their maxima, minima, and zeros at different times.
That difference in cycle position is described by phase.
Phase Shift
In the form
\[ y=A\sin(B(x-C))+D, \]the quantity
\[ \boxed{C} \]is the phase shift, or horizontal shift.
If
\[ C>0, \]the graph shifts right.
If
\[ C<0, \]the graph shifts left.
Why \(x-C\) Shifts Right
Consider
\[ y=\sin(x-C). \]The parent sine function begins its familiar zero-to-positive-rise pattern when its internal input equals zero.
Set
\[ x-C=0. \]Then
\[ x=C. \]So the characteristic starting point moves from \(x=0\) to \(x=C\).
Phase-Angle Form
A sinusoid may also be written as
\[ y=A\sin(Bx+\phi)+D. \]Here
\[ \phi \]is a phase angle.
Factoring \(B\) from the internal expression gives
\[ Bx+\phi = B\left(x+\frac{\phi}{B}\right). \]Therefore the horizontal phase shift is
\[ \boxed{ C=-\frac{\phi}{B} }. \]Phase Angle and Phase Shift Are Different Quantities
In
\[ A\sin(Bx+\phi), \]\(\phi\) is an angular offset inside the function.
The actual horizontal displacement of the graph is
\[ -\frac{\phi}{B}. \]When \(B\neq1\), simply calling \(\phi\) the horizontal shift gives the wrong distance.
Example: Find the Phase Shift
Consider
\[ y=\sin(2x-\pi). \]Factor the \(2\):
\[ 2x-\pi = 2\left(x-\frac{\pi}{2}\right). \]Therefore,
\[ \boxed{ \text{phase shift}=\frac{\pi}{2}\text{ to the right} }. \]Example: A Left Shift
Consider
\[ y=\cos\left(3x+\frac{\pi}{2}\right). \]Factor \(3\):
\[ 3x+\frac{\pi}{2} = 3\left(x+\frac{\pi}{6}\right). \]Therefore the phase shift is
\[ \boxed{-\frac{\pi}{6}}, \]meaning a shift left by
\[ \boxed{\frac{\pi}{6}}. \]Vertical Shift
The parameter \(D\) moves the graph vertically.
For
\[ y=A\sin(B(x-C))+D, \]every parent-function output is increased by \(D\).
The midline becomes
\[ \boxed{y=D}. \]Example: Reading All Four Parameters
Consider
\[ y=3\sin\left(2\left(x-\frac{\pi}{4}\right)\right)+5. \]Here,
\[ A=3,\qquad B=2,\qquad C=\frac{\pi}{4},\qquad D=5. \]Therefore:
\[ \boxed{\text{amplitude}=3}, \] \[ \boxed{\text{period}=\frac{2\pi}{2}=\pi}, \] \[ \boxed{\text{phase shift}=\frac{\pi}{4}\text{ right}}, \]and
\[ \boxed{\text{midline }y=5}. \]Maximum and Minimum in the Same Example
For
\[ y=3\sin\left(2\left(x-\frac{\pi}{4}\right)\right)+5, \]the amplitude is \(3\) and the midline is \(5\).
Thus,
\[ \text{maximum}=5+3=8 \]and
\[ \text{minimum}=5-3=2. \]The range is
\[ \boxed{[2,8]}. \]Quarter-Cycle Length
Sine and cosine graphs are especially easy to sketch by dividing one period into four equal parts.
If the period is \(T\), then the quarter-cycle length is
\[ \boxed{\frac{T}{4}}. \]Since
\[ T=\frac{2\pi}{|B|}, \]the quarter-cycle length is
\[ \boxed{\frac{\pi}{2|B|}}. \]Five Key Sine Positions
For a positive-amplitude sine function beginning at its phase shift \(x=C\), one cycle follows the midline pattern
\[ D,\quad D+|A|,\quad D,\quad D-|A|,\quad D. \]These values are separated horizontally by one quarter of the period.
Five Key Cosine Positions
For a positive-amplitude cosine function beginning at \(x=C\), one cycle follows
\[ D+|A|,\quad D,\quad D-|A|,\quad D,\quad D+|A|. \]Again, consecutive key points are separated by one quarter-period.
Effect of a Negative Amplitude
If \(A<0\), the graph is reflected across its midline.
For example,
\[ y=-2\sin x \]begins at the midline but initially moves downward rather than upward.
Its amplitude is still
\[ \boxed{2}. \]Example: Sketching a Transformed Sine Function
Consider
\[ y=2\sin\left(x-\frac{\pi}{2}\right)+1. \]The amplitude is
\[ 2, \]the period is
\[ 2\pi, \]the phase shift is
\[ \frac{\pi}{2}\text{ right}, \]and the midline is
\[ y=1. \]The maximum is \(3\), and the minimum is \(-1\).
Key Points for the Example
The quarter-period is
\[ \frac{2\pi}{4}=\frac{\pi}{2}. \]Beginning at
\[ x=\frac{\pi}{2}, \]the five key points are
\[ \left(\frac{\pi}{2},1\right), \] \[ (\pi,3), \] \[ \left(\frac{3\pi}{2},1\right), \] \[ (2\pi,-1), \]and
\[ \left(\frac{5\pi}{2},1\right). \]Example: Sketching a Transformed Cosine Function
Consider
\[ y=-4\cos(2x)+3. \]The amplitude is
\[ 4, \]the period is
\[ \frac{2\pi}{2}=\pi, \]there is no horizontal shift, and the midline is
\[ y=3. \]The negative coefficient reflects the cosine wave across the midline.
Range of the Transformed Cosine Example
Since the midline is \(3\) and the amplitude is \(4\),
\[ \text{maximum}=3+4=7 \]and
\[ \text{minimum}=3-4=-1. \]Therefore,
\[ \boxed{\operatorname{range}=[-1,7]}. \]Amplitude Is Not Used for Every Trigonometric Function
The term amplitude is naturally defined for bounded oscillations such as sine and cosine.
Tangent and cotangent are unbounded, so they do not have an amplitude.
Secant and cosecant are also unbounded and do not have amplitude in the ordinary sinusoidal sense.
Their vertical scale factors still affect their graphs, but calling those factors amplitudes is misleading.
Period of Tangent and Cotangent
The parent tangent and cotangent functions have period
\[ \pi. \]Therefore, for
\[ y=A\tan(B(x-C))+D \]or
\[ y=A\cot(B(x-C))+D, \]the period is
\[ \boxed{ T=\frac{\pi}{|B|} }. \]Period of Secant and Cosecant
Secant follows the repeating structure of cosine, while cosecant follows sine.
Therefore, for transformed secant and cosecant functions with internal coefficient \(B\),
\[ \boxed{ T=\frac{2\pi}{|B|} }. \]Frequency for Any Repeating Model
Once a period \(T\) has been identified, ordinary frequency can be defined by
\[ \boxed{f=\frac1T}. \]The exact period formula depends on the parent function.
For sine and cosine,
\[ T=\frac{2\pi}{|B|}, \]while for tangent and cotangent,
\[ T=\frac{\pi}{|B|}. \]Phase Difference
Two waves can have the same period and amplitude but different horizontal placements.
For example,
\[ y_1=\sin x \]and
\[ y_2=\sin\left(x-\frac{\pi}{3}\right) \]have the same amplitude and period.
The second is shifted right by
\[ \frac{\pi}{3}. \]This horizontal offset represents a phase difference between the waves.
In Phase
Two periodic signals are in phase when corresponding features of their cycles occur at the same horizontal positions.
For example,
\[ y=\sin x \]and
\[ y=3\sin x \]have different amplitudes but the same phase.
Their zeros, maxima, and minima occur at the same inputs.
Out of Phase
If corresponding cycle features occur at different horizontal positions, the signals are out of phase.
For example,
\[ \sin x \]and
\[ \cos x \]have the same amplitude and period but are shifted relative to one another by a quarter-cycle:
\[ \boxed{\frac{\pi}{2}}. \]Half-Cycle Phase Difference
A horizontal shift by half a sine or cosine period is
\[ \pi. \]This produces
\[ \sin(x-\pi)=-\sin x \]and
\[ \cos(x-\pi)=-\cos x. \]A half-cycle shift therefore produces the same graph as a vertical reflection of the parent sinusoid.
Equivalent Sinusoidal Forms
The same periodic graph can often be represented in more than one way.
For example,
\[ \sin x = \cos\left(x-\frac{\pi}{2}\right). \]Thus a sine model can often be rewritten as a cosine model by changing the phase.
Different formulas may therefore describe the same graph.
Choosing Sine or Cosine for a Model
Both sine and cosine can model the same sinusoidal behavior with an appropriate phase shift.
Cosine is often convenient when the model begins at a maximum or minimum.
Sine is often convenient when the model begins at the midline and moves upward or downward.
The best form is usually the one that makes the given information easiest to encode.
Building a Sinusoidal Model from Extremes
Suppose a periodic quantity ranges from
\[ 20 \]to
\[ 80. \]The amplitude is
\[ \frac{80-20}{2}=30. \]The midline is
\[ \frac{80+20}{2}=50. \]Any corresponding sine or cosine model therefore has
\[ \boxed{|A|=30,\qquad D=50}. \]Building a Model from the Period
Suppose the same quantity repeats every
\[ 12 \]time units.
Then
\[ T=12. \]For a sine or cosine model,
\[ |B|=\frac{2\pi}{T} = \frac{2\pi}{12} = \frac{\pi}{6}. \]Thus a possible internal coefficient is
\[ \boxed{B=\frac{\pi}{6}}. \]Example: Model Beginning at a Maximum
Suppose a quantity ranges from \(20\) to \(80\), repeats every \(12\) units, and begins at its maximum when \(t=0\).
A convenient cosine model is
\[ \boxed{ y=30\cos\left(\frac{\pi}{6}t\right)+50 }. \]At \(t=0\),
\[ y=30(1)+50=80, \]as required.
Example: Model Beginning at the Midline
Suppose instead that the same quantity begins at its midline and initially increases.
A convenient model is
\[ \boxed{ y=30\sin\left(\frac{\pi}{6}t\right)+50 }. \]At \(t=0\),
\[ y=50, \]and the sine pattern initially moves upward.
Periodic Models Need Units
In applications, amplitude has the same units as the dependent variable.
Period has the units of the independent variable.
Frequency has reciprocal units.
For example, if vertical displacement is measured in centimeters and time in seconds, then:
- amplitude may be measured in centimeters;
- period is measured in seconds;
- frequency is measured in cycles per second or hertz.
Example: Oscillating Height
Suppose
\[ h(t)=4\cos(\pi t)+10, \]where \(h\) is measured in meters and \(t\) in seconds.
The amplitude is
\[ \boxed{4\text{ m}}. \]The period is
\[ T=\frac{2\pi}{\pi}=2\text{ s}. \]Therefore the frequency is
\[ \boxed{f=\frac12\text{ Hz}}. \]Example: Temperature Cycle
Suppose a simplified temperature model is
\[ T(t)=8\sin\left(\frac{\pi}{12}(t-6)\right)+18. \]The amplitude is
\[ 8, \]the period is
\[ \frac{2\pi}{\pi/12}=24, \]the phase shift is \(6\) units to the right, and the midline is
\[ T=18. \]The model therefore oscillates between
\[ 10 \]and
\[ 26. \]Example: High-Frequency Motion
Consider
\[ y=2\sin(20\pi t). \]Here
\[ \omega=20\pi. \]Therefore,
\[ f=\frac{20\pi}{2\pi}=10. \]If \(t\) is in seconds, the system completes
\[ \boxed{10\text{ cycles per second}}. \]Amplitude versus Frequency
Amplitude and frequency describe different aspects of an oscillation.
Increasing amplitude makes the vertical excursion larger.
Increasing frequency makes cycles occur more rapidly.
A wave can have:
- large amplitude and low frequency;
- small amplitude and high frequency;
- large values of both;
- small values of both.
Period versus Phase
Period determines the width of an entire cycle.
Phase shift determines where that cycle is positioned horizontally.
Changing phase does not change how long a cycle lasts.
Changing period does not merely slide the graph; it changes the horizontal scale of the repeating pattern.
Amplitude versus Vertical Shift
Amplitude measures distance from the midline.
Vertical shift determines where the midline itself lies.
In
\[ y=5\sin x+12, \]the amplitude is
\[ 5, \]while the vertical shift and midline value are
\[ 12. \]Common Error: Calling \(A\) the Amplitude without Absolute Value
For
\[ y=-6\cos x, \]the amplitude is
\[ \boxed{6}, \]not \(-6\).
The negative sign represents reflection.
Common Error: Multiplying the Parent Period by \(B\)
For sine and cosine, the period is not
\[ 2\pi|B|. \]It is
\[ \boxed{ \frac{2\pi}{|B|} }. \]A larger \(|B|\) means more cycles fit into the same horizontal interval, so the period must become smaller.
Common Error: Confusing Period and Frequency
Period measures horizontal length per cycle.
Frequency measures cycles per unit input.
They are reciprocals:
\[ \boxed{f=\frac1T}. \]A high-frequency signal therefore has a short period.
Common Error: Calling \(B\) the Ordinary Frequency
In
\[ y=A\sin(Bt), \]when \(t\) is time and the internal angle is measured in radians, \(B\) is angular frequency.
Ordinary frequency is
\[ \boxed{ f=\frac{|B|}{2\pi} }. \]The distinction matters whenever units are being interpreted.
Common Error: Reading the Phase Shift before Factoring \(B\)
Consider
\[ y=\sin(2x-\pi). \]The phase shift is not \(\pi\).
Factor the internal coefficient:
\[ 2x-\pi = 2\left(x-\frac{\pi}{2}\right). \]Therefore the actual phase shift is
\[ \boxed{\frac{\pi}{2}}. \]Common Error: Reversing the Direction of a Phase Shift
In
\[ \sin(x-C), \]positive \(C\) shifts the graph right.
Thus
\[ \sin(x-4) \]shifts right by \(4\), while
\[ \sin(x+4) \]shifts left by \(4\).
The visible sign inside the parentheses is opposite the direction word.
Common Error: Giving Tangent an Amplitude
A function such as
\[ y=3\tan x \]has a vertical scale factor of \(3\), but tangent is unbounded.
It therefore does not have amplitude in the usual sense.
Common Error: Ignoring the Midline
For
\[ y=4\sin x+7, \]the oscillation is centered on
\[ \boxed{y=7}, \]not on the \(x\)-axis.
Its maximum is \(11\) and minimum is \(3\).
A Reliable Procedure for Analyzing a Sinusoid
- Rewrite the function in the form \(A\sin(B(x-C))+D\) or \(A\cos(B(x-C))+D\).
- Find the amplitude \(|A|\).
- Find the period \(2\pi/|B|\).
- Find the phase shift \(C\).
- Find the midline \(y=D\).
- Find the maximum \(D+|A|\) and minimum \(D-|A|\).
- Divide the period into four equal parts.
- Plot five key points for one cycle.
- Extend the pattern periodically.
A Reliable Procedure for Finding Frequency
- Identify the period \(T\).
- Compute \(f=1/T\).
- If the equation uses angular frequency \(\omega\), use \(f=|\omega|/(2\pi)\).
- Attach appropriate reciprocal units.
- Check that higher frequency corresponds to shorter period.
A Reliable Procedure for Building a Periodic Model
- Find the maximum and minimum values.
- Compute amplitude as half their difference.
- Compute the midline as their average.
- Determine the period from repeated features.
- Compute \(B=2\pi/T\) for a sine or cosine model.
- Choose sine or cosine based on a convenient starting feature.
- Determine the needed phase shift.
- Write the model.
- Check important inputs against the original situation.
Practice 1: Amplitude
Find the amplitude of
\[ y=-7\sin x. \]Practice 2: Period
Find the period of
\[ y=\cos(4x). \]Practice 3: Horizontal Stretch
Find the period of
\[ y=\sin\left(\frac{x}{2}\right). \]Practice 4: Phase Shift
Find the phase shift of
\[ y=\sin\left(3x-\frac{\pi}{2}\right). \]Practice 5: Midline and Range
For
\[ y=5\cos x-2, \]find the midline, maximum, minimum, and range.
Practice 6: Frequency from Period
A periodic motion has period
\[ 0.25\text{ s}. \]Find its frequency in hertz.
Practice 7: Frequency from Angular Frequency
For
\[ y=3\sin(12\pi t), \]find the ordinary frequency if \(t\) is measured in seconds.
Practice 8: Analyze a Sinusoid
Analyze
\[ y=2\cos\left(3\left(x+\frac{\pi}{6}\right)\right)+4. \]Find the amplitude, period, phase shift, midline, maximum, and minimum.
Practice 9: Build a Model
A periodic quantity has maximum \(14\), minimum \(6\), period \(10\), and begins at a maximum when \(t=0\).
Write a cosine model.
Practice 10: Find Amplitude and Midline from Extremes
A periodic graph has maximum value \(25\) and minimum value \(9\).
Find its amplitude and midline.
Practice 11: Tangent Period
Find the period of
\[ y=\tan(5x). \]Practice 12: Identify the Error
A student says that
\[ y=\sin(4x) \]has period
\[ 8\pi \]because the coefficient \(4\) multiplies the parent period \(2\pi\).
Explain the error.
Practice Answers
Practice 1
\[ \boxed{\text{amplitude}=|-7|=7}. \]Practice 2
\[ T=\frac{2\pi}{4} = \boxed{\frac{\pi}{2}}. \]Practice 3
Here
\[ B=\frac12. \]Therefore,
\[ T=\frac{2\pi}{1/2} = \boxed{4\pi}. \]Practice 4
Factor \(3\):
\[ 3x-\frac{\pi}{2} = 3\left(x-\frac{\pi}{6}\right). \]Therefore,
\[ \boxed{ \text{phase shift}=\frac{\pi}{6}\text{ right} }. \]Practice 5
The amplitude is \(5\), and
\[ D=-2. \]Therefore the midline is
\[ \boxed{y=-2}. \]The maximum is
\[ -2+5=3, \]and the minimum is
\[ -2-5=-7. \]Thus,
\[ \boxed{\operatorname{range}=[-7,3]}. \]Practice 6
\[ f=\frac1{0.25} = \boxed{4\text{ Hz}}. \]Practice 7
Here
\[ \omega=12\pi. \]Therefore,
\[ f=\frac{12\pi}{2\pi} = \boxed{6\text{ Hz}}. \]Practice 8
From
\[ y=2\cos\left(3\left(x+\frac{\pi}{6}\right)\right)+4, \]we have
\[ A=2,\qquad B=3,\qquad C=-\frac{\pi}{6},\qquad D=4. \]Therefore,
\[ \boxed{\text{amplitude}=2}, \] \[ \boxed{\text{period}=\frac{2\pi}{3}}, \] \[ \boxed{\text{phase shift}=\frac{\pi}{6}\text{ left}}, \] \[ \boxed{\text{midline }y=4}, \] \[ \boxed{\text{maximum}=6}, \]and
\[ \boxed{\text{minimum}=2}. \]Practice 9
The amplitude is
\[ \frac{14-6}{2}=4. \]The midline is
\[ \frac{14+6}{2}=10. \]Since
\[ T=10, \]we have
\[ B=\frac{2\pi}{10}=\frac{\pi}{5}. \]Beginning at a maximum suggests cosine:
\[ \boxed{ y=4\cos\left(\frac{\pi}{5}t\right)+10 }. \]Practice 10
Amplitude:
\[ \frac{25-9}{2}=8. \]Midline:
\[ \frac{25+9}{2}=17. \]Therefore,
\[ \boxed{\text{amplitude}=8,\qquad \text{midline }y=17}. \]Practice 11
Tangent has parent period \(\pi\), so
\[ T=\frac{\pi}{5}. \]Therefore,
\[ \boxed{T=\frac{\pi}{5}}. \]Practice 12
The coefficient multiplying \(x\) changes the horizontal scale reciprocally.
For sine,
\[ T=\frac{2\pi}{|B|}. \]With \(B=4\),
\[ T=\frac{2\pi}{4} = \boxed{\frac{\pi}{2}}. \]A larger internal coefficient produces a shorter period, not a longer one.
What to Remember
For
\[ y=A\sin(B(x-C))+D \]or
\[ y=A\cos(B(x-C))+D, \]the major features are
\[ \boxed{\text{amplitude}=|A|}, \] \[ \boxed{\text{period}=\frac{2\pi}{|B|}}, \] \[ \boxed{\text{phase shift}=C}, \]and
\[ \boxed{\text{midline }y=D}. \]- Amplitude measures distance from the midline.
- Maximum and minimum are \(D\pm|A|\).
- Frequency is the reciprocal of period: \(f=1/T\).
- Angular frequency satisfies \(\omega=2\pi f\).
- Phase describes position within a cycle.
- In \(A\sin(Bx+\phi)+D\), the horizontal phase shift is \(-\phi/B\).
- Negative \(A\) reflects a sinusoid across its midline.
- Tangent and cotangent have period \(\pi/|B|\), not \(2\pi/|B|\).
- Amplitude is not ordinarily assigned to unbounded tangent, cotangent, secant, or cosecant graphs.
- Periodic models should preserve the physical units of amplitude, period, and frequency.
Where We Go Next
Amplitude, period, frequency, phase, and midline describe how periodic graphs are positioned and scaled.
The next lesson, Trigonometric Identities, shifts attention from graph structure to algebraic relationships among trigonometric functions and develops tools for rewriting and verifying equivalent expressions.