9.2 Graphs of Sine and Cosine & Sinusoidal Modeling

9.2 Graphs of Sine and Cosine & Sinusoidal Modeling

Quick Answer: Transformed sinusoidal functions follow $y = A\sin(B(x - C)) + D$ or $y = A\cos(B(x - C)) + D$. The amplitude is $|A| = \frac{\text{max} - \text{min}}{2}$, the midline is $y = D = \frac{\text{max} + \text{min}}{2}$, the period is $P = \frac{2\pi}{|B|}$ (so $B = \frac{2\pi}{P}$), and the phase shift is $C$. Modeling cyclic real-world scenarios requires identifying maximum and minimum values, calculating $A, D,$ and $B$, and selecting cosine if the cycle begins at a peak/trough or sine if it begins at the midline.

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1. Parent Graphs of Sine and Cosine (F-IF.7e)

The graphs of the parent trigonometric functions $f(x) = \sin(x)$ and $g(x) = \cos(x)$ are continuous, smooth, undulating waves called sinusoids. Their wave shape reflects the cyclical variation of $y$ and $x$ coordinates as a point rotates around the unit circle.

   y ^
   1 +       * (π/2, 1)                  * (5π/2, 1)
     |      / \                         / \
   0 +----*----+----*---------*-------*----+----*---> x
     |  (0,0)   \  (π,0)     /     (2π,0)   \  (3π,0)
  -1 +           * (3π/2, -1)                * (7π/2, -1)
       Graph of y = sin(x): Zeros at integer multiples of π; Peak at π/2; Trough at 3π/2

The Five-Point Method for One Full Cycle ($[0, 2\pi]$)

Graphing one complete period of a sinusoid requires plotting five critical points: the start point, quarter-cycle, half-cycle, three-quarter cycle, and end point.

Fundamental Properties Comparison

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2. The General Transformed Sinusoidal Equation (F-TF.5)

Every periodic wave in Algebra II can be represented by the transformed sinusoidal function:

$$y = A\sin(B(x - C)) + D \quad \text{or} \quad y = A\cos(B(x - C)) + D$$

Each parameter controls a specific geometric transformation of the parent curve:

1. Amplitude ($|A|$): Vertical Stretch / Reflection

• The amplitude measures the vertical distance from the horizontal midline to the maximum peak (or to the minimum trough): $$\text{Amplitude} = |A| = \frac{\text{Maximum Value} - \text{Minimum Value}}{2}$$
• Range: The range of the function is $[D - |A|, D + |A|]$.
• Sign of $A$: Amplitude is strictly a non-negative distance ($|A| > 0$). If $A < 0$, the wave is vertically reflected across the midline $y = D$.

2. Midline ($y = D$): Vertical Shift

• The midline is the horizontal line running halfway between the maximum and minimum values: $$\text{Midline: } y = D = \frac{\text{Maximum Value} + \text{Minimum Value}}{2}$$
• It represents the equilibrium or baseline state in physical modeling.

3. Frequency Parameter ($B$) & Period ($P$): Horizontal Stretch / Compression

• The period ($P$) is the horizontal distance required for the function to complete one full cycle: $$P = \frac{2\pi}{|B|}$$
• Conversely, when given the period $P$ from data or a graph, the frequency coefficient $B$ is found using: $$B = \frac{2\pi}{P}$$
• Frequency ($f$): The number of complete cycles executed per unit of $x$: $f = \frac{1}{P} = \frac{|B|}{2\pi}$.

4. Phase Shift ($C$): Horizontal Translation

• The parameter $C$ represents the horizontal shift of the wave.
• In the standard factored form $B(x - C)$, a value of $C > 0$ shifts the wave right by $C$ units, while $C < 0$ (written as $(x + |C|)$) shifts the wave left by $|C|$ units.

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3. Selecting Sine vs. Cosine for Periodic Real-World Scenarios

Because sine and cosine are horizontal translations of each other, any periodic sinusoid can be written using either function. However, strategic selection simplifies model construction based on the initial condition at $t = 0$:

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4. Step-by-Step Protocol for Constructing Sinusoidal Models

When given a table of values, a graph, or a contextual description of a cyclic process:

+--------------------------------------------------------------------------------+
|                      Sinusoidal Modeling Workflow                              |
+--------------------------------------------------------------------------------+
| 1. Identify Extrema    --> Determine Maximum and Minimum values.               |
| 2. Calculate Midline   --> D = (Max + Min) / 2.                                |
| 3. Calculate Amplitude --> |A| = (Max - Min) / 2.                              |
| 4. Determine Period    --> P = Time for 1 full cycle (or 2 × time from min to max)|
| 5. Solve for B         --> B = 2\pi / P.                                         |
| 6. Choose Sine/Cosine  --> Align start condition at t = 0 to establish C and sign.|
+--------------------------------------------------------------------------------+

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5. Worked Examples

Worked Problem 1: Ferris Wheel Height Over Time

Problem: A passenger boards a ferris wheel from a platform $6 \text{ feet}$ above the ground. The ferris wheel has a radius of $40 \text{ feet}$ and rotates continuously, completing one full revolution every $80 \text{ seconds}$. Construct a sinusoidal function $h(t)$ modeling the passenger's height in feet above the ground as a function of time $t$ in seconds after boarding.

• Step 1: Identify maximum and minimum heights.

  • Minimum height (boarding platform): $\text{Min} = 6 \text{ ft}$.
  • Maximum height: $\text{Max} = \text{Min} + 2(\text{radius}) = 6 + 2(40) = 86 \text{ ft}$.

• Step 2: Calculate the vertical midline $D$.

  $$D = \frac{\text{Max} + \text{Min}}{2} = \frac{86 + 6}{2} = \frac{92}{2} = 46 \text{ ft}$$

• Step 3: Calculate the amplitude $|A|$.

  $$|A| = \frac{\text{Max} - \text{Min}}{2} = \frac{86 - 6}{2} = \frac{80}{2} = 40 \text{ ft}$$

  (Note: For a circular wheel, amplitude always equals the wheel's radius).

• Step 4: Calculate the period $P$ and frequency parameter $B$. The wheel completes one revolution in $80 \text{ seconds}$, so $P = 80$.

  $$B = \frac{2\pi}{P} = \frac{2\pi}{80} = \frac{\pi}{40}$$

• Step 5: Determine function type and sign based on $t = 0$. At time $t = 0$, the passenger is at the minimum height ($6 \text{ ft}$). An unshifted inverted cosine function ($-\cos$) begins at the minimum value:

  $$h(t) = -40\cos\left(\frac{\pi}{40}t\right) + 46$$

Worked Problem 2: Tidal Water Depth Prediction

Problem: The water depth in a coastal harbor oscillates sinusoidally with the tide. On a particular day, high tide occurs at 2:00 AM with a depth of $18 \text{ feet}$, and low tide occurs $6.2 \text{ hours}$ later at 8:12 AM with a depth of $6 \text{ feet}$. Let $t$ represent elapsed hours since midnight ($t = 0$ at 12:00 AM).

1. Write a sinusoidal equation $D(t)$ modeling the water depth.
2. Determine the predicted water depth at 11:00 AM ($t = 11$).

• Step 1: Calculate amplitude and midline.

  • $\text{Max} = 18 \text{ ft}$, $\text{Min} = 6 \text{ ft}$.
  • Midline: $D = \frac{18 + 6}{2} = 12 \text{ ft}$.
  • Amplitude: $A = \frac{18 - 6}{2} = 6 \text{ ft}$.

• Step 2: Determine period and frequency parameter $B$. The time from a maximum to the consecutive minimum represents half a period:

  $$\frac{1}{2}P = 6.2 \implies P = 12.4 \text{ hours}$$ $$B = \frac{2\pi}{12.4} = \frac{2\pi}{\frac{62}{5}} = \frac{10\pi}{62} = \frac{5\pi}{31}$$

• Step 3: Identify the phase shift $C$. High tide occurs at 2:00 AM, so a crest occurs at $t = 2$. Using a positive cosine model, shift right by $C = 2$:

  $$D(t) = 6\cos\left(\frac{5\pi}{31}(t - 2)\right) + 12$$

• Step 4: Evaluate depth at 11:00 AM ($t = 11$).

  $$D(11) = 6\cos\left(\frac{5\pi}{31}(11 - 2)\right) + 12 = 6\cos\left(\frac{45\pi}{31}\right) + 12$$

  Using a calculator in Radian mode:

  $$\cos\left(\frac{45\pi}{31}\right) \approx -0.1514$$ $$D(11) = 6(-0.1514) + 12 = -0.908 + 12 \approx 11.09 \text{ feet}$$

Worked Problem 3: Analyzing an Algebraic Sinusoidal Function

Problem: Given the function $f(x) = -5\cos\left(\frac{\pi}{6}(x - 2)\right) + 11$, state the amplitude, midline, period, range, and horizontal phase shift.

• Step 1: Amplitude ($|A|$).

  $$|A| = |-5| = 5$$

• Step 2: Midline ($y = D$).

  $$y = 11$$

• Step 3: Period ($P$).

  $$P = \frac{2\pi}{|B|} = \frac{2\pi}{\frac{\pi}{6}} = 2\pi \times \frac{6}{\pi} = 12$$

• Step 4: Range.

  $$\text{Min} = D - |A| = 11 - 5 = 6 \quad \text{and} \quad \text{Max} = D + |A| = 11 + 5 = 16$$ $$\text{Range: } [6, 16]$$

• Step 5: Phase shift ($C$). The factor inside is $(x - 2)$, corresponding to a phase shift of $2$ units to the right.

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6. Common Regents Pitfalls & Exam Strategies

• Pitfall 1: Reporting negative amplitude. Amplitude is defined as a distance, meaning it must always be expressed as a positive number. If $A = -5$, the amplitude is $5$, not $-5$.
• Pitfall 2: Equating frequency parameter $B$ directly to period. Writing $P = B$ or $P = \frac{\pi}{6}$ is a common rubric violation. Remember: $P = \frac{2\pi}{B}$ and $B = \frac{2\pi}{P}$.
• Pitfall 3: Failing to double the half-period. When given the time from minimum to maximum, students often mistake this for the full period. The time from low to high is only half of the complete wave cycle ($P = 2 \times \Delta t$).