Trigonometry

Sinusoidal Modeling

Last updated: March 2026 · Advanced
Before you start

You should be comfortable with:

Real-world applications
📐
Carpentry

Measurements, material estimation, cutting calculations

Electrical

Voltage drop, wire sizing, load balancing

In Graphs of Sine and Cosine, you learned the general form y=Asin(B(xC))+Dy = A\sin(B(x - C)) + D and what each parameter controls. This page teaches you to go the other direction: given real-world data that repeats in a pattern, find AA, BB, CC, and DD to create a mathematical model.

The Four Parameters from Data

Every sinusoidal model requires four values. Here is how to extract each one from data:

The Four Parameters of a Sinusoidal Model

D (midline)maxminAPeriod = 2π/BC
ParameterFormulaWhat It Means
DD (vertical shift)D=max+min2D = \dfrac{\text{max} + \text{min}}{2}The midline — the average of the highest and lowest values
AA (amplitude)A=maxmin2A = \dfrac{\text{max} - \text{min}}{2}Half the distance between peak and trough
BB (angular frequency)B=2πperiodB = \dfrac{2\pi}{\text{period}}Controls how fast the wave cycles
CC (phase shift)Find the xx-value where the cycle startsHorizontal offset from the origin

Worked Example 1: Daylight Hours

In a certain city, June 21 (day 172) has the most daylight at 15.2 hours, and December 21 (day 355) has the least at 9.0 hours. Model the daylight as a function of day number.

Step 1 — Find DD (midline):

D=15.2+9.02=24.22=12.1 hoursD = \frac{15.2 + 9.0}{2} = \frac{24.2}{2} = 12.1 \text{ hours}

Step 2 — Find AA (amplitude):

A=15.29.02=6.22=3.1 hoursA = \frac{15.2 - 9.0}{2} = \frac{6.2}{2} = 3.1 \text{ hours}

Step 3 — Find BB (from the period). The cycle repeats every 365 days.

B=2π3650.01721B = \frac{2\pi}{365} \approx 0.01721

Step 4 — Find CC (phase shift). The maximum occurs on day 172. For a cosine function, the peak is at x=Cx = C, so:

C=172C = 172

Using cosine (since we identified the maximum first):

y=3.1cos ⁣(2π365(x172))+12.1\boxed{y = 3.1\cos\!\left(\frac{2\pi}{365}(x - 172)\right) + 12.1}

Verify: At x=172x = 172 (June 21): y=3.1cos(0)+12.1=3.1+12.1=15.2y = 3.1\cos(0) + 12.1 = 3.1 + 12.1 = 15.2

At x=355x = 355 (Dec 21): y=3.1cos ⁣(2π365(183))+12.13.1cos(π)+12.1=3.1+12.1=9.0y = 3.1\cos\!\left(\frac{2\pi}{365}(183)\right) + 12.1 \approx 3.1\cos(\pi) + 12.1 = -3.1 + 12.1 = 9.0

Worked Example 2: Tide Height

A coastal harbor records the following tide data:

Time (hours)Height (feet)
0:004.2
3:007.8
6:004.2
9:000.6
12:004.2

A dock worker needs to know: when is the tide at least 6 feet?

Step 1 — Identify max and min. Max =7.8= 7.8 ft (at t=3t = 3), min =0.6= 0.6 ft (at t=9t = 9).

Step 2 — Find DD:

D=7.8+0.62=4.2 ftD = \frac{7.8 + 0.6}{2} = 4.2 \text{ ft}

Step 3 — Find AA:

A=7.80.62=3.6 ftA = \frac{7.8 - 0.6}{2} = 3.6 \text{ ft}

Step 4 — Find the period. From max at t=3t = 3 to the next time the pattern would reach the same max: the data shows a half-period from t=3t = 3 (max) to t=9t = 9 (min) is 6 hours. So the full period is 12 hours.

B=2π12=π6B = \frac{2\pi}{12} = \frac{\pi}{6}

Step 5 — Find CC. The maximum is at t=3t = 3. Using cosine:

C=3C = 3

Model:

h(t)=3.6cos ⁣(π6(t3))+4.2h(t) = 3.6\cos\!\left(\frac{\pi}{6}(t - 3)\right) + 4.2

Solving the dock worker’s question: When is h(t)6h(t) \ge 6?

3.6cos ⁣(π6(t3))+4.263.6\cos\!\left(\frac{\pi}{6}(t - 3)\right) + 4.2 \ge 6

cos ⁣(π6(t3))1.83.6=0.5\cos\!\left(\frac{\pi}{6}(t - 3)\right) \ge \frac{1.8}{3.6} = 0.5

Cosine is 0.5\ge 0.5 when the angle is in [π3,π3][-\frac{\pi}{3}, \frac{\pi}{3}] (within one period).

π3π6(t3)π3-\frac{\pi}{3} \le \frac{\pi}{6}(t - 3) \le \frac{\pi}{3}

2t32-2 \le t - 3 \le 2

1t51 \le t \le 5

Answer: The tide is at least 6 feet from 1:00 AM to 5:00 AM and again from 1:00 PM to 5:00 PM (each 12-hour cycle).

Choosing Sine vs. Cosine

Both model the same wave — the only difference is the default starting position:

  • Cosine starts at a maximum: cos(0)=1\cos(0) = 1. Use cosine when you can easily identify where the max occurs.
  • Sine starts at the midline going up: sin(0)=0\sin(0) = 0. Use sine when you can identify where the data crosses the midline while increasing.

Either function works for any data set — one just gives a simpler phase shift CC than the other.

Checking Your Model

After building a model, always plug in at least two known data points to verify:

  1. Check the maximum: does yy equal the max value at the right xx?
  2. Check the minimum: does yy equal the min value at the right xx?
  3. Check a midline crossing: does yy equal DD at a point between max and min?

If any check fails, the most common errors are:

  • Phase shift sign error (positive CC shifts right)
  • Period error (using the half-period instead of the full period)
  • Mixing up AA and DD

Common Mistakes

  1. Using the wrong period. The time from max to min is a half-period, not a full period. The full period is twice that distance.
  2. Sign errors in the phase shift. In y=Asin(B(xC))+Dy = A\sin(B(x - C)) + D, the CC is subtracted. A positive CC shifts the graph right. Students often get the direction backwards.
  3. Forgetting that BB multiplies the entire (xC)(x - C). The phase shift is CC, not C/BC/B. If you write y=Asin(BxC)+Dy = A\sin(Bx - C) + D instead of y=Asin(B(xC))+Dy = A\sin(B(x - C)) + D, the shift is C/BC/B, not CC.
  4. Not converting time units consistently. If the period is in hours, make sure all inputs are in hours. Mixing hours and minutes (or days and months) breaks the model.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Problem 1: A city’s average monthly temperature has a maximum of 84°F in July (month 7) and a minimum of 32°F in January (month 1). Write a cosine model for temperature as a function of month.

D=(84+32)/2=58°FD = (84 + 32)/2 = 58°F

A=(8432)/2=26°FA = (84 - 32)/2 = 26°F

Period = 12 months, so B=2π/12=π/6B = 2\pi/12 = \pi/6

Max at month 7: C=7C = 7

T(m)=26cos ⁣(π6(m7))+58T(m) = 26\cos\!\left(\frac{\pi}{6}(m - 7)\right) + 58

Verify: T(7)=26cos(0)+58=84T(7) = 26\cos(0) + 58 = 84 ✓ and T(1)=26cos(π)+58=26+58=32T(1) = 26\cos(-\pi) + 58 = -26 + 58 = 32

Answer: T(m)=26cos ⁣(π6(m7))+58T(m) = 26\cos\!\left(\frac{\pi}{6}(m - 7)\right) + 58

Problem 2: A Ferris wheel has a diameter of 60 feet, and its center is 35 feet above the ground. It completes one revolution every 3 minutes. A rider boards at the lowest point. Write a function for the rider’s height above the ground as a function of time in minutes.

Radius = 30 ft. Lowest point = 3530=535 - 30 = 5 ft. Highest point = 35+30=6535 + 30 = 65 ft.

D=35D = 35 (center height), A=30A = 30 (radius)

Period = 3 minutes, so B=2π/3B = 2\pi/3

The rider starts at the bottom (minimum). A negative cosine starts at a minimum:

h(t)=30cos ⁣(2π3t)+35h(t) = -30\cos\!\left(\frac{2\pi}{3}t\right) + 35

Verify: h(0)=30(1)+35=5h(0) = -30(1) + 35 = 5 ft (bottom) ✓ and h(1.5)=30cos(π)+35=30+35=65h(1.5) = -30\cos(\pi) + 35 = 30 + 35 = 65 ft (top) ✓

Answer: h(t)=30cos ⁣(2π3t)+35h(t) = -30\cos\!\left(\frac{2\pi}{3}t\right) + 35

Problem 3: Using the tide model h(t)=3.6cos ⁣(π6(t3))+4.2h(t) = 3.6\cos\!\left(\frac{\pi}{6}(t - 3)\right) + 4.2, find the tide height at 4:30 AM (t=4.5t = 4.5).

h(4.5)=3.6cos ⁣(π6(4.53))+4.2h(4.5) = 3.6\cos\!\left(\frac{\pi}{6}(4.5 - 3)\right) + 4.2

=3.6cos ⁣(π6(1.5))+4.2=3.6cos ⁣(π4)+4.2= 3.6\cos\!\left(\frac{\pi}{6}(1.5)\right) + 4.2 = 3.6\cos\!\left(\frac{\pi}{4}\right) + 4.2

=3.6×0.7071+4.2=2.546+4.2=6.7466.7 ft= 3.6 \times 0.7071 + 4.2 = 2.546 + 4.2 = 6.746 \approx 6.7 \text{ ft}

Answer: The tide is approximately 6.7 feet at 4:30 AM.

Problem 4: An AC voltage source has the equation V(t)=170sin(120πt)V(t) = 170\sin(120\pi t), where tt is in seconds. Find the frequency (cycles per second) and the period.

B=120πB = 120\pi

Period =2πB=2π120π=160= \frac{2\pi}{B} = \frac{2\pi}{120\pi} = \frac{1}{60} seconds

Frequency =1period=60= \frac{1}{\text{period}} = 60 Hz (cycles per second)

This is the standard 60 Hz AC power in North America. The peak voltage is 170 V, which gives an RMS voltage of 170/2120170/\sqrt{2} \approx 120 V. For more on AC circuits and trig, see Trig for Electricians.

Answer: Frequency = 60 Hz, Period = 1/60 second

Problem 5: Given the model y=Acos(B(xC))+Dy = A\cos(B(x - C)) + D with D=50D = 50, A=20A = 20, and period = 8, find BB. Then, if the maximum occurs at x=2x = 2, write the complete model.

B=2π8=π4B = \frac{2\pi}{8} = \frac{\pi}{4}

Max at x=2x = 2: C=2C = 2

y=20cos ⁣(π4(x2))+50y = 20\cos\!\left(\frac{\pi}{4}(x - 2)\right) + 50

Answer: y=20cos ⁣(π4(x2))+50y = 20\cos\!\left(\frac{\pi}{4}(x - 2)\right) + 50

Key Takeaways

  • To build a sinusoidal model: find DD (midline), AA (amplitude), BB (from the period), and CC (phase shift from where the max or midline crossing occurs)
  • Use cosine when you identify the maximum first; use sine when you identify a midline crossing
  • The time from maximum to minimum is a half-period — double it for the full period
  • Always verify the model by plugging in at least two known data points
  • Sinusoidal models apply to any repeating phenomenon: daylight hours, tides, temperature cycles, Ferris wheels, and AC voltage

Return to Trigonometry for more topics in this section.

Last updated: March 28, 2026