AP Precalculus 3.4 Sine & Cosine Graphs — Notes

〰️

Graph of f(θ) = sin θ

Input = angle θ on horizontal axis · output = sin θ on vertical axis

💡

Building the sine graph from the unit circle

Since angle measures in standard position are periodic, f(θ) = sin θ is periodic too. We use θ on the horizontal axis and sin θ (the y-coordinate on the unit circle) on the vertical axis. Every angle from the unit circle gives us a point on the graph.

f(θ) = sin θ (red) and g(θ) = cos θ (blue dashed) — one full period 0 to 2π

θ	0	π/6	π/3	π/2	2π/3	5π/6	π	7π/6	3π/2	5π/3	11π/6	2π
sin θ	0	½	√3/2	1	√3/2	½	0	−½	−1	−√3/2	−½	0

Properties of f(θ) = sin θ
Midline Halfway between max and min	Amplitude Distance from midline to max/min	Period Length of one full cycle	Frequency Reciprocal of the period
y = 0	a = 1	P = 2π	1/(2π)

🔑

Key shape facts for sin θ

Starts at (0, 0) · rises to max (π/2, 1) · returns to zero at (π, 0) · falls to min (3π/2, −1) · returns to zero at (2π, 0).
The graph oscillates between concave down (first half) and concave up (second half).

〰️

Graph of g(θ) = cos θ

Same shape as sin θ, but shifted left by π/2 · starts at the maximum

Properties of g(θ) = cos θ
Midline	Amplitude	Period	Frequency
y = 0	a = 1	P = 2π	1/(2π)

🔑

Key shape facts for cos θ

Starts at maximum (0, 1) · crosses zero at (π/2, 0) · reaches min at (π, −1) · crosses zero at (3π/2, 0) · returns to max at (2π, 1).
The graph also oscillates between concave down and concave up.

📖

Key Definitions

Midline · Amplitude · Period · Frequency · Sinusoidal Functions

Term	Definition	Formula
Midline	Horizontal line halfway between the maximum and minimum values of the function	y = (max + min) / 2
Amplitude	Distance from the midline to the maximum (or minimum) value	a = (max − min) / 2
Period	The length of one full cycle of the function	P = (distance from max to next max)
Frequency	The reciprocal of the period — how many cycles occur per unit	f = 1/P

🎯

Max to min = HALF a period

The distance from a maximum to the very next minimum is only half a period. So if max is at θ = π and min is at θ = 3π, the half-period is 2π → full period = 4π. Always double the max-to-min distance to get the full period!

📌 Sinusoidal Functions

A sinusoidal function is any function that involves additive and multiplicative transformations of f(θ) = sin θ.

Both sine and cosine are sinusoidal functions because: g(θ) = cos θ = sin(θ + π/2)

Cosine is just a sine function shifted left by π/2. This means any sinusoidal function can be written as either a sine or cosine transformation.

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Reading Properties from Graphs — Examples 1 & 2

Find max and min → compute midline, amplitude · find repeat distance → period

Graph of f — Example 1 (max=6, min=−6, period=π)

📌 Example 1 — Find period, frequency, amplitude, and midline for f(θ).

1

Period: One full cycle from 0 to π (the graph completes a full wave by θ=π). P = π

2

Frequency: 1/P = 1/π

3

Max = 6, Min = −6 · Midline = (6 + (−6))/2 = y = 0

4

Amplitude: (max − min)/2 = (6 − (−6))/2 = 12/2 = 3... wait: max = 6 so amplitude = 6 − 0 = 3? No — max is 6, midline is 0 (not 3). Amplitude = max − midline = 6 − 0 = 3. Hmm, but graph goes to 6...

Period

π

Frequency

1/π

Midline

y = 0

Amplitude

3

From the graph: maximum = 6, minimum = 0. Midline = (6+0)/2 = 3. Amplitude = 6−3 = 3.

✏️

Corrected reading: max=6, min=0

The graph of f oscillates between max = 6 and min = 0 (it never goes below the x-axis). So: midline = (6+0)/2 = y = 3. Amplitude = 6 − 3 = 3. Period = π (one complete cycle seen). Frequency = 1/π.

📌 Example 2 — Find period, frequency, amplitude, and midline for g(θ). (max = 3, min = −1)

1

Period: One full cycle completes in π/2 (two cycles visible from 0 to π). P = π/2

2

Frequency: 1/P = 1/(π/2) = 2/π

3

Midline: (max + min)/2 = (3 + (−1))/2 = 2/2 = y = 1

4

Amplitude: max − midline = 3 − 1 = 2

Period

π/2

Frequency

2/π

Midline

y = 1

Amplitude

2

🔢

Finding Properties from Max/Min Points — Example 3

Max to min = half period · midline = average of max and min values

📌 Example 3 — h(θ) has a maximum at (π, 8) and the first minimum after is at (3π, −2). Find all properties.

1

Half-period: Distance from max to min = 3π − π = 2π → this is HALF the period.

2

Period: Full period = 2 × (half period) = 2 × 2π = 4π

3

Frequency: 1/P = 1/(4π)

4

Midline: (max + min)/2 = (8 + (−2))/2 = 6/2 = y = 3

5

Amplitude: max − midline = 8 − 3 = 5

Period

4π

Frequency

1/(4π)

Midline

y = 3

Amplitude

5

⚙️ Step-by-step derivation

Max at θ=π (y=8) · Min at θ=3π (y=−2)

Half period = 3π − π = 2π → Full period P = 4π

Midline = (8 + (−2))/2 = 6/2 = y = 3

Amplitude = 8 − 3 = 5

🕐

Real-World Application — Example 4 (FRQ Model)

Clock problem · identify 5 key points · find all properties · intervals for decreasing and concave up

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Setup

A clock is mounted on a wall. Center of clock = 120 inches above the floor. Minute hand = 8 inches long. Clock runs twice as fast as normal, so one revolution = 30 minutes. At t=0 the hand points straight up (12 o'clock). h(t) = distance from endpoint of hand to floor.

Part (A) — Five Key Points

t=0: Hand points up → endpoint = 120 + 8 = 128 in
t=7.5: Hand points right (¼ revolution) → endpoint at midline height = 120 in
t=15: Hand points down (½ revolution) → endpoint = 120 − 8 = 112 in
t=22.5: Hand points left (¾ revolution) → endpoint at midline = 120 in
t=30: Back to top (full revolution) → endpoint = 128 in

F: (0, 128)
G: (7.5, 120)
J: (15, 112)
K: (22.5, 120)
P: (30, 128)

Part (B) — Properties of h(t)

Max = 128 (hand points up)
Min = 112 (hand points down)
Midline = (128 + 112)/2 = y = 120
Amplitude = 128 − 120 = 8
Period = 30 minutes (twice as fast)
Frequency = 1/30

Period

30

Frequency

1/30

Midline

y = 120

Amplitude

8

Part (C) — Find two intervals where h(t) is both decreasing AND concave up.

1

Decreasing: h is decreasing when the hand moves from the top (max) toward the bottom (min): from t=0 to t=15, and again from t=30 to t=45 in the second cycle.

2

Concave up: A sine curve is concave up in its lower half — from the midline crossing (going down) to the midline crossing (coming back up). That's from t=7.5 to t=22.5 for each cycle.

3

Intersection: Decreasing (0,15) ∩ Concave up (7.5,22.5) = (7.5, 15). In second cycle: (37.5, 45).

h is decreasing AND concave up on: (7.5, 15) and (37.5, 45)

💡

How to find "decreasing and concave up" on a sinusoidal graph

Sketch one full cycle. The graph is decreasing from max to min. It is concave up in the bottom half (below the midline) — from the descending midline crossing to the ascending midline crossing. The intersection of these two behaviors is the second quarter of each cycle (from the descending midline crossing to the minimum).

⚡

Quick Reference

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🔑 Key Formulas

Midline = (max + min) / 2

Amplitude = max − midline = (max − min)/2

Period P = 2 × (distance from max to min)

Frequency = 1/P

cos θ = sin(θ + π/2) — sinusoidal!

⚠️ Common Mistakes

❌ max − min = amplitude

(max − min) = twice the amplitude. Divide by 2: amplitude = (max − min)/2. Or: amplitude = max − midline.

❌ Max-to-min distance = full period

Max to min is only HALF the period. Always multiply by 2 to get the full period.

❌ sin starts at max; cos starts at 0

It's the other way: sin(0)=0 (starts at zero), cos(0)=1 (starts at MAX). sin begins at the midline, cos begins at the peak.

❌ Frequency = period

Frequency = 1/Period. They are reciprocals. Larger period → smaller frequency (fewer cycles per unit).

Sine & CosineFunction Graphs

Building the sine graph from the unit circle

Key shape facts for sin θ

Key shape facts for cos θ

Max to min = HALF a period

📌 Sinusoidal Functions

Corrected reading: max=6, min=0

Setup

Part (A) — Five Key Points

Part (B) — Properties of h(t)

How to find "decreasing and concave up" on a sinusoidal graph

Sine & Cosine
Function Graphs