Watch one rivet on a Ferris wheel from the side, and plot its height against time. As the wheel turns steadily, the rivet's height rises, crests, falls, bottoms out, rises again — tracing the smooth, endless wave you know as the sine curve.
That's literally what the sine graph is: the y-coordinate of a point moving around a circle, unrolled along a time axis. The circle from Lesson 17 and the wave in this one are the same object in two costumes — and switching fluently between costumes is the skill Unit 3 grades.
Let θ increase steadily and plot the unit-circle coordinates against θ:
y = sin θ (the height of P(θ)): starts at 0, rises to 1 at π/2, back through 0 at π, down to −1 at 3π/2, home at 2π.y = cos θ (the horizontal coordinate): starts at 1 (its maximum — the point starts at (1,0)), falls through 0 at π/2, bottoms at −1 at π, returns to 1 at 2π.[GRAPH: Two aligned panels over one period [0, 2π]. Top: y = sin θ — zeros at 0, π, 2π marked; max (π/2, 1); min (3π/2, −1). Bottom: y = cos θ — max (0, 1); zeros at π/2, 3π/2; min (π, −1). Dashed vertical gridlines at multiples of π/2 connecting the panels; midline y = 0 dashed on both.]
Same wave, offset start: cos θ = sin(θ + π/2) — cosine is sine slid left a quarter turn. Either curve, and every vertical/horizontal dilation or translation of them, is called sinusoidal.
For a sinusoidal graph:
d = (max + min)/2a = (max − min)/2 (always positive as a distance)Example: a sinusoid oscillating between 3 and 11 has midline y = (11+3)/2 = 7 and amplitude (11−3)/2 = 4.
For the parents sin θ and cos θ: midline y = 0, amplitude 1, period 2π, frequency 1/(2π).
Over one period, the sine curve's story (cosine's is the same story starting at a peak):
| feature | where (for sin θ on [0, 2π]) |
|---|---|
| increasing | (0, π/2) ∪ (3π/2, 2π) |
| decreasing | (π/2, 3π/2) |
| concave down | (0, π) — the crest half, above the midline |
| concave up | (π, 2π) — the trough half, below the midline |
| zeros (= crossing the midline) | 0, π, 2π |
| extrema | max at π/2, min at 3π/2 |
Two linked facts the exam loves:
Sinusoids have no end behavior in the limit sense — lim θ→∞ sin θ does not exist; the graph oscillates forever between its bounds. Say "oscillates between −1 and 1," never "approaches."
sin is odd (sin(−θ) = −sin θ: origin symmetry); cos is even (cos(−θ) = cos θ: y-axis symmetry) — inherited directly from the unit circle (Lesson 17) and visible in the graphs.
Problem: State the amplitude, midline, period, and range of y = 2 cos θ − 5.
Solution: Amplitude 2; midline y = −5; period 2π (no horizontal dilation); range: midline ± amplitude → [−7, −3].
Interpretation: Outside constants (2, −5) set the vertical anatomy; the period waits for inside constants (next lesson).
Problem: A sinusoidal graph has consecutive maxima at (1, 11) and (9, 11), with a minimum of 3 between them. Find the period, midline, amplitude, and the input of that minimum.
Solution: Period: peak-to-peak = 9 − 1 = 8. Midline: (11 + 3)/2 = y = 7. Amplitude: (11 − 3)/2 = 4. The minimum sits midway between consecutive maxima: input (1 + 9)/2 = 5 → minimum point (5, 3).
Interpretation: Symmetry does the locating: min halfway between maxes, midline crossings halfway between each extreme pair (at inputs 3 and 7 here).
Problem: For y = sin θ on [0, 2π], find where the function is simultaneously decreasing and concave up, and describe that piece in Ferris-wheel language.
Solution: Decreasing: (π/2, 3π/2). Concave up: (π, 2π). Overlap: (π, 3π/2). Ferris wheel: the rivet is falling (decreasing) but the fall is flattening as it nears the bottom (concave up) — the last quarter of the descent.
Interpretation: Sinusoids cycle through the four direction/concavity combinations each period, one quarter each — a favorite MC pattern-check.
Problem: A sinusoid has midline y = 6, amplitude 2.5, and period 12, with a maximum at x = 3. Without writing a formula, find its value and behavior at x = 9 and at x = 6.
Solution: Max at x = 3 → min half a period later at x = 3 + 6 = 9: value 6 − 2.5 = 3.5 at x = 9. Midline crossings occur a quarter period (3 units) from each extreme: at x = 6 the curve crosses the midline (value 6), heading downward (it's between a max and the following min).
Interpretation: Quarter-period bookkeeping (max → midline↓ → min → midline↑ → max) answers value questions with no equation at all. Build this reflex now — FRQ 2 rewards it.
1. (A). |3| = 3. (B) is the full max-to-min swing.
2. (B). One trip around the unit circle: 2π. (D) confuses radians with degrees.
3. (C). The −5 drops the midline to y = −5; amplitude 2 doesn't move it. (B) subtracts amplitude from shift.
4. (B). Midline 1 ± amplitude 4: [−3, 5]. (C) forgets the downward reach; (A) ignores the shift.
5. (D). cos starts at (0, 1), its maximum. (A) describes sin.
6. (C). Rising from −1 to 1 as θ runs (−π/2, π/2) (through the origin). (A) spans the crest — half rising, half falling.
7. (D). (11 − 3)/2 = 4. (A) is the full swing; (B) the midline.
8. (A). (11 + 3)/2 = y = 7.
9. (B). Crest half — above the midline: (0, π). (A) is the concave-up trough.
10. (C). Frequency = 1/period = 1/(2π). (D) forgets the 2π.
11. (D). Slide sine left a quarter period: sin(θ + π/2) = cos θ. Check θ = 0: sin(π/2) = 1 = cos 0 ✓. (A) shifts the wrong way (it equals −cos θ).
12. (FRQ-style, 6 points) (i) [2 pts] Min at x = 2 → max at x = 2 + 5 = 7 (half period): h(7) = 28. Quarter period after the min, at x = 4.5, h crosses the midline: h(4.5) = 20 (heading upward). (ii) [2 pts] Increasing from min to max: (2, 7); decreasing from max to next min: (7, 12). (iii) [2 pts] Inflections at the midline crossings: x = 4.5 (rising) and x = 9.5 (falling). A sinusoid's concavity flips exactly where it crosses its midline — below the midline it's concave up, above it concave down.
12. (FRQ-style) 🚫 A sinusoidal function h has period 10, midline y = 20, amplitude 8, and a minimum at x = 2. (i) Find the value of h at x = 7 and at x = 4.5, using quarter-period reasoning (no formula). (ii) On what interval(s) between x = 2 and x = 12 is h increasing? Decreasing? (iii) Identify the x-values in [2, 12] where h has points of inflection, and justify with the midline.
The figure shows one cycle of a sinusoidal function g: a maximum at (0, 9), a midline crossing at (4, 5) heading downward, a minimum at (8, 1), a midline crossing at (12, 5) heading upward, and a return to a maximum at (16, 9).
(a) (i) State the period, midline, and amplitude of g. (ii) State the range of g.
(b) (i) On what interval in (0, 16) is g decreasing? (ii) Justify using the graph's extrema.
(c) (i) State the intervals in (0, 16) where g is concave up. (ii) Explain the connection between the midline crossings and the concavity changes.
(a) [2 pts] (i) [1 pt] Period 16 (max at 0 to max at 16); midline y = 5; amplitude 4. (ii) [1 pt] Range [1, 9] (midline ± amplitude).
(b) [2 pts] (i) [1 pt] g is decreasing on (0, 8). (ii) [1 pt] g falls from its maximum at x = 0 to its minimum at x = 8; between a maximum and the next minimum, outputs strictly fall — so g decreases on the whole interval, then increases on (8, 16).
(c) [2 pts] (i) [1 pt] Concave up on (4, 12) — the trough half of the cycle, where the graph lies below its midline. (ii) [1 pt] A sinusoid changes concavity exactly where it crosses its midline (x = 4 and x = 12): these crossings are the points of inflection, separating the concave-down crest (above the midline) from the concave-up trough (below it).
1. (A). |3| = 3. (B) is the full max-to-min swing.
2. (B). One trip around the unit circle: 2π. (D) confuses radians with degrees.
3. (C). The −5 drops the midline to y = −5; amplitude 2 doesn't move it. (B) subtracts amplitude from shift.
4. (B). Midline 1 ± amplitude 4: [−3, 5]. (C) forgets the downward reach; (A) ignores the shift.
5. (D). cos starts at (0, 1), its maximum. (A) describes sin.
6. (C). Rising from −1 to 1 as θ runs (−π/2, π/2) (through the origin). (A) spans the crest — half rising, half falling.
7. (D). (11 − 3)/2 = 4. (A) is the full swing; (B) the midline.
8. (A). (11 + 3)/2 = y = 7.
9. (B). Crest half — above the midline: (0, π). (A) is the concave-up trough.
10. (C). Frequency = 1/period = 1/(2π). (D) forgets the 2π.
11. (D). Slide sine left a quarter period: sin(θ + π/2) = cos θ. Check θ = 0: sin(π/2) = 1 = cos 0 ✓. (A) shifts the wrong way (it equals −cos θ).
12. (FRQ-style, 6 points) (i) [2 pts] Min at x = 2 → max at x = 2 + 5 = 7 (half period): h(7) = 28. Quarter period after the min, at x = 4.5, h crosses the midline: h(4.5) = 20 (heading upward). (ii) [2 pts] Increasing from min to max: (2, 7); decreasing from max to next min: (7, 12). (iii) [2 pts] Inflections at the midline crossings: x = 4.5 (rising) and x = 9.5 (falling). A sinusoid's concavity flips exactly where it crosses its midline — below the midline it's concave up, above it concave down.
🎯 Exam tip: Draw the "quarter-period ruler" for any sinusoid: mark the given extremum, then step T/4 at a time, cycling max → midline↓ → min → midline↑ → max. Nearly every graph-reading question in Unit 3 is answered by this ruler before any formula appears.