What is the period of y = sin x, and at what x in [0, 2π] is its maximum? [1 point]

Is y = cos x concave up or concave down on ((π)/(2), (3π)/(2))? [1 point]

College Board AP 2023 (style)-style practice: Section I, Part A (multiple choice, no calculator). On the interval [0, 2π], where does the graph of y = cos x attain its minimum value? (A) x = 0 (B) x = (π)/(2) (C) x = π (D) x = (3π)/(2)

The correct answer is (C), x = π. The cosine graph starts at its maximum (0, 1), falls to 0 at (π)/(2), reaches its minimum -1 at π, returns to 0 at (3π)/(2), and back to 1 at 2π. So the minimum occurs at x = π.

College Board AP 2024 (style)-style practice: Section II (free response, no calculator). Consider y = sin x on [0, 2π]. (a) State all zeros, the maximum point, and the minimum point. (b) Describe the concavity of the graph on (0, π) and justify using the rate of change.

A 3-point question on reading the sine graph. (a) Key points (2 points): zeros at x = 0, π, 2π; maximum at ((π)/(2), 1); minimum at ((3π)/(2), -1). (b) Concavity (1 point): on (0, π) the graph is concave down. It rises to the peak at (π)/(2) then falls, so its rate of change is decreasing throughout (0, π), which means the graph bends downward (concave down) on that interval.

United StatesPrecalculusSyllabus dot point

What do the graphs of sine and cosine look like, and how do their period, amplitude, midline and concavity arise from the unit circle?

Topic 3.4 Sine and Cosine Function Graphs: construct and interpret the graphs of sine and cosine, identifying period, amplitude, midline, zeros and concavity.

A focused answer to AP Precalculus Topic 3.4, covering how the sine and cosine graphs are generated from the unit circle, their period, amplitude, midline, zeros, maxima and minima, and concavity within a cycle.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

As an angle sweeps around the unit circle, the $y$ -coordinate traces $\sin x$ and the $x$ -coordinate traces $\cos x$ , producing two smooth waves. Both have period $2\pi$ , amplitude $1$ and midline $y = 0$ . The sine graph passes through the origin, rising to a maximum $1$ at $\frac{\pi}{2}$ , back to $0$ at $\pi$ , down to $-1$ at $\frac{3\pi}{2}$ , and back to $0$ at $2\pi$ . The cosine graph starts at its maximum $1$ at $x = 0$ and is the sine graph shifted left by $\frac{\pi}{2}$ . Within a cycle the curves change concavity: concave down across each peak and concave up across each trough, with the inflection points at the midline crossings. Knowing these landmark points lets you sketch either graph by hand and read features off it.

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What this topic is asking
From the circle to the wave
Landmark points
Concavity within a cycle
Sine and cosine as shifts of each other
Try this

What this topic is asking

The College Board (Topic 3.4) wants you to construct and read the graphs of $y = \sin x$ and $y = \cos x$ . You should know their period ( $2\pi$ ), amplitude ( $1$ ), midline ( $y = 0$ ), the locations of their zeros, maxima and minima, and the concavity within each cycle, all of which come directly from the unit circle.

From the circle to the wave

The repeating wave shape is the direct visual record of going around the circle again and again.

Landmark points

For $y = \sin x$ on $[0, 2\pi]$ : zeros at $0, \pi, 2\pi$ ; maximum $\left(\frac{\pi}{2}, 1\right)$ ; minimum $\left(\frac{3\pi}{2}, -1\right)$ . For $y = \cos x$ : zeros at $\frac{\pi}{2}, \frac{3\pi}{2}$ ; maximum $(0, 1)$ and $(2\pi, 1)$ ; minimum $(\pi, -1)$ . Plotting these five-per-cycle landmark points and joining them smoothly reproduces either graph.

Sketching one cycle of cosine

Sketch $y = \cos x$ on $[0, 2\pi]$ using its landmark points, and state where it is increasing.

step 1 Start at the maximum

At $x = 0$ , $\cos 0 = 1$ : the graph begins at its peak $(0, 1)$ .

step 2 Mark the quarter points

At $\frac{\pi}{2}$ , $\cos = 0$ (a zero, falling). At $\pi$ , $\cos = -1$ (the minimum). At $\frac{3\pi}{2}$ , $\cos = 0$ (a zero, rising). At $2\pi$ , $\cos = 1$ (back to the peak).

step 3 Join smoothly

Connect $(0,1) \to \left(\frac{\pi}{2},0\right) \to (\pi,-1) \to \left(\frac{3\pi}{2},0\right) \to (2\pi,1)$ with a smooth wave.

step 4 State the increasing interval

The graph falls from $0$ to $\pi$ then rises from $\pi$ to $2\pi$ , so $y = \cos x$ is increasing on $(\pi, 2\pi)$ .

Concavity within a cycle

The concavity of a sine or cosine wave alternates. Across each peak the curve bends downward (concave down); across each trough it bends upward (concave up); the changeovers (inflection points) happen exactly at the midline crossings. For $y = \sin x$ , the interval $(0, \pi)$ is concave down (it contains the peak) and $(\pi, 2\pi)$ is concave up (it contains the trough). This is the change-in-tandem language of Topic 1.1 applied to the wave: where the rate of change is decreasing, the graph is concave down.

Sine and cosine as shifts of each other

A point worth stating once is that cosine is sine shifted left by $\frac{\pi}{2}$ : $\cos x = \sin\left(x + \frac{\pi}{2}\right)$ . The two graphs have identical shape, period, amplitude and midline; they differ only in horizontal position. Recognizing this means you only ever need to remember one wave and where it starts. This single shift relationship is the seed of the general sinusoidal model in Topics 3.5 and 3.6, where amplitude, period, phase and vertical shift are all adjusted at once.

Try this

Q1. What is the period of $y = \sin x$ , and at what $x$ in $[0, 2\pi]$ is its maximum? [1 point]

Cue. Period $2\pi$ ; maximum at $x = \frac{\pi}{2}$ .

Q2. Is $y = \cos x$ concave up or concave down on $\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$ ? [1 point]

Cue. Concave up: this interval contains the minimum at $x = \pi$ , so the curve bends upward across it.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)1 marksSection I, Part A (multiple choice, no calculator). On the interval

[0, 2\pi]

, where does the graph of

y = \cos x

attain its minimum value? (A)

x = 0

(B)

x = \frac{\pi}{2}

(C)

x = \pi

(D)

x = \frac{3\pi}{2}

Show worked answer →

The correct answer is (C), $x = \pi$ .

The cosine graph starts at its maximum $(0, 1)$ , falls to $0$ at $\frac{\pi}{2}$ , reaches its minimum $-1$ at $\pi$ , returns to $0$ at $\frac{3\pi}{2}$ , and back to $1$ at $2\pi$ . So the minimum occurs at $x = \pi$ .

AP 2024 (style)3 marksSection II (free response, no calculator). Consider

y = \sin x

[0, 2\pi]

. (a) State all zeros, the maximum point, and the minimum point. (b) Describe the concavity of the graph on

(0, \pi)

and justify using the rate of change.

Show worked answer →

A 3-point question on reading the sine graph.

(a) Key points (2 points): zeros at $x = 0, \pi, 2\pi$ ; maximum at $\left(\frac{\pi}{2}, 1\right)$ ; minimum at $\left(\frac{3\pi}{2}, -1\right)$ .
(b) Concavity (1 point): on $(0, \pi)$ the graph is concave down. It rises to the peak at $\frac{\pi}{2}$ then falls, so its rate of change is decreasing throughout $(0, \pi)$ , which means the graph bends downward (concave down) on that interval.

Related dot points

Sources & how we know this

AP Precalculus Course and Exam Description — College Board (2023)