Unit 3 · Topic 3.4

Sine and Cosine Function Graphs

The graph of y = sin θ is the height of the point at angle θ on the unit circle; y = cos θ is its horizontal position.

Practice 2: Multiple Representations

Part 1 · Hook

Why this matters

Watch a reflector on a turning bike wheel from the side: it rises, slows at the top, falls, slows at the bottom and rises again. Plot its height against the angle the wheel has turned and you get the most important periodic graph there is. That graph, and its twin for the horizontal position, is what every model in the rest of this unit is built from.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is sin⁡7π6\sin\frac{7\pi}{6}?

  1. −12-\frac{1}{2}
  2. 12\frac{1}{2}
  3. −32-\frac{\sqrt{3}}{2}
  4. 32\frac{\sqrt{3}}{2}
Show the answer

Quadrant III, reference angle π6\frac{\pi}{6}: the height is −12-\frac{1}{2}.

  • Correct: −12-\frac{1}{2}:
  • 12\frac{1}{2}:
  • −32-\frac{\sqrt{3}}{2}:
  • 32\frac{\sqrt{3}}{2}:

2. A function has period PP. Over any interval of length PP, what is its average rate of change?

  1. 00
  2. PP
  3. 1P\frac{1}{P}
  4. It depends on the interval
Show the answer

The output after one period equals the output at the start, so the change in output is 00.

  • Correct: 00:
  • PP:
  • 1P\frac{1}{P}:
  • It depends on the interval:

Part 4 · See it

See it first

Left: a circle of radius 1 with the point P at angle 2 pi over 3, whose height above the x-axis is sine of 2 pi over 3, about 0.87. Right: the graph of y = sin theta for theta from 0 to 2 pi in green, with the graph of y = cos theta dashed in blue for comparison. A dashed horizontal line carries the height of P across to the point (2 pi over 3, sin 2 pi over 3) on the sine graph. The sine graph starts at 0, rises to 1 at pi/2, returns to 0 at pi, falls to −1 at 3 pi over 2 and returns to 0 at 2 pi.
The height of the point at angle θ\theta on the unit circle is the height of the sine graph at θ\theta. The dashed blue curve is cosine, the horizontal position. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Let the angle θ\theta be the input and the height of the point on the unit circle the outputyou get y=sin⁡θy = \sin\theta: 00 at θ=0\theta = 0, up to 11 at π2\frac{\pi}{2}, back to 00 at π\pi, down to −1-1 at 3π2\frac{3\pi}{2}, back to 00 at 2π2\pi
  2. Use the horizontal position as the output insteadyou get y=cos⁡θy = \cos\theta: 1,0,−1,0,11, 0, -1, 0, 1 at the same five inputs, the sine graph moved π2\frac{\pi}{2} to the left
  3. One more full turn brings the point backboth graphs repeat every 2π2\pi and stay between −1-1 and 11
  4. Near the top or bottom of the circle the height changes slowly; near the xx-axis it changes fastthe rate of change of sin⁡θ\sin\theta is largest in size at the zeros, and the graph is concave down where sin⁡θ>0\sin\theta > 0 and concave up where sin⁡θ<0\sin\theta < 0

Part 6 · Key ideas

Key ideas

  • Both graphs have domain all real numbers, range [−1,1][-1, 1] and period 2π2\pi.
  • Five key points per cycle: sine (0,0),(0, 0),(π2,1),\left(\frac{\pi}{2}, 1\right),(π,0),(\pi, 0),(3π2,−1),\left(\frac{3\pi}{2}, -1\right),(2π,0)(2\pi, 0); cosine (0,1),(0, 1),(π2,0),\left(\frac{\pi}{2}, 0\right),(π,−1),(\pi, -1),(3π2,0),\left(\frac{3\pi}{2}, 0\right),(2π,1)(2\pi, 1).
  • Sine is zero at every multiple of π\pi; cosine is zero at π2\frac{\pi}{2} plus every multiple of π\pi.
  • Each graph is concave down above the θ\theta-axis and concave up below it, so its rate of change is decreasing where the graph is above the axis.

Part 7 · Misconception

A common mistake

The wrong idea: The sine graph is made of curved arches that are steepest at the top, like a hill.

What actually happens: At a highest point the graph is flat: the point on the circle is moving sideways, not up. The graph is steepest where it crosses the θ\theta-axis, where the point on the circle moves straight up or down.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The graph of y=cos⁡θy = \cos\theta

π/2π3π/22π−1.5−1−0.50.511.5θyy = cos θ

1. On which interval is cos⁡θ\cos\theta decreasing?

  1. (0,π)(0, \pi)
  2. (π,2π)(\pi, 2\pi)
  3. (π2,3π2)\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)
  4. (π2,2π)\left(\frac{\pi}{2}, 2\pi\right)
Show the answer

The graph falls from 11 at θ=0\theta = 0 to −1-1 at θ=π\theta = \pi, then rises.

  • Correct: (0,π)(0, \pi): From the highest point to the lowest point: (0,π)(0, \pi).
  • (π,2π)(\pi, 2\pi): On (π,2π)(\pi, 2\pi) the graph rises from −1-1 back to 11.
  • (π2,3π2)\left(\frac{\pi}{2}, \frac{3\pi}{2}\right): This is where the graph is below the axis, not where it falls; it rises on (π,3π2)\left(\pi, \frac{3\pi}{2}\right).
  • (π2,2π)\left(\frac{\pi}{2}, 2\pi\right): Cosine falls on (π2,π)\left(\frac{\pi}{2}, \pi\right) but rises on (π,2π)(\pi, 2\pi), so it is not decreasing on this whole interval.

2. On which interval is the graph of y=cos⁡θy = \cos\theta concave up?

  1. (π2,3π2)\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)
  2. (0,π2)\left(0, \frac{\pi}{2}\right)
  3. (π,2π)(\pi, 2\pi)
  4. (0,π)(0, \pi)
Show the answer

The graph bends upward where it is below the axis, between the zeros π2\frac{\pi}{2} and 3π2\frac{3\pi}{2}.

  • Correct: (π2,3π2)\left(\frac{\pi}{2}, \frac{3\pi}{2}\right): Between the two zeros, around the lowest point, the graph is concave up.
  • (0,π2)\left(0, \frac{\pi}{2}\right): Near the highest point (0,1)(0, 1) the graph bends downward: concave down.
  • (π,2π)(\pi, 2\pi): This is where cosine increases. On (3π2,2π)\left(\frac{3\pi}{2}, 2\pi\right) it is concave down again.
  • (0,π)(0, \pi): This is where cosine decreases; the bend changes at π2\frac{\pi}{2}.

3. What is the average rate of change of cos⁡θ\cos\theta on [0,π2]\left[0, \frac{\pi}{2}\right]?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

cos⁡(π/2)−cos⁡0π/2−0\frac{\cos(\pi/2) - \cos 0}{\pi/2 - 0}=0−1π/2= \frac{0 - 1}{\pi/2}=−2π= -\frac{2}{\pi}, the slope of the secant line through (0,1)(0, 1) and (π2,0)\left(\frac{\pi}{2}, 0\right).

  • Answer: −2π\frac{-2}{\pi} ≈ -0.637

4. The average rates of change of cos⁡θ\cos\theta on [0,π4]\left[0, \frac{\pi}{4}\right] and on [π4,π2]\left[\frac{\pi}{4}, \frac{\pi}{2}\right] are about −0.373-0.373 and −0.900-0.900. Which describes cos⁡θ\cos\theta on (0,π2)\left(0, \frac{\pi}{2}\right)?

  1. Decreasing, and the rate of change is decreasing
  2. Decreasing, and the rate of change is increasing
  3. Increasing, and the rate of change is decreasing
  4. Decreasing, and the rate of change is constant
Show the answer

Both rates are negative, so cosine is decreasing. Over equal intervals the rate goes from −0.373-0.373 to −0.900-0.900, a smaller number, so the rate of change is decreasing: concave down.

  • Correct: Decreasing, and the rate of change is decreasing: Negative rates that get more negative: decreasing, concave down.
  • Decreasing, and the rate of change is increasing: −0.900<−0.373-0.900 < -0.373, so the rate went down, not up.
  • Increasing, and the rate of change is decreasing: The rates are negative, so the outputs fall.
  • Decreasing, and the rate of change is constant: The two rates differ, so the rate is not constant.

5. What is the range of y=sin⁡θy = \sin\theta?

  1. [−1,1][-1, 1]
  2. [0,1][0, 1]
  3. [0,2π][0, 2\pi]
  4. (−∞,∞)(-\infty, \infty)
Show the answer

Sine is the height of a point on a circle of radius 1, so it goes from −1-1 at the bottom to 11 at the top.

  • Correct: [−1,1][-1, 1]: From the bottom of the unit circle to the top.
  • [0,1][0, 1]: Below the xx-axis the height is negative, down to −1-1.
  • [0,2π][0, 2\pi]: This is one period of inputs, not the outputs.
  • (−∞,∞)(-\infty, \infty): This is the domain: any angle can go in, but the outputs stay between −1-1 and 11.

6. On which intervals is sin⁡θ\sin\theta increasing? Select all that apply.

  1. (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)
  2. (3π2,5π2)\left(\frac{3\pi}{2}, \frac{5\pi}{2}\right)
  3. (π2,3π2)\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)
  4. (0,π)(0, \pi)
Show the answer

Sine rises from its lowest point to its highest point: from −π2-\frac{\pi}{2} to π2\frac{\pi}{2}, and again one period later, from 3π2\frac{3\pi}{2} to 5π2\frac{5\pi}{2}.

  • Correct: (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right): From −1-1 at −π2-\frac{\pi}{2} up to 11 at π2\frac{\pi}{2}.
  • Correct: (3π2,5π2)\left(\frac{3\pi}{2}, \frac{5\pi}{2}\right): The same interval shifted by 2π2\pi.
  • (π2,3π2)\left(\frac{\pi}{2}, \frac{3\pi}{2}\right): Here sine falls from 11 to −1-1.
  • (0,π)(0, \pi): Sine rises on (0,π2)\left(0, \frac{\pi}{2}\right) but falls on (π2,π)\left(\frac{\pi}{2}, \pi\right).

Part 9 · Summary

Summary

The graph of y=sin⁡θy = \sin\theta is the height of the point at angle θ\theta on the unit circle; y=cos⁡θy = \cos\theta is its horizontal position. Both repeat every 2π2\pi and stay between −1-1 and 11, and cosine is sine shifted π2\frac{\pi}{2} to the left. Each graph is concave down above the axis and concave up below it, steepest at its zeros and flat at its highest and lowest points.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

Question set for this topic