Precalculus · F.TF.A.4

Graphing sine and cosine

A reflector on a turning bicycle wheel rises, falls and rises again. If its height is plotted against the angle the wheel has turned, what shape comes out, and why that shape?

Where you are 8 in Trigonometric functions and their graphs

When you reach for it

On the unit circle a sine is found one angle at a time: choose the angle, find its terminal point, read the y-coordinate (Trig functions on the unit circle). Many questions are about all angles at once. Where is the sine positive? Where is it growing? Which angles between 0 and 20 put the terminal point at the very top of the circle? Which angles have \displaystyle \sin x = \frac{1}{2}? A single point answers none of these quickly. A graph answers them all by looking, the way the graph of any function shows its whole behavior at once.

Measured in radians, an angle is a real number (Sine, cosine and tangent as functions of a real number), so sine and cosine are functions from real numbers to real numbers and can be drawn on ordinary axes. This lesson draws y = \sin x and y = \cos x themselves, with no stretching or shifting. It runs over two days. Part 1 (Examples 1 to 4) builds the sine graph from the circle. Part 2 (Examples 5 to 7) builds the cosine graph and the five key points that pin down one cycle of either.

One point about letters. On these graphs x is the input, an angle in radians, exactly as for any function graphed in the plane, and y is the output. The x on the axis is not the x-coordinate of the terminal point.

The procedure

Part 1: unrolling the circle into the sine graph

Let the angle x grow from 0 to 2\pi. Its terminal point starts at (1, 0) and travels once counterclockwise around the unit circle. At every moment the point has a height, its y-coordinate, which is \sin x. Plot each pair (x, \sin x): the angle across, the height up. That is the whole idea. The graph is a record of the height of a point going around a circle, with the circle’s turn laid out flat along the horizontal axis.

\displaystyle \frac{\pi}{3}
\displaystyle \frac{5\pi}{3}
\displaystyle \frac{\pi}{2}
\pi
\displaystyle \frac{3\pi}{2}
2\pi
-1
1
x
y
Figure 1Each terminal point is carried across at its own height. The graph is the record of the height as the angle grows from 0 to 2\pi.

The special angles give exact heights to plot:

x\sin x
00
\displaystyle \frac{\pi}{6}\displaystyle \frac{1}{2}
\displaystyle \frac{\pi}{4}\displaystyle \frac{\sqrt{2}}{2}
\displaystyle \frac{\pi}{3}\displaystyle \frac{\sqrt{3}}{2}
\displaystyle \frac{\pi}{2}1
\displaystyle \frac{2\pi}{3}\displaystyle \frac{\sqrt{3}}{2}
\displaystyle \frac{3\pi}{4}\displaystyle \frac{\sqrt{2}}{2}
\displaystyle \frac{5\pi}{6}\displaystyle \frac{1}{2}
\pi0

Here \displaystyle \frac{\sqrt{2}}{2} \approx 0.71 and \displaystyle \frac{\sqrt{3}}{2} \approx 0.87. Read quadrant by quadrant:

  • From 0 to \displaystyle \frac{\pi}{2} the point climbs from (1, 0) to (0, 1), so the graph rises from height 0 to height 1.
  • From \displaystyle \frac{\pi}{2} to \pi it comes down the left side to (-1, 0), so the graph falls back to 0. The terminal points of x and \pi - x are mirror images across the y-axis (Unit circle symmetry and even-odd identities), so they have the same height, and this piece of the graph is the mirror image of the first piece across the line \displaystyle x = \frac{\pi}{2}.
  • From \pi to 2\pi the point is below the x-axis, falling to (0, -1) at \displaystyle \frac{3\pi}{2} and rising back to (1, 0). The graph dips below the axis to -1 and returns to 0. Example 1 fills in these heights.

Why the curve is rounded. The heights do not change at a steady rate. Over the first sixth of a half turn, from 0 to \displaystyle \frac{\pi}{6}, the average rate of change of the height is

\frac{\frac{1}{2} - 0}{\frac{\pi}{6}} = \frac{3}{\pi} \approx 0.95

Over an input interval of the same width just below the top, from \displaystyle \frac{\pi}{3} to \displaystyle \frac{\pi}{2}, it is

\frac{1 - \frac{\sqrt{3}}{2}}{\frac{\pi}{6}} = \frac{3(2 - \sqrt{3})}{\pi} \approx 0.26

The same amount of turning gains almost four times as much height near the start. The reason is the direction the point is moving. At (1, 0) it moves straight up, so all of its motion goes into height. Near (0, 1) it moves almost sideways, so its height barely changes. The graph leaves the origin climbing steeply and flattens into a rounded top at \displaystyle \left(\frac{\pi}{2}, 1\right). It is not made of straight pieces, and it has no corners.

Beyond one turn. Adding 2\pi to an angle adds one full turn and lands on the same terminal point (Periodicity and the periodicity identities), so \sin(x + 2\pi) = \sin x. The piece of graph over [0, 2\pi] is therefore copied over [2\pi, 4\pi], over [-2\pi, 0], and so on forever in both directions. Negative x means clockwise turning. One repetition is a cycle, and its width 2\pi is the period (Periodic functions).

y = \sin x
-2\pi
\displaystyle -\frac{3\pi}{2}
-\pi
\displaystyle -\frac{\pi}{2}
\displaystyle \frac{\pi}{2}
\pi
\displaystyle \frac{3\pi}{2}
2\pi
-1
1
x
y
Figure 2The sine graph over two full turns. The cycle on [0, 2\pi] repeats in both directions.

Facts about y = \sin x, each read from the circle:

  • Domain all real numbers, since every angle has a terminal point.
  • Range [-1, 1], since heights on the unit circle run from -1 to 1. Each value strictly between -1 and 1 is taken twice in every cycle; 1 and -1 are taken once each.
  • Zeros at x = n\pi for every integer n, where the point is on the x-axis.
  • Highest points (x, 1) at \displaystyle x = \frac{\pi}{2} + 2\pi n; lowest points (x, -1) at \displaystyle x = \frac{3\pi}{2} + 2\pi n, for integers n.
  • Period 2\pi.
  • Symmetric about the origin. The angle -x is x reflected across the x-axis, so its height is the opposite: \sin(-x) = -\sin x. Each point (x, y) of the graph comes with (-x, -y), so sine is an odd function (Even and odd functions).

Part 2: cosine, and the five key points

The cosine graph comes from the same moving point, recording its x-coordinate instead of its height. The point starts at (1, 0), so the graph starts at 1. As the point moves up and to the left through Quadrant I its x-coordinate falls to 0 at \displaystyle \frac{\pi}{2}; it keeps falling to -1 at \pi, then climbs back through 0 at \displaystyle \frac{3\pi}{2} to 1 at 2\pi.

x\cos x
01
\displaystyle \frac{\pi}{6}\displaystyle \frac{\sqrt{3}}{2}
\displaystyle \frac{\pi}{4}\displaystyle \frac{\sqrt{2}}{2}
\displaystyle \frac{\pi}{3}\displaystyle \frac{1}{2}
\displaystyle \frac{\pi}{2}0
\displaystyle \frac{2\pi}{3}\displaystyle -\frac{1}{2}
\displaystyle \frac{3\pi}{4}\displaystyle -\frac{\sqrt{2}}{2}
\displaystyle \frac{5\pi}{6}\displaystyle -\frac{\sqrt{3}}{2}
\pi-1

The same shape, a quarter turn ahead. Turning a point a quarter turn counterclockwise about the origin sends (a, b) to (-b, a): the new point is just as far from the origin and at right angles to the old one, as (1, 0) \to (0, 1) and (0, 1) \to (-1, 0) show. So if the terminal point of x is (a, b), the terminal point of \displaystyle x + \frac{\pi}{2} is (-b, a), and its height is a:

\sin\left(x + \frac{\pi}{2}\right) = \cos x

The height of the sine graph at \displaystyle x + \frac{\pi}{2} equals the height of the cosine graph at x. Every point of the sine graph, slid \displaystyle \frac{\pi}{2} to the left, lands on the cosine graph. The cosine graph is the same wave, shifted, with the same rounded tops and steep crossings.

y = \cos x
-2\pi
\displaystyle -\frac{3\pi}{2}
-\pi
\displaystyle -\frac{\pi}{2}
\displaystyle \frac{\pi}{2}
\pi
\displaystyle \frac{3\pi}{2}
2\pi
-1
1
x
y
Figure 3The cosine graph. It is the sine graph slid \displaystyle \frac{\pi}{2} to the left, and it is symmetric about the y-axis.

Facts about y = \cos x: domain all real numbers; range [-1, 1]; zeros at \displaystyle x = \frac{\pi}{2} + \pi n; highest points (x, 1) at x = 2\pi n; lowest points (x, -1) at x = \pi + 2\pi n; period 2\pi. It is symmetric about the y-axis, because the terminal points of x and -x are mirror images across the x-axis and have the same x-coordinate: \cos(-x) = \cos x, so cosine is an even function.

The five key points. Split one cycle into four quarter turns. Starting from 0, each quarter turn carries the terminal point from one axis to the next:

(1, 0) \to (0, 1) \to (-1, 0) \to (0, -1) \to (1, 0)

At these five angles every coordinate is 0, 1 or -1, so they are exactly where each graph crosses the axis or reaches its highest or lowest value. Between two neighbors the point stays in one quadrant, where both coordinates move steadily one way, so the graph only rises or only falls there, with the rounded shape found in Part 1.

x\sin x\cos x
001
\displaystyle \frac{\pi}{2}10
\pi0-1
\displaystyle \frac{3\pi}{2}-10
2\pi01

Sine runs zero, top, zero, bottom, zero. Cosine runs top, zero, bottom, zero, top. To sketch a cycle by hand, plot the five points and join them with the rounded wave, never with straight segments. A cycle may start at any multiple of \displaystyle \frac{\pi}{2}: take that start and four more steps of \displaystyle \frac{\pi}{2}, and the five points are again the axis crossings and extremes.

y = \sin x
y = \cos x
\displaystyle \frac{\pi}{2}
\pi
\displaystyle \frac{3\pi}{2}
2\pi
-1
1
x
y
Figure 4The five key points of one cycle of each graph, at the quarter turns. Sine is solid, cosine dashed.

Worked examples

Example 1

Part 1: the second half of the sine cycle

Find \sin x exactly at \displaystyle x = \frac{7\pi}{6}, \displaystyle \frac{5\pi}{4}, \displaystyle \frac{4\pi}{3}, \displaystyle \frac{3\pi}{2}, \displaystyle \frac{5\pi}{3}, \displaystyle \frac{7\pi}{4}, \displaystyle \frac{11\pi}{6} and 2\pi, and describe the graph on [\pi, 2\pi].

Solution

Each of these angles is \pi more than an angle in the table for [0, \pi]. Adding \pi is a half turn, which sends a terminal point (a, b) to the opposite point (-a, -b), so

\sin x = -\sin(x - \pi)

For example \displaystyle \frac{7\pi}{6} - \pi = \frac{\pi}{6}, so \displaystyle \sin\frac{7\pi}{6} = -\frac{1}{2}. Doing the same for each angle:

x\sin x
\pi0
\displaystyle \frac{7\pi}{6}\displaystyle -\frac{1}{2}
\displaystyle \frac{5\pi}{4}\displaystyle -\frac{\sqrt{2}}{2}
\displaystyle \frac{4\pi}{3}\displaystyle -\frac{\sqrt{3}}{2}
\displaystyle \frac{3\pi}{2}-1
\displaystyle \frac{5\pi}{3}\displaystyle -\frac{\sqrt{3}}{2}
\displaystyle \frac{7\pi}{4}\displaystyle -\frac{\sqrt{2}}{2}
\displaystyle \frac{11\pi}{6}\displaystyle -\frac{1}{2}
2\pi0

Each entry is the opposite of the entry for the angle \pi smaller in the first table. So the graph on [\pi, 2\pi] is the arch over [0, \pi] turned upside down: a trough below the axis with its lowest point at \displaystyle \left(\frac{3\pi}{2}, -1\right). Check one value directly: \displaystyle \frac{5\pi}{3} is in Quadrant IV with reference angle \displaystyle \frac{\pi}{3}, and sine is negative there, so \displaystyle \sin\frac{5\pi}{3} = -\frac{\sqrt{3}}{2}, as listed.

Example 2

Part 1: where sine climbs

On [-2\pi, 2\pi], where is y = \sin x increasing, where is it positive, and where is it both?

Solution

The graph rises while the terminal point moves upward. Going counterclockwise, the point moves up while it is on the right half of the circle, from the bottom (0, -1) through (1, 0) to the top (0, 1), and moves down on the left half. In angles that is from \displaystyle -\frac{\pi}{2} to \displaystyle \frac{\pi}{2}, and every copy of that stretch shifted by 2\pi. Inside [-2\pi, 2\pi] the copies give

\left[-2\pi, -\tfrac{3\pi}{2}\right] \cup \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right] \cup \left[\tfrac{3\pi}{2}, 2\pi\right]

The sine is positive where the point is above the x-axis, strictly between 0 and \pi, and on the copy 2\pi to the left: (-2\pi, -\pi) \cup (0, \pi).

Both hold on the overlap. The first increasing piece meets (-2\pi, -\pi) in \displaystyle \left(-2\pi, -\frac{3\pi}{2}\right]; the middle piece meets (0, \pi) in \displaystyle \left(0, \frac{\pi}{2}\right]; the last piece is in neither positive interval. So the sine is increasing and positive on \displaystyle \left(-2\pi, -\frac{3\pi}{2}\right] \cup \left(0, \frac{\pi}{2}\right]: the climbing half of each arch. Every one of these can be read straight off the sine graph above.

Example 3

Part 1: reading off where the sine is one half

Find every x in [-2\pi, 2\pi] with \displaystyle \sin x = \frac{1}{2}.

Solution

The solutions are where the horizontal line \displaystyle y = \frac{1}{2} meets the graph. Since \displaystyle \frac{1}{2} is strictly between 0 and 1, the line crosses each arch twice and misses each trough. In [0, 2\pi] the only arch is over (0, \pi), and the table gives the two crossings: \displaystyle \frac{\pi}{6} and \displaystyle \frac{5\pi}{6}, mirror images across \displaystyle \frac{\pi}{2}. On (\pi, 2\pi) the sine is negative, so there are no others there.

The arch over (-2\pi, -\pi) is the same arch shifted 2\pi left, so its crossings are \displaystyle \frac{\pi}{6} - 2\pi = -\frac{11\pi}{6} and \displaystyle \frac{5\pi}{6} - 2\pi = -\frac{7\pi}{6}. Shifting right instead gives \displaystyle \frac{13\pi}{6}, which is past 2\pi. The solutions are

x = -\frac{11\pi}{6},\ -\frac{7\pi}{6},\ \frac{\pi}{6},\ \frac{5\pi}{6}
Example 4

Part 1: Vihaan’s reflector

Vihaan’s bicycle is on a repair stand, and seen from his side the back wheel turns counterclockwise. A reflector is clipped to a spoke 1 foot from the hub. Put the hub at the origin, with the x-axis pointing forward and the y-axis up, in feet. The reflector starts level with the hub, directly in front of it, at (1, 0).

x
1\ \text{ft}
\sin x
\text{reflector}
\text{hub}
Figure 5The reflector in Example 4. It moves on a circle of radius 1 foot, so after a turn of x radians its height above the hub is \sin x feet.

(a) How high is the reflector after the wheel turns 1 radian, and after 3.5 radians, to the nearest hundredth of a foot? (b) During the first full turn, when is the reflector above the hub, and when is it highest? (c) Vihaan turns the wheel through 3\pi radians. How many times is the reflector at the top?

Solution

The reflector moves on a circle of radius 1 about the origin, starting at (1, 0). After a turn of x radians it is at the terminal point of x, so its height above the hub is h = \sin x feet. The questions are about the graph of y = \sin x.

(a) With the calculator in radian mode, \sin 1 \approx 0.8415, so the reflector is about 0.84 feet above the hub. And \sin 3.5 \approx -0.3508, so it is about 0.35 feet below the hub. That sign fits the graph: 3.5 lies between \pi \approx 3.14 and \displaystyle \frac{3\pi}{2} \approx 4.71, where the graph is in its trough.

(b) It is above the hub when \sin x > 0, which in the first turn is 0 < x < \pi. It is highest when \sin x = 1, at \displaystyle x = \frac{\pi}{2}, a height of 1 foot.

(c) The tops are at \displaystyle x = \frac{\pi}{2} + 2\pi n. Those in [0, 3\pi] are \displaystyle \frac{\pi}{2} and \displaystyle \frac{5\pi}{2}; the next, \displaystyle \frac{9\pi}{2}, is past 3\pi. The reflector is at the top 2 times.

Example 5

Part 2: a cycle of sine that starts at minus pi

Meera wants one full cycle of y = \sin x that starts at x = -\pi. Give its five key points and describe the sketch.

Solution

A cycle has width 2\pi, so this one is [-\pi, \pi], and its quarter steps of \displaystyle \frac{\pi}{2} are \displaystyle -\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi. Each of these sends the terminal point to an axis. Turning \pi clockwise ends at (-1, 0), height 0. Turning \displaystyle \frac{\pi}{2} clockwise ends at the bottom (0, -1), height -1. Then 0, \displaystyle \frac{\pi}{2} and \pi give heights 0, 1 and 0. The key points are

(-\pi, 0),\ \left(-\tfrac{\pi}{2}, -1\right),\ (0, 0),\ \left(\tfrac{\pi}{2}, 1\right),\ (\pi, 0)

The sketch falls from (-\pi, 0) into a trough, rises through the origin, and runs over an arch back down to (\pi, 0). This cycle opens with its trough, while the cycle on [0, 2\pi] opens with its arch; they are the same graph seen through different windows. The points also show the symmetry about the origin: \displaystyle \left(-\frac{\pi}{2}, -1\right) is paired with \displaystyle \left(\frac{\pi}{2}, 1\right).

Example 6

Part 2: where sine and cosine agree

On [0, 2\pi], solve \sin x = \cos x, and find where \sin x > \cos x.

Solution

The two coordinates of a terminal point are equal exactly when the point lies on the line y = x. That line meets the unit circle at \displaystyle \left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right), the point of \displaystyle \frac{\pi}{4}, and at \displaystyle \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right), the point of \displaystyle \frac{5\pi}{4}. So \sin x = \cos x at \displaystyle x = \frac{\pi}{4} and \displaystyle x = \frac{5\pi}{4}, and these are where the two curves in the key-point figure cross.

The curves cross nowhere else in [0, 2\pi], so on each of the three pieces between crossings one curve stays above the other, and one test value decides. At \displaystyle \frac{\pi}{2}, \sin = 1 and \cos = 0, so sine is higher on \displaystyle \left(\frac{\pi}{4}, \frac{5\pi}{4}\right). At 0, \sin = 0 < 1 = \cos, and at \displaystyle \frac{3\pi}{2}, \sin = -1 < 0 = \cos, so cosine is higher on the two outer pieces. Therefore \sin x > \cos x exactly on \displaystyle \left(\frac{\pi}{4}, \frac{5\pi}{4}\right).

Example 7

Part 2: twenty radians of the reflector

Dhruv spins the same wheel through 20 radians from the start in Example 4. After the start, how many times is the reflector level with the hub, how many times is it at the top, and how many times is it farthest forward?

Solution

Level with the hub means height 0: \sin x = 0 at x = n\pi. The values with 0 < x \le 20 need \displaystyle n \le \frac{20}{\pi} \approx 6.37, so n = 1, 2, \dots, 6: 6 times.

At the top means \sin x = 1, at \displaystyle x = \frac{\pi}{2} + 2\pi n: \displaystyle \frac{\pi}{2} \approx 1.57, \displaystyle \frac{5\pi}{2} \approx 7.85, \displaystyle \frac{9\pi}{2} \approx 14.14, and the next, \displaystyle \frac{13\pi}{2} \approx 20.42, is too far. That is 3 times.

Farthest forward means the x-coordinate is 1, so \cos x = 1, at x = 2\pi n: 2\pi \approx 6.28, 4\pi \approx 12.57, 6\pi \approx 18.85, with 8\pi \approx 25.13 too far. That is 3 times. As a check, \displaystyle \frac{20}{2\pi} \approx 3.18 full turns pass each single position three times, and pass the level position, which comes twice a turn, six times.

Boundaries

  • Radians. The input is in radians. In degrees the graph has the same up and down pattern, but the axis is marked 90^\circ, 180^\circ, \dots and the period is 360^\circ, so the curve is stretched sideways by a factor \displaystyle \frac{180}{\pi}. One axis cannot mix the two.
  • Only these two graphs. Everything here is for y = \sin x and y = \cos x exactly. Steps of \displaystyle \frac{\pi}{2} between key points belong to period 2\pi; a number inside, as in \sin 2x, or in front, as in 3\cos x, changes the step or the heights, and later lessons handle each change.
  • The letter x. On the graph, x is the angle. The x-coordinate of the terminal point is \cos x, a different number.
  • Heights outside [-1, 1]. The graph never goes above 1 or below -1, so an equation such as \sin x = c or \cos x = c has no solution when |c| > 1.

Where this goes next

The two graphs drawn here are the parents of every sinusoid. Multiplying by a number stretches the wave up and down, which is Amplitude and reflection. A number multiplying x squeezes or spreads the cycle, the subject of Finding the period from the equation. Sliding the wave sideways is a phase shift, read correctly only after factoring the inside, and sliding it up or down moves the midline in Midline and vertical shift. Each of those moves the same five key points found here. Read in reverse, they let you write an equation from a wave, and they are the starting point for modeling things that repeat, from tides to the turning wheel. The same unrolling, done with the slope of the terminal side, gives the tangent graph and its asymptotes, and reading where one graph sits above a line is the heart of Trigonometric inequalities.

Exercises

Part 1: points on the sine graph

Give the exact point of the graph of y = \sin x at each x, and the quadrant of the terminal point.

  1. \displaystyle x = -\frac{3\pi}{4}
  2. \displaystyle x = \frac{13\pi}{6}
  3. \displaystyle x = -\frac{5\pi}{3}

Part 1: reading the sine graph

  1. Where on [0, 4\pi] is \sin x negative?
  2. Where on [0, 4\pi] is y = \sin x decreasing?
  3. Find every x in [-2\pi, 4\pi] with \sin x = -1.
  4. Find every x in [0, 4\pi] with \displaystyle \sin x = -\frac{\sqrt{3}}{2}.
  5. List the zeros of y = \sin x on [-3\pi, 3\pi]. How many are there?

Part 1: the reflector

Use Vihaan’s wheel from Example 4: the reflector is 1 foot from the hub, starts at (1, 0), and has height \sin x feet after a counterclockwise turn of x radians.

  1. How high or low is the reflector after the wheel turns 4 radians, to the nearest hundredth of a foot, and in which quadrant is it?
  2. For which x in [0, 4\pi] is the reflector at least \displaystyle \frac{\sqrt{2}}{2} foot above the hub?

Part 2: the cosine graph

  1. Give the exact point of the graph of y = \cos x at \displaystyle x = \frac{7\pi}{6}.
  2. Give the exact point of the graph of y = \cos x at \displaystyle x = -\frac{4\pi}{3}.
  3. Where on [0, 2\pi] is y = \cos x increasing?
  4. Find every x in [-2\pi, 2\pi] with \cos x = 0.
  5. Find every x in [-2\pi, 2\pi] with \displaystyle \cos x = \frac{\sqrt{2}}{2}.

Part 2: key points of a cycle

  1. Give the five key points of the cycle of y = \cos x that starts at x = \pi.
  2. Give the five key points of the cycle of y = \sin x that starts at \displaystyle x = \frac{\pi}{2}, and say how this cycle begins.

Part 2: sine and cosine together

  1. Where on [0, 2\pi] are \sin x and \cos x both negative?
  2. Find every x in [0, 2\pi] with \sin x = -\cos x.
  3. Meera turns Vihaan’s wheel through 15 radians from the start. After the start, how many times is the reflector level with the hub, how many times is it at the bottom, and how many times is it farthest back?

Reasoning

  1. Using the motion of the terminal point, explain why the sine graph is steepest where it crosses the x-axis and flat at its highest and lowest points.
  2. Show that \displaystyle \cos\left(x - \frac{\pi}{2}\right) = \sin x, and say what it means for the two graphs.
  3. Show that the graph of y = \cos x is symmetric about the vertical line x = \pi.
  4. Dhruv sketches y = \sin x on [0, 2\pi] by joining its five key points with straight segments. Meera says the sketch is wrong between the points. Use \displaystyle x = \frac{\pi}{6} to decide who is right.
  5. Explain why the line y = c never meets the sine graph when |c| > 1, and why it meets it exactly twice in every cycle [2\pi n, 2\pi n + 2\pi) when -1 < c < 1.
Answers
  1. \displaystyle \left(-\frac{3\pi}{4}, -\frac{\sqrt{2}}{2}\right), Quadrant III. A clockwise turn of \displaystyle \frac{3\pi}{4} lands on \displaystyle \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right).
  2. \displaystyle \left(\frac{13\pi}{6}, \frac{1}{2}\right), Quadrant I. \displaystyle \frac{13\pi}{6} - 2\pi = \frac{\pi}{6}.
  3. \displaystyle \left(-\frac{5\pi}{3}, \frac{\sqrt{3}}{2}\right), Quadrant I. \displaystyle -\frac{5\pi}{3} + 2\pi = \frac{\pi}{3}.
  4. (\pi, 2\pi) \cup (3\pi, 4\pi). The troughs of the first two cycles, where the point is below the x-axis.
  5. \displaystyle \left[\frac{\pi}{2}, \frac{3\pi}{2}\right] \cup \left[\frac{5\pi}{2}, \frac{7\pi}{2}\right]. The point moves down while on the left half of the circle, from the top to the bottom.
  6. \displaystyle x = -\frac{\pi}{2}, \frac{3\pi}{2}, \frac{7\pi}{2}. The lowest points are at \displaystyle \frac{3\pi}{2} + 2\pi n; n = -1, 0, 1 fall in the interval.
  7. \displaystyle x = \frac{4\pi}{3}, \frac{5\pi}{3}, \frac{10\pi}{3}, \frac{11\pi}{3}. In [0, 2\pi] the trough crosses the line at \displaystyle \pi + \frac{\pi}{3} and \displaystyle 2\pi - \frac{\pi}{3}; add 2\pi to each.
  8. -3\pi, -2\pi, -\pi, 0, \pi, 2\pi, 3\pi: 7 zeros, one at every multiple of \pi.
  9. About 0.76 feet below the hub, in Quadrant III. \sin 4 \approx -0.7568, and 4 lies between \pi \approx 3.14 and \displaystyle \frac{3\pi}{2} \approx 4.71.
  10. \displaystyle \left[\frac{\pi}{4}, \frac{3\pi}{4}\right] \cup \left[\frac{9\pi}{4}, \frac{11\pi}{4}\right]. On each arch the height is at least \displaystyle \frac{\sqrt{2}}{2} between the two crossings \displaystyle \frac{\pi}{4} and \displaystyle \frac{3\pi}{4}; the second arch is the first shifted 2\pi.
  11. \displaystyle \left(\frac{7\pi}{6}, -\frac{\sqrt{3}}{2}\right). Quadrant III, reference angle \displaystyle \frac{\pi}{6}, cosine negative.
  12. \displaystyle \left(-\frac{4\pi}{3}, -\frac{1}{2}\right). Cosine is even, so \displaystyle \cos\left(-\frac{4\pi}{3}\right) = \cos\frac{4\pi}{3}, in Quadrant III with reference angle \displaystyle \frac{\pi}{3}.
  13. [\pi, 2\pi]. The x-coordinate grows while the point travels the bottom half from (-1, 0) back to (1, 0).
  14. \displaystyle x = -\frac{3\pi}{2}, -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}. The point is on the y-axis at every odd multiple of \displaystyle \frac{\pi}{2}.
  15. \displaystyle x = -\frac{7\pi}{4}, -\frac{\pi}{4}, \frac{\pi}{4}, \frac{7\pi}{4}. In [0, 2\pi]: \displaystyle \frac{\pi}{4} and \displaystyle \frac{7\pi}{4}; subtract 2\pi from each for the rest.
  16. (\pi, -1), \displaystyle \left(\frac{3\pi}{2}, 0\right), (2\pi, 1), \displaystyle \left(\frac{5\pi}{2}, 0\right), (3\pi, -1). The point starts at (-1, 0) and visits the axes in turn.
  17. \displaystyle \left(\frac{\pi}{2}, 1\right), (\pi, 0), \displaystyle \left(\frac{3\pi}{2}, -1\right), (2\pi, 0), \displaystyle \left(\frac{5\pi}{2}, 1\right). It begins at a highest point, so it opens falling, with the same pattern as the cosine cycle on [0, 2\pi].
  18. \displaystyle \left(\pi, \frac{3\pi}{2}\right). Both coordinates are negative only in Quadrant III.
  19. \displaystyle x = \frac{3\pi}{4}, \frac{7\pi}{4}. The coordinates are opposites on the line y = -x, which meets the circle at \displaystyle \left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right) and \displaystyle \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right).
  20. Level 4 times, at the bottom 2 times, farthest back 2 times. Level at n\pi \le 15: n = 1 to 4, since \displaystyle \frac{15}{\pi} \approx 4.77. Bottom at \displaystyle \frac{3\pi}{2} \approx 4.71 and \displaystyle \frac{7\pi}{2} \approx 11.00; \displaystyle \frac{11\pi}{2} \approx 17.28 is too far. Farthest back, \cos x = -1, at \pi and 3\pi \approx 9.42; 5\pi \approx 15.71 is too far.
  21. Where the graph crosses the axis the point is at (1, 0) or (-1, 0) and moves straight up or down, so all of its motion changes the height. At the top or bottom it moves sideways, so a small turn barely changes the height and the graph is flat there.
  22. A quarter turn clockwise sends (a, b) to (b, -a). If the terminal point of x is (a, b), that of \displaystyle x - \frac{\pi}{2} is (b, -a), whose x-coordinate is b = \sin x. So the cosine graph at \displaystyle x - \frac{\pi}{2} has the height of the sine graph at x: the sine graph is the cosine graph slid \displaystyle \frac{\pi}{2} to the right.
  23. For any h, the angles \pi + h and \pi - h have terminal points that are mirror images across the x-axis, since \pi + h and -(\pi - h) differ by 2\pi. Mirror images across the x-axis share their x-coordinate, so \cos(\pi + h) = \cos(\pi - h): the heights at equal distances on either side of x = \pi agree.
  24. Meera is right. The segment from (0, 0) to \displaystyle \left(\frac{\pi}{2}, 1\right) is \displaystyle y = \frac{2}{\pi}x, which at \displaystyle \frac{\pi}{6} gives \displaystyle \frac{1}{3}, but \displaystyle \sin\frac{\pi}{6} = \frac{1}{2}. The true graph bulges above the segment.
  25. Heights of points on the unit circle lie in [-1, 1], so no angle has sine c when |c| > 1. For -1 < c < 1 the horizontal line y = c meets the unit circle at two points, and each point is the terminal point of exactly one angle in a turn [2\pi n, 2\pi n + 2\pi).
kg1128 · updated 2026-10-20 · CC-BY-SA-4.0