1. Turn a circle's height into the sine graph

Explain it like I am five

Picture a Ferris wheel of radius 1. A rider boards on the right, level with the hub (the center), and the wheel turns counterclockwise by an amount x. Track the rider's height: + above the hub, − below. The height is the second number of the rider's point, the y-coordinate, and on this wheel it equals sin x, said "sine of x".

Follow one lap, measuring x in radians, where π (pi, about 3.14) is half a turn. At x = 0 the height is 0. A quarter turn later, at x = $π over 2$ (about 1.57), the rider is at the top: 1. At x = π, level again: 0. At x = $3 π over 2$ , the bottom: −1. At x = 2π, back at the start: 0.

Now unroll the record. Plot (0, 0), ( $π over 2$ , 1), (π, 0), ( $3 π over 2$ , −1) and (2π, 0), and join them with a smooth curve: that is the sine graph. It stays between −1 and 1, because the rider never leaves the wheel, and it repeats every 2π, because every lap is the same trip.

In plain words

Picture a dot riding around a circular wheel with radius 1. You watch how high the dot sits above or below the wheel's center. Start at the right edge. The dot begins at height 0, rises to 1, returns to 0, drops to −1, and returns to 0 again. Now unroll your record onto a page: distance traveled around the circle goes across, and height goes up or down. That record is the sine graph. Its input x measures the turn in radians. Its output y is the dot's height. You draw each input and output together as a point (x, y).

The sine record continues through every real input and stays between heights −1 and 1.

Reminder

Input and output coordinates. At input $π over 2$ , sine gives output 1, so the graph point is ( $π over 2$ , 1). Input goes first in (x, y).
Radians and the unit circle. One full turn is 2π radians, so a quarter turn is $2 π over 4$ = $π over 2$ radians, or 90°. On radius 1, radian input equals signed rim distance; ordinary traveled arc length equals its magnitude.
Special-angle exact values. At 30°, or $π over 6$ , the circle height is $1 over 2$ ; at 45°, or $π over 4$ , it is $square root of 2 over 2$ ; at 60°, or $π over 3$ , it is $square root of 3 over 2$ .
Sign of a circle height. Above the horizontal axis sine is positive; below it sine is negative. Thus sin( $7 π over 6$ ) = − $1 over 2$ .
Common denominators. 2π = $12 π over 6$ , so $13 π over 6$ − 2π = $π over 6$ by subtracting the numerators.
Exact and rounded numbers. $square root of 2 over 2$ ≈ 0.71. The ≈ sign says the decimal is rounded; the fraction with the root stays exact.
Domain and range notation. (−∞, ∞) has no finite endpoints. [−1, 1] includes its two endpoints, the same statement as −1 ≤ y ≤ 1.

Why it works. On the unit circle, sine is the vertical coordinate of the moving point. The point moves without jumping, so its height changes without jumping. No height can exceed the radius 1 or fall below −1. A full lap travels 2π and returns to the same point, so the height record repeats every 2π. Walking the same distance backward puts the dot at the opposite height, which explains sin(−x) = −sin x and the graph's symmetry about the origin.

RuleRule: For y = sin x: Domain is all real numbers; Range is [−1, 1]; the graph is Continuous.
Period (T) = 2π; sin(x + 2kπ) = sin x for any Integer k; sin(−x) = −sin x.

The same idea, five ways

Say it

Say: sine of x is the circle point's height after a turn of x radians.

Write it

The sine graph records every radian input and the height that input produces.

In math

y = sin x
Graph points: (x, sin x)
Domain: (−∞, ∞)
Range: [−1, 1], or −1 ≤ y ≤ 1
Domain: −∞ < x < ∞
Domain: {x | x is a real number}
Range: {y | −1 ≤ y ≤ 1}

Track a wheel rider's height as the rider circles the center.

See it

The horizontal direction records the turn; the vertical direction records height.

The same idea, other ways

As a picture

A moving point on a Unit circle supplies the height. At a quarter turn its vertical coordinate is 1, so the sine graph contains ( $π over 2$ , 1).

At the top of the circle, sine is the height 1.

As a story

Imagine writing the wheel rider's height on a strip of paper that moves steadily sideways. Each full lap makes another copy of the same rise and fall.

Two laps produce two matching height records.

With small numbers

At the right, top, left, bottom, and right positions, the heights are 0, 1, 0, −1, 0. Those five numbers give the shape before you fill in extra points.

Read down each column to pair an input with its height.

Why it must repeat

A full lap returns to the same physical point. Its height cannot change merely because you reached it after another lap. Therefore sin(x + 2π) = sin x.

The longer turn finishes at the same point and has the same sine height.

x exact, in radians	sin x exact	x rounded to 2 decimal places	sin x rounded to 2 decimal places
0	0	0.00	0.00
$π over 6$	$1 over 2$	0.52	0.50
$π over 4$	$square root of 2 over 2$	0.79	0.71
$π over 3$	$square root of 3 over 2$	1.05	0.87
$π over 2$	1	1.57	1.00
$2 π over 3$	$square root of 3 over 2$	2.09	0.87
$3 π over 4$	$square root of 2 over 2$	2.36	0.71
$5 π over 6$	$1 over 2$	2.62	0.50
π	0	3.14	0.00
$7 π over 6$	− $1 over 2$	3.67	−0.50
$5 π over 4$	− $square root of 2 over 2$	3.93	−0.71
$4 π over 3$	− $square root of 3 over 2$	4.19	−0.87
$3 π over 2$	−1	4.71	−1.00
$5 π over 3$	− $square root of 3 over 2$	5.24	−0.87
$7 π over 4$	− $square root of 2 over 2$	5.50	−0.71
$11 π over 6$	− $1 over 2$	5.76	−0.50
2π	0	6.28	0.00

.1Domain and Continuous

A Domain is the list of inputs you are allowed to give a function. For sine, you can turn the wheel any distance forward or backward, so every Real number is allowed. Continuous means the drawn record has no break or jump. Think of following a road without teleporting to a different height. These are two different facts: allowing every input does not, by itself, guarantee a smooth trip between inputs.

Rule: Domain of sin x: all real numbers, written (−∞, ∞).
Rule: Continuous: sine has no breaks or jumps anywhere in its domain.
Rule: The circle point and its height move continuously; that is why sine is continuous.

Every horizontal input reaches a point on the unbroken sine curve.

Reminder

Real number and infinity. 7, −2, and $π over 6$ are real inputs. ∞ describes endless extent; it is not an input or endpoint you reach.

The same idea, five ways

Say it

Say: every real input works, and the graph has no jumps.

Write it

Sine has all real numbers as its domain and is continuous everywhere.

In math

Domain: (−∞, ∞)
For every real x, sin x exists.
−∞ < x < ∞
{x | x is a real number}

You can keep riding the circular track without falling into a gap.

See it

The horizontal shadow includes every input.

Worked exampleAn input beyond the first lap still works

This asks whether 7 is an allowed input for sin x, rather than asking for its exact output.

The graph continues beyond the first cycle to input 7 and farther.

Read the Domain statement: every Real number is allowed.You can walk any real signed distance around the unit circle.
7 is a Real number, so 7 belongs to the domain.The allowed inputs are not limited to the first drawn cycle.

Answer

Yes. sin 7 exists
the input 7 is in the domain.

Check After one lap of length 2π, the rider can continue traveling to distance 7. The circle still supplies a height.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: sine only accepts 0 ≤ x ≤ 2π because that is the first graph shown.

The drawing shows one useful portion, not a restriction on inputs.

✓ Instead: Its domain is (−∞, ∞); more turns repeat the pattern.

Tips and tricks

Tip: Domain asks about the horizontal inputs; Continuous asks whether the curve breaks.

.2Range

The Range is the collection of outputs the function can make. The rider's wheel has radius 1, so the rider can get 1 unit above the center or 1 unit below it, but no farther. Every height in between occurs somewhere on the lap. You can give sine a very large input, meaning many turns, without making its output very large. More travel makes another lap, not a taller wheel.

Rule: Range of sin x: [−1, 1].
Rule: Equivalent statements: −1 ≤ sin x ≤ 1 and |sin x| ≤ 1.
Rule: The endpoints are included because sine reaches both 1 and −1.

The curve reaches heights 1 and −1 and stays between them.

Reminder

Absolute value and closed intervals. |−1| = 1 is distance from zero. [−1, 1] includes both ends, as do the signs in −1 ≤ y ≤ 1.

The same idea, five ways

Say it

Say: sine's output is between minus one and one, including both.

Write it

Every sine output lies in the closed interval from −1 to 1.

In math

Range: [−1, 1]
−1 ≤ y ≤ 1
|sin x| ≤ 1
{y | −1 ≤ y ≤ 1}

A rider on a radius-one wheel cannot be more than one unit above its center.

See it

The vertical shadow is exactly the interval from −1 to 1.

Worked exampleTell an output from an input

This asks whether $3 over 2$ can be a sine output, rather than whether $3 over 2$ is an allowed input.

Height 1.5 is above every point of the sine graph.

Compare $3 over 2$ with the highest allowed output 1: $3 over 2$ = 1.5 > 1.Range is an output question, and the radius-one circle has top height 1.
Reject $3 over 2$ as an output, while keeping it as a permitted input.Every real input is allowed, but outputs stay between −1 and 1.

Answer

$3 over 2$ is not in the range of sine.
$3 over 2$ is in its domain.

Check The graph has no point at height 1.5, but the vertical line at input 1.5 does meet the curve.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: input x = 10 means sine's output is 10.

The input tells how far the rider turns; the output tells height. They measure different things.

✓ Instead: sin 10 exists and lies between −1 and 1.

Tips and tricks

Tip: Memory device: Domain goes across; Range rises and falls. Point to the horizontal and vertical axes as you say it.

.3Periodic function, Period (T), and Cycle

A repeating wallpaper strip has a whole design that returns again and again. A Periodic function does the same thing with its outputs. A positive shift p is a period if moving every input p units to the right gives the same output. When we say Period (T), we mean the smallest positive shift that repeats the entire design. A Cycle is one complete copy of that design. For sine, one cycle comes from one full circle lap. This least positive repeat distance is also called the Fundamental period.

Rule: A Periodic function has a positive p such that x + p is also in its domain and f(x + p) = f(x) for every x in its domain.
Rule: For sine, the smallest positive period is T = 2π.
Rule: sin(x + 2kπ) = sin x for every Integer k; negative k repeats the pattern to the left.
Rule: An Integer is a whole number with a possible minus sign, such as −2, 0, or 3.

Matching full wave patterns are separated by 2π horizontally.

Reminder

Adding fractions with π. $π over 2$ + π = $π over 2$ + $2 π over 2$ = $3 π over 2$ because the pieces must have the same denominator.

The same idea, five ways

Say it

Say: sine's whole pattern repeats every two pi units.

Write it

A sine cycle has horizontal length 2π, and the same cycle continues both ways.

In math

f(x + p) = f(x), p > 0
T = 2π for sin x
sin(x + 2kπ) = sin x, k an integer

One complete wallpaper design repeats after the same width each time.

See it

The second full wave copies the first.

Worked exampleA matching pair of zeros is not a whole period

This asks whether shifting sine by π repeats every output. Try the simple input 0 and the harder check input $π over 2$ .

Only a full-cycle shift takes every peak to another matching peak.

At 0, sin 0 = 0 and sin(0 + π) = sin π = 0.These are both circle points level with the center.
At $π over 2$ , sin( $π over 2$ ) = 1, but sin( $π over 2$ + π) = sin( $3 π over 2$ ) = −1.A half turn carries the top point to the bottom point.
Reject π as a period and use 2π.A period must repeat every output, and the peak test has failed for π.

Answer

π is not a period of sine.
The smallest positive period is 2π.

Check Shifting the top point (, 1) by 2π gives (, 1), another top point rather than a bottom point.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: the zeros at 0 and π prove the period is π.

The output at changes from 1 to −1 under that same shift.

✓ Instead: Check the whole pattern, or matching peaks; sine's period is 2π.

Tips and tricks

Tip: A period repeats the entire design. Measure from a peak to the next peak or from an upward middle crossing to the next upward middle crossing.

.4Odd function and symmetry about the origin

Imagine turning the wheel the same amount in opposite directions from the right edge. One rider goes up while the other goes down by the same amount. An Odd function has opposite outputs at opposite inputs: changing x to −x changes the output's sign too. Symmetry about the origin means a half turn of the drawn page around (0, 0) leaves the graph in the same place. The matching points are (x, y) and (−x, −y).

Rule: Sine is an Odd function: sin(−x) = −sin x.
Rule: Symmetry about the origin matches (x, y) with (−x, −y).
Rule: With respect to means measured or described in relation to; here the center of the symmetry is the Origin.

Opposite horizontal inputs have opposite sine heights.

Reminder

A minus sign in front of an output. −( $1 over 2$ ) = − $1 over 2$ , and −(−1) = 1 because reversing a negative direction gives a positive direction.

The same idea, five ways

Say it

Say: reverse the input and sine reverses the output.

Write it

Sine is odd, so its graph has symmetry about the origin.

In math

sin(−x) = −sin x
(x, y) matches (−x, −y)

Equal turns forward and backward lift one rider and lower the other by equal amounts.

See it

Rotating the sine graph half a turn around (0, 0) preserves it.

Worked exampleReverse an input and then extend beyond a full turn

This asks for sine heights at negative inputs. Find sin(− $π over 6$ ) and sin(− $13 π over 6$ ) using the positive heights already taught.

The paired points reverse both input and output signs.

sin(− $π over 6$ ) = −sin( $π over 6$ ) = − $1 over 2$ .Oddness reverses the height when the input is reversed.
sin( $13 π over 6$ ) = $1 over 2$ , so sin(− $13 π over 6$ ) = − $1 over 2$ .The earlier repeat example found the positive height, and oddness supplies the negative one.

Answer

sin(− $π over 6$ ) = − $1 over 2$ .
sin(− $13 π over 6$ ) = − $1 over 2$ .

Check − + 2π = −, so both negative inputs finish at the same circle point and have the same height.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: origin symmetry changes (x, y) to (−x, y).

That changes only the input; sine also changes the height's sign.

✓ Instead: Use (−x, −y). At x = , the partner of height 1 has height −1.

Tips and tricks

Tip: Memory device: Odd means opposite input, opposite output. Check it with sin( $π over 2$ ) = 1 and sin(− $π over 2$ ) = −1.

.5Increasing and Decreasing: follow the height, not the angle alone

Picture the dot riding around the unit circle while you record its height. From the right edge to the top, the dot climbs. From the top around the left side to the bottom, it descends. From the bottom back to the right edge, it climbs again. Increasing means the output rises whenever the input moves right within the interval you named. Decreasing means the output falls there. A larger input therefore does not always mean a larger sine output. First locate both inputs within the same rising or falling stretch, the way you would check which side of a hill you are walking on.

Rule: Increasing on an interval means u < v gives sin u < sin v for any two inputs u and v in that interval.
Rule: Decreasing on an interval means u < v gives sin u > sin v for any two inputs u and v in that interval.
Rule: Sine is increasing on [0, $π over 2$ ] and [ $3 π over 2$ , 2π ], because the circle dot climbs on those arcs. These intervals are [0°, 90°] and [270°, 360°] when the inputs use degrees.
Rule: Sine is decreasing on [ $π over 2$ , $3 π over 2$ ], because the dot descends from the top through the left side to the bottom. This interval is [90°, 270°] for degree inputs.
Rule: These intervals repeat after every whole turn. For any Integer k, sine increases on [2kπ, $π over 2$ + 2kπ] and [ $3 π over 2$ + 2kπ, 2π + 2kπ], and decreases on [ $π over 2$ + 2kπ, $3 π over 2$ + 2kπ].

Read left to right: sine climbs to the top, descends to the bottom, then climbs back to its starting height.

Reminder

Sine is circle height. At the rightmost dot sin 0 = 0; at the top sin( $π over 2$ ) = 1; at the bottom sin( $3 π over 2$ ) = −1.
Degrees and radians. The top is 90° = $π over 2$ , the bottom is 270° = $3 π over 2$ , and a full lap is 360° = 2π. Compare angles using one unit.
Closed intervals. [0, $π over 2$ ] includes the start and top. Increasing still compares distinct inputs: 0 < $π over 2$ and sin 0 < sin( $π over 2$ ).
A whole-turn repeat. sin(x + 2kπ) = sin x for Integer k. Shifting both inputs by the same whole turn keeps their output comparison.

The same idea, five ways

Say it

Say: first find the rising or falling stretch, then compare heights.

Write it

On an increasing interval, a larger input has a larger output; on a decreasing interval, a larger input has a smaller output.

In math

Increasing: if u < v in the same increasing interval, sin u < sin v.
Decreasing: if u < v in the same decreasing interval, sin u > sin v.
Sine increases on [0, $π over 2$ ] and [ $3 π over 2$ , 2π ].
Sine decreases on [ $π over 2$ , $3 π over 2$ ].
Graph words: rising left to right means increasing; falling left to right means decreasing.

Check which side of the hill holds both stops before comparing their heights.

See it

Read left to right: sine climbs to the top, descends to the bottom, then climbs back to its starting height.

Worked exampleCompare two heights before a peak, then across the peak

Compare sin 35° with sin 55°, then compare sin 75° with sin 105°. This asks which circle dot is higher in each pair. Find the order or equality without calculating decimal outputs.

The graph's inputs are radians, with the two degree angles marked at their matching positions on its rising stretch.

The 75° dot sits a short distance to the right of the vertical axis.

The 105° dot is the vertical-mirror partner of the 75° dot, so their heights agree.

Locate 35° and 55°: 0° < 35° < 55° < 90°.Both dots lie on the right-to-top arc, where the dot's height increases as its angle increases.
Write sin 35° < sin 55°.The later position on this rising arc is higher, so the sine outputs keep the order of the inputs.
Locate the other pair: 75° < 90° < 105°. Calculate 90° − 75° = 15° and 105° − 90° = 15°.The inputs lie on opposite sides of the top, equally far from its vertical radius. They are not together in one rising interval.
Reflect the 75° circle point across the vertical axis. It lands at 105° with the same vertical coordinate. Write sin 75° = sin 105°.Reflection across a vertical mirror changes sideways position but keeps height. Sine reads that unchanged height.

Answer

sin 35° < sin 55°.
sin 75° = sin 105°.

Check On the graph, 35° and 55° lie on the rising approach to the peak, so the second point is higher. The inputs 75° and 105° are each 15° from the peak at 90°. The circle picture pairs them at one common height on opposite sides of the vertical axis, confirming equality.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: 75° < 105°, so sin 75° < sin 105°.

These inputs cross the top at 90°. The circle points are reflections across the vertical axis and have equal heights.

✓ Instead: sin 75° = sin 105°. Use input order only after identifying one increasing or decreasing interval containing both inputs.

✗ Not this: Counterexample: an increasing sine output must be positive.

Increasing describes the direction of change. From the bottom at to the right edge at 2π, sine rises from −1 to 0 while remaining negative before the endpoint.

✓ Instead: A value can be negative while the graph increases. Separate output sign from rising or falling direction.

Tips and tricks

Tip: Trace the sine graph left to right before comparing two outputs. Name the interval you are using.
Tip: Memory cue: rising keeps order, falling reverses order. Check the location of both inputs before applying it.
Tip: Rebuild the rise and fall intervals from the circle's top and bottom. Do not memorize a separate interval for every repeated lap.
Tip: An angle's quadrant determines coordinate signs; it does not make sine increasing throughout every positive-height region. The top splits the upper half into a rising side and a falling side.

Strategy: step by step

1. Treat x as the Input, also called the Independent variable, and y = sin x as the Output, also called the Dependent variable. The output depends on which input you choose.
2. Use radians on the horizontal axis. On a Unit circle, a Radian input has the same numerical value as signed rim distance along the circle. Ordinary traveled Arc length equals the input's magnitude. The Central angle is the turn at the center.
3. Read each column of the value table: its top entry is x and its bottom entry is sin x. Point plotting means putting the Ordered pair (x, sin x) on the Graph in the xy-plane.
4. Locate the five Key points (0, 0), ( $π over 2$ , 1), (π, 0), ( $3 π over 2$ , −1), and (2π, 0). They record the start, top, middle, bottom, and return. A Key point is an anchor at a quarter-cycle position; five such anchors divide a cycle into four gaps. A quarter turn is one fourth of a full turn. A peak is a top height and a valley, also called a trough, is a bottom height.
5. Draw a smooth curve through the points. An Intercept is a crossing or touching of an axis; here x = 0, π, and 2π give output 0.
6. Continue the same pattern left and right. One Cycle has horizontal length 2π, because a full circle lap repeats the entire height record.

Strategy

Strategy: build a sine graph from circle heights

Is the input outside the cycle from 0 to 2π?

YesAdd or subtract whole multiples of 2π until you reach the matching input in that cycle.

NoIf its input has a listed column, read the height there. Between listed inputs the circle still supplies a height; the short table does not list every real input.

↓

Is the input negative?

YesYou can also use sin(−x) = −sin x: read the positive input and reverse the height's sign.

NoKeep the height's sign shown by the circle position.

Confirm that the horizontal inputs are radians.
Read down each input column to get its sine height.
Plot the five quarter-turn points first, then the intermediate source points.
Connect smoothly and repeat by adding or subtracting 2π from inputs.

Worked exampleOne cycle of y = sin x on [−2π, 0] from clockwise trips around the unit circle

A point P starts at (1, 0) on the unit circle. For an input x in radians, P travels a rim distance equal to the magnitude of x, counterclockwise when x > 0 and clockwise when x < 0. The output y = sin x is the height (y-coordinate) of the point where P stops. The figure marks P after a counterclockwise quarter turn of 90°, at (0, 1). (a) For x = −2π, − $3 π over 2$ , −π, − $π over 2$ and 0, find where P stops and make a value table of sin x. (b) Plot the ordered pairs (x, sin x), using π ≈ 3.14 to place the inputs, and draw one cycle of y = sin x on [−2π, 0]. Name its intercepts, its peak and its valley. (c) Continue the pattern to find sin( $21 π over 2$ ), and state the domain and range of y = sin x.

Unit circle with P at (0, 1), reached from (1, 0) by a counterclockwise quarter turn of 90°, the input x = . The inputs x = − (three quarter turns clockwise) and x = (five full laps plus a quarter turn) stop at this same point, so each has height 1.

Name the variables. The input x is the independent variable: the signed rim distance P travels. The output y = sin x is the dependent variable: the height where P stops.Where P stops, and so how high it is, depends on how far and which way it travels. So x is chosen first and y is read from it.
Turn each input into quarter turns. The unit circle's circumference is 2π × 1 = 2π, so a full lap covers 2π of rim. A quarter turn, a central angle of 90°, covers $2 π over 4$ = $π over 2$ . Divide each magnitude by $π over 2$ : $π over 2$ is 1 quarter turn, π is 2, $3 π over 2$ is 3 and 2π is 4, one full lap. The nonzero inputs are negative, so those trips go clockwise. For x = 0, P does not move.On a unit circle a radian input has the same value as the signed rim distance. The arc length traveled equals the input's magnitude, and the sign gives the direction.
Walk clockwise from (1, 0) one quarter turn at a time: P reaches (0, −1), then (−1, 0), then (0, 1), then (1, 0) again. So sin(− $π over 2$ ) = −1, sin(−π) = 0, sin(− $3 π over 2$ ) = 1 and sin(−2π) = 0. Also sin 0 = 0, since P stays at (1, 0).Clockwise from (1, 0) the path heads down first, passing the bottom, the left side and the top of the circle in that order. sin x is the y-coordinate of the stopping point.
Write the value table with the inputs in increasing order. Top row x: −2π, − $3 π over 2$ , −π, − $π over 2$ , 0. Bottom row sin x: 0, 1, 0, −1, 0. Each column is one ordered pair (x, sin x).A column's top entry is the input and its bottom entry is the output for that input, so every column gives exactly one point to plot.
Plot (−2π, 0), (− $3 π over 2$ , 1), (−π, 0), (− $π over 2$ , −1) and (0, 0). With π ≈ 3.14: 2 × 3.14 = 6.28, 1.5 × 3.14 = 4.71 and 0.5 × 3.14 = 1.57. So the inputs sit at about −6.28, −4.71, −3.14, −1.57 and 0 on the horizontal axis.The horizontal axis is measured in radians, which are real numbers, so each multiple of π has its own place on the number line.
Name the five key points: start (−2π, 0), top (− $3 π over 2$ , 1), middle (−π, 0), bottom (− $π over 2$ , −1) and return (0, 0). Neighboring inputs are $π over 2$ apart, so the five anchors split the cycle into four equal gaps.The inputs are one quarter turn apart, which is the spacing of the key points. Height 1 belongs to the highest point of the circle (the peak), and height −1 to the lowest (the valley).
Draw a smooth curve through the points. It rises from (−2π, 0) to the peak (− $3 π over 2$ , 1), falls through (−π, 0) to the valley (− $π over 2$ , −1), and rises again to (0, 0). The intercepts are at x = −2π, −π and 0, where sin x = 0, and the curve meets the y-axis at the origin.P slides along the rim without jumping, so its height changes gradually and the graph is continuous. The curve stays between −1 and 1 because no point of the unit circle is higher than 1 or lower than −1.
Compare with the cycle on [0, 2π]. Adding 2π to the inputs gives 0, $π over 2$ , π, $3 π over 2$ , 2π with the same outputs 0, 1, 0, −1, 0. So the cycle on [−2π, 0] is the cycle on [0, 2π] moved 2π to the left. Repeating it left and right covers every real x: the domain is all real numbers and the range is [−1, 1].Adding 2π adds one full counterclockwise lap, which ends at the same point, so sin(x + 2π) = sin x. One cycle has horizontal length 2π.
For x = $21 π over 2$ , take away 6 full laps, 12π = $24 π over 2$ : $21 π over 2$ − $24 π over 2$ = − $3 π over 2$ . So sin( $21 π over 2$ ) = sin(− $3 π over 2$ ) = 1, the peak value in the table.sin(x + 2kπ) = sin x for any integer k. Here k = −6, which moves the input into the cycle already drawn on [−2π, 0].

Answer

(a) For x = −2π, −, −π, − and 0, P stops at (1, 0), (0, 1), (−1, 0), (0, −1) and (1, 0). The value table is x: −2π, −, −π, −, 0 with sin x: 0, 1, 0, −1, 0. (b) The cycle is a smooth curve through (−2π, 0), (−, 1), (−π, 0), (−, −1) and (0, 0). Its intercepts are at x = −2π, −π and 0, its peak is (−, 1) and its valley is (−, −1). (c) sin() = 1. The domain is all real numbers and the range is [−1, 1].

Check Use sin(−x) = −sin x with the key points on [0, 2π]: sin(−) = −sin() = −1, sin(−π) = −sin(π) = 0, sin(−) = −sin() = −(−1) = 1 and sin(−2π) = −sin(2π) = 0. All four match the table. For (c), use k = −5 instead: − 10π = − = . That is five laps plus a counterclockwise quarter turn, which stops at the top point (0, 1) marked in the figure. So sin() = sin() = 1 again.

Work to write

Quarter turn: $2 π over 4$ = $π over 2$ of rim; x < 0 means clockwise
x = − $π over 2$ : P = (0, −1), sin(− $π over 2$ ) = −1
x = −π: P = (−1, 0), sin(−π) = 0
x = − $3 π over 2$ : P = (0, 1), sin(− $3 π over 2$ ) = 1
x = −2π: P = (1, 0), sin(−2π) = 0; x = 0: sin 0 = 0
x: −2π, − $3 π over 2$ , −π, − $π over 2$ , 0 and sin x: 0, 1, 0, −1, 0
Key points: (−2π, 0), (− $3 π over 2$ , 1), (−π, 0), (− $π over 2$ , −1), (0, 0)
Intercepts: x = −2π, −π, 0; peak (− $3 π over 2$ , 1); valley (− $π over 2$ , −1)
$21 π over 2$ − 12π = − $3 π over 2$ , so sin( $21 π over 2$ ) = sin(− $3 π over 2$ ) = 1
Domain: all real numbers; range: [−1, 1]; period 2π

(a) For x = −2π, − $3 π over 2$ , −π, − $π over 2$ and 0, P stops at (1, 0), (0, 1), (−1, 0), (0, −1) and (1, 0). The value table is x: −2π, − $3 π over 2$ , −π, − $π over 2$ , 0 with sin x: 0, 1, 0, −1, 0. (b) The cycle is a smooth curve through (−2π, 0), (− $3 π over 2$ , 1), (−π, 0), (− $π over 2$ , −1) and (0, 0). Its intercepts are at x = −2π, −π and 0, its peak is (− $3 π over 2$ , 1) and its valley is (− $π over 2$ , −1). (c) sin( $21 π over 2$ ) = 1. The domain is all real numbers and the range is [−1, 1].

Ladder: from easy to exam-hard. Press Try it first on any rung to hide its steps and use them as hints.

Rung 1Rung 1: use a reference angle above the left axis

Find sin( $3 π over 4$ ) exactly. This asks for a circle height after a 135° turn; first find the small angle to the horizontal axis.

The dot is above the horizontal axis, so sine remains positive.

$3 π over 4$ is between $π over 2$ = $2 π over 4$ and π = $4 π over 4$ , so its terminal ray is in Quadrant II.A turn between 90° and 180° finishes above and left of the center.
Reference angle = π − $3 π over 4$ = $4 π over 4$ − $3 π over 4$ = $π over 4$ . Add it back: $3 π over 4$ + $π over 4$ = π.Subtracting from the left-axis angle finds and checks the acute tilt of this ray from that axis.
The 45° triangle gives height magnitude $square root of 2 over 2$ . Keep the positive sign, so sin( $3 π over 4$ ) = $square root of 2 over 2$ .The triangle measures size, and a point above the horizontal axis has positive height.

Answer

Reference angle = $π over 4$ .
sin( $3 π over 4$ ) = $square root of 2 over 2$ .

Check Reflect the 45° circle point across the vertical axis. Its angle becomes 180° − 45° = 135°, while its height stays . The height agrees with the reference-angle calculation.

Rung 2Rung 2: keep size and sign separate below the left axis

Find sin( $5 π over 4$ ) exactly. This asks for the circle height at 225°, where the reference triangle supplies size and the region supplies sign.

Below the axis, the same reference triangle gives a negative sine height.

π = $4 π over 4$ < $5 π over 4$ < $6 π over 4$ = $3 π over 2$ , so the ray is in Quadrant III.The turn finishes between the left side and the bottom, below the horizontal axis.
Reference angle = $5 π over 4$ − π = $5 π over 4$ − $4 π over 4$ = $π over 4$ . Check π + $π over 4$ = $5 π over 4$ .Subtracting the half turn finds and verifies the small tilt below the left horizontal axis.
Use magnitude $square root of 2 over 2$ and attach a minus sign: sin( $5 π over 4$ ) = − $square root of 2 over 2$ .The 45° triangle gives an unsigned leg length, but the circle point's height is negative below the axis.

Answer

Reference angle = $π over 4$ .
sin( $5 π over 4$ ) = − $square root of 2 over 2$ .

Check A half turn carries the 45° unit-circle point to the 225° point and reverses both coordinates. Its former height becomes −, matching the answer.

Rung 3Rung 3: remove several full turns before reading the height

Find sin( $17 π over 4$ ) exactly. This asks for a height after more than two full turns; use the repeats to reach a known first-lap angle.

Two extra laps preserve the 45° circle position and sine height.

Two full turns have angle 4π = $16 π over 4$ . Subtract them: $17 π over 4$ − 4π = $π over 4$ .Removing complete laps finds a smaller angle that finishes at the same circle point.
Verify the decomposition: 4π + $π over 4$ = $16 π over 4$ + $π over 4$ = $17 π over 4$ .Adding back confirms that only whole turns were removed.
Use sin(x + 2kπ) = sin x with k = 2: sin( $17 π over 4$ ) = sin( $π over 4$ ) = $square root of 2 over 2$ .The two extra laps preserve the height, and the first rung rebuilt the exact 45° value.

Answer

Matching first-lap angle = $π over 4$ .
sin( $17 π over 4$ ) = $square root of 2 over 2$ .

Check Convert the angle using π radians = 180°: radians = = = 765°. Removing 720°, two full turns, leaves 765° − 720° = 45°. That positive first-quadrant point has equal coordinates and height , agreeing with the radian calculation.

Counterexamples: what it is not, and the tempting wrong moves

✗ Not this: Counterexample: sine is continuous because its domain is all real numbers.

A function can allow every input and still jump. The domain statement alone does not control how neighboring outputs change.

✓ Instead: Sine is continuous because the circle point's height changes without jumps as the point moves.

✗ Not this: Counterexample: sin() = 0 because sine starts at height 0.

The start is input 0. The input is a quarter turn later.

✓ Instead: sin 0 = 0, but sin() = 1 at the wheel's top.

✗ Not this: Counterexample: the rounded input 1.57 is exactly .

π is not a terminating decimal, so that table label is only an approximation.

✓ Instead: ≈ 1.57; keep in exact work.

Tips and tricks

Tip: Know cold: sine starts at the middle and rises. Memory device: start, top, middle, bottom, start gives heights 0, 1, 0, −1, 0.
Tip: Understand, then rebuild: the five points come from the wheel's right, top, left, bottom, and right positions. You can rebuild them in one lap rather than memorize seventeen pairs.
Tip: Put on the cheat sheet: Domain (−∞, ∞), Range [−1, 1], T = 2π, sin(x + 2kπ) = sin x, and sin(−x) = −sin x.
Tip: The source's rounded row is a plotting aid. Exact fractions and roots are the answers to use when exact values are requested.

Trap. Do not turn the exact radian input into a degree input of °. It means a quarter turn, or 90°. The decimal 1.57 is only a rounded label for that exact input.

Keep in mind

Sine is a number, the point's height, not the y-axis itself: a rider 0.6 below the hub gives a sine of −0.6.
The input is in radians: $π over 2$ means a quarter turn (90°), and 1.57 is only its rounded label.
Far inputs repeat, so remove whole laps of 2π first: sin( $13 π over 6$ ) = sin( $π over 6$ ) = $1 over 2$ .
√ alone means the nonnegative (principal) root, so $square root of 2$ ≈ 1.41, while the equation $z^{2}$ = 2 has two answers, $square root of 2$ and − $square root of 2$ ; a height below the hub gets its minus sign written in front, as in − $square root of 2 over 2$ .

Memory hookSine is the rider's height: start in the middle and go up first. Middle, top, middle, bottom, middle: 0, 1, 0, −1, 0.

Flash cards: say the answer out loud, then flip

What are the domain and range of y = sin x?

Domain (inputs allowed): all real numbers
Range (outputs produced): [−1, 1]

What is the period of y = sin x?

2π: the graph repeats every full lap.

List the five key points of one sine cycle.

(0, 0), (, 1), (π, 0), (, −1), (2π, 0)

Find sin().

1, because − 6π = : three full laps removed

What does sin(−x) equal?

−sin x: turning backward gives the opposite height.

Can sin x equal 1.2?

No. The height never passes the radius 1, so sine stays in [−1, 1].