Definition 3: the unit circle

Explain it like I am five

Picture a running track of radius 1 around the origin, the unit circle, with its start line at (1, 0). A number t says how far to walk: counterclockwise when t is positive, clockwise when t is negative. Where you stop is the terminal point P(t), read "P of t".

With r = 1, the rules become sin t = $y over 1$ = y and cos t = $x over 1$ = x. So the stopping point's address is (cos t, sin t), cosine first like x. A question such as "P is on the unit circle, find cos t and sin t" asks you to read that address.

Say P(t) = (− $15 over 17$ , $8 over 17$ ). It is on the track, since $x^{2}$ + $y^{2}$ = $225 over 289$ + $64 over 289$ = 1. Read cos t = − $15 over 17$ and sin t = $8 over 17$ , so tan t = $\frac{y}{x}$ = − $8 over 15$ . Any circle around the origin shrinks to this one: the map point (−15, 8), 17 from home, divided by 17 lands here on the same ray, which is what sin = $\frac{y}{r}$ and cos = $\frac{x}{r}$ do.

In plain words

Picture a round running track with radius 1 and the starting mark at (1, 0). This is the Unit circle. A real number t gives signed travel around it: walk a distance |t|, counterclockwise for t > 0 and clockwise for t < 0. The bars |t| mean the size without its sign; for t = −3 the distance is 3. At t = 0 you stay at the start. Your stopping place is the Terminal point P(t). If it is (x, y), cosine reads x and sine reads y. These trig functions with real-number inputs are also called Circular functions. A half lap changes both coordinates' signs; reversing the travel changes only the up or down coordinate.

Cosine is the leftward x-coordinate; sine is the downward y-coordinate.

Reminder

Absolute value. |t| is the nonnegative size: |− $3 π over 2$ | = $3 π over 2$ . The original minus sign still sets clockwise travel.
Circle equation. A radius-1 circle centered at the origin satisfies $x^{2}$ + $y^{2}$ = 1: $144 over 169$ + $25 over 169$ = 1.
Dividing fractions. $5 over 13$ ÷ $12 over 13$ = $5 over 12$ , since their equal denominators cancel.
Radians. π is a half turn and 2π is a full turn. In a diagram, π radians is drawn as 180°.

Why it works. The distance r from the origin is 1, so the point definition becomes sin t = = y and cos t = = x. Radian magnitude is arc length divided by radius; on this track, a distance |t| turns you through |t| radians. Direction supplies its sign. Thus signed travel t matches the angle t radians. A diameter, a straight line through the center from one circle point to another, connects opposite points, so a half turn reverses both coordinates. Reversing travel reflects across the x-axis, keeping x and reversing y. A full turn returns to the same point.

Rule + = 1 and P(t) = (cos t, sin t). tan t = and sec t = require x ≠ 0; cot t = and csc t = require y ≠ 0.
P(t ± π) = (−x, −y); P(−t) = (x, −y); P(t + 2nπ) = (x, y), where n is any integer.

The same idea, five ways

Say it

Cosine is the first coordinate; sine is the second. Signed travel includes both distance and direction.

Write it

A real number t names the point reached by walking |t| along the unit circle in the direction given by its sign.

In math

P(t) = (cos t, sin t)
$x^{2}$ + $y^{2}$ = 1
P(t ± π) = (−x, −y)
P(−t) = (x, −y)
P(t + 2nπ) = P(t), n any integer

Walk around a track, then report how far right or left and up or down you stopped.

See it

Cosine is the leftward x-coordinate; sine is the downward y-coordinate.

The same idea, other ways

As a track walk

The circle starts at (1, 0). Positive t sends you counterclockwise; negative t reverses that direction. If t = −π, walk distance π clockwise. You reach (−1, 0), the same point reached by a positive half turn.

The negative sign supplies direction; the distance walked is π.

As a triangle shrunk to one

A point (12, 5) is 13 from the origin because $12 squared$ + $5 squared$ = 169 = $13 squared$ . Divide both coordinates and the radius by 13. The new point ( $12 over 13$ , $5 over 13$ ) has radius 1. Sine becomes its y-coordinate because y divided by 1 is y.

On the unit circle the hypotenuse is 1, so the leg ratios are the coordinates.

As a coordinate address

Remember the order P(t) = (cos t, sin t). Cosine is horizontal and comes first, as x does. Sine is vertical and comes second, as y does. At P(t) = (− $3 over 5$ , − $4 over 5$ ), write cos t = − $3 over 5$ and sin t = − $4 over 5$ before any division.

Cosine is the leftward x-coordinate; sine is the downward y-coordinate.

As three picture moves

A half turn moves across the center: (x, y) becomes (−x, −y). A full turn returns home: (x, y) remains (x, y). Reversing the signed travel reflects across the x-axis: (x, y) becomes (x, −y). These work even when t already includes several laps.

Half turn: (x, y) becomes (−x, −y)

Full turn: (x, y) becomes (x, y)

Reverse travel: (x, y) becomes (x, −y)

Decide which move was requested before changing any coordinate.

	Right triangle	Point P(x, y), distance r	Unit circle point (x, y)
Works for	acute angles	any angle	any real number t
sin	$\frac{o p p}{h y p}$	$\frac{y}{r}$	y
cos	$\frac{a d j}{h y p}$	$\frac{x}{r}$	x
tan	$\frac{o p p}{a d j}$	$\frac{y}{x}$	$\frac{y}{x}$
csc, sec, cot	$\frac{h y p}{o p p}$ , $\frac{h y p}{a d j}$ , $\frac{a d j}{o p p}$	$\frac{r}{y}$ , $\frac{r}{x}$ , $\frac{x}{y}$	$1 over y$ , $1 over x$ , $\frac{x}{y}$

.1Read the coordinates of a unit point

On a radius-1 track, the horizontal coordinate already is cosine and the vertical coordinate already is sine. There is no extra division for these two because their usual denominator is 1. Check that a given point really belongs on that track, then build the remaining ratios from its coordinates.

P(t) = (x, y) gives cos t = x and sin t = y only on the unit circle.
Check $x^{2}$ + $y^{2}$ = 1.
tan = $\frac{y}{x}$ , cot = $\frac{x}{y}$ , sec = $1 over x$ and csc = $1 over y$ whenever the denominator is nonzero.

The first coordinate is cosine and the second is sine.

Worked exampleRead a new unit point

P(u) = ( $12 over 13$ , $5 over 13$ ). Find sin u, cos u and tan u. In words, verify the point and read its coordinates.

Read y for sine, x for cosine, then for tangent.

( $12 over 13$ $) squared$ + ( $5 over 13$ $) squared$ = $144 over 169$ + $25 over 169$ = 1.The sum confirms radius 1.
sin u = $5 over 13$ and cos u = $12 over 13$ .On the unit circle sine is y and cosine is x.
tan u = $5 over 13$ ÷ $12 over 13$ = $5 over 12$ .The common denominators cancel.

Answer

sin u = $5 over 13$
cos u = $12 over 13$
tan u = $5 over 12$

Check All are positive because both coordinates place the point in I.

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

✗ Not this: sin u = .

That is the horizontal coordinate.

✓ Instead: sin u = , the vertical coordinate.

Tips and tricks

Write cos = x first, then sin = y, to preserve the coordinate order.

.2A half turn: change both coordinate signs

Walk half a lap forward or backward from any stopping point. You reach the point across the center of the track. Left becomes right and up becomes down, so both coordinates change sign. Adding π or subtracting π gives that same opposite point because those two trips differ by one full lap.

P(t + π) = P(t − π) = (−x, −y).
A half turn negates sine, cosine, secant and cosecant wherever defined. A zero value remains zero.
Tangent and cotangent keep their values when defined because both coordinates reverse signs.

A half turn puts the original point directly across the center.

Worked exampleMove across the center

P(u) = ( $12 over 13$ , $5 over 13$ ). Find P(u + π), P(u − π) and tan(u + π). In words, walk half a lap and form the new ratio.

The new point lies in III, across from the original point in I.

P(u + π) = (− $12 over 13$ , − $5 over 13$ ).A half turn reverses both coordinates.
P(u − π) = (− $12 over 13$ , − $5 over 13$ ) too.Forward and backward half turns reach the same opposite point.
tan(u + π) = (− $5 over 13$ ) ÷ (− $12 over 13$ ) = $5 over 12$ .Both signs reverse and cancel in the quotient.

Answer

P(u + π) = (− $12 over 13$ , − $5 over 13$ )
P(u − π) = (− $12 over 13$ , − $5 over 13$ )
tan(u + π) = $5 over 12$

Check The new point has the same squared coordinates and radius 1; its positive tangent fits III.

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

✗ Not this: P(u + π) changes only y.

Changing only y is reflection across the x-axis, not a half turn.

✓ Instead: A half turn changes both x and y.

Tips and tricks

Half turn means both signs. Check that the point changed to the opposite quadrant.

.3Full turns: keep both coordinates

A complete lap returns you to your starting spot. Two laps, or a clockwise lap, do the same. The distance and direction of the trip can change while the stopping point remains fixed. Adding any whole-number multiple of 2π therefore keeps both coordinates and every defined trig value.

P(t + 2nπ) = P(t) for every integer n.
A full turn is 2π radians = 360°.
The values and undefined locations repeat after full turns; tangent also repeats after half turns.

A full extra turn returns to the 60° terminal side.

Worked exampleOne extra lap after a special angle

Find P( $7 π over 3$ ) and sin $7 π over 3$ . In words, remove a full turn and read the remaining 60° point.

The extra lap changes travel, but the final coordinates stay the same.

$7 π over 3$ − 2π = $7 π over 3$ − $6 π over 3$ = $π over 3$ .Subtracting one full turn preserves the stopping point.
P( $π over 3$ ) = ( $1 over 2$ , $square root of 3 over 2$ ).The 60° special-angle values give cosine first and sine second.
P( $7 π over 3$ ) is the same point, so sin $7 π over 3$ = $square root of 3 over 2$ .A full lap keeps both coordinates.

Answer

P( $7 π over 3$ ) = ( $1 over 2$ , $square root of 3 over 2$ )
sin $7 π over 3$ = $square root of 3 over 2$

Check 420° − 360° = 60°, which gives the same ray in degrees.

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

✗ Not this: Adding 2π changes both signs.

Two π half turns reverse the signs twice and return them to their original values.

✓ Instead: Adding 2π changes neither coordinate.

Tips and tricks

Remove complete laps before working with a large input.

.4Reverse the signed travel: reflect across the x-axis

Start again at the rightmost point and reverse your entire trip. A counterclockwise trip becomes clockwise, or a clockwise trip becomes counterclockwise. The two stopping points are mirror images above and below the horizontal axis. Their horizontal coordinate stays the same, and their vertical coordinate changes sign. This describes P(−t), even when t includes several laps.

P(−t) = (x, −y).
cos(−t) = cos t and sin(−t) = −sin t.
Where defined, sec keeps its value, while tan, cot and csc change signs.

Reversing the trip reflects the original upper-left point to the lower left.

Worked exampleReverse a trip to the upper left

P(v) = (− $5 over 13$ , $12 over 13$ ). Find P(−v), cos(−v) and sin(−v). In words, reverse the signed walk and read its mirror point.

Reflection changes up to down while the point stays left.

P(−v) = (− $5 over 13$ , − $12 over 13$ ).Reflection across the x-axis leaves x fixed and reverses y.
cos(−v) = − $5 over 13$ and sin(−v) = − $12 over 13$ .Cosine reads the unchanged horizontal coordinate; sine reads the reversed vertical one.

Answer

P(−v) = (− $5 over 13$ , − $12 over 13$ )
cos(−v) = − $5 over 13$
sin(−v) = − $12 over 13$

Check The original is in II and the mirror is in III; both are left of the y-axis.

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

✗ Not this: P(−v) = (, −).

That reverses both coordinates and represents a half turn.

✓ Instead: Keep x = − and reverse only y.

Tips and tricks

The minus inside P(−t) reverses travel; it is not a command to negate both coordinates.

Strategy: step by step

For travel t, begin at (1, 0); travel distance |t| in the direction given by its sign.
Read x and y from the Terminal point P(t), and check $x^{2}$ + $y^{2}$ = 1 if a point is supplied.
Write cos t = x and sin t = y.
Use tan = $\frac{y}{x}$ , cot = $\frac{x}{y}$ , sec = $1 over x$ and csc = $1 over y$ , checking each denominator.
For t ± π, flip both signs. For −t, flip only y. For a full-turn change 2nπ, keep both coordinates.
For several moves, apply them one at a time and check the resulting quadrant.

Strategy

Read a unit-circle point and move it

Are you given a point?

YesCheck + = 1; read cos = x and sin = y.

NoConvert the signed travel to a familiar angle and locate the point.

↓

Is the input changed by a full-turn multiple 2nπ?

YesKeep both coordinates.

NoCheck for a half turn or reversed travel.

↓

Does the change reverse the travel, replacing t with −t?

YesReflect across the x-axis: keep x and negate y.

NoIf the requested change is t + π or t − π, negate both coordinates. For a different change, locate the new point by its stated travel.

↓

Is a requested function's denominator zero?

YesWrite undefined.

NoForm and simplify the ratio.

For travel t, begin at (1, 0); travel distance |t| in the direction given by its sign.
Read x and y from the Terminal point P(t), and check $x^{2}$ + $y^{2}$ = 1 if a point is supplied.
Write cos t = x and sin t = y.
Use tan = $\frac{y}{x}$ , cot = $\frac{x}{y}$ , sec = $1 over x$ and csc = $1 over y$ , checking each denominator.
For t ± π, flip both signs. For −t, flip only y. For a full-turn change 2nπ, keep both coordinates.
For several moves, apply them one at a time and check the resulting quadrant.

Worked exampleClockwise travel to t = −: terminal point by symmetry and the six functions

A point starts at (1, 0) on the unit circle $x^{2}$ + $y^{2}$ = 1 and travels clockwise a distance of $19 π over 6$ , so it marks the real number t = − $19 π over 6$ . You may use P( $π over 6$ ) = ( $square root of 3 over 2$ , $1 over 2$ ). Describe where the travel ends. Then find the terminal point P(t) by applying the symmetry rules one move at a time, name its quadrant and confirm that it lies on the unit circle. Finally, find the exact values of sin t, cos t, tan t, cot t, sec t and csc t, rationalizing every denominator.

Unit circle showing the terminal point P of t = − at (−, ) in Quadrant II. This is the same point as for , an angle of 150°.

Locate the end of the travel. Split the distance: $19 π over 6$ = 2π + π + $π over 6$ . Going clockwise from (1, 0), the 2π is one full turn back to (1, 0). The π is a half turn to (−1, 0). The last $π over 6$ carries the point on from (−1, 0) toward (0, 1), so P(t) is in Quadrant II.t is negative, so the point travels clockwise a distance |t| = $19 π over 6$ . Clockwise motion passes (1, 0), (0, −1), (−1, 0) and (0, 1) in that order, so a point just past (−1, 0) is in Quadrant II.
Write t as moves applied to $π over 6$ : t = − $π over 6$ − π − 2π.− $π over 6$ − $6 π over 6$ − $12 π over 6$ = − $19 π over 6$ . Each move (change the sign, subtract π, subtract 2π) has a known effect on the coordinates of P( $π over 6$ ).
Move 1: P(− $π over 6$ ) = ( $square root of 3 over 2$ , − $1 over 2$ ), in Quadrant IV.P(−t) = (x, −y): changing the sign of t flips only the y-coordinate of P( $π over 6$ ) = ( $square root of 3 over 2$ , $1 over 2$ ).
Move 2: P(− $π over 6$ − π) = (− $square root of 3 over 2$ , $1 over 2$ ), in Quadrant II.P(t − π) = (−x, −y): subtracting π is a half turn, which flips both signs of ( $square root of 3 over 2$ , − $1 over 2$ ).
Move 3: P(− $π over 6$ − π − 2π) = (− $square root of 3 over 2$ , $1 over 2$ ), still in Quadrant II. So P(− $19 π over 6$ ) = (− $square root of 3 over 2$ , $1 over 2$ ).P(t + 2nπ) = (x, y) with n = −1: a full turn keeps both coordinates. Quadrant II agrees with the travel in the first step.
Check the point: (− $square root of 3 over 2$ $) squared$ + ( $1 over 2$ $) squared$ = $3 over 4$ + $1 over 4$ = 1.Every terminal point lies on $x^{2}$ + $y^{2}$ = 1. Sign flips do not change $x^{2}$ or $y^{2}$ , so this also confirms the supplied point P( $π over 6$ ).
Read x = − $square root of 3 over 2$ and y = $1 over 2$ . Write cos t = − $square root of 3 over 2$ and sin t = $1 over 2$ .The unit-circle definition is P(t) = (cos t, sin t). Cosine is the x-coordinate and sine is the y-coordinate.
Check the denominators: x = − $square root of 3 over 2$ ≠ 0 and y = $1 over 2$ ≠ 0.tan t and sec t divide by x, and cot t and csc t divide by y. Both are nonzero, so all four functions are defined here.
tan t = $\frac{y}{x}$ = $1 over 2$ · (− $2 over square root of 3$ ) = − $1 over square root of 3$ = − $square root of 3 over 3$ , and cot t = $\frac{x}{y}$ = (− $square root of 3 over 2$ ) · 2 = − $square root of 3$ .Dividing by a fraction is the same as multiplying by its reciprocal. Multiplying the top and bottom of − $1 over square root of 3$ by $square root of 3$ removes the root from the denominator.
sec t = $1 over x$ = − $2 over square root of 3$ = − $2 square root of 3 over 3$ , and csc t = $1 over y$ = 2.The reciprocal of − $square root of 3 over 2$ is − $2 over square root of 3$ , which is rationalized by multiplying top and bottom by $square root of 3$ . The reciprocal of $1 over 2$ is 2.

Answer

P(−) = (−, ), in Quadrant II. sin t = , cos t = −, tan t = −, cot t = −, sec t = −, csc t = 2.

Check Add two full turns (n = 2): t + 4π = − + = . This lies between and π, so P(t) = P() is in Quadrant II with x < 0 and y > 0, as found. The signs fit Quadrant II: sin t and csc t are positive, and cos t, tan t, cot t and sec t are negative. Each reciprocal pair multiplies to 1: ()(2) = 1, (−)(−) = = 1 and (−)(−) = = 1.

Work to write

$19 π over 6$ = 2π + π + $π over 6$ clockwise from (1, 0), ending in Quadrant II
t = − $π over 6$ − π − 2π
P(− $π over 6$ ) = ( $square root of 3 over 2$ , − $1 over 2$ )
P(− $π over 6$ − π) = (− $square root of 3 over 2$ , $1 over 2$ )
P(− $19 π over 6$ ) = (− $square root of 3 over 2$ , $1 over 2$ ), Quadrant II
(− $square root of 3 over 2$ $) squared$ + ( $1 over 2$ $) squared$ = $3 over 4$ + $1 over 4$ = 1
cos t = − $square root of 3 over 2$ , sin t = $1 over 2$
tan t = − $square root of 3 over 3$ , cot t = − $square root of 3$
sec t = − $2 square root of 3 over 3$ , csc t = 2

P(− $19 π over 6$ ) = (− $square root of 3 over 2$ , $1 over 2$ ), in Quadrant II. sin t = $1 over 2$ , cos t = − $square root of 3 over 2$ , tan t = − $square root of 3 over 3$ , cot t = − $square root of 3$ , sec t = − $2 square root of 3 over 3$ , csc t = 2.

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: the starting point after laps

Find sin 4π and cos 4π. In words, read the point after two full counterclockwise laps.

Two full laps return to the start.

4π = 2 × 2π, so P(4π) = (1, 0).Two full laps return to the rightmost starting point.
sin 4π = 0 and cos 4π = 1.Sine is y and cosine is x.

Answer

sin 4π = 0
cos 4π = 1

Check 4π radians is 720°, the axis value already rebuilt.

Rung 2Rung 2: six values from a new point

P(u) = ( $8 over 17$ , $15 over 17$ ). Find all six functions. In words, check the unit point, read its coordinates and form four ratios.

The higher y-coordinate gives the larger sine.

$64 over 289$ + $225 over 289$ = $289 over 289$ = 1.The squared coordinates confirm radius 1.
sin u = $15 over 17$ and cos u = $8 over 17$ .Read y and x respectively.
tan u = $15 over 8$ , cot u = $8 over 15$ , sec u = $17 over 8$ , csc u = $17 over 15$ .The shared 17s cancel in quotients; dividing 1 by a nonzero fraction flips it.

Answer

sin u = $15 over 17$
cos u = $8 over 17$
tan u = $15 over 8$
cot u = $8 over 15$
sec u = $17 over 8$
csc u = $17 over 15$

Check Every value is positive in I; sec u × cos u = × = 1.

Rung 3Rung 3: half turn and reverse travel

P(u) = ( $12 over 13$ , $5 over 13$ ). Find P(u − π), P(−u) and their tangents. In words, distinguish the opposite point from the mirror point.

Original: (, )

Half turn: (, )

Reverse: (, )

Both moves reverse y, but only a half turn reverses x.

P(u − π) = (− $12 over 13$ , − $5 over 13$ ).A backward half turn reverses both coordinates.
P(−u) = ( $12 over 13$ , − $5 over 13$ ).Reversing the original signed walk reflects only y.
tan(u − π) = $5 over 12$ ; tan(−u) = − $5 over 12$ .The half-turn ratio reverses both signs; the reflection ratio reverses only its numerator.

Answer

P(u − π) = (− $12 over 13$ , − $5 over 13$ )
P(−u) = ( $12 over 13$ , − $5 over 13$ )
tan(u − π) = $5 over 12$
tan(−u) = − $5 over 12$

Check The half-turn point is in III and its tangent is positive; the reflection is in IV and its tangent is negative.

Rung 4Rung 4: combine reflection, half turn and full turn

P(v) = (− $5 over 13$ , $12 over 13$ ). Find P(−v + 3π) and all six functions there. In words, reverse the walk, then add one half turn and one full turn.

Reflection and a half turn send the original upper-left point to the upper right; the full lap then returns here.

First P(−v) = (− $5 over 13$ , − $12 over 13$ ).Reflection across the x-axis keeps x and reverses y.
Write 3π = π + 2π.This separates a half turn, which changes signs, from a full turn, which keeps them.
P(−v + π) = ( $5 over 13$ , $12 over 13$ ), and adding 2π keeps that point.The half turn reverses both coordinates; the final full lap changes neither.
sin(−v + 3π) = $12 over 13$ , cos(−v + 3π) = $5 over 13$ .Read y and x from the final point.
tan = $12 over 5$ , cot = $5 over 12$ , sec = $13 over 5$ , csc = $13 over 12$ .Form the quotients and reciprocals using the nonzero final coordinates.

Answer

P(−v + 3π) = ( $5 over 13$ , $12 over 13$ )
sin = $12 over 13$
cos = $5 over 13$
tan = $12 over 5$
cot = $5 over 12$
sec = $13 over 5$
csc = $13 over 12$

Check The final point lies in I, so all six values are positive. Its squared coordinates still sum to 1.

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

✗ Not this: P(−t) = (−x, −y).

Both sign changes describe a half turn. Reversing travel gives an x-axis reflection instead.

✓ Instead: P(−t) = (x, −y); P(t + π) = (−x, −y).

✗ Not this: t = −π means walking a negative distance.

Distance cannot be negative. The sign records direction; |−π| = π is the distance.

✓ Instead: Walk distance π clockwise from (1, 0).

✗ Not this: Any supplied point can be read as (cos t, sin t).

That shortcut needs radius 1. For (2, 2), + = 8, so it is not on the unit circle.

✓ Instead: Use r = = 2 and divide by r, or normalize to (, ).

Tips and tricks

Know P(t) = (cos t, sin t) cold: c before s, x before y is the memory cue.
Sketch the starting point on the right before a signed walk. A negative input reverses travel.
Rebuild half-turn and reflection rules from the picture. Put the coordinate formulas on the cheat sheet.
A point may have infinitely many real-number inputs t because full laps return to it. You need its coordinates, not one unique t.
Read reference tables by columns: keep the input label and its output in the same vertical column.

Trap. Swapping x and y. The point is P(t) = (cos t, sin t): cosine comes first, as x does. Alphabetical order helps: c before s, x before y.

Keep in mind

The first number is always cos t: P(t) = (0.28, 0.96) means cos t = 0.28 and sin t = 0.96.
A point is on the unit circle only if $x^{2}$ + $y^{2}$ = 1: (0.5, 0.5) is not, since 0.25 + 0.25 = 0.5.
Half a lap flips both signs and walking backward flips only y: if P(t) = (0.28, 0.96), then P(t + π) = (−0.28, −0.96) and P(−t) = (0.28, −0.96).
tan and sec need x ≠ 0, and cot and csc need y ≠ 0: at P(t) = (0, 1), tan t = $1 over 0$ is undefined.

Memory hookThe point is (cos t, sin t): alphabetical, c before s as x before y. Positive t walks counterclockwise, negative t walks clockwise.

Flash cards: say the answer out loud, then flip

What is the unit circle?

The circle of radius 1 centered at the origin.

What is the terminal point P(t)?

Where you stop after traveling t along the unit circle from (1, 0). It is (cos t, sin t).

What equation does every point (x, y) on the unit circle satisfy?

+ = 1

P(t) = (−, −) is on the unit circle. Find cos t, sin t and tan t.

cos t = − $40 over 41$
sin t = − $9 over 41$
tan t = $9 over 40$

P(t) = (−, ). Find P(t + π) and P(−t).

P(t + π) = ( $15 over 17$ , − $8 over 17$ ), since half a lap flips both signs
P(−t) = (− $15 over 17$ , − $8 over 17$ ), since walking backward flips only y

P(t) = (0.96, 0.28). Is sin t = 0.96?

No. The first number is cos t = 0.96. sin t is the second number, 0.28.