Definition 2: a point on the terminal side

Explain it like I am five

Picture a map with home at (0, 0), the origin. The address P(−15, 8) means walk 15 blocks left, then 8 up. Those numbers are coordinates, signed distances, not the axes: x = −15 is negative because it points left, y = 8 positive because it points up.

Put θ in standard position: corner at home, starting side pointing right, and a positive angle turning counterclockwise, against a clock's hands. Its terminal side, the ray (half-line) where the turn stops, runs from home through P. The distance home is r = $square root of (−15 ) squared + 8 squared$ = $square root of 225 + 64$ = $square root of 289$ = 17. A distance is never negative, and you never pick home, so r > 0.

Why sine uses y: for an angle under 90°, dropping a line from P to the x-axis makes a right triangle whose up-down leg y is opposite θ and sideways leg x is adjacent. So sin θ = $\frac{y}{r}$ and cos θ = $\frac{x}{r}$ . Larger angles keep them, signs included: sin θ = $8 over 17$ , cos θ = − $15 over 17$ , tan θ = $\frac{y}{x}$ = − $8 over 15$ .

In plain words

Think of a map with your home at (0, 0), the origin. An address P(x, y) tells you how far right or left to walk, then how far up or down. The straight route home has positive length r. To describe an angle such as 120°, put its corner at home and start its rotation pointing right. This is standard position. Its ending ray is the terminal side. Choose any address on that ray except home. Sine compares the up-down coordinate with r, and cosine compares the left-right coordinate with r. Coordinates can carry minus signs even though distance cannot. That extends the triangle ratios to every angle.

The drawing uses (−2.5, 6), half of the note point (−5, 12); they lie on the same ray and give the same ratios.

Reminder

Squaring negatives. (−5 $) squared$ = (−5) × (−5) = 25. Keep parentheses around a negative coordinate before squaring.
Signed division. A positive over a negative is negative: $8 over −6$ = − $4 over 3$ . Two negatives give a positive ratio.
Positive square root. $square root of 169$ = 13. Although $z^{2}$ = 169 has two solutions, a distance selects the positive one.
Division by zero. $0 over 4$ = 0 because 0 × 4 = 0. $4 over 0$ has no value because no number times 0 gives 4.

Why it works. For an acute angle, drop a perpendicular from P to the x-axis. The resulting right triangle has adjacent leg x, opposite leg y and hypotenuse r. The fractions therefore agree with the triangle definition. A positive enlargement multiplies x, y and r by the same factor, which cancels in every ratio, so choosing another point on the same ray changes nothing. Larger angles keep the signed coordinates. Pythagorean distance gives r = ; because the origin is excluded, r > 0. A zero x or y remains possible, so check each denominator.

Ruler = > 0. sin θ = , cos θ = , tan θ = (x ≠ 0), cot θ = (y ≠ 0), sec θ = (x ≠ 0), csc θ = (y ≠ 0).

The same idea, five ways

Say it

Say sine is y over r and cosine is x over r. Say r is the positive distance to the origin.

Write it

A point other than the origin (0, 0) on an angle's terminal side supplies its six signed coordinate ratios.

In math

P(x, y) ≠ (0, 0), r = $square root of x squared + y squared$ > 0
sin θ = $\frac{y}{r}$ , cos θ = $\frac{x}{r}$
tan θ = $\frac{y}{x}$ , sec θ = $\frac{r}{x}$ , x ≠ 0
cot θ = $\frac{x}{y}$ , csc θ = $\frac{r}{y}$ , y ≠ 0

A signed map address and the positive straight route home.

See it

The drawing uses (−2.5, 6), half of the note point (−5, 12); they lie on the same ray and give the same ratios.

The same idea, other ways

As a signed map address

Coordinates tell direction: negative x means left, negative y means down. Distance tells length, so r stays positive. Write the coordinate signs before you calculate.

Four right and five down gives a negative y but a positive distance.

As an enlarged photograph

Moving from (−5, 12) to (−10, 24) doubles the signed coordinates and the distance, from 13 to 26. Every numerator and denominator doubles, so the fractions agree.

(−5, 12), r = 13

(−10, 24), r = 26

Positive scaling leaves the final ray and the ratios unchanged.

As the old triangle on a map

For an acute angle, x is adjacent, y is opposite and r is hypotenuse. Signed coordinates let the same formulas continue when the angle no longer fits inside a right triangle.

In this positive-coordinate example, the map legs are the triangle's adjacent and opposite sides.

As checking the bottom first

Each function asks for a division. Its denominator decides whether an answer exists. For P(0, −4), r is 4: sine divides by 4, but tangent would divide by 0.

No number multiplied by zero can give a nonzero numerator.

Why the squared identity still holds with signs

For any non-origin point, r > 0 and $r^{2}$ = $x^{2}$ + $y^{2}$ . The definitions give sin θ = $\frac{y}{r}$ and cos θ = $\frac{x}{r}$ . Squaring and adding gives $\frac{y^{2}}{r^{2}}$ + $\frac{x^{2}}{r^{2}}$ = $y squared + x squared over r squared$ = $\frac{r^{2}}{r^{2}}$ = 1. Negative coordinate signs disappear when squared, so the same Pythagorean identity holds in every direction.

= + , with r > 0

θ + θ =

= 1

The signed coordinate definitions derive the identity for every angle.

Function	Domain (θ in radians)	Undefined when
sin θ, cos θ	All real θ: −∞ < θ < ∞, or (−∞, ∞)	Never: the point you chose (never the origin) has r > 0.
tan θ, sec θ	θ ≠ $π over 2$ ± nπ, equivalently θ ≠ $π over 2$ + nπ, with n any integer; in degrees θ ≠ 90° + n·180°	Terminal side on the y-axis: x = 0 and cos θ = 0.
cot θ, csc θ	θ ≠ ± nπ, equivalently θ ≠ nπ, with n any integer; in degrees θ ≠ n·180°	Terminal side on the x-axis: y = 0 and sin θ = 0.

.1r (distance to the origin)

Coordinates are directions and lengths combined. A negative coordinate sends you left or down. The straight route back to the origin is a length, so find it from squared coordinates and take the positive root.

Rule: exclude P(0, 0).
Rule: r = $square root of x squared + y squared$ > 0.
Rule: only positive scaling keeps a point on the same terminal ray.

The legs are 4 and 5 long even though the signed y-coordinate is −5.

Worked exampleFind the radius from a signed point

Find the distance from the origin to P(4, −5). In words, turn the signed address into two positive leg lengths.

The slanted length is .

$x^{2}$ = 16 and $y^{2}$ = 25.A negative times a negative is positive.
r = $square root of 16 + 25$ = $square root of 41$ .The Pythagorean theorem gives the straight distance; keep the positive root.

Answer

r =

Check Squaring gives 41, equal to the sum of the two squared leg lengths.

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

✗ Not this: r = 4 + 5 = 9.

Adding legs gives the walking route, not the straight distance.

✓ Instead: The straight route is = .

Tips and tricks

Square the coordinates before adding them.

.2Sine from y and r

Use the map address to compare up-down position with straight distance. Write the signed coordinates before taking the ratio. The denominator's condition tells you when that division is possible.

Rule: sin θ = $\frac{y}{r}$ .
When it works: always, because r (the distance from the origin to your point) is never 0 for every point except the origin.

Use the requested coordinate ratio for P(4, −5), whose positive distance is .

Worked exampleSine from y and r: one address

For P(4, −5) on a terminal side, find sin θ. In words, compare up-down position with straight distance.

Compare only the two quantities named in this function's ratio.

r = $square root of 4 squared + (−5 ) squared$ = $square root of 41$ .Distance uses the squared coordinates and the positive root.
sin θ = $\frac{y}{r}$ = $−5 over square root of 41$ .Sine compares the signed y-coordinate with positive r, so this value is negative.
$−5 over square root of 41$ = $−5 square root of 41 over 41$ = − $5 square root of 41 over 41$ .Multiply top and bottom by $square root of 41$ ; its square gives a rational denominator of 41.

Answer

sin θ = −

Check Its reciprocal csc θ = − gives a product of = 1 with the sine. The negative signs cancel, agreeing with the reciprocal identity.

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

✗ Not this: sin θ = for P(4, −5).

This drops the negative y-coordinate. A positive radius does not erase the numerator's sign.

✓ Instead: sin θ = −.

Tips and tricks

The sine follows y because its denominator is positive.

.3Cosine from x and r

Use the map address to compare left-right position with straight distance. Write the signed coordinates before taking the ratio. The denominator's condition tells you when that division is possible.

Rule: cos θ = $\frac{x}{r}$ .
When it works: always, because r (the distance from the origin to your point) is never 0 for every point except the origin.

Use the requested coordinate ratio for P(4, −5), whose positive distance is .

Worked exampleCosine from x and r: one address

For P(4, −5) on a terminal side, find cos θ. In words, compare left-right position with straight distance.

Compare only the two quantities named in this function's ratio.

r = $square root of 4 squared + (−5 ) squared$ = $square root of 41$ .Distance uses the squared coordinates and the positive root.
cos θ = $\frac{x}{r}$ = $4 over square root of 41$ .Cosine compares the signed x-coordinate with positive r; both are positive.
$4 over square root of 41$ = $4 square root of 41 over 41$ .Multiply top and bottom by $square root of 41$ , whose square is 41.

Answer

cos θ =

Check Its reciprocal sec θ = gives a product of = 1 with the cosine.

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

✗ Not this: cos θ = − for P(4, −5).

That uses y over r, which is sine. Cosine uses the x-coordinate 4.

✓ Instead: cos θ = .

Tips and tricks

The cosine follows x because its denominator is positive.

.4Tangent from y and x

Use the map address to compare up-down position with left-right position. Write the signed coordinates before taking the ratio. The denominator's condition tells you when that division is possible.

Rule: tan θ = $\frac{y}{x}$ .
It only works when x ≠ 0: x is on the bottom, and you cannot divide by 0.

Use the requested coordinate ratio for P(4, −5), whose positive distance is .

Worked exampleTangent from y and x: one address

For P(4, −5) on a terminal side, find tan θ. In words, compare up-down position with left-right position.

Compare only the two quantities named in this function's ratio.

x = 4 ≠ 0 and y = −5.Tangent's denominator is x, so this address permits the division.
tan θ = $\frac{y}{x}$ = $−5 over 4$ = − $5 over 4$ .Compare signed rise with run; a negative over a positive is negative. Tangent does not need r.

Answer

tan θ = −

Check Use the quotient identity as a second route: ÷ = −, because the equal nonzero bottoms cancel.

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

✗ Not this: tan θ = − for P(4, −5).

That reverses y and x and gives cotangent.

✓ Instead: tan θ = = −.

Tips and tricks

Opposite coordinate signs give a negative tangent.

.5Cotangent from x and y

Use the map address to compare left-right position with up-down position. Write the signed coordinates before taking the ratio. The denominator's condition tells you when that division is possible.

Rule: cot θ = $\frac{x}{y}$ .
It only works when y ≠ 0: y is on the bottom, and you cannot divide by 0.

Use the requested coordinate ratio for P(4, −5), whose positive distance is .

Worked exampleCotangent from x and y: one address

For P(4, −5) on a terminal side, find cot θ. In words, compare left-right position with up-down position.

Compare only the two quantities named in this function's ratio.

y = −5 ≠ 0 and x = 4.Cotangent's denominator is y, so this address permits the division.
cot θ = $\frac{x}{y}$ = $4 over −5$ = − $4 over 5$ .Compare signed run with rise; a positive over a negative is negative. Cotangent does not need r.

Answer

cot θ = −

Check tan θ = −, so cot θ × tan θ = (−) × (−) = 1. Both are defined because x and y are nonzero.

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

✗ Not this: cot θ = − for P(4, −5).

That reverses x and y and gives tangent.

✓ Instead: cot θ = = −.

Tips and tricks

Use x over y directly, including when tangent has no value.

.6Secant from r and x

Use the map address to compare straight distance with left-right position. Write the signed coordinates before taking the ratio. The denominator's condition tells you when that division is possible.

Rule: sec θ = $\frac{r}{x}$ .
It only works when x ≠ 0: x is on the bottom, and you cannot divide by 0.

Use the requested coordinate ratio for P(4, −5), whose positive distance is .

Worked exampleSecant from r and x: one address

For P(4, −5) on a terminal side, find sec θ. In words, compare straight distance with left-right position.

Compare only the two quantities named in this function's ratio.

r = $square root of 4 squared + (−5 ) squared$ = $square root of 41$ .Distance comes from squared coordinates and is positive.
sec θ = $square root of 41 over 4$ .Secant is distance over x. Here x = 4 ≠ 0, and the root is already on top, so no rationalization is needed.

Answer

sec θ =

Check cos θ = , so sec θ × cos θ = × = 1.

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

✗ Not this: sec θ = for P(4, −5).

That is cosine. Secant uses the distance over x, the reciprocal order.

✓ Instead: sec θ = .

Tips and tricks

The secant has the same sign as the nonzero cosine.

.7Cosecant from r and y

Use the map address to compare straight distance with up-down position. Write the signed coordinates before taking the ratio. The denominator's condition tells you when that division is possible.

Rule: csc θ = $\frac{r}{y}$ .
It only works when y ≠ 0: y is on the bottom, and you cannot divide by 0.

Use the requested coordinate ratio for P(4, −5), whose positive distance is .

Worked exampleCosecant from r and y: one address

For P(4, −5) on a terminal side, find csc θ. In words, compare straight distance with up-down position.

Compare only the two quantities named in this function's ratio.

r = $square root of 4 squared + (−5 ) squared$ = $square root of 41$ .Distance comes from squared coordinates and is positive.
csc θ = $square root of 41 over −5$ = − $square root of 41 over 5$ .Cosecant is distance over y. Here y = −5 ≠ 0, so the positive distance divided by negative y gives a negative value.

Answer

csc θ = −

Check sin θ = , so csc θ × sin θ = (−) × (−) = 1.

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

✗ Not this: csc θ = for P(4, −5).

The y denominator is −5, so its sign cannot be removed.

✓ Instead: csc θ = −.

Tips and tricks

The cosecant has the same sign as the nonzero sine.

.8Domains: all real inputs and the skipped axis families

A domain is the list of inputs a function can accept. Sine and cosine divide by positive r, so every real angle works. Other functions divide by x or y and skip the angles where that coordinate is zero. There are infinitely many skipped angles because another half turn lands on the same axis.

Rule: sine and cosine have domain (−∞, ∞), also written −∞ < θ < ∞.
Notation: ∞ means without bound. It is not a real endpoint, so interval notation uses parentheses.
Rule: tan and sec exclude θ = $π over 2$ + nπ, or 90° + n·180°.
Rule: cot and csc exclude θ = nπ, or n·180°.
Notation: n is any integer, ..., −2, −1, 0, 1, 2, ... . The notes' ± notation lists the same families.
Rule: an original zero denominator means undefined, never zero.

sin, cos: (−∞, ∞)

tan, sec: θ ≠ + nπ

cot, csc: θ ≠ nπ

n is any integer

The skipped families come from zero x or zero y.

Worked exampleCheck an angle against an excluded family

At θ = $7 π over 2$ , which functions are undefined? In words, determine the final axis and see which denominators become zero.

The terminal side is vertical, forcing x = 0.

$7 π over 2$ = $π over 2$ + 3π.This matches the tangent/secant excluded family with the integer n = 3.
The terminal side is on the y-axis, so x = 0 and y ≠ 0.Each extra π is a half turn and stays on the same axis.
Tangent $\frac{y}{x}$ and secant $\frac{r}{x}$ are undefined.Both have zero x in their denominator; the other four have nonzero r or y bottoms.

Answer

tan $7 π over 2$ : undefined
sec $7 π over 2$ : undefined
The other four functions are defined.

Check − 2π = , a downward vertical direction; it again has x = 0.

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

✗ Not this: Infinity is included as the last sine input.

Infinity is not a real number and the real line has no last value.

✓ Instead: Write (−∞, ∞); every finite real angle is accepted.

Tips and tricks

Identify the coordinate in the denominator, then identify the axis that makes it zero.

Strategy: step by step

Write down x and y with their signs.
Find r = $square root of x squared + y squared$ and simplify the root. r is always positive.
sin θ = $\frac{y}{r}$ and cos θ = $\frac{x}{r}$ : reduce, then rationalize.
tan θ = $\frac{y}{x}$ and cot θ = $\frac{x}{y}$ : r is not involved.
sec θ = $\frac{r}{x}$ and csc θ = $\frac{r}{y}$ : the flips of cos and sin.
Check each sign directly: r > 0; a ratio is negative when its numerator and denominator have opposite signs. We will organize these signs by quadrant in a later lesson.

Strategy

Find signed trig values from a terminal point

Is P the origin (0, 0)?

YesChoose a different point on the terminal side; r = 0 supplies no ratios.

NoCompute r > 0.

↓

Is x = 0?

YesTangent and secant are undefined; the other formulas still use their own denominators.

NoTangent and secant can be evaluated.

↓

Is y = 0?

YesCotangent and cosecant are undefined.

NoCotangent and cosecant can be evaluated.

↓

Is a denominator an unsimplified nonzero root?

YesReduce, then rationalize without changing the sign.

NoKeep the reduced ratio.

Check that the point is not the origin and keep both coordinate signs.
Compute the positive distance r from the squared coordinates.
Before dividing, identify which of x and y are zero.
Use each coordinate ratio; write undefined whenever its denominator is zero.
Reduce and rationalize; check signs from the coordinates and check $\sin^{2}$ θ + $\cos^{2}$ θ = 1.

Worked exampleAll six values from the point P(-6, -9)

The terminal side of an angle θ in standard position passes through the point P(-6, -9). Find the exact values of sin θ, cos θ, tan θ, cot θ, sec θ and csc θ. Simplify every square root and leave no square root in a denominator.

The point P(-6, -9) and the line through the origin and P. The terminal side of θ is only the ray from the origin through P, which is the Quadrant III half of the line. r is the distance from the origin to P.

Write the coordinates of P with their signs: x = -6, y = -9.Definition 2 works with any point (x, y) on the terminal side other than the origin. The minus signs stay attached to the numbers because they decide the signs of the answers.
r = $square root of x squared + y squared$ = $square root of (-6 ) squared + (-9 ) squared$ = $square root of 36 + 81$ = $square root of 117$ = $square root of 9 · 13$ = 3 $square root of 13$ .r is the distance from the origin to P, so it is the positive root, and squaring removes both minus signs. 9 is the largest perfect square that divides 117, so $square root of 117$ = $square root of 9$ · $square root of 13$ = 3 $square root of 13$ .
sin θ = $\frac{y}{r}$ = $-9 over 3 square root of 13$ = $-3 over square root of 13$ = $-3 over square root of 13$ · $square root of 13 over square root of 13$ = - $3 square root of 13 over 13$ . cos θ = $\frac{x}{r}$ = $-6 over 3 square root of 13$ = $-2 over square root of 13$ = $-2 over square root of 13$ · $square root of 13 over square root of 13$ = - $2 square root of 13 over 13$ .Cancel the common factor 3 first so the numbers stay small. Then multiply by $square root of 13 over square root of 13$ , which equals 1, to clear the root from the denominator, since $square root of 13$ · $square root of 13$ = 13.
tan θ = $\frac{y}{x}$ = $-9 over -6$ = $3 over 2$ and cot θ = $\frac{x}{y}$ = $-6 over -9$ = $2 over 3$ .These two use only the coordinates, so r is not involved and no root appears. Both are defined because x = -6 ≠ 0 and y = -9 ≠ 0. A negative divided by a negative is positive, and the common factor 3 cancels.
sec θ = $\frac{r}{x}$ = $3 square root of 13 over -6$ = - $square root of 13 over 2$ and csc θ = $\frac{r}{y}$ = $3 square root of 13 over -9$ = - $square root of 13 over 3$ .sec θ and csc θ are the flips of cos θ = $-2 over square root of 13$ and sin θ = $-3 over square root of 13$ . With r on top, the root is already in the numerator, so after cancelling the 3 there is nothing to rationalize.
Check each sign: r = 3 $square root of 13$ > 0, x = -6 < 0, y = -9 < 0. sin θ, cos θ, sec θ and csc θ each pair r with one negative coordinate, so all four are negative. tan θ and cot θ divide one negative coordinate by the other, so both are positive. The values found agree.A ratio is negative exactly when its numerator and denominator have opposite signs. r is always positive, so a ratio that uses r takes the sign of the coordinate in it.

Answer

sin θ = -, cos θ = -, tan θ = , cot θ = , sec θ = -, csc θ = -

Check Rebuild P from r: r · cos θ = 3 · (-) = - = -6 = x and r · sin θ = 3 · (-) = - = -9 = y. Also θ + θ = ( + ( = + = 1, as it must be because + = . Each flip pair multiplies to 1: (-) · (-) = = 1, (-) · (-) = = 1 and · = 1.

Work to write

x = -6, y = -9
r = $square root of (-6 ) squared + (-9 ) squared$ = $square root of 117$ = 3 $square root of 13$
sin θ = $-9 over 3 square root of 13$ = - $3 square root of 13 over 13$
cos θ = $-6 over 3 square root of 13$ = - $2 square root of 13 over 13$
tan θ = $-9 over -6$ = $3 over 2$
cot θ = $-6 over -9$ = $2 over 3$
sec θ = $3 square root of 13 over -6$ = - $square root of 13 over 2$
csc θ = $3 square root of 13 over -9$ = - $square root of 13 over 3$
x < 0 and y < 0 with r > 0: sin, cos, sec, csc negative; tan, cot positive

sin θ = - $3 square root of 13 over 13$ , cos θ = - $2 square root of 13 over 13$ , tan θ = $3 over 2$ , cot θ = $2 over 3$ , sec θ = - $square root of 13 over 2$ , csc θ = - $square root of 13 over 3$

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: positive coordinates and a whole-number distance

P(4, 3) is on a terminal side. Find all six functions. In words, turn its positive address into six ratios.

All coordinates and the distance are positive.

r = $square root of 16 + 9$ = 5.Square the coordinates, add and take the positive root.
sin = $3 over 5$ , cos = $4 over 5$ , tan = $3 over 4$ .Use $\frac{y}{r}$ , $\frac{x}{r}$ and $\frac{y}{x}$ .
cot = $4 over 3$ , sec = $5 over 4$ , csc = $5 over 3$ .Use $\frac{x}{y}$ , $\frac{r}{x}$ and $\frac{r}{y}$ ; none of these bottoms is zero.

Answer

sin = $3 over 5$
cos = $4 over 5$
tan = $3 over 4$
cot = $4 over 3$
sec = $5 over 4$
csc = $5 over 3$

Check The squared sine and cosine sum to + = 1.

Rung 2Rung 2: one negative coordinate and reducible ratios

P(−6, 8) is on a terminal side. Find all six functions. In words, keep the negative x while reducing the six ratios.

This half-sized point lies on the same ray as (−6, 8).

r = $square root of 36 + 64$ = 10.The negative coordinate squares to a positive number.
sin = $8 over 10$ = $4 over 5$ , cos = $−6 over 10$ = − $3 over 5$ , tan = $8 over −6$ = − $4 over 3$ .Substitute with signs, then reduce.
cot = $−6 over 8$ = − $3 over 4$ , sec = $10 over −6$ = − $5 over 3$ , csc = $10 over 8$ = $5 over 4$ .The reciprocal ratios keep their coordinate signs.

Answer

sin = $4 over 5$
cos = − $3 over 5$
tan = − $4 over 3$
cot = − $3 over 4$
sec = − $5 over 3$
csc = $5 over 4$

Check The half-sized point (−3, 4) has distance 5 and gives exactly the reduced ratios.

Rung 3Rung 3: a large point and a small equivalent point

P(9, −12) is on a terminal side. Find all six functions. In words, find the distance and reduce ratios without dropping the negative y.

Half of (9, −12) fits the drawing and keeps the same ray.

r = $square root of 81 + 144$ = $square root of 225$ = 15.Squares erase the coordinate signs; distance uses the positive root.
sin = $−12 over 15$ = − $4 over 5$ , cos = $9 over 15$ = $3 over 5$ , tan = $−12 over 9$ = − $4 over 3$ .Use $\frac{y}{r}$ , $\frac{x}{r}$ and $\frac{y}{x}$ and divide common factors.
cot = − $3 over 4$ , sec = $5 over 3$ , csc = − $5 over 4$ .Use $\frac{x}{y}$ , $\frac{r}{x}$ and $\frac{r}{y}$ with the original signs.

Answer

sin = − $4 over 5$
cos = $3 over 5$
tan = − $4 over 3$
cot = − $3 over 4$
sec = $5 over 3$
csc = − $5 over 4$

Check Divide both coordinates by 3 to obtain (3, −4), whose distance is 5; its six ratios match.

Rung 4Rung 4: two negatives and an irrational distance

P(−5, −7) is on a terminal side. Find all six functions. In words, preserve both negative coordinates and rationalize the ratios with r on the bottom.

The picture uses half of (−5, −7) to show the same signed ray.

r = $square root of 25 + 49$ = $square root of 74$ .Distance stays positive, and 74 has no perfect-square factor above 1.
sin = $−7 over square root of 74$ = − $7 square root of 74 over 74$ , cos = $−5 over square root of 74$ = − $5 square root of 74 over 74$ .Multiply top and bottom by $square root of 74$ to rationalize.
tan = $−7 over −5$ = $7 over 5$ , cot = $−5 over −7$ = $5 over 7$ .Two negatives give a positive ratio.
sec = $square root of 74 over −5$ = − $square root of 74 over 5$ , csc = $square root of 74 over −7$ = − $square root of 74 over 7$ .A positive distance over a negative coordinate gives a negative value.

Answer

sin = − $7 square root of 74 over 74$
cos = − $5 square root of 74 over 74$
tan = $7 over 5$
cot = $5 over 7$
sec = − $square root of 74 over 5$
csc = − $square root of 74 over 7$

Check + = + = 1; tangent and cotangent alone are positive because they divide two negative coordinates.

Rung 5Rung 5: a zero coordinate changes which values exist

P(0, −4) is on a terminal side. Find all six functions. In words, check every bottom before deciding whether to write a value or undefined.

The point has x = 0, so only fractions with x on the bottom fail.

r = $square root of 0 squared + (−4 ) squared$ = 4.The point is not the origin, so its distance is positive.
sin = $−4 over 4$ = −1 and cos = $0 over 4$ = 0.Their denominator r is nonzero; a zero numerator gives zero.
tan = $−4 over 0$ and sec = $4 over 0$ are undefined.These two divide by x = 0.
cot = $0 over −4$ = 0 and csc = $4 over −4$ = −1.Their denominator y is nonzero.

Answer

sin = −1
cos = 0
tan: undefined
cot = 0
sec: undefined
csc = −1

Check + = 1 + 0 = 1. Direct denominator checks distinguish the two zeros from the two undefined values.

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

✗ Not this: For a point left of the origin, r is negative.

Left is a direction encoded by x. Distance is length and is never negative.

✓ Instead: For P(−5, 12), r = 13 while x = −5.

✗ Not this: P(0, 0) works because it lies on every terminal side.

It has r = 0, making sine and cosine divide by zero; it cannot define the ratios.

✓ Instead: Pick a point other than the origin (0, 0) on the requested ray.

✗ Not this: P(−4, −3) is interchangeable with P(4, 3).

Those points lie on opposite rays. Only a positive scale factor preserves the same terminal side.

✓ Instead: Use points reached by multiplying both coordinates by the same positive number.

✗ Not this: A zero numerator and a zero denominator both mean the answer is zero.

Zero over a nonzero number is 0; dividing by zero has no value.

✓ Instead: = 0, while is undefined.

Tips and tricks

Write x, y and r separately before forming any fractions.
Check denominators before rationalizing; no rewriting repairs division by zero.
Choose a small convenient point on the same ray when the problem lets you.
Read signs directly from the coordinate fractions here; the next sign lesson collects them into a quadrant table.
Read reference tables by columns: keep the input label and its output in the same vertical column.

Trap. Giving r a sign. r is a distance, so it is positive even when x and y are negative. Every minus sign in the answers comes from x or y, never from r.

Keep in mind

r is a distance, so it is always positive: for P(−9, −12), r = $square root of 81 + 144$ = 15, never −15.
Every minus sign in an answer comes from x or y: for P(−15, 8), cos θ = − $15 over 17$ is negative because x is negative.
Any point on the same ray gives the same six values: P(−30, 16) has r = 34 and sin θ = $16 over 34$ = $8 over 17$ , the same as P(−15, 8).
A zero coordinate can still put a 0 on the bottom: for P(0, 5), tan θ = $5 over 0$ is undefined.

Memory hookAlphabetical: c comes before s as x comes before y, so cosine takes x and sine takes y: cos θ = , sin θ = .

Flash cards: say the answer out loud, then flip

What is standard position?

The angle's corner sits at the origin, its starting side points right along the positive x-axis, and positive angles turn counterclockwise.

What is the terminal side of an angle?

The ray where the turn stops. With a point P given, it is the ray from the origin through P.

What is r, and can it be negative?

The distance from the origin to P, r = . It is always positive.

P(−24, −7) is on the terminal side of θ. Find r, sin θ and cos θ.

r = $square root of 576 + 49$ = 25
sin θ = − $7 over 25$
cos θ = − $24 over 25$

Why is sin θ = and not ?

For an angle under 90°, y is the leg opposite θ and r is the hypotenuse, so SOH gives .

P(−7, −24) is on the terminal side. Is r = −25?

No. r = = 25. A distance is always positive, and the minus signs stay in x and y.