All six functions from one value and a quadrant

Explain it like I am five

Picture a kite on a string exactly 1 unit long. Its height and its distance out from you are tied together by the Pythagorean theorem (in a right triangle, le $g^{2}$ + le $g^{2}$ = long sid $e^{2}$ ): heigh $t^{2}$ + distanc $e^{2}$ = 1. On the unit circle that reads $\sin^{2}$ θ + $\cos^{2}$ θ = 1. Here θ (theta) names the angle, and $\sin^{2}$ θ means (sin θ $) squared$ , sine squared.

Example: sin θ = $24 over 25$ and θ is in Quadrant II, the upper left. Find cos θ.
Step 1: $\cos^{2}$ θ = 1 − ( $24 over 25$ $) squared$ = $625 over 625$ − $576 over 625$ = $49 over 625$ .
Step 2, the size: $square root of 49 over 625$ = $7 over 25$ . The symbol √ alone always means the positive root, the principal square root, even though − $7 over 25$ also squares to $49 over 625$ .
Step 3, the sign: Quadrant II is left of center, so x is negative and cos θ = − $7 over 25$ .

Then divide and flip: tan θ = $24 over 25$ ÷ (− $7 over 25$ ) = − $24 over 7$ , cot θ = − $7 over 24$ , sec θ = − $25 over 7$ , csc θ = $25 over 24$ .

In plain words

Picture a ladder of length 1 reaching from the center of a circular track to its rim. Its height and horizontal reach must fit together to make that length. Knowing one tells you the size of the other. You still need to know whether the endpoint is above or below the center, and right or left. The quadrant supplies those directions and therefore the signs. Sine is the signed height, and cosine is the signed horizontal reach. Once you know those two, division and reciprocals produce the other four outputs. The same relationships also let you simplify expressions whose unfamiliar angles would otherwise block you.

Scaled up by 5, the point (3, −4) has r = 5, so cos θ = and sin θ = − in Quadrant IV.

Reminder

Squaring a negative fraction. (− $1 over 3$ $) squared$ = $1 over 9$ , because multiplying two negative numbers gives a positive number.
Common denominators. 1 − $9 over 25$ = $25 over 25$ − $9 over 25$ = $16 over 25$ .
Square roots of fractions. $square root of 16 over 25$ = $square root of 16 over square root of 25$ = $4 over 5$ . This gives a size, before the quadrant chooses a sign.
Dividing fractions. (− $4 over 5$ ) ÷ $3 over 5$ = (− $4 over 5$ ) × $5 over 3$ = − $4 over 3$ .
Rationalizing. $1 over square root of 5$ = $square root of 5 over 5$ , because multiplying top and bottom by $square root of 5$ is multiplying by 1.

Scaled up by 5, the point (3, −4) has r = 5, so cos θ = and sin θ = − in Quadrant IV.

Read the output entry in the column under its input or category.

Why it works. The radius and the two coordinate legs form a right triangle. The Pythagorean theorem gives + = 1, so θ + θ = 1. Subtracting the known square gives the missing coordinate's square. Taking a square root gives its size, and the quadrant fixes its sign. Tangent, cotangent, secant and cosecant are defined from these coordinates by quotients and reciprocals. Dividing the Pythagorean identity by a nonzero cosine square or sine square gives the other two identities, with their denominator restrictions.

Ruleθ + θ = 1. Thus cos θ = ± and sin θ = ±, with the sign chosen by the quadrant. Then tan θ = , cot θ = , sec θ = , csc θ = , wherever the denominator is nonzero.

The same idea, five ways

Say it

The square of sine plus the square of cosine is one. Knowing one coordinate and the quadrant lets you recover the other.

Write it

A point one unit from the origin has coordinate squares that add to one; its location chooses the signs.

In math

$\sin^{2}$ θ + $\cos^{2}$ θ = 1.
cos θ = ± $square root of 1 − sin squared θ$ ; sin θ = ± $square root of 1 − cos squared θ$ .
$\tan^{2}$ θ + 1 = $\sec^{2}$ θ when cos θ ≠ 0.
$\cot^{2}$ θ + 1 = $\csc^{2}$ θ when sin θ ≠ 0.

A ladder's height and horizontal reach must fit its fixed length; its location decides the coordinate signs.

See it

The coordinate squares are and , which add to 1.

The same idea, other ways

As a ladder

A ladder of length 1 has horizontal reach cos θ and height sin θ, interpreted as signed coordinates. The physical side lengths are their magnitudes. The Pythagorean theorem connects their squares. The quadrant chooses the coordinate directions.

The triangle's lengths are magnitudes, while the circle coordinates carry signs.

With small numbers

Suppose the horizontal coordinate is $3 over 5$ . Its square is $9 over 25$ . The remaining height square is 1 − $9 over 25$ = $16 over 25$ , so the height has size $4 over 5$ . The point may be above or below the center; the quadrant settles that final choice.

The scaled triangle has legs 3 and 4 and radius 5.

As two possible locations

One coordinate usually leaves two points on the circle. A horizontal position $3 over 5$ gives heights $4 over 5$ and − $4 over 5$ . The quadrant is the instruction that selects one of those points.

Quadrant IV selects the lower of the two possible points.

As a rebuilding kit

Keep sine and cosine as the two basic coordinates. Build tangent by dividing height by horizontal position, cotangent by reversing that division, and secant and cosecant by taking reciprocals. This rebuilds the six values without six separate guesses.

tan θ = sin θ ÷ cos θ

cot θ = cos θ ÷ sin θ

sec θ = 1 ÷ cos θ

csc θ = 1 ÷ sin θ

Four outputs are rebuilt from the two coordinates.

Family	Identities and restrictions
Reciprocal	csc θ = $1 over sin θ$ when sin θ ≠ 0; sec θ = $1 over cos θ$ when cos θ ≠ 0; cot θ = $1 over tan θ$ when sin θ ≠ 0 and cos θ ≠ 0.
Quotient	tan θ = $\frac{\sin θ}{\cos θ}$ when cos θ ≠ 0; cot θ = $\frac{\cos θ}{\sin θ}$ when sin θ ≠ 0.
Pythagorean	$\sin^{2}$ θ + $\cos^{2}$ θ = 1 for every real θ; $\tan^{2}$ θ + 1 = $\sec^{2}$ θ when cos θ ≠ 0; $\cot^{2}$ θ + 1 = $\csc^{2}$ θ when sin θ ≠ 0.

.1Reciprocal identities and their allowed inputs

A reciprocal is one divided by a nonzero number, like changing halves per whole into wholes per half. The sine and cosecant pair and the cosine and secant pair are reciprocals wherever the relevant coordinate is not zero. Tangent and cotangent are reciprocals only where both are defined and nonzero.

csc θ = $1 over sin θ$ , with sin θ ≠ 0.
sec θ = $1 over cos θ$ , with cos θ ≠ 0.
cot θ = $1 over tan θ$ , with sin θ ≠ 0 and cos θ ≠ 0.
sin θ · csc θ = 1 and cos θ · sec θ = 1 on their respective common domains.
tan θ · cot θ = 1 when sin θ ≠ 0 and cos θ ≠ 0.

sin θ · csc θ = 1, sin θ ≠ 0

cos θ · sec θ = 1, cos θ ≠ 0

tan θ · cot θ = 1, sin θ cos θ ≠ 0

A reciprocal pair multiplies to 1 only when both expressions exist.

Reminder

A reciprocal keeps the sign. The reciprocal of − $1 over 5$ is −5, and their product is 1.

Worked exampleA reciprocal identity with a negative value

Suppose cos θ = − $1 over 5$ . The question gives a nonzero cosine and asks for secant and the product of this reciprocal pair.

cos θ =

sec θ = −5

()(−5) = 1

The reciprocal has the same sign and multiplies with cosine to give 1.

sec θ = 1 ÷ (− $1 over 5$ ) = −5.Secant is the reciprocal of cosine, and the given cosine is nonzero.
(− $1 over 5$ ) × (−5) = 1.The two negative signs cancel, and the factors 5 cancel.

Answer

sec θ = −5
cos θ · sec θ = 1

Check The secant and cosine signs match, as reciprocals must.

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

✗ Not this: cot θ = can be used at every input where cotangent exists.

At θ = , cotangent is 0 but tangent is undefined, so 1 divided by tangent cannot be used there.

✓ Instead: Use cot θ = on its full domain. The reciprocal formula requires both sine and cosine to be nonzero.

Tips and tricks

Before writing 1 divided by a function, check that the function exists and is not 0.

.2Quotient identities rebuild tangent and cotangent

A quotient is a division result. Picture a ramp: tangent compares the signed rise with the signed horizontal run. Cotangent reverses that comparison. These are definitions, so using them replaces a function name with the coordinates that give its value.

tan θ = $\frac{\sin θ}{\cos θ}$ when cos θ ≠ 0.
cot θ = $\frac{\cos θ}{\sin θ}$ when sin θ ≠ 0.
A zero top with a nonzero bottom gives 0. A zero bottom gives an undefined expression.

Tangent compares rise with run; the quadrant supplies each coordinate's sign.

Reminder

Different signs in a quotient. $3 over −4$ = − $3 over 4$ .

Worked exampleAll six functions from a secant in Quadrant IV

Suppose sec θ = $7 over 4$ and θ lies in Quadrant IV. Find sin θ, cos θ, tan θ, cot θ and csc θ exactly. Rationalize any denominators.

Name the given value and the targets. Given: sec θ = $7 over 4$ , θ in Quadrant IV. Find: sin θ, cos θ, tan θ, cot θ, csc θ.One trig value and a quadrant fix all six functions. Listing the targets keeps each of the five in view.
Check the given value against Quadrant IV. In Quadrant IV, cos θ > 0 and sin θ < 0, so sec θ > 0. Also |sec θ| = $7 over 4$ > 1.Inside a quadrant, 0 < |cos θ| < 1, so |sec θ| must be greater than 1. Its sign must match cosine's sign, which is positive in Quadrant IV. Both conditions hold, so such a θ exists.
Take the reciprocal of secant: cos θ = $1 over sec θ$ = $\frac{1}{\frac{7}{4}}$ = $4 over 7$ .Secant is the reciprocal of cosine. Dividing 1 by $7 over 4$ means multiplying by $4 over 7$ .
Substitute into the identity: $\sin^{2}$ θ + ( $4 over 7$ $) squared$ = 1, so $\sin^{2}$ θ = 1 − $16 over 49$ = $49 over 49$ − $16 over 49$ = $33 over 49$ . $\sin^{2}$ θ + $\cos^{2}$ θ = 1. Solving for $\sin^{2}$ θ gives the size of sine without choosing its sign yet.
Take the nonnegative root: $square root of 33 over 49$ = $square root of 33 over 7$ . In Quadrant IV sine is negative, so sin θ = − $square root of 33 over 7$ .The square root gives only the magnitude. The quadrant decides the sign, and below the x-axis the y-coordinate is negative.
tan θ = $\frac{\sin θ}{\cos θ}$ = $\frac{- \frac{\sqrt{33}}{7}}{\frac{4}{7}}$ = − $square root of 33 over 7$ · $7 over 4$ = − $square root of 33 over 4$ .Tangent is sine divided by cosine. Dividing by $4 over 7$ means multiplying by $7 over 4$ , and the sevens cancel.
cot θ = $1 over tan θ$ = − $4 over square root of 33$ = − $4 over square root of 33$ · $square root of 33 over square root of 33$ = − $4 square root of 33 over 33$ .Cotangent is the reciprocal of tangent. Multiplying the top and bottom by $square root of 33$ rationalizes the denominator.
csc θ = $1 over sin θ$ = − $7 over square root of 33$ = − $7 over square root of 33$ · $square root of 33 over square root of 33$ = − $7 square root of 33 over 33$ .Cosecant is the reciprocal of sine. Rationalize by multiplying the top and bottom by $square root of 33$ .

Answer

sin θ = −, cos θ = , tan θ = −, cot θ = −, csc θ = − (and sec θ = as given).

Check Identity: θ + θ = + = = 1. Signs in Quadrant IV: sin −, cos +, tan −, cot −, sec +, csc −, all as found. Given value: = = = sec θ.

Work to write

sec θ = $7 over 4$ > 1 and is positive, so it is valid in Quadrant IV.
cos θ = $1 over sec θ$ = $4 over 7$
$\sin^{2}$ θ = 1 − $16 over 49$ = $33 over 49$
sin θ = − $square root of 33 over 7$ (negative in Quadrant IV)
tan θ = $\frac{- \frac{\sqrt{33}}{7}}{\frac{4}{7}}$ = − $square root of 33 over 4$
cot θ = − $4 over square root of 33$ = − $4 square root of 33 over 33$
csc θ = − $7 over square root of 33$ = − $7 square root of 33 over 33$
Check: $33 over 49$ + $16 over 49$ = 1

sin θ = − $square root of 33 over 7$ , cos θ = $4 over 7$ , tan θ = − $square root of 33 over 4$ , cot θ = − $4 square root of 33 over 33$ , csc θ = − $7 square root of 33 over 33$ (and sec θ = $7 over 4$ as given).

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

✗ Not this: tan θ = .

That division reverses the definition and gives cotangent instead.

✓ Instead: tan θ = ; cot θ = .

Tips and tricks

Tangent is height divided by horizontal position. Write y ÷ x before converting to trig names.

.3The three Pythagorean identities and why they hold

The radius and coordinate legs form a triangle. Its two leg squares add to the radius square. Because the radius is 1, the total is 1. The other two identities come from dividing every term by one nonzero coordinate square, like changing all measurements in an equation to the same unit.

$\sin^{2}$ θ means (sin θ $) squared$ . It squares the output, rather than the angle.
$\sin^{2}$ θ + $\cos^{2}$ θ = 1 for every real θ.
$\tan^{2}$ θ + 1 = $\sec^{2}$ θ when cos θ ≠ 0.
$\cot^{2}$ θ + 1 = $\csc^{2}$ θ when sin θ ≠ 0.
The identities $\sec^{2}$ θ − $\tan^{2}$ θ = 1 and $\csc^{2}$ θ − $\cot^{2}$ θ = 1 are rearrangements of the same facts.

The coordinate squares add to the radius square: + = 1.

Reminder

Dividing an equation. Divide every term on both sides by the same nonzero number. Dividing 4 + 5 = 9 by 3 gives $4 over 3$ + $5 over 3$ = 3.

Worked exampleDerive all three identities

Explain why the three Pythagorean identities hold, and state when each formula is defined.

θ + θ = 1

Divide by θ: θ + 1 = θ

The last two identities come from nonzero division of the first.

For a unit-circle point away from the axes, the coordinate legs and radius form a right triangle, giving $x^{2}$ + $y^{2}$ = $1 squared$ = 1. On an axis, the coordinates are 0 and ±1, so the same equality holds directly.The coordinate legs and radius form a right triangle, with radius 1.
Substitute x = cos θ and y = sin θ: $\sin^{2}$ θ + $\cos^{2}$ θ = 1.Sine and cosine are the coordinate definitions.
When cos θ ≠ 0, divide every term by $\cos^{2}$ θ: (sin θ ÷ cos θ $) squared$ + 1 = (1 ÷ cos θ $) squared$ .Dividing both sides by the same nonzero quantity preserves equality and turns the cosine square term into 1.
Use the quotient and reciprocal definitions: $\tan^{2}$ θ + 1 = $\sec^{2}$ θ.Sine divided by cosine is tangent; 1 divided by cosine is secant.
When sin θ ≠ 0, divide the original identity by $\sin^{2}$ θ: 1 + (cos θ ÷ sin θ $) squared$ = (1 ÷ sin θ $) squared$ .The same nonzero division turns the sine square term into 1.
Thus $\cot^{2}$ θ + 1 = $\csc^{2}$ θ.Cosine divided by sine is cotangent; 1 divided by sine is cosecant.

Answer

$\sin^{2}$ θ + $\cos^{2}$ θ = 1 for every real θ.
$\tan^{2}$ θ + 1 = $\sec^{2}$ θ when cos θ ≠ 0.
$\cot^{2}$ θ + 1 = $\csc^{2}$ θ when sin θ ≠ 0.

Check At θ = , sine and cosine squares each equal , while tangent and cotangent squares each equal 1 and secant and cosecant squares each equal 2. Each identity checks.

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

✗ Not this: θ + θ = (sin θ + cos θ.

Squaring a sum also produces the cross term 2 sin θ cos θ. The Pythagorean identity contains no such cross term.

✓ Instead: Add the two separate output squares: (sin θ + (cos θ = 1.

✗ Not this: θ + 1 = θ can be evaluated on the y-axis.

Both tangent and secant divide by cosine. On the y-axis cosine is 0.

✓ Instead: Apply this identity only when cos θ ≠ 0.

Tips and tricks

Rebuild the other two identities from $\sin^{2}$ θ + $\cos^{2}$ θ = 1 by dividing by the appropriate nonzero square.

.4The algebra behind verifying and simplifying an identity

Verifying an identity means turning one side into the other without changing its value, like exchanging a coin for equivalent smaller coins. Work on one side. Replace trig names by sine and cosine, use equal-sized fraction pieces, and cancel only a common nonzero factor that multiplies an entire top and bottom.

LHS means left side. RHS means right side.
Multiplication distributes: a(b + c) = ab + ac. Reversing this rule factors a common factor: ab + ac = a(b + c).
With a nonzero bottom b, a = $\frac{a b}{b}$ . This gives a whole term the same denominator as a nearby fraction.
A factor multiplies an entire expression. You may cancel such a factor from a fraction only when it is nonzero.
A numerical substitution checks a result at one input. It does not prove an identity for all allowed inputs.

A verification rewrites one side with equal expressions until the two sides match.

Reminder

Common denominators. $2 over 5$ + $3 over 5$ = $5 over 5$ = 1, because both terms count fifths.
A factor of 1. For b ≠ 0, $\frac{b}{b}$ = 1. This is why multiplying top and bottom by the same nonzero quantity preserves a fraction.

Worked exampleMake denominators match before combining

For sin t ≠ 0, simplify ( $\cos^{2}$ t ÷ sin t) + sin t. In words, combine two terms into one expression.

t ÷ sin t + sin t

= (t + t) ÷ sin t

= 1 ÷ sin t

= csc t

A common denominator lets the two coordinate squares combine.

Rewrite sin t as $\frac{\sin^{2} t}{\sin t}$ .Multiplying sin t by $\frac{\sin t}{\sin t}$ multiplies by 1, since sin t ≠ 0. This gives both terms the same bottom.
$\frac{\cos^{2} t}{\sin t}$ + $\frac{\sin^{2} t}{\sin t}$ = $cos squared t + sin squared t over sin t$ .Fractions with equal nonzero bottoms combine by adding their tops.
$cos squared t + sin squared t over sin t$ = $1 over sin t$ = csc t.The Pythagorean identity turns the top into 1, and the reciprocal definition gives cosecant.

Answer

csc t, with sin t ≠ 0

Check At t = , the left expression is ( ÷ ) + = + = 2. The right expression is csc = 2.

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

✗ Not this: Cancel a term from a sum: = b.

The top is a sum, so a does not multiply the whole top. For a = 4 and b = 8, = 3, rather than 8.

✓ Instead: If a ≠ 0, = 1 + . Cancel only whole common factors.

✗ Not this: A test at one angle proves an identity.

One matching input cannot establish equality at every allowed input.

✓ Instead: Use a chain of justified algebraic equalities for proof; use a numerical test only as a check.

✗ Not this: The original fraction equals sin θ even when tan θ = 7.

At that input its entire bottom is 0, so the original expression is undefined. Cancelling a zero factor would be division by zero.

✓ Instead: The simplification equals sin θ only with the original restrictions, including tan θ ≠ 7.

Tips and tricks

Write restrictions before changing fractions. Keep them even after a denominator cancels.
When proving, keep the target side visible but work on the other side.

.5When tangent or cotangent is the given value

A ratio tells you how two coordinates compare, like knowing a ramp's rise for each unit of run. It does not directly give their sizes on a circle of radius 1. Write one coordinate as a multiple of the other. The sum-of-squares identity then fixes the scale, and the quadrant fixes the signs.

If tan θ = q, then sin θ = q cos θ when cos θ ≠ 0.
If cot θ = q, then cos θ = q sin θ when sin θ ≠ 0.
Substitution means replacing a quantity by an equal expression.
For tan θ = q, ( $q^{2}$ + 1) $\cos^{2}$ θ = 1. For cot θ = q, ( $q^{2}$ + 1) $\sin^{2}$ θ = 1.

A rise-to-run magnitude ratio of gives legs 4 and 7 before scaling to radius 1.

Reminder

Solving for a square. If $65 over 49$ $z^{2}$ = 1, divide by $65 over 49$ to get $z^{2}$ = $49 over 65$ . This finds the size that the final square root uses.

Worked exampleRecover the scale from a tangent ratio

tan θ = − $4 over 7$ and θ is in Quadrant II. Find sine and cosine first. The ratio gives their comparison, while the identity gives their sizes.

Divide both leg sizes by , then give cosine the Quadrant II negative sign.

sin θ = (− $4 over 7$ ) cos θ.Multiply the quotient equation sin θ ÷ cos θ = − $4 over 7$ by the nonzero cosine. This expresses sine in terms of cosine.
(− $4 over 7$ cos θ $) squared$ + $\cos^{2}$ θ = 1, so ( $16 over 49$ + 1) $\cos^{2}$ θ = 1.Substitute the equal expression for sine into $\sin^{2}$ θ + $\cos^{2}$ θ = 1 and square the factors.
$65 over 49$ $\cos^{2}$ θ = 1, hence $\cos^{2}$ θ = $49 over 65$ .Combine $16 over 49$ + $49 over 49$ , then divide by $65 over 49$ . This isolates the cosine square and finds its size.
cos θ = − $7 over square root of 65$ = − $7 square root of 65 over 65$ .The square root gives size $7 over square root of 65$ . Quadrant II requires negative cosine. Rationalizing preserves the value.
sin θ = (− $4 over 7$ ) × (− $7 over square root of 65$ ) = $4 over square root of 65$ = $4 square root of 65 over 65$ .Use the original ratio relation. The two negative signs cancel, giving positive sine in Quadrant II.

Answer

sin θ = $4 square root of 65 over 65$
cos θ = − $7 square root of 65 over 65$

Check Their quotient is −, and their squares add to + = 1.

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

✗ Not this: tan θ = − means sin θ = −4 and cos θ = 7.

Those coordinates do not lie on a circle of radius 1, and their signs contradict Quadrant II.

✓ Instead: The ratio fixes proportional leg sizes. Normalize using the radius , and use the quadrant to choose signed coordinates.

Tips and tricks

A tangent or cotangent ratio gives a shape before it gives unit-circle coordinates.

Strategy: step by step

1. Translate the request: one trig value and a quadrant are given; find each of the five other outputs.
2. Check the sign and range against the stated quadrant. Inside a quadrant both coordinates are nonzero: require 0 < |sin θ| < 1 and 0 < |cos θ| < 1, or |sec θ| > 1 and |csc θ| > 1. A given tangent or cotangent must be nonzero with the quadrant's sign. Reject an endpoint or zero that forces an axis.
3. If secant or cosecant is given, take its reciprocal to get cosine or sine. If tangent or cotangent is given, use its quotient formula to express sine in terms of cosine or cosine in terms of sine.
4. Substitute into $\sin^{2}$ θ + $\cos^{2}$ θ = 1. Solve for the missing square; this finds the coordinate's size without yet choosing its sign.
5. Take the nonnegative square root, then choose the coordinate's sign from the quadrant.
6. Use quotients and reciprocals for the other functions. Show division by fractions and any rationalizing.
7. Check the sum of the sine and cosine squares, every quadrant sign and the original given value.
Read a reference-table entry in the column under its input or category in the matching picture.

Strategy

Find five trig outputs from one given output

Does the given value have the quadrant's sign and fit its interior values: sine or cosine size strictly between 0 and 1, secant or cosecant size greater than 1, or nonzero tangent or cotangent?

YesThe given value is compatible with an interior quadrant. Continue, then verify that both rebuilt coordinates are nonzero.

NoState that the information is inconsistent with the quadrant. An axis value cannot lie inside a quadrant.

↓

Is secant or cosecant given?

YesTake its reciprocal to recover cosine or sine.

NoKeep a given sine or cosine, or use the quotient identity if tangent or cotangent is given.

↓

Is tangent or cotangent given?

YesFor tan θ = q, use sin θ = q cos θ. For cot θ = q, use cos θ = q sin θ. Substitute into θ + θ = 1 to find a coordinate square.

NoSubtract the known coordinate square from 1 to find the missing coordinate square.

↓

Does a denominator become zero?

YesMark that function undefined; do not take its reciprocal.

NoCompute the quotient or reciprocal and simplify its exact form.

1. Translate the request: one trig value and a quadrant are given; find each of the five other outputs.
2. Check the sign and range against the stated quadrant. Inside a quadrant both coordinates are nonzero: require 0 < |sin θ| < 1 and 0 < |cos θ| < 1, or |sec θ| > 1 and |csc θ| > 1. A given tangent or cotangent must be nonzero with the quadrant's sign. Reject an endpoint or zero that forces an axis.
3. If secant or cosecant is given, take its reciprocal to get cosine or sine. If tangent or cotangent is given, use its quotient formula to express sine in terms of cosine or cosine in terms of sine.
4. Substitute into $\sin^{2}$ θ + $\cos^{2}$ θ = 1. Solve for the missing square; this finds the coordinate's size without yet choosing its sign.
5. Take the nonnegative square root, then choose the coordinate's sign from the quadrant.
6. Use quotients and reciprocals for the other functions. Show division by fractions and any rationalizing.
7. Check the sum of the sine and cosine squares, every quadrant sign and the original given value.

Worked exampleAll six functions from cos θ = in Quadrant IV

Suppose cos θ = $2 over 7$ and the terminal side of θ lies in Quadrant IV. Find the exact values of sin θ, tan θ, cot θ, sec θ and csc θ. Rationalize any denominators.

Unit circle with the terminal point P of θ ≈ 287° in Quadrant IV. P has x-coordinate cos θ = and y-coordinate sin θ = −.

Given: cos θ = $2 over 7$ , θ in Quadrant IV. Find: sin θ, tan θ, cot θ, sec θ, csc θ.One trig value and a quadrant fix the point on the unit circle. The other five functions come from it.
Check the given value. It is positive, and 0 < $2 over 7$ < 1. In Quadrant IV, cosine is positive and sine is negative.Inside a quadrant both coordinates are nonzero with absolute value less than 1. Cosine is positive in Quadrant IV, so the data are consistent.
Cosine is given directly, so no reciprocal or quotient step is needed. Substitute it into $\sin^{2}$ θ + $\cos^{2}$ θ = 1: $\sin^{2}$ θ + ( $2 over 7$ $) squared$ = 1, so $\sin^{2}$ θ = 1 − $4 over 49$ = $45 over 49$ .The Pythagorean identity gives the size of the missing coordinate. Its sign is not chosen yet.
Take the nonnegative root: $square root of 45 over 49$ = $square root of 45 over 7$ = $3 square root of 5 over 7$ . Sine is negative in Quadrant IV, so sin θ = − $3 square root of 5 over 7$ .The quadrant decides the sign. Points in Quadrant IV lie below the x-axis, so the y-coordinate is negative.
tan θ = $\frac{\sin θ}{\cos θ}$ = (− $3 square root of 5 over 7$ ) ÷ $2 over 7$ = − $3 square root of 5 over 7$ · $7 over 2$ = − $3 square root of 5 over 2$ .Dividing by a fraction means multiplying by its reciprocal. The 7s cancel.
cot θ = $1 over tan θ$ = − $2 over 3 square root of 5$ = − $2 square root of 5 over 15$ .Cotangent is the reciprocal of tangent. Multiply top and bottom by $square root of 5$ to rationalize: 3 $square root of 5$ · $square root of 5$ = 15.
sec θ = $1 over cos θ$ = $7 over 2$ .Secant is the reciprocal of cosine.
csc θ = $1 over sin θ$ = − $7 over 3 square root of 5$ = − $7 square root of 5 over 15$ .Cosecant is the reciprocal of sine. Rationalize by multiplying top and bottom by $square root of 5$ .

Answer

sin θ = −, tan θ = −, cot θ = −, sec θ = , csc θ = −

Check Pythagorean identity: θ + θ = + = = 1. Signs in Quadrant IV: cos and sec are positive; sin, csc, tan and cot are negative. All six match. Given value: = = cos θ. Products: tan θ · cot θ = (−)(−) = = 1, and sin θ · csc θ = (−)(−) = = 1.

Work to write

$\sin^{2}$ θ = 1 − $4 over 49$ = $45 over 49$
sin θ = − $3 square root of 5 over 7$ (negative in Quadrant IV)
tan θ = − $3 square root of 5 over 2$
cot θ = − $2 square root of 5 over 15$
sec θ = $7 over 2$
csc θ = − $7 square root of 5 over 15$
Check: $45 over 49$ + $4 over 49$ = 1

sin θ = − $3 square root of 5 over 7$ , tan θ = − $3 square root of 5 over 2$ , cot θ = − $2 square root of 5 over 15$ , sec θ = $7 over 2$ , csc θ = − $7 square root of 5 over 15$

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

Rung 1All six functions from a positive sine in Quadrant II

Suppose sin θ = $8 over 17$ and the terminal side of θ lies in Quadrant II. Find the exact values of cos θ, tan θ, cot θ, sec θ and csc θ.

An angle of about 152° in standard position. Its terminal side lies in Quadrant II, where the y-coordinate is positive (sin θ = ) and the x-coordinate is negative (cos θ = −).

Given: sin θ = $8 over 17$ , with θ in Quadrant II. Find cos θ, tan θ, cot θ, sec θ and csc θ.One trig value and a quadrant are given, so the other five functions can be found.
Check the given value: 0 < $8 over 17$ < 1. In Quadrant II, sine is positive, so the sign matches.Inside a quadrant both coordinates are nonzero. That requires 0 < |sin θ| < 1 and the quadrant's sign. In Quadrant II, x < 0 and y > 0.
Sine is already given, so no reciprocal or quotient conversion is needed.Step 3 applies only when secant, cosecant, tangent or cotangent is given.
Substitute into $\sin^{2}$ θ + $\cos^{2}$ θ = 1: ( $8 over 17$ $) squared$ + $\cos^{2}$ θ = 1. This gives $64 over 289$ + $\cos^{2}$ θ = 1, so $\cos^{2}$ θ = 1 − $64 over 289$ = $289 over 289$ − $64 over 289$ = $225 over 289$ .The Pythagorean identity gives the square of the missing coordinate before its sign is chosen.
Take the nonnegative root: $square root of 225 over 289$ = $15 over 17$ . In Quadrant II, cosine is negative, so cos θ = − $15 over 17$ .The square root gives the size of the coordinate. The quadrant decides its sign.
tan θ = $\frac{\sin θ}{\cos θ}$ = $8 over 17$ ÷ (− $15 over 17$ ) = $8 over 17$ · (− $17 over 15$ ) = − $8 over 15$ . cot θ = $\frac{\cos θ}{\sin θ}$ = − $15 over 17$ · $17 over 8$ = − $15 over 8$ .Dividing by a fraction means multiplying by its reciprocal, and the 17s cancel.
sec θ = $1 over cos θ$ = 1 ÷ (− $15 over 17$ ) = − $17 over 15$ . csc θ = $1 over sin θ$ = 1 ÷ $8 over 17$ = $17 over 8$ .Secant and cosecant are the reciprocals of cosine and sine. No denominators contain roots, so nothing needs rationalizing.
Check: ( $8 over 17$ $) squared$ + (− $15 over 17$ $) squared$ = $64 over 289$ + $225 over 289$ = $289 over 289$ = 1. In Quadrant II, sine and cosecant are positive while cosine, tangent, cotangent and secant are negative. sin θ = $8 over 17$ matches the given value.The identity, every quadrant sign and the original value must all agree.

Answer

cos θ = −, tan θ = −, cot θ = −, sec θ = −, csc θ =

Check ( + (− = + = 1. Also tan θ · cot θ = (−)(−) = 1 and cos θ · sec θ = (−)(−) = 1. The signs fit Quadrant II, where only sine and cosecant are positive.

Work to write

$\cos^{2}$ θ = 1 − $64 over 289$ = $225 over 289$
cos θ = − $15 over 17$ (Quadrant II, cosine negative)
tan θ = $8 over 17$ · (− $17 over 15$ ) = − $8 over 15$
cot θ = − $15 over 8$
sec θ = − $17 over 15$
csc θ = $17 over 8$
Check: $64 over 289$ + $225 over 289$ = 1

cos θ = − $15 over 17$ , tan θ = − $8 over 15$ , cot θ = − $15 over 8$ , sec θ = − $17 over 15$ , csc θ = $17 over 8$

Rung 2All six functions from a cotangent in Quadrant III

Given cot θ = $20 over 21$ and θ lies in Quadrant III, find sin θ, cos θ, tan θ, sec θ and csc θ.

Translate the request. Known: cot θ = $20 over 21$ , with θ in Quadrant III. Wanted: sin θ, cos θ, tan θ, sec θ, csc θ.One trig value and a quadrant fix the point on the unit circle, so they fix all five other functions.
Check the sign and the range. In Quadrant III, x < 0 and y < 0, so sin θ < 0 and cos θ < 0. That makes tan θ and cot θ positive. The given cot θ = $20 over 21$ is positive and nonzero.A given cotangent must be nonzero and must have the quadrant's sign. It passes, so no axis point is forced.
Use the quotient formula: cot θ = $\frac{\cos θ}{\sin θ}$ = $20 over 21$ , so cos θ = $20 over 21$ sin θ.This writes cosine in terms of sine, which leaves one unknown for the identity.
Substitute into $\sin^{2}$ θ + $\cos^{2}$ θ = 1: $\sin^{2}$ θ + $400 over 441$ $\sin^{2}$ θ = 1. Combine the terms: $441 over 441$ $\sin^{2}$ θ + $400 over 441$ $\sin^{2}$ θ = $841 over 441$ $\sin^{2}$ θ = 1. So $\sin^{2}$ θ = $441 over 841$ .The Pythagorean identity gives the size of the sine before its sign is chosen.
Take the nonnegative root: $square root of 441 over 841$ = $21 over 29$ . In Quadrant III sine is negative, so sin θ = − $21 over 29$ .Both 441 = $21 squared$ and 841 = $29 squared$ are perfect squares. The quadrant decides the sign.
Find cosine: cos θ = $20 over 21$ · (− $21 over 29$ ) = − $20 over 29$ . This value is negative, as Quadrant III requires.Substituting back into the quotient relation gives cosine, with its sign already correct.
tan θ = $\frac{\sin θ}{\cos θ}$ = (− $21 over 29$ ) ÷ (− $20 over 29$ ) = (− $21 over 29$ ) · (− $29 over 20$ ) = $21 over 20$ . This equals $1 over cot θ$ , as it should.Dividing by a fraction means multiplying by its reciprocal. The 29s cancel and the two negatives give a positive result.
sec θ = $1 over cos θ$ = 1 ÷ (− $20 over 29$ ) = − $29 over 20$ , and csc θ = $1 over sin θ$ = 1 ÷ (− $21 over 29$ ) = − $29 over 21$ . No rationalizing is needed because there are no radicals.Secant and cosecant are the reciprocals of cosine and sine. They keep the same signs.

Answer

sin θ = −, cos θ = −, tan θ = , sec θ = −, csc θ = − (and cot θ = as given).

Check Sum of squares: + = = 1. Signs: sin, cos, sec and csc are negative, and tan and cot are positive, which fits Quadrant III. Original value: = (−) ÷ (−) = ✓.

Work to write

Quadrant III: sin θ < 0, cos θ < 0, cot θ > 0 ✓
cos θ = $20 over 21$ sin θ
$\sin^{2}$ θ + $400 over 441$ $\sin^{2}$ θ = 1 ⇒ $841 over 441$ $\sin^{2}$ θ = 1 ⇒ $\sin^{2}$ θ = $441 over 841$
sin θ = − $21 over 29$ (Quadrant III)
cos θ = $20 over 21$ · (− $21 over 29$ ) = − $20 over 29$
tan θ = (− $21 over 29$ ) · (− $29 over 20$ ) = $21 over 20$
sec θ = − $29 over 20$ , csc θ = − $29 over 21$
Check: $441 over 841$ + $400 over 841$ = 1

sin θ = − $21 over 29$ , cos θ = − $20 over 29$ , tan θ = $21 over 20$ , sec θ = − $29 over 20$ , csc θ = − $29 over 21$ (and cot θ = $20 over 21$ as given).

Rung 3Rung 3: a root that must be simplified

sin θ = − $2 over 7$ and θ is in Quadrant IV. Find the other five exact outputs.

The horizontal coordinate is positive and the height is negative.

Cosine and secant are positive; sine, cosecant, tangent and cotangent are negative.Quadrant IV is right of the y-axis and below the x-axis.
$\cos^{2}$ θ = 1 − (− $2 over 7$ $) squared$ = $49 over 49$ − $4 over 49$ = $45 over 49$ .This subtraction isolates the missing coordinate's square.
cos θ = $square root of 45 over 49$ = $square root of 45 over 7$ = $3 square root of 5 over 7$ .Cosine is positive here. The perfect-square factor 9 in 45 gives $square root of 45$ = 3 $square root of 5$ .
tan θ = (− $2 over 7$ ) ÷ $3 square root of 5 over 7$ = − $2 over 3 square root of 5$ = − $2 square root of 5 over 15$ .The quotient definition gives tangent; multiplying top and bottom by $square root of 5$ rationalizes it.
cot θ = $3 square root of 5 over 7$ ÷ (− $2 over 7$ ) = − $3 square root of 5 over 2$ .Use cosine divided by sine.
sec θ = $7 over 3 square root of 5$ = $7 square root of 5 over 15$ ; csc θ = − $7 over 2$ .Take the coordinate reciprocals and rationalize secant.

Answer

cos θ = $3 square root of 5 over 7$
tan θ = − $2 square root of 5 over 15$
cot θ = − $3 square root of 5 over 2$
sec θ = $7 square root of 5 over 15$
csc θ = − $7 over 2$

Check The coordinate squares and add to 1. Their signs fit Quadrant IV.

Rung 4Rung 4: begin with secant

sec θ = −3 and θ is in Quadrant II. Find the other five exact outputs.

Quadrant II chooses positive height and negative horizontal position.

cos θ = 1 ÷ (−3) = − $1 over 3$ .Taking secant's reciprocal recovers cosine.
$\sin^{2}$ θ = 1 − (− $1 over 3$ $) squared$ = $9 over 9$ − $1 over 9$ = $8 over 9$ .The identity isolates sine's square, which finds its magnitude.
sin θ = $square root of 8 over 9$ = $2 square root of 2 over 3$ .Quadrant II has positive sine; $square root of 8$ = 2 $square root of 2$ .
tan θ = $2 square root of 2 over 3$ ÷ (− $1 over 3$ ) = −2 $square root of 2$ .The quotient definition and the fraction reciprocal produce a negative tangent.
cot θ = − $1 over 2 square root of 2$ = − $square root of 2 over 4$ ; csc θ = $3 over 2 square root of 2$ = $3 square root of 2 over 4$ .Take the nonzero reciprocals and rationalize.

Answer

sin θ = $2 square root of 2 over 3$
cos θ = − $1 over 3$
tan θ = −2 $square root of 2$
cot θ = − $square root of 2 over 4$
csc θ = $3 square root of 2 over 4$

Check The coordinate squares add to + = 1, and 1 ÷ cos θ = −3 recovers the given secant.

Rung 5Rung 5: begin with a fractional cosecant

csc θ = $7 over 2$ and θ is in Quadrant II. Find the other five exact outputs.

The positive sine and negative cosine place the point in Quadrant II.

sin θ = 1 ÷ $7 over 2$ = $2 over 7$ .Taking cosecant's reciprocal recovers sine.
$\cos^{2}$ θ = 1 − ( $2 over 7$ $) squared$ = $49 over 49$ − $4 over 49$ = $45 over 49$ .Subtract the known sine square to isolate the unknown cosine square.
cos θ = − $square root of 45 over 49$ = − $3 square root of 5 over 7$ .The square root gives the size, and Quadrant II chooses negative cosine.
tan θ = $2 over 7$ ÷ (− $3 square root of 5 over 7$ ) = − $2 over 3 square root of 5$ = − $2 square root of 5 over 15$ .Use the quotient formula and rationalize.
cot θ = − $3 square root of 5 over 2$ ; sec θ = − $7 over 3 square root of 5$ = − $7 square root of 5 over 15$ .Use cosine divided by sine, then cosine's reciprocal.

Answer

sin θ = $2 over 7$
cos θ = − $3 square root of 5 over 7$
tan θ = − $2 square root of 5 over 15$
cot θ = − $3 square root of 5 over 2$
sec θ = − $7 square root of 5 over 15$

Check The coordinate squares add to 1, only sine and cosecant are positive, and 1 ÷ sin θ = recovers the given value.

Rung 6Rung 6: a quotient is given instead of a coordinate

tan θ = − $4 over 7$ and θ is in Quadrant II. Find the other five exact outputs.

Normalize the legs by the hypotenuse, then give cosine its Quadrant II negative sign.

sin θ = (− $4 over 7$ ) cos θ.Multiply the tangent quotient equation by nonzero cosine to express sine in terms of cosine.
(− $4 over 7$ cos θ $) squared$ + $\cos^{2}$ θ = 1, so $65 over 49$ $\cos^{2}$ θ = 1.Substitute into the Pythagorean identity, square the coefficient and combine $16 over 49$ + $49 over 49$ .
$\cos^{2}$ θ = $49 over 65$ .Divide by $65 over 49$ to isolate the unknown square. This determines cosine's size.
cos θ = − $7 over square root of 65$ = − $7 square root of 65 over 65$ .The square root gives size $7 over square root of 65$ and Quadrant II chooses negative cosine.
sin θ = (− $4 over 7$ ) × (− $7 over square root of 65$ ) = $4 over square root of 65$ = $4 square root of 65 over 65$ .Substitute into the ratio relation; two negative factors give positive sine.
cot θ = − $7 over 4$ , sec θ = − $square root of 65 over 7$ , csc θ = $square root of 65 over 4$ .Take tangent's, cosine's and sine's reciprocals respectively.

Answer

sin θ = $4 square root of 65 over 65$
cos θ = − $7 square root of 65 over 65$
cot θ = − $7 over 4$
sec θ = − $square root of 65 over 7$
csc θ = $square root of 65 over 4$

Check The quotient sin θ ÷ cos θ is −, and the coordinate squares and add to 1.

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

✗ Not this: A positive square root means the missing coordinate is positive.

The square root gives a nonnegative size. Both coordinate signs have that square; the quadrant chooses the sign.

✓ Instead: In Quadrant IV with cos θ = , sin θ = − = −.

✗ Not this: θ means sin().

The exponent applies to the output of sine, not to its input. At θ = , (sin θ = 1, while squaring the angle would be a different input.

✓ Instead: θ means (sin θ.

✗ Not this: Cancellation makes a simplified formula valid at every input.

A cancelled factor may have been zero, making the original expression undefined. Algebra cannot create an original output there.

✓ Instead: Carry every restriction from the original expression through the final answer.

✗ Not this: sin θ = 1 and θ is in Quadrant I is consistent because the sign is positive.

Sine 1 forces cosine 0, so the terminal point is (0, 1) on an axis.

✓ Instead: No angle inside Quadrant I has sine 1; inside a quadrant, both coordinates are nonzero and sine's size is less than 1.

Tips and tricks

Memory device: coordinate squares add to the radius square. On the unit circle, that radius square is 1.
Write the quadrant signs above the calculations. Use them again after you take each square root.
Check both the identity and the original given value. A correct sum of squares alone cannot catch every wrong sign.
For an identity, begin with one side and preserve its allowed inputs through each change.
Know cold:
Sine reads y and cosine reads x. Memory device: cosine names the horizontal coordinate; sine names height.
A zero denominator means undefined. Memory device: zero on the bottom blocks division.
The quadrant sign pattern. Memory device: All Students Take Calculus.
The period pattern. Memory device: ratios repeat after half a lap; coordinates and their reciprocals need a full lap.
The even pair. Memory device: cosine and its reciprocal secant are even.
The unit-circle square rule. Memory device: coordinate squares add to the radius square.
Understand, then rebuild when needed:
Rebuild the tangent and cotangent signs by dividing the coordinate signs.
Rebuild the other two Pythagorean identities by dividing the unit-circle identity by a nonzero coordinate square.
Rebuild the missing coordinate from the given square and the quadrant sign.
Rebuild the special-angle reciprocals from sine and cosine. These take a few steps; you do not need six unrelated memorized lists.
Put on the study reference card or look up while studying:
The six-function domain and range table.
The special-angle and axis-point tables.
The sine and cosine direction-of-change intervals.
For a closed-book exam, use the card to practice rebuilding the facts, then practice without it.

Trap. Taking the sign from the square root. = is positive, but in Quadrant IV sine is negative, so sin θ = −. The quadrant decides the sign, never the square root.

Keep in mind

√ alone is never negative, but squaring hides the sign: $square root of 49 over 625$ = $7 over 25$ , yet $7 over 25$ and − $7 over 25$ both square to $49 over 625$ .
Check that the given value is possible before starting: sin θ = $9 over 8$ has no angle, because $9 over 8$ = 1.125 > 1.
$\sin^{2}$ θ means (sin θ $) squared$ , the output squared, not sin( $θ^{2}$ ): with sin θ = 0.6, $\sin^{2}$ θ = 0.36.
Finish with a check: (− $7 over 25$ $) squared$ + ( $24 over 25$ $) squared$ = $49 over 625$ + $576 over 625$ = 1, and only sine and cosecant came out positive, as Quadrant II requires.

Memory hookRoot for the size, quadrant for the sign: √ says how big, All Students Take Calculus says plus or minus; then divide and flip for the other four.

Flash cards: say the answer out loud, then flip

What is the principal square root?

The nonnegative root: = 9, never −9.

State the Pythagorean identity.

θ + θ = 1.

cos θ = − and θ is in Quadrant II. Find sin θ.

: θ = 1 − = , the root is , and sine is positive in II.

sin θ = and cos θ = −. Find tan θ.

−.

Which decides the sign of the missing coordinate: the square root or the quadrant?

The quadrant. The square root gives only the size.

True or false: = ±.

False: √ alone means the positive root, . The ± appears only when you solve θ = .