The fundamental identities

Explain it like I am five

Picture a ladder 1 unit long. Its height up the wall is sin θ and its distance along the floor is cos θ, since opposite ÷ 1 and adjacent ÷ 1 are those lengths. The Pythagorean theorem, le $g^{2}$ + le $g^{2}$ = hypotenus $e^{2}$ , says $\sin^{2}$ θ + $\cos^{2}$ θ = 1. Read $\sin^{2}$ θ as "sine of theta, squared": find sin θ, then square it.

That is an identity: an equation true for every allowed input, as one $5 bill always equals five $1 bills. Say θ is acute (under 90°) and sin θ = $7 over 25$ . Then $\cos^{2}$ θ = 1 − ( $7 over 25$ $) squared$ = $625 over 625$ − $49 over 625$ = $576 over 625$ . Since $24 squared$ = 576 and $25 squared$ = 625, cos θ = $24 over 25$ , positive because θ is acute.

Next, tan θ = $\frac{\sin θ}{\cos θ}$ = $7 over 25$ ÷ $24 over 25$ = $7 over 24$ , since the 25s cancel. Flip for the rest: csc θ = $25 over 7$ , sec θ = $25 over 24$ , cot θ = $24 over 7$ . Dividing $\sin^{2}$ θ + $\cos^{2}$ θ = 1 by $\cos^{2}$ θ gives $\tan^{2}$ θ + 1 = $\sec^{2}$ θ, and dividing by $\sin^{2}$ θ gives $\cot^{2}$ θ + 1 = $\csc^{2}$ θ.

In plain words

Think of changing a $5 bill into five $1 bills. The amount stays the same even though it looks different. An identity is an equation that stays true for every allowed input, meaning every input for which its expressions exist. The six side ratios come from the same triangle, so they cannot vary independently. Reciprocal identities flip matching fractions. Quotient identities build tangent or cotangent by dividing sine and cosine. The Pythagorean identity ties their squares together. The notation $\sin^{2}$ θ means find sine first, then square that answer. For an acute angle, one known ratio lets you rebuild the other five. Absolute value bars give size without a sign: |−3| = 3.

θ means (sin θ

θ + θ = 1

cos θ ≠ 0: θ + 1 = θ

sin θ ≠ 0: θ + 1 = θ

One square-sum identity generates two others when the chosen divisor is nonzero.

Reminder

Balance an equation. Subtract from both sides: $u^{2}$ + $4 over 25$ = 1 gives $u^{2}$ = 1 − $4 over 25$ = $21 over 25$ .
Square a fraction. ( $2 over 7$ $) squared$ = $4 over 49$ . A minus sign inside parentheses also disappears when squared.
Subtract from 1. 1 − $4 over 49$ = $49 over 49$ − $4 over 49$ = $45 over 49$ . Match the bottoms first.
Principal root and two solutions. $square root of 9$ = 3 and $square root of 0$ = 0. Solving $u^{2}$ = 9 gives u = ±3 because both signs square to 9.
Divide fractions. $3 over 5$ ÷ $4 over 5$ = $3 over 5$ × $5 over 4$ = $3 over 4$ .
Absolute value. |−3| = 3 and |3| = 3: the bars report size, or distance from zero, without a sign.

Why it works. Let the adjacent leg be a, the opposite leg b, and the hypotenuse c. Divide + = by nonzero : + = 1. The two fractions are θ and θ, giving θ + θ = 1. Flipping and dividing side ratios gives the other families. The parts show every divided term and its reason. The triangle proves these connections for acute angles; the later point definition preserves them for every angle where their expressions are defined. Know the first square-sum identity cold and rebuild the other two by division.

Ruleθ + θ = 1 for every angle. tan θ = and sec θ = require cos θ ≠ 0; cot θ = and csc θ = require sin θ ≠ 0.
Where both reciprocal expressions exist, cot θ = . θ + 1 = θ requires cos θ ≠ 0; θ + 1 = θ requires sin θ ≠ 0.

The same idea, five ways

Say it

Sine squared plus cosine squared equals one.

Write it

The square of the sine of an angle plus the square of its cosine is always one.

In math

$\sin^{2}$ θ + $\cos^{2}$ θ = 1
(sin θ $) squared$ + (cos θ $) squared$ = 1
$\cos^{2}$ θ = 1 − $\sin^{2}$ θ
$\sin^{2}$ θ = 1 − $\cos^{2}$ θ

Two square tiles share a fixed total area; one tile's area determines the other.

See it

The hypotenuse-one triangle shows why the two squared legs total 1.

The same idea, other ways

Shrink the hypotenuse to one

Divide every side by c. The hypotenuse becomes 1, and the legs become $\frac{b}{c}$ , the sine, and $\frac{a}{c}$ , the cosine. The squared legs still add to the squared hypotenuse, now 1.

After scaling, the leg lengths are the sine and cosine values.

Concrete numbers

For sin θ = $3 over 5$ and cos θ = $4 over 5$ , their squares add to $9 over 25$ + $16 over 25$ = 1. Their ordinary sum is $7 over 5$ , so the squares matter.

θ =

+ = 1

Square both values before adding.

Same amount, different form

$\sin^{2}$ θ and 1 − $\cos^{2}$ θ are the same amount, like five $1 bills and one $5 bill. You can replace one with the other inside an expression without changing its value.

θ = 1 − θ

Subtract one square from both sides to isolate the other.

A rebuilding recipe

Know sine? Square it, subtract from 1, take the root and choose the sign to get cosine. Divide the two for tangent, reverse the division for cotangent, and flip sine and cosine for cosecant and secant.

sine then square then 1 minus square

root and sign then cosine

divide and flip then other functions

One given value begins a sequence of small arithmetic moves.

Identity	Why it is true
csc θ = $1 over sin θ$	$\frac{h y p}{o p p}$ is $\frac{o p p}{h y p}$ upside down
sec θ = $1 over cos θ$	$\frac{h y p}{a d j}$ is $\frac{a d j}{h y p}$ upside down
cot θ = $1 over tan θ$	$\frac{a d j}{o p p}$ flips $\frac{o p p}{a d j}$ where tan exists and is nonzero; cot = cos ÷ sin applies more generally
tan θ = $\frac{\sin θ}{\cos θ}$	$\frac{o p p}{h y p}$ ÷ $\frac{a d j}{h y p}$ = $\frac{o p p}{h y p}$ × $\frac{h y p}{a d j}$ = $\frac{o p p}{a d j}$
cot θ = $\frac{\cos θ}{\sin θ}$	$\frac{a d j}{h y p}$ ÷ $\frac{o p p}{h y p}$ = $\frac{a d j}{o p p}$ , provided sin θ ≠ 0
$\sin^{2}$ θ + $\cos^{2}$ θ = 1	divide $a^{2}$ + $b^{2}$ = $c^{2}$ by $c^{2}$
$\tan^{2}$ θ + 1 = $\sec^{2}$ θ	divide every term by nonzero $\cos^{2}$ θ
$\cot^{2}$ θ + 1 = $\csc^{2}$ θ	divide every term by nonzero $\sin^{2}$ θ
\|sin θ\| ≤ 1 and \|cos θ\| ≤ 1	Absolute value bars give size. Both nonnegative squares add to 1, so each square is at most 1 and each value has size at most 1. In a triangle a leg cannot exceed hyp.
\|csc θ\| ≥ 1 and \|sec θ\| ≥ 1	A defined reciprocal flips a size between 0 and 1 to a size at least 1, as 1 ÷ $1 over 2$ = 2. Zero has no reciprocal.

.1Read $\sin^{2}$ θ and $\cos^{2}$ θ

A vending machine gives a number, and then you square that number. $\sin^{2}$ θ follows this order: find sine of θ, then multiply its output by itself. It never squares θ first.

$\sin^{2}$ θ = (sin θ $) squared$ ; $\cos^{2}$ θ = (cos θ $) squared$ .
The same convention applies to $\tan^{2}$ θ, $\cot^{2}$ θ, $\sec^{2}$ θ and $\csc^{2}$ θ.

θ = (sin θ

sin θ = then θ =

The square acts after the function produces its output.

The same idea, five ways

Say it

Sine squared means the sine value multiplied by itself.

Write it

A superscript square on a trig name squares its output, not its input angle.

In math

$\sin^{2}$ θ = (sin θ $) squared$

Square the number the machine gives you, like squaring a score after receiving it.

See it

θ = (sin θ

sin θ = then θ =

The square acts after the function produces its output.

Worked exampleSquare the output

sin θ = $2 over 5$ . Find $\sin^{2}$ θ. In words, square the given output, not the angle.

θ = (sin θ

sin θ = then θ =

The square acts after the function produces its output.

$\sin^{2}$ θ = ( $2 over 5$ $) squared$ .The notation squares the sine output.
( $2 over 5$ $) squared$ = $2 × 2 over 5 × 5$ = $4 over 25$ .A fraction squared squares its numerator and denominator.

Answer

θ = .

Check is less than , matching the fact that multiplying a positive fraction below 1 by itself makes it smaller.

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

✗ Not this: Square only the bottom to get .

Squaring multiplies the entire fraction by itself.

✓ Instead: × = .

Tips and tricks

Say square the sine value aloud.

.2Reciprocal identities

A reciprocal reverses a fraction, like changing miles per hour into hours per mile. Sine pairs with cosecant, cosine with secant, and tangent with cotangent. A zero output has no reciprocal.

csc θ = $1 over sin θ$ when sin θ ≠ 0.
sec θ = $1 over cos θ$ when cos θ ≠ 0.
cot θ = $1 over tan θ$ only where tan is defined and nonzero; cot = cos ÷ sin is the general quotient.

csc θ = , sin θ ≠ 0

sec θ = , cos θ ≠ 0

cot θ = where both exist

The flip pairs require nonzero defined values.

The same idea, five ways

Say it

Cosecant is one divided by sine; secant is one divided by cosine.

Write it

The reciprocal identities reverse a nonzero defined ratio into its paired function.

In math

csc θ = $1 over sin θ$ when sin θ ≠ 0.
sec θ = $1 over cos θ$ when cos θ ≠ 0.
cot θ = $1 over tan θ$ only where tan is defined and nonzero; cot = cos ÷ sin is the general quotient.

Miles per hour and hours per mile reverse the same comparison.

See it

csc θ = , sin θ ≠ 0

sec θ = , cos θ ≠ 0

cot θ = where both exist

The flip pairs require nonzero defined values.

Worked exampleFlip one cosine

cos θ = $2 over 5$ . Find sec θ. In words, reverse the given nonzero cosine ratio.

csc θ = , sin θ ≠ 0

sec θ = , cos θ ≠ 0

cot θ = where both exist

The flip pairs require nonzero defined values.

sec θ = 1 ÷ $2 over 5$ = $5 over 2$ .Cosine is nonzero, so its reciprocal exists. Dividing by a fraction flips it.

Answer

sec θ = .

Check × = = 1, checking the reciprocal pair.

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

✗ Not this: The reciprocal of zero is zero.

No number multiplied by 0 gives 1.

✓ Instead: 1 ÷ 0 is undefined.

Tips and tricks

Check for zero before taking a reciprocal.

.3Quotient identities

Dividing sine by cosine cancels the hypotenuse shared by both fractions and leaves opposite over adjacent. Reverse the division for cotangent, like changing height per floor distance into floor distance per height.

tan θ = $\frac{\sin θ}{\cos θ}$ when cos θ ≠ 0.
cot θ = $\frac{\cos θ}{\sin θ}$ when sin θ ≠ 0.
$\frac{o p p}{h y p}$ ÷ $\frac{a d j}{h y p}$ = $\frac{o p p}{h y p}$ × $\frac{h y p}{a d j}$ = $\frac{o p p}{a d j}$ .

tan θ = , cos θ ≠ 0

cot θ = , sin θ ≠ 0

Tangent and cotangent divide the same two values in reverse orders.

The same idea, five ways

Say it

Tangent is sine divided by cosine; cotangent reverses that order.

Write it

The quotient identities express tangent and cotangent as ratios of sine and cosine.

In math

tan θ = $\frac{\sin θ}{\cos θ}$ when cos θ ≠ 0.
cot θ = $\frac{\cos θ}{\sin θ}$ when sin θ ≠ 0.
$\frac{o p p}{h y p}$ ÷ $\frac{a d j}{h y p}$ = $\frac{o p p}{h y p}$ × $\frac{h y p}{a d j}$ = $\frac{o p p}{a d j}$ .

Height per floor distance and floor distance per height reverse the same ladder measurements.

See it

tan θ = , cos θ ≠ 0

cot θ = , sin θ ≠ 0

Tangent and cotangent divide the same two values in reverse orders.

Worked exampleDivide two known outputs

sin θ = $5 over 13$ and cos θ = $12 over 13$ . Find tan θ and cot θ. In words, divide the outputs in both possible orders.

tan θ = , cos θ ≠ 0

cot θ = , sin θ ≠ 0

Tangent and cotangent divide the same two values in reverse orders.

tan θ = $5 over 13$ ÷ $12 over 13$ = $5 over 13$ × $13 over 12$ = $5 over 12$ .The nonzero cosine may divide sine; flip the second fraction and cancel factor 13.
cot θ = $12 over 13$ ÷ $5 over 13$ = $12 over 13$ × $13 over 5$ = $12 over 5$ .The nonzero sine may divide cosine; reverse the order and cancel 13.

Answer

tan θ = $5 over 12$
cot θ = $12 over 5$

Check × = 1. The givens also satisfy + = 1.

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

✗ Not this: Equal bottoms cancel when adding too.

They cancel here because division flips the second fraction. Addition retains the bottom.

✓ Instead: + = , while ÷ = .

Tips and tricks

Name the division order before calculating.

.4Pythagorean identity

Shrink a right triangle until hyp is 1. Its leg lengths become sine and cosine. The two small square areas still total the large square area, now $1 squared$ = 1.

$\sin^{2}$ θ + $\cos^{2}$ θ = 1.
$\cos^{2}$ θ = 1 − $\sin^{2}$ θ and $\sin^{2}$ θ = 1 − $\cos^{2}$ θ.
The later point definition extends the triangle result to every angle.

The two squared leg lengths add to = 1.

The same idea, five ways

Say it

Sine squared plus cosine squared equals one.

Write it

After scaling the hypotenuse to 1, the two squared leg lengths add to 1.

In math

$\sin^{2}$ θ + $\cos^{2}$ θ = 1.
$\cos^{2}$ θ = 1 − $\sin^{2}$ θ and $\sin^{2}$ θ = 1 − $\cos^{2}$ θ.
The later point definition extends the triangle result to every angle.

The two leg-square areas share the fixed total area of the hypotenuse square.

See it

The two squared leg lengths add to = 1.

Worked exampleCheck two squared values

sin θ = $8 over 17$ and cos θ = $15 over 17$ . Check the identity. In words, square the outputs and add.

Name the sides from the marked angle before making a fraction.

$\sin^{2}$ θ = $64 over 289$ and $\cos^{2}$ θ = $225 over 289$ .Square both numerator and denominator of each fraction.
$64 over 289$ + $225 over 289$ = $289 over 289$ = 1.Equal denominators let you add the numerators.

Answer

θ + θ = 1.

Check The original side lengths satisfy + = , so the side-length check agrees.

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

✗ Not this: Add sine and cosine without squaring.

The identity relates square areas, not lengths.

✓ Instead: Square first; + = , not 1.

Tips and tricks

Write both superscript squares before substituting.

.5Rebuild the other two Pythagorean identities

Changing units can make the same total look different, like measuring money in dollars or quarters. Divide every term by the same nonzero square, then use the quotient and reciprocal definitions to rename the fractions.

For cos θ ≠ 0: $\frac{\sin^{2} θ}{\cos^{2} θ}$ + $\frac{\cos^{2} θ}{\cos^{2} θ}$ = $1 over cos squared θ$ , giving $\tan^{2}$ θ + 1 = $\sec^{2}$ θ.
For sin θ ≠ 0: $\frac{\sin^{2} θ}{\sin^{2} θ}$ + $\frac{\cos^{2} θ}{\sin^{2} θ}$ = $1 over sin squared θ$ , giving 1 + $\cot^{2}$ θ = $\csc^{2}$ θ.

+ =

θ + 1 = θ

+ =

1 + θ = θ

Divide every term before renaming the fractions.

The same idea, five ways

Say it

Tangent squared plus one equals secant squared; one plus cotangent squared equals cosecant squared.

Write it

Dividing every term of the sine-cosine identity by a nonzero square gives another equivalent identity.

In math

For cos θ ≠ 0: $\frac{\sin^{2} θ}{\cos^{2} θ}$ + $\frac{\cos^{2} θ}{\cos^{2} θ}$ = $1 over cos squared θ$ , giving $\tan^{2}$ θ + 1 = $\sec^{2}$ θ.
For sin θ ≠ 0: $\frac{\sin^{2} θ}{\sin^{2} θ}$ + $\frac{\cos^{2} θ}{\sin^{2} θ}$ = $1 over sin squared θ$ , giving 1 + $\cot^{2}$ θ = $\csc^{2}$ θ.

Changing from dollars to quarters changes every amount by the same factor, preserving the equality.

See it

+ =

θ + 1 = θ

+ =

1 + θ = θ

Divide every term before renaming the fractions.

Worked exampleDerive both divided identities

Start with $\sin^{2}$ θ + $\cos^{2}$ θ = 1. Derive the tangent-secant and cotangent-cosecant identities. In words, show every term after division.

+ =

θ + 1 = θ

+ =

1 + θ = θ

Divide every term before renaming the fractions.

For cos θ ≠ 0, divide every term by $\cos^{2}$ θ: $\frac{\sin^{2} θ}{\cos^{2} θ}$ + $\frac{\cos^{2} θ}{\cos^{2} θ}$ = $1 over cos squared θ$ .The divisor is nonzero, and equal operations on both sides preserve equality.
$\frac{\sin^{2} θ}{\cos^{2} θ}$ = ( $\frac{\sin θ}{\cos θ}$ $) squared$ = $\tan^{2}$ θ.A quotient squared squares its top and bottom; sine divided by cosine is tangent.
$\frac{\cos^{2} θ}{\cos^{2} θ}$ = 1 and $1 over cos squared θ$ = ( $1 over cos θ$ $) squared$ = $\sec^{2}$ θ. Hence $\tan^{2}$ θ + 1 = $\sec^{2}$ θ.A nonzero number divided by itself is 1, and the reciprocal of cosine is secant.
For sin θ ≠ 0, divide every term by $\sin^{2}$ θ: $\frac{\sin^{2} θ}{\sin^{2} θ}$ + $\frac{\cos^{2} θ}{\sin^{2} θ}$ = $1 over sin squared θ$ .Again the same nonzero divisor must act on every term.
$\frac{\sin^{2} θ}{\sin^{2} θ}$ = 1 and $\frac{\cos^{2} θ}{\sin^{2} θ}$ = ( $\frac{\cos θ}{\sin θ}$ $) squared$ = $\cot^{2}$ θ.Cosine divided by sine is cotangent.
$1 over sin squared θ$ = ( $1 over sin θ$ $) squared$ = $\csc^{2}$ θ. Hence 1 + $\cot^{2}$ θ = $\csc^{2}$ θ.The reciprocal of sine is cosecant.

Answer

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

Check With sin = and cos = , the checks are + 1 = = and 1 + = = .

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

✗ Not this: Divide only the first term by θ.

Equality is preserved only when every term receives the division.

✓ Instead: Display three fractions, then rename them.

Tips and tricks

Memorize the original square-sum identity and rebuild these two when needed.

.6Bounds: the allowed sizes

Absolute value bars give size without direction, like distance from home: |−3| and |3| both equal 3. Sine and cosine have size at most 1. Their defined reciprocals, cosecant and secant, have size at least 1.

|sin θ| ≤ 1 and |cos θ| ≤ 1.
Both squares are nonnegative and total 1, so neither square can exceed 1. A value with size greater than 1 would have a square greater than 1.
Where defined, |csc θ| ≥ 1 and |sec θ| ≥ 1. Flipping a nonzero size at most 1 gives size at least 1.
Equivalently, −1 ≤ sin θ ≤ 1 and −1 ≤ cos θ ≤ 1.

Sine and cosine remain between −1 and 1, including the endpoints.

The same idea, five ways

Say it

Sine and cosine lie between negative one and one, inclusive.

Write it

Sine and cosine have absolute value at most one; defined secant and cosecant have absolute value at least one.

In math

|sin θ| ≤ 1
|cos θ| ≤ 1
−1 ≤ sin θ ≤ 1
−1 ≤ cos θ ≤ 1
|csc θ| ≥ 1 where defined
|sec θ| ≥ 1 where defined

The height reached by a ladder cannot exceed the ladder's length.

See it

Sine and cosine remain between −1 and 1, including the endpoints.

Worked exampleReject impossible values

Could sin θ = $6 over 5$ ? Could sec θ = $3 over 5$ ? In words, check each proposed output against the identity and reciprocal definitions.

Sine and cosine remain between −1 and 1, including the endpoints.

$\sin^{2}$ θ would be $36 over 25$ > 1.A single square already exceeds the total 1, leaving no room for the other nonnegative square.
$\cos^{2}$ θ would be 1 − $36 over 25$ = − $11 over 25$ , which is impossible.No real square is negative.
sec θ = $3 over 5$ would give cos θ = $5 over 3$ , whose square $25 over 9$ also exceeds 1.Taking this nonzero reciprocal is valid, but its cosine violates the bound.

Answer

sin θ = $6 over 5$ is impossible.
sec θ = $3 over 5$ is impossible.

Check A valid comparison is cos θ = , sec θ = . The cosine has size below 1 and its reciprocal size above 1.

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

✗ Not this: All six functions must have size at most 1.

The bound concerns leg lengths after the hypotenuse is scaled to 1, sine and cosine; their reciprocals can be larger.

✓ Instead: Sine and cosine have size at most 1; defined sec and csc have size at least 1. Tan and cot have no such bound.

Tips and tricks

A bound checks size, not sign.

Strategy: step by step

Read squared notation first: $\sin^{2}$ θ = (sin θ $) squared$ . Square the trig value, not the angle.
If the given value is csc or sec, flip it to obtain sin or cos. If it is cot, its reciprocal gives tan only when cot is nonzero and tan exists.
If sin or cos is known, substitute into $\sin^{2}$ θ + $\cos^{2}$ θ = 1. Subtract its square from both sides to isolate the missing square.
Take the nonnegative square root to find the size. For an acute angle the missing sine or cosine is positive, so choose that sign. $square root of 0$ is 0, not positive.
If tan is known, use $\tan^{2}$ θ + 1 = $\sec^{2}$ θ, choose positive sec for an acute angle, then flip for cos. Multiply tan by cos for sin.
Use tan = sin ÷ cos and cot = cos ÷ sin when their denominators are nonzero. Flip cosine for secant and sine for cosecant.
Reduce and rationalize. Check that the squares add to 1 and each reciprocal pair multiplies to 1.

Strategy

Recover values from one acute-angle function

Is the given function sec or csc?

YesFlip it to get cos or sin.

NoKeep sine, cosine or tangent; check nonzero cot before flipping it.

↓

Do you know sine or cosine?

YesSquare it, subtract from 1, take the root and select the positive acute-angle value.

NoWith tangent, get positive sec from = + 1, then cos = 1 ÷ sec and sin = tan × cos.

↓

Does the next division have zero on the bottom?

YesThat expression is undefined. For cot, check cos ÷ sin directly; undefined tan does not force undefined cot.

NoDivide or flip, reduce and rationalize.

↓

Is the angle acute?

YesEvery triangle ratio must be positive.

NoUse the later quadrant information to choose a missing sine or cosine sign.

Read squared notation first: $\sin^{2}$ θ = (sin θ $) squared$ . Square the trig value, not the angle.
If the given value is csc or sec, flip it to obtain sin or cos. If it is cot, its reciprocal gives tan only when cot is nonzero and tan exists.
If sin or cos is known, substitute into $\sin^{2}$ θ + $\cos^{2}$ θ = 1. Subtract its square from both sides to isolate the missing square.
Take the nonnegative square root to find the size. For an acute angle the missing sine or cosine is positive, so choose that sign. $square root of 0$ is 0, not positive.
If tan is known, use $\tan^{2}$ θ + 1 = $\sec^{2}$ θ, choose positive sec for an acute angle, then flip for cos. Multiply tan by cos for sin.
Use tan = sin ÷ cos and cot = cos ÷ sin when their denominators are nonzero. Flip cosine for secant and sine for cosecant.
Reduce and rationalize. Check that the squares add to 1 and each reciprocal pair multiplies to 1.

Worked exampleRecover five functions from an acute cosecant

In a right triangle, csc θ = 7. Find the other five functions. In words, you know the hypotenuse divided by the opposite leg; recover the remaining ratios.

The given cosecant fixes hypotenuse over opposite as seven over one.

Reference table: read the reason entry directly under its identity label.

sin θ = 1 ÷ 7 = $1 over 7$ .Cosecant and sine are reciprocals.
$\sin^{2}$ θ + $\cos^{2}$ θ = 1 becomes $1 over 49$ + $\cos^{2}$ θ = 1. Subtract $1 over 49$ from both sides: $\cos^{2}$ θ = $49 over 49$ − $1 over 49$ = $48 over 49$ .This isolates the unknown cosine square. Subtracting the same fraction from both sides keeps the equation balanced.
cos θ = $square root of 48 over 49$ = $square root of 48 over 7$ = $4 square root of 3 over 7$ . $square root of 48$ = $square root of 16 × 3$ = 4 $square root of 3$ . Cosine is positive for an acute angle.
tan θ = $1 over 7$ ÷ $4 square root of 3 over 7$ = $1 over 7$ × $7 over 4 square root of 3$ = $1 over 4 square root of 3$ = $square root of 3 over 12$ .The 7s cancel; multiplying top and bottom by $square root of 3$ gives bottom 4 × 3 = 12.
cot θ = 1 ÷ $1 over 4 square root of 3$ = 4 $square root of 3$ ; sec θ = 1 ÷ $4 square root of 3 over 7$ = $7 over 4 square root of 3$ = $7 square root of 3 over 12$ .Both known values are nonzero, so their reciprocals exist. For secant multiply top and bottom by $square root of 3$ ; the bottom becomes 4 × 3 = 12.

Answer

sin θ = $1 over 7$
cos θ = $4 square root of 3 over 7$
tan θ = $square root of 3 over 12$
cot θ = 4 $square root of 3$
sec θ = $7 square root of 3 over 12$

Check A triangle with opp = 1 and hyp = 7 has adj = = 4. Its cosine is , independently confirming the identity method. The square check is + = 1.

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

Rung 1Rung 1: start from cosine

θ is acute and cos θ = $8 over 17$ . Find the other five functions. In words, recover sine's square first.

Name the sides from the marked angle before making a fraction.

$\sin^{2}$ θ = 1 − $64 over 289$ = $289 over 289$ − $64 over 289$ = $225 over 289$ .Subtract cosine squared from both sides.
sin θ = $square root of 225 over 289$ = $15 over 17$ .The top and bottom roots are 15 and 17, and the acute sine is positive.
tan θ = $15 over 17$ ÷ $8 over 17$ = $15 over 8$ ; cot θ = $8 over 15$ .Divide in each order.
sec θ = $17 over 8$ ; csc θ = $17 over 15$ .Flip each nonzero partner.

Answer

sin θ = $15 over 17$
tan θ = $15 over 8$
cot θ = $8 over 15$
sec θ = $17 over 8$
csc θ = $17 over 15$

Check + = 1. A triangle with adj = 8 and hyp = 17 has opp = 15.

Rung 2Rung 2: the given value is a reciprocal

θ is acute and sec θ = $5 over 4$ . Find the other five functions. In words, obtain cosine first.

Name the sides from the marked angle before making a fraction.

cos θ = 1 ÷ $5 over 4$ = $4 over 5$ .Secant is the nonzero reciprocal of cosine.
$\sin^{2}$ θ = 1 − $16 over 25$ = $9 over 25$ , so sin θ = $3 over 5$ .Subtract the cosine square, then take the positive acute sine.
tan θ = $3 over 5$ ÷ $4 over 5$ = $3 over 4$ , cot θ = $4 over 3$ .Divide the outputs in the required orders.
csc θ = $5 over 3$ .Flip sine; secant was already given.

Answer

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

Check × = 1, and + = 1.

Rung 3Rung 3: a partner with a root

θ is acute and cos θ = $1 over 3$ . Find the other five functions. In words, recover a sine with a root and simplify its reciprocal.

Name the sides from the marked angle before making a fraction.

$\sin^{2}$ θ = 1 − $1 over 9$ = $8 over 9$ , so sin θ = $square root of 8 over 3$ = $2 square root of 2 over 3$ .Subtract the cosine square and take the positive root; $square root of 8$ = $square root of 4 × 2$ .
tan θ = $2 square root of 2 over 3$ ÷ $1 over 3$ = 2 $square root of 2$ .The equal bottoms cancel during division.
cot θ = $1 over 2 square root of 2$ = $square root of 2 over 4$ .Flip tangent and rationalize: the bottom becomes 2 × 2 = 4.
sec θ = 3; csc θ = $3 over 2 square root of 2$ = $3 square root of 2 over 4$ .Flip cosine and sine, rationalizing cosecant.

Answer

sin θ = $2 square root of 2 over 3$
tan θ = 2 $square root of 2$
cot θ = $square root of 2 over 4$
sec θ = 3
csc θ = $3 square root of 2 over 4$

Check + = 1, and 3 × = 1 checks secant.

Rung 4Rung 4: two rationalizations

θ is acute and sin θ = $2 over 7$ . Find the other five functions. In words, keep the root manipulations visible after finding cosine.

Name the sides from the marked angle before making a fraction.

$\cos^{2}$ θ = 1 − $4 over 49$ = $49 over 49$ − $4 over 49$ = $45 over 49$ .Subtract the known sine square from both sides.
cos θ = $square root of 45 over 7$ = $3 square root of 5 over 7$ .The acute cosine is positive, and $square root of 45$ = $square root of 9 × 5$ .
tan θ = $2 over 7$ ÷ $3 square root of 5 over 7$ = $2 over 3 square root of 5$ = $2 square root of 5 over 15$ .Cancel 7 during division, then multiply top and bottom by $square root of 5$ .
cot θ = $3 square root of 5 over 2$ .Reverse the quotient; the root is already on top.
sec θ = $7 over 3 square root of 5$ = $7 square root of 5 over 15$ , csc θ = $7 over 2$ .Flip cosine and sine, rationalizing the root denominator.

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 + = 1. With opp = 2 and hyp = 7, the Pythagorean theorem gives adj = 3, confirming cosine independently.

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

✗ Not this: θ means sin().

The superscript squares the function output, not the input angle.

✓ Instead: θ = (sin θ. If sin θ = , then θ = .

✗ Not this: sin θ + cos θ = 1.

The identity adds squares, not the original values. + = .

✓ Instead: ( + ( = + = 1.

✗ Not this: cot is undefined whenever tan is undefined.

If sin θ = 1 and cos θ = 0, then tan = 1 ÷ 0 is undefined but cot = 0 ÷ 1 = 0.

✓ Instead: Use cot = cos ÷ sin for sin ≠ 0; use cot = 1 ÷ tan only where both expressions exist.

✗ Not this: A principal square root is always positive.

The principal root is nonnegative, so it includes = 0.

✓ Instead: = 3, but the equation = 9 has solutions u = ±3.

✗ Not this: A sine of 1.4 is possible because tangent may exceed 1.

= 1.96 alone exceeds the total 1 allowed for the two nonnegative squares.

✓ Instead: Sine and cosine have size at most 1; tangent is not subject to that bound.

✗ Not this: An acute angle has csc γ = .

Cosecant is hypotenuse over opposite. The hypotenuse is longer, so an acute cosecant is greater than 1; is less than 1.

✓ Instead: There is no such acute angle. The worked example uses csc γ = .

Tips and tricks

Know $\sin^{2}$ θ + $\cos^{2}$ θ = 1 cold. Rebuild the two divided identities instead of memorizing three unrelated equations.
Remember the flip pairs: sin with csc, cos with sec, tan with cot. The reciprocal shortcut needs both expressions to exist.
Write the missing square before taking a root; the equation step and sign step do different jobs.
Sine or cosine outside −1 to 1 signals an error. Defined secant or cosecant with size below 1 signals an error too.
For these acute-angle questions all six values are positive. In later problems a quadrant clue chooses the sign.
Read reference tables by columns: keep the input label and its output in the same vertical column.

Trap. Reading θ as sin(). θ means (sin θ: find sine first, then square its value. Also, θ + θ = 1 relates squares; sin θ + cos θ is usually not 1. With sine and cosine , the ordinary sum is .

Keep in mind

$\sin^{2}$ θ means (sin θ $) squared$ , the sine squared, not the sine of $θ^{2}$ : if sin θ = $2 over 5$ , then $\sin^{2}$ θ = $4 over 25$ .
The squares add to 1, but the plain values usually do not: $3 over 5$ + $4 over 5$ = $7 over 5$ , while $9 over 25$ + $16 over 25$ = 1.
An identity with a fraction needs a nonzero bottom: tan θ = $\frac{\sin θ}{\cos θ}$ has no value when cos θ = 0, as for a unit ladder standing straight up, where it would be $1 over 0$ .
One line on your cheat sheet, $\sin^{2}$ θ + $\cos^{2}$ θ = 1, rebuilds the other two in seconds: dividing by $\cos^{2}$ θ gives $\tan^{2}$ θ + 1 = $\sec^{2}$ θ.

Memory hookUnit ladder: heigh + floo = is θ + θ = 1. Divide by θ for the tan and sec one, by θ for the cot and csc one.

Flash cards: say the answer out loud, then flip

What is an identity?

An equation that is true for every allowed input.

Write the three reciprocal identities.

csc θ = $1 over sin θ$
sec θ = $1 over cos θ$
cot θ = $1 over tan θ$

How do you get θ + 1 = θ from θ + θ = 1?

Divide every term by θ, where cos θ ≠ 0.

Is sin θ + cos θ = 1 for every angle?

No. The squares add to 1. At 45°, sin θ + cos θ = ≈ 1.414.

The fundamental identities

.1Read sin2sin2θ and cos2cos2θ

.2Reciprocal identities

.3Quotient identities

.4Pythagorean identity

.5Rebuild the other two Pythagorean identities

.6Bounds: the allowed sizes

Recover values from one acute-angle function

.1Read $\sin^{2}$ θ and $\cos^{2}$ θ