5. Cofunctions: change the partner and subtract the angle

Explain it like I am five

Picture a ladder leaning against a wall. With the floor it makes a right triangle, a triangle with one square corner of 90°. The two pointed corners share the other 90°: if the floor corner is 35°, the top corner is 90° − 35° = 55°.

Stand at the floor corner. The wall height is opposite you, so sin 35° = $\frac{w a l l h e i g h t}{l a d d e r l e n g t h}$ , by SOH: opposite over hypotenuse, the ladder. Now stand at the top corner. The same wall height is adjacent, next to you, so cos 55° = $\frac{w a l l h e i g h t}{l a d d e r l e n g t h}$ , by CAH. Same fraction, so sin 35° = cos 55°; a calculator gives about 0.5736 for both. This swap gives the cofunction pairs: sine and cosine, tangent and cotangent, secant and cosecant. The co means complement: the two angles add to 90°.

The rule: switch to the partner and replace the angle x by $π over 2$ − x in radians, or 90° − x in degrees. Example: tan( $π over 7$ ) = cot( $π over 2$ − $π over 7$ ). With a common bottom, $7 π over 14$ − $2 π over 14$ = $5 π over 14$ , so tan( $π over 7$ ) = cot( $5 π over 14$ ).

In plain words

Imagine looking at the same ladder from its two pointed corners. From the floor corner, the wall is across from you. From the wall corner, the floor is across from you. The corners add to 90°, and the two legs trade jobs. That trade creates Cofunctions, three pairs of trig functions that give the same value at matching angles. Sine pairs with cosine, tangent with cotangent, and secant with cosecant. To change partners, subtract the old angle from a quarter turn. Use 90° for degrees or $π over 2$ for radians. The result may be negative; the identity works wherever both functions are defined.

sin x = cos( − x); cos x = sin( − x)

tan x = cot( − x); cot x = tan( − x)

sec x = csc( − x); csc x = sec( − x)

Use 90° in degrees. Both sides must be defined.

Each pair uses a quarter turn minus the old argument.

Reminder

Degrees and radians. 90° and $π over 2$ name the same quarter turn, but each subtraction needs one unit: $π over 2$ − $π over 6$ uses radians.
Common denominators. $π over 2$ = $3 π over 6$ and $2 π over 3$ = $4 π over 6$ , giving difference − $π over 6$ .
Signed subtraction. 90 − 110 = −20. Subtracting a negative reverses that move: 90 − (−20) = 110.
SOH CAH TOA. Sine is opposite over hypotenuse, cosine is adjacent over hypotenuse, and tangent is opposite over adjacent. sin 30° = $1 over 2$ .
Quotients and reciprocals. tan θ = $\frac{\sin θ}{\cos θ}$ , cot θ = $\frac{\cos θ}{\sin θ}$ , sec θ = $1 over cos θ$ , and csc θ = $1 over sin θ$ . At 45°, sec and csc both equal $square root of 2$ .
Rationalizing. $1 over square root of 3$ = $square root of 3 over 3$ after multiplying top and bottom by $square root of 3$ .
Quadrant signs. In Quadrant II sine is positive and cosine negative, so cotangent is negative; $−1 over 0$ at an axis is undefined.

Why it works. A triangle's angles total 180°. The right angle uses 90°, leaving 90° for the two acute angles. Changing corners exchanges opposite and adjacent while keeping the hypotenuse. Thus opposite over hypotenuse at one corner equals adjacent over hypotenuse at the other. The other pairs follow from the same exchange. Beyond triangle angles, the unit circle gives the same exchange: the point at − x has coordinates (sin x, cos x). Swapping those coordinates produces the identities. A zero denominator still has no value, so an identity cannot turn an undefined expression into a number.

RuleRule: Wherever both sides are defined: sin x = cos( − x), cos x = sin( − x), tan x = cot( − x), cot x = tan( − x), sec x = csc( − x), and csc x = sec( − x). For degree angles, replace with 90° throughout.

The same idea, five ways

Say it

Say: switch to the cofunction and subtract the old angle from a quarter turn.

Write it

A function of an angle equals its cofunction at the complementary argument, wherever both expressions have values.

In math

α + β = 90°
β = 90° − α
β = $π over 2$ − α
sin α = cos β
tan α = cot β
sec α = csc β

From the other ladder corner, across from and next to exchange jobs.

See it

The two arguments together fill one quarter turn.

The same idea, other ways

As two corners of one triangle

From the floor corner, sine is wall height over ladder length. From the wall corner, cosine is the same wall height over ladder length. The same fraction has two names at complementary angles.

The highlighted sides give sine from one acute corner and cosine from the other.

With small exact values

30° and 60° add to 90°. sin 30° = cos 60° = $1 over 2$ . tan 30° = cot 60° = $square root of 3 over 3$ . The same triangle sides supply each matching ratio.

Changing acute corners exchanges opposite and adjacent.

As a coordinate exchange

At angle x the unit circle point is (cos x, sin x). Reflect it across the diagonal where the two coordinates are equal. This changes angle x to $π over 2$ − x and swaps the coordinates. Its horizontal coordinate is now sin x and its vertical coordinate is now cos x. This argument includes negative and larger angles. Reflection across the diagonal sends the rightward axis to the upward axis and reverses counterclockwise turning to clockwise turning. Starting at $π over 2$ and turning clockwise by x therefore gives the reflected signed angle $π over 2$ − x. This also applies to negative or larger turns. The argument follows from the changed direction as well as the swapped coordinates.

At x: (cos x, sin x)

At − x: (sin x, cos x)

The coordinates exchange jobs.

The coordinate exchange extends the identities beyond acute triangle angles.

.1Sine and cosine

A right triangle's two acute corners are Complementary angles because their sum is 90°. A leg opposite one corner is adjacent to the other. Dividing the same leg by the unchanged hypotenuse gives sine from the first corner and cosine from the second. At general angles, the unit circle's coordinate exchange gives the same relation.

Rule: sin x = cos( $π over 2$ − x) and cos x = sin( $π over 2$ − x).
Rule: In degrees the new argument is 90° − x.
Rule: Sine and cosine exist at every real angle.

Changing corners swaps the legs while leaving the hypotenuse c.

Reminder

Complement arithmetic. 27° + β = 90° becomes β = 63° after subtracting 27°; 27° + 63° checks it.

The same idea, five ways

Say it

Say: sine becomes cosine at the quarter turn minus the old angle.

Write it

Sine and cosine exchange when their arguments total a quarter turn.

In math

sin α = cos β
cos α = sin β
α + β = $π over 2$

One leg is across from one corner and beside the other.

See it

The two arguments together fill one quarter turn.

Worked exampleChange cosine to sine in degrees

Rewrite cos 27° as sine of another angle. Find an equal partner expression, not a decimal.

The two arguments together fill one quarter turn.

Choose sine and set 27° + β = 90°.The partner argument must complete the quarter turn.
β = 90° − 27° = 63°.Subtract the old angle to find the new one.
Put it back: 27° + 63° = 90°.This verifies the argument.
cos 27° = sin 63°.The cofunction identity applies, and both expressions are defined.

Answer

cos 27° = sin 63°.

Check Reverse the identity: sin 63° = cos(90° − 63°) = cos 27°.

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

✗ Not this: Counterexample: sin 30° = cos 30°.

Keeping the old angle gives on one side and on the other.

✓ Instead: sin 30° = cos 60°.

Tips and tricks

Tip: switch both the function name and the argument.

.2Tangent and cotangent

Tangent compares opposite with adjacent. Cotangent reverses that comparison at the same corner. Moving to the other acute corner also swaps the legs' names. The two reversals cancel, making tangent at one corner equal to cotangent at the complementary corner. At general angles use the coordinate fractions and remember that a zero denominator gives an Undefined expression.

Rule: tan x = cot( $π over 2$ − x) and cot x = tan( $π over 2$ − x), wherever defined.
Rule: tan x = $\frac{\sin x}{\cos x}$ and cot x = $\frac{\cos x}{\sin x}$ .
Rule: Tangent needs nonzero cosine; cotangent needs nonzero sine.

The highlighted ratio is tangent at one corner and cotangent at the other.

Reminder

Division by zero. $0 over 5$ = 0, but $5 over 0$ is undefined. Check the bottom.

The same idea, five ways

Say it

Say: tangent pairs with cotangent at the complementary argument.

Write it

Changing corners and reversing the leg ratio gives the same value.

In math

tan α = cot β
cot α = tan β
α + β = $π over 2$

Swapping labels and reversing the comparison returns the original ratio.

See it

tan α = opposite ÷ adjacent

The other corner exchanges the legs.

cot β uses the original ratio.

Two reversals return the same leg ratio.

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

✗ Not this: Counterexample: cot(3π) = tan(−) is an equality of real values.

cot(3π) divides −1 by 0, and tan(−) also divides −1 by 0. Both are undefined.

✓ Instead: The arguments sum to , but neither expression has a number value.

Tips and tricks

Tip: at an axis angle, check the quotient denominator before writing a value.

.3Secant and cosecant

Secant is the reciprocal of cosine, and cosecant is the reciprocal of sine. A reciprocal divides 1 by the original number. Equal nonzero cosine and sine values have equal reciprocals. The triangle gives the same explanation: hypotenuse over adjacent at one corner becomes hypotenuse over opposite at the other. If the original value is zero, its reciprocal is undefined.

Rule: sec x = csc( $π over 2$ − x) and csc x = sec( $π over 2$ − x), wherever defined.
Rule: sec x = $1 over cos x$ and csc x = $1 over sin x$ .
Rule: Zero has no reciprocal.

Hypotenuse over b changes names when the corner changes.

Reminder

Reciprocals. 1 ÷ $1 over 2$ = 2, but zero has no reciprocal.

The same idea, five ways

Say it

Say: secant pairs with cosecant at the quarter turn minus the old angle.

Write it

Equal nonzero sine and cosine values have equal reciprocals.

In math

sec α = csc β
csc α = sec β
α + β = $π over 2$

Turn equal nonzero fractions upside down and they stay equal.

See it

cos α = sin β

sec α = csc β, when defined

Taking reciprocals preserves equality of nonzero values.

Worked exampleA negative cosecant partner argument

Rewrite csc( $3 π over 5$ ) as secant of another angle. A negative complementary argument is allowed.

−

= −

The partner argument is negative because the original argument exceeds a quarter turn.

Set $3 π over 5$ + β = $π over 2$ .This finds the missing secant argument.
β = $π over 2$ − $3 π over 5$ = $5 π over 10$ − $6 π over 10$ = − $π over 10$ .Common tenths allow subtraction, giving a negative result.
Put it back: $3 π over 5$ − $π over 10$ = $6 π over 10$ − $π over 10$ = $π over 2$ .This verifies the argument.
csc( $3 π over 5$ ) = sec(− $π over 10$ ).The first angle has nonzero sine in Quadrant II, and the second has nonzero cosine in Quadrant IV.

Answer

csc() = sec(−).

Check The underlying identity sin() = cos(−) equates two nonzero values. Their reciprocals therefore match.

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

✗ Not this: Counterexample: sec() = csc 0 = 0.

Both reciprocal denominators are zero, so neither expression is zero.

✓ Instead: Both are undefined. Numerical cofunction equalities require both sides to be defined.

Tips and tricks

Tip: check the underlying cosine for secant and sine for cosecant before taking a reciprocal.

Strategy: step by step

1. Choose the partner: sine with cosine, tangent with cotangent, or secant with cosecant.
2. Use 90° for degree angles and $π over 2$ for radian angles.
3. Subtract the old angle from that quarter turn. For radian fractions, make the denominators match first.
4. Write the partner at the resulting angle. Keep any negative result.
5. Add the old and new arguments to check their quarter turn sum. Check that both functions are defined.

Strategy

Strategy: fill a cofunction angle blank

Does the original angle have a degree sign?

YesSubtract from 90° and retain degree signs.

NoUse the radian quarter turn .

↓

Do the radian fractions have different denominators?

YesRewrite both over a common denominator before subtraction.

NoFor radians already over the same denominator, subtract numerators and keep that denominator. For degree measures, subtract the numbers directly and keep the degree sign.

↓

Is the old argument greater than a quarter turn?

YesKeep the negative new argument.

NoThe new argument is zero or positive.

↓

Does either expression divide by zero?

YesReport undefined, not a numerical equality.

NoWrite the equal cofunction expressions.

Choose the function's partner.
Subtract the old angle from the quarter turn in the same unit.
Keep the sign, check the sum, and check the denominator.

Worked exampleSecant to cosecant when the angle is past a quarter turn

The figure shows the angle $4 π over 5$ radians, which is 144°, in standard position. Use a cofunction identity to rewrite sec( $4 π over 5$ ) as the cosecant of a single angle in radians. Write that angle as a fraction of π in lowest terms. Show the subtraction with matching denominators. Then check that the old and new angles add to a quarter turn and that both functions are defined.

The angle radians (144°) drawn in standard position. Its terminal side lies in Quadrant II, already past a quarter turn, which is why − comes out negative.

Choose the partner. Secant pairs with cosecant, so the answer will have the form csc(new angle).The cofunction pairs are sine with cosine, tangent with cotangent, and secant with cosecant. The rule sec x = csc( $π over 2$ − x) turns a secant into a cosecant.
Use the quarter turn $π over 2$ : sec( $4 π over 5$ ) = csc( $π over 2$ − $4 π over 5$ ). $4 π over 5$ is a radian measure, so the quarter turn must be in radians too. The 144° in the figure is the same angle, but 90° − $4 π over 5$ would mix degrees with radians.
Make the denominators match: $π over 2$ = $5 π over 10$ and $4 π over 5$ = $8 π over 10$ .Fractions can be subtracted only over a common denominator. 10 is the least common multiple of 2 and 5, so multiply the top and bottom of $π over 2$ by 5, and the top and bottom of $4 π over 5$ by 2.
Subtract the old angle from the quarter turn: $π over 2$ − $4 π over 5$ = $5 π over 10$ − $8 π over 10$ = − $3 π over 10$ .The rule is quarter turn minus old angle, in that order. $8 π over 10$ is larger than $5 π over 10$ , and the figure shows the angle already past a quarter turn, so the difference is negative. Reversing the order would give $3 π over 10$ , which has the wrong sign. $3 π over 10$ is in lowest terms because 3 and 10 have no common factor.
Write the partner at the new angle and keep the sign: sec( $4 π over 5$ ) = csc(− $3 π over 10$ ).The partner is evaluated at exactly the result of the subtraction, minus sign included. Dropping the sign would change the value. sec( $4 π over 5$ ) is negative because cosine is negative in Quadrant II. csc( $3 π over 10$ ) is positive because $3 π over 10$ lies in Quadrant I, where sine is positive.
Check the quarter-turn sum: $4 π over 5$ + (− $3 π over 10$ ) = $8 π over 10$ − $3 π over 10$ = $5 π over 10$ = $π over 2$ .The old and new arguments of a cofunction pair always add to a quarter turn. Getting exactly $π over 2$ confirms both the subtraction and its sign.
Check that both functions are defined. Because $π over 2$ < $4 π over 5$ < π, the angle $4 π over 5$ ends inside Quadrant II, off the y-axis, so cos( $4 π over 5$ ) ≠ 0. Because − $π over 2$ < − $3 π over 10$ < 0, the angle − $3 π over 10$ ends inside Quadrant IV, off the x-axis, so sin(− $3 π over 10$ ) ≠ 0.The identity holds only where both sides are defined. sec x = $1 over cos x$ needs cos x ≠ 0, and csc x = $1 over sin x$ needs sin x ≠ 0. An angle whose terminal side lies strictly inside a quadrant has nonzero sine and nonzero cosine.

Answer

sec() = csc(−)

Check Sum: + (−) = − = = , a quarter turn. Defined: neither angle ends on an axis, so cos() ≠ 0 and sin(−) ≠ 0. Calculator in radian mode: cos() ≈ −0.8090, so sec() ≈ −1.2361. Also sin(−) ≈ −0.8090, so csc(−) ≈ −1.2361. The two sides agree, sign included.

Work to write

sec( $4 π over 5$ ) = csc( $π over 2$ − $4 π over 5$ )
$π over 2$ − $4 π over 5$ = $5 π over 10$ − $8 π over 10$ = − $3 π over 10$
sec( $4 π over 5$ ) = csc(− $3 π over 10$ )
Check: $4 π over 5$ + (− $3 π over 10$ ) = $8 π over 10$ − $3 π over 10$ = $5 π over 10$ = $π over 2$
cos( $4 π over 5$ ) ≠ 0 and sin(− $3 π over 10$ ) ≠ 0, so both sides are defined

sec( $4 π over 5$ ) = csc(− $3 π over 10$ )

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: degree subtraction

Rewrite sin 18° as cosine. Find the angle in the partner expression.

The two arguments together fill one quarter turn.

Choose cosine and set 18° + β = 90°.Degrees need a 90° quarter turn.
β = 90° − 18° = 72°.Subtracting isolates the new angle.
Put it back: 18° + 72° = 90°.The sum checks the complement.
sin 18° = cos 72°.Sine and cosine are the correct pair and both are defined.

Answer

sin 18° = cos 72°.

Check The reverse subtraction 90° − 72° = 18° returns the original argument.

Rung 2Rung 2: reciprocal functions in degrees

Rewrite sec 32° as cosecant. Find and check its argument.

The two arguments together fill one quarter turn.

Choose cosecant and set 32° + β = 90°.Secant pairs with cosecant.
β = 90° − 32° = 58°.This finds the partner angle.
Put it back: 32° + 58° = 90°.The sum verifies the argument.
sec 32° = csc 58°.Cosine at 32° and sine at 58° are equal and positive, so their reciprocals are defined.

Answer

sec 32° = csc 58°.

Check The reverse subtraction 90° − 58° = 32° recovers the secant argument.

Rung 3Rung 3: a radian common denominator

Rewrite cot( $π over 10$ ) as tangent. The partner argument must be in radians.

The two arguments together fill one quarter turn.

Choose tangent and set $π over 10$ + β = $π over 2$ .The arguments total a radian quarter turn.
β = $5 π over 10$ − $π over 10$ = $4 π over 10$ = $2 π over 5$ .Rewrite the quarter turn in tenths, subtract, and reduce by 2.
Put it back: $π over 10$ + $2 π over 5$ = $5 π over 10$ = $π over 2$ .This verifies the missing argument.
cot( $π over 10$ ) = tan( $2 π over 5$ ).Both acute arguments have nonzero denominators.

Answer

cot() = tan().

Check Reverse the subtraction: − = .

Rung 4Rung 4: unequal denominators and a larger numerator

Rewrite cos( $2 π over 5$ ) as sine. Carry the numerator 2 through the subtraction.

The two arguments together fill one quarter turn.

Choose sine and set $2 π over 5$ + β = $π over 2$ .This finds the complementary sine argument.
β = $π over 2$ − $2 π over 5$ = $5 π over 10$ − $4 π over 10$ = $π over 10$ .Both fractions must describe equal sized tenths before subtraction.
Put it back: $2 π over 5$ + $π over 10$ = $π over 2$ .The sum verifies the new argument.
cos( $2 π over 5$ ) = sin( $π over 10$ ).Sine and cosine are defined at every real angle.

Answer

cos() = sin().

Check The reverse subtraction − returns .

Rung 5Rung 5: a negative radian result

Rewrite cot( $7 π over 12$ ) as tangent. The old angle is greater than a quarter turn.

−

= −

Keep the negative partner argument.

Choose tangent and set $7 π over 12$ + β = $π over 2$ .The identity still uses a quarter turn beyond acute triangle angles.
β = $π over 2$ − $7 π over 12$ = $6 π over 12$ − $7 π over 12$ = − $π over 12$ .Six twelfths minus seven twelfths leaves a negative twelfth.
Put it back: $7 π over 12$ − $π over 12$ = $6 π over 12$ = $π over 2$ .The negative sign is necessary for the required sum.
cot( $7 π over 12$ ) = tan(− $π over 12$ ).The first angle has nonzero sine in Quadrant II, and the second has nonzero cosine in Quadrant IV.

Answer

cot() = tan(−).

Check Both functions are negative in their respective quadrants. Reversing the subtraction returns .

Rung 6Rung 6: a negative reciprocal partner

Rewrite csc 143° as secant. Explain why its negative argument is allowed.

90° − 143° = −53°

143° + (−53°) = 90°

csc 143° = sec(−53°)

A negative argument is valid even though it is not an acute corner.

Choose secant and set 143° + β = 90°.Cosecant pairs with secant at a quarter turn sum.
β = 90° − 143° = −53°.The old angle is 53° beyond the quarter turn.
Put it back: 143° + (−53°) = 90°.This verifies the negative partner argument.
csc 143° = sec(−53°).Sine in Quadrant II and cosine in Quadrant IV are positive and nonzero here, so both reciprocals exist.

Answer

csc 143° = sec(−53°).

Check 90° − (−53°) = 143° recovers the original argument. The underlying identity sin 143° = cos(−53°) confirms the reciprocals match.

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

✗ Not this: Counterexample: tan() = cot().

The values are and . At the same angle, cotangent is tangent's reciprocal, not usually its equal.

✓ Instead: tan() = cot().

✗ Not this: Counterexample: For csc 110°, subtract from 180° to get sec 70°.

180° is a half turn; cofunctions use a quarter turn.

✓ Instead: 90° − 110° = −20°, so csc 110° = sec(−20°).

Tips and tricks

Tip: remember the three pairs, then rebuild the angle by subtraction.
Tip: check that the two arguments add to a quarter turn instead of evaluating unfamiliar trig values.
Tip: keep the six identities on the cheat sheet; individual complement answers can be rebuilt.
Tip: a negative partner argument is expected when the original argument exceeds a quarter turn.

Trap. Changing the function name but keeping the old angle, or subtracting radians from 90°. Change name and argument, using one angle unit throughout.

Keep in mind

Change both the name and the angle: sin 35° = cos 55°, not cos 35°.
Stay in one unit: subtract degrees from 90° and radians from $π over 2$ , never $π over 7$ from 90°.
LHS and RHS mean the left-hand side and the right-hand side of the equals sign: in sin 35° = cos 55°, the LHS is sin 35° and the RHS is cos 55°.
A cofunction identity, an equation true for every allowed angle, holds only where both sides are defined, because a fraction with 0 on the bottom has no value: 6 ÷ 0 would need a number that times 0 gives 6, and none exists.

Memory hookCo means complement: a cofunction trades sine for cosine, tan for cot, sec for csc, and trades the angle x for 90° − x.

Flash cards: say the answer out loud, then flip

What are the three cofunction pairs?

sine and cosine
tangent and cotangent
secant and cosecant

What does the co in cofunction stand for?

Complement: the two angles add to 90°, or .

Fill the blank: sin 28° = cos ___

62°

Fill the blank: sec() = csc ___

, since − = −

Fill the blank: cos 125° = sin ___. Is a negative answer allowed?

−35°, from 90° − 125°
yes, the identity works for negative angles.