Defines the remaining four trigonometric functions as ratios and reciprocals of the cosine and sine. Establishes where each is undefined by asking which denominator vanishes, reads the tangent as the slope of the terminal side, and derives the two further Pythagorean identities by dividing the original one through.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 5 — Trigonometric Functions
§5.3 The Other Trigonometric Functions, pp. 674-692
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §5.3 The Other Trigonometric Functions §5.3, pp. 674-692 — the pages these objectives are drawn from
Warm-up
Six are named. The question is how many are independent.
Discussion prompt
If you knew the cosine and the sine of an angle, could you compute the other four?
Hint: What are the other four defined as?
Answer:
Yes, all four. The tangent is the sine over the cosine, the cotangent is the reverse, and the secant and cosecant are the reciprocals of the cosine and the sine.
So there are really only two independent functions. The other four are abbreviations for combinations that turn up often enough to deserve names.
That is worth knowing because it means there is nothing new to learn about their values, signs or special angles. Everything follows by arithmetic from §5.2, and only the undefined points are genuinely new behaviour.
Concept
The tangent, cotangent, secant and cosecant are all defined from the cosine and sine. Their values, signs and identities follow by arithmetic, and their undefined points come from zero denominators.
\[ \tan=\frac{\sin}{\cos}, \; \cot=\frac{\cos}{\sin}, \; \sec=\frac{1}{\cos}, \; \csc=\frac{1}{\sin} \]
The one thing worth memorising is the pairing, because it is counterintuitive: the secant is the reciprocal of the cosine, not of the sine. The prefix co- in the name does not match the co- in the reciprocal, and this trips people up more than anything else in the section.
Figure (svg): The six trigonometric functions arranged to show that four of them are defined as ratios or reciprocals of the cosine and sine
OpenStax, Precalculus, §5.3 The Other Trigonometric Functions §5.3, pp. 674-679
Section
Section 1
Concept
The tangent and cotangent are the two ratios of sine and cosine; the secant and cosecant are their reciprocals, paired crosswise.
The crosswise pairing of secant with cosine is a historical accident and is genuinely the section's main memorisation burden. The mnemonic that works is that the co- names pair with the non-co- ones: cosecant with sine, secant with cosine.
Figure (svg): The six trigonometric functions arranged to show that four of them are defined as ratios or reciprocals of the cosine and sine
OpenStax, Precalculus, §5.3 The Other Trigonometric Functions §5.3, pp. 674-680
Picture it
The left box is defined; the right box is arithmetic.
Figure (svg): The six trigonometric functions arranged to show that four of them are defined as ratios or reciprocals of the cosine and sine
The note at the bottom is the only thing here worth memorising. Everything else can be reconstructed from the two definitions on the left.
Worked example
Find the two coordinates, then do arithmetic.
\[ \text{Find all six functions at } 30^\circ. \]
Recall the coordinates
Why: From §5.2's special values.
\[ \cos \sqrt{3} / 2, \sin 1 / 2 \]
Compute the tangent
Why: Sine over cosine.
\[ 1 / \sqrt{3} = \sqrt{3} / 3 \]
Compute the reciprocals of those two
Why: Flip each.
\[ \cot \sqrt{3}, \csc 2 \]
Compute the secant
Why: Reciprocal of cosine.
\[ 2 / \sqrt{3} = 2 \sqrt{3} / 3 \]
Figure (svg): The six trigonometric functions arranged to show that four of them are defined as ratios or reciprocals of the cosine and sine
\[ \cos=\tfrac{\sqrt3}{2}, \sin=\tfrac12, \tan=\tfrac{\sqrt3}{3}, \cot=\sqrt3, \sec=\tfrac{2\sqrt3}{3}, \csc=2 \]
Verify: check the reciprocal pairs
Why: The tangent times the cotangent should be 1: root 3 over 3 times root 3 is 1 — correct. The cosine times the secant should be 1 as well, and the sine times the cosecant, which they are. Three reciprocal checks confirm all six values at once.
OpenStax, Precalculus, §5.3 The Other Trigonometric Functions §5.3, pp. 675-678
Matching
Two ratios and two reciprocals, paired crosswise.
Match the pairs
Why: The two ratios are reciprocals of each other, and the two reciprocal functions pair crosswise with the originals. The last two are the ones to fix in memory, since the naming actively misleads.
Worked example
Everything reduces to the two coordinates.
\[ \text{If } \sec\theta=3 \text{ and } \theta \text{ is in quadrant IV, find } \tan\theta. \]
Convert the secant to a cosine
Why: Take the reciprocal.
\[ \cos = \frac{1}{3} \]
Find the sine's size
Why: From the Pythagorean identity.
\[ \sin ^{2} = \frac{8}{9} \]
Apply the quadrant's sign
Why: Sine is negative in quadrant IV.
\[ \sin = -2 \sqrt{2} / 3 \]
Form the tangent
Why: Sine over cosine.
\[ \tan = -2 \sqrt{2} \]
Figure (svg): The solution to Worked example find one from another shown as a ladder of expressions, one row per legal move
\[ \tan\theta=-2\sqrt2 \]
Verify: check with the derived identity
Why: One plus the tangent squared should equal the secant squared: 1 plus 8 is 9, and 3 squared is 9 — agreeing. Using the derived identity as a check is worth doing, since it tests the whole chain rather than any single step.
OpenStax, Precalculus, §5.3 The Other Trigonometric Functions §5.3, pp. 678-680
Trap
\[ \sec\theta=\frac{1}{\sin\theta} \]
Pair the co- prefixes
Why: Secant has no co-, so it is paired with sine, which has none either.
The secant is defined as the reciprocal of the sine.
The pairing is crosswise. The secant is the reciprocal of the cosine and the cosecant is the reciprocal of the sine.
Check at 60 degrees: the cosine is one half, so the secant is 2. The sine is root 3 over 2, so the cosecant is 2 over root 3 — a different number.
The co- names pair with the non-co- ones. It is a historical accident rather than a pattern, and it is the one thing in this section that has to be memorised rather than derived.
Faded example
An angle has cosine 3 over 5 and sine 4 over 5.
Fill in the blanks
\tan = \frac43 = \frac___}___, \qquad \sec = \frac______}
Why: The fifths cancel in the tangent, leaving four thirds. The secant is the reciprocal of the cosine, so it is five thirds. Both are arithmetic on the two coordinates, with nothing new required.
Prediction
Six functions are named.
Predict first
How many independent quantities are there?
Correct: Two: everything follows from the cosine and sine.
Why: The other four are defined as ratios and reciprocals, so knowing the two coordinates determines all six. And even those two are linked by the Pythagorean identity, so given one and a quadrant the other follows — which means there is really only one independent quantity plus a sign.
Socratic
They add no information beyond the cosine and sine.
Discussion prompt
If they are just combinations, why do they have names?
Hint: Which combinations turn up most often in formulas?
Answer:
Because those particular combinations appear constantly, and a name is shorter than a ratio. The tangent especially: it is the slope of a line, which is one of the most-used quantities in the whole course.
The secant and cosecant earn their names in calculus and integration, where they appear in derivative formulas and in the identities that make certain integrals tractable.
Historically several of them were tabulated separately for navigation, before calculators made a division cheap. The names survive because the formulas that use them are cleaner, not because they add anything the cosine and sine could not express.
Section
Section 2
Concept
Each of the four derived functions has a denominator, and it is undefined wherever that denominator is zero — which happens at the quadrantal angles.
There is nothing special about trigonometry in these failures. A ratio with a zero denominator is undefined, and the only new content is knowing which angles make each coordinate vanish — which §5.2's quadrantal values already supply.
Figure (svg): The unit circle with the four quadrantal angles marked, showing which trigonometric functions are undefined at each because a denominator vanishes there
OpenStax, Precalculus, §5.3 The Other Trigonometric Functions §5.3, pp. 680-685
Picture it
Each undefined point corresponds to one coordinate being zero.
Figure (svg): The unit circle with the four quadrantal angles marked, showing which trigonometric functions are undefined at each because a denominator vanishes there
The two failures alternate around the circle, since the cosine and sine vanish at different quadrantal angles. That alternation is what §6.2's graphs will show as asymptotes.
Worked example
Set its denominator to zero.
\[ \text{Where is } \tan\theta \text{ undefined?} \]
Identify the denominator
Why: The tangent is sine over cosine.
Set it to zero
Why: Where the horizontal coordinate vanishes.
\[ \cos \theta = 0 \]
Locate those angles
Why: On the vertical axis.
\[ 90\text{ and } 270 ^\circ \]
Add full turns
Why: The functions are periodic.
\[ \text{every } 180 ^\circ \]
Figure (svg): The unit circle with the four quadrantal angles marked, showing which trigonometric functions are undefined at each because a denominator vanishes there
\[ \theta=90^\circ+180^\circ k \]
Verify: check with the slope reading
Why: The tangent is the slope of the terminal side, and at 90 and 270 degrees that side is vertical. A vertical line has no slope, as §2.1 established, so the tangent has no value — the two explanations agree and reinforce each other.
OpenStax, Precalculus, §5.3 The Other Trigonometric Functions §5.3, pp. 681-683
Sorting
Only a zero denominator causes a failure.
Sort into buckets
Sort each function.
Worked example
Ask which coordinate is zero there.
\[ \text{Which of the six are undefined at } 180^\circ? \]
Find the coordinates
Why: The point is one unit left.
\[ \cos - 1, \sin 0 \]
Check the sine's zero
Why: It is zero, so its reciprocal fails.
Check what else divides by sine
Why: The cotangent does.
Check the rest
Why: Cosine is negative 1, so nothing else fails.
Figure (svg): The solution to Worked example which functions fail at 180 degrees shown as a ladder of expressions, one row per legal move
\[ \cot 180^\circ \text{ and } \csc 180^\circ \text{ are undefined} \]
Verify: compute the four that are defined
Why: The cosine is negative 1, the sine is 0, the tangent is 0 over negative 1 which is 0, and the secant is 1 over negative 1 which is negative 1. All four are perfectly ordinary numbers, which shows that a zero sine only affects the functions that divide by it.
OpenStax, Precalculus, §5.3 The Other Trigonometric Functions §5.3, pp. 683-685
Error analysis
A student reasons about a quadrantal angle.
Annotate
On: \( \text{at } 90^\circ, \cos=0, \text{ so all six functions are undefined} \)
A zero value and an undefined value are different things. The cosine is zero at 90 degrees, which is a perfectly good number, and only its reciprocals suffer.
Prediction
The tangent is undefined at 90 degrees.
Predict first
Where else is it undefined?
Correct: Every 180 degrees from there.
Why: The cosine is zero at 90 and again at 270, which are 180 apart, and then repeats every full turn. So the tangent's failures come every half turn rather than every full one — which is why §6.2 will find that the tangent has period 180 degrees rather than 360.
Faded example
The cosecant is the reciprocal of the sine.
Fill in the blanks
\sin\theta = 0 \text180 0^\circ \text180 ___^\circ, \text___ \csc \text___ ___^\circ
Why: The sine vanishes where the point is on the horizontal axis, which is at 0 and 180 degrees. Those are 180 apart and the pattern repeats, so the cosecant is undefined at every multiple of 180 degrees.
Analogy
The undefined points are not new behaviour.
Match the pairs
Why: All three are one phenomenon: division by zero. Recognising the trigonometric case as familiar rather than new means the asymptotes in §6.2's graphs arrive as expected, and it explains why the tangent's failures coincide exactly with the vertical terminal sides.
Section
Section 3
Concept
The tangent is the sine over the cosine, which is the vertical coordinate over the horizontal one — exactly the slope of the terminal side.
The last point is worth noticing because it explains a feature that would otherwise be surprising. Angles differing by a half turn point in opposite directions but lie on the same line, so they have the same slope — which is why the tangent repeats twice as often as the sine and cosine.
Figure (svg): A terminal side drawn through the origin with its slope marked, showing that the tangent of the angle is the slope of that line
OpenStax, Precalculus, §5.3 The Other Trigonometric Functions §5.3, pp. 685-689
Picture it
Rise over run for the line through the origin.
Figure (svg): A terminal side drawn through the origin with its slope marked, showing that the tangent of the angle is the slope of that line
Reading the tangent as a slope explains its undefined points, its unbounded range and its shorter period, all from §2.1's facts about lines.
Worked example
The slope is the tangent, so the angle is the inverse tangent.
\[ \text{A line through the origin has slope } 1. \text{ What angle does it make with the horizontal?} \]
Recognise the slope as a tangent
Why: The terminal side's slope.
\[ \tan \theta = 1 \]
Find where the tangent is 1
Why: Sine equals cosine.
Identify the angle
Why: The diagonal is at 45 degrees.
\[ 45 ^\circ \]
Note the other solution
Why: A half turn later lies on the same line.
\[ \text{also } 225 ^\circ \]
Figure (svg): A terminal side drawn through the origin with its slope marked, showing that the tangent of the angle is the slope of that line
\[ \theta=45^\circ \text{ (or } 225^\circ) \]
Verify: check both give the same line
Why: At 45 degrees the point is up and to the right; at 225 it is down and to the left. Both lie on the same line through the origin, so both have slope 1 — which is exactly why the tangent repeats every half turn rather than every full one.
OpenStax, Precalculus, §5.3 The Other Trigonometric Functions §5.3, pp. 686-688
Prediction
The cosine approaches zero as the angle approaches 90 degrees.
Predict first
What happens to the tangent?
Correct: It grows without bound.
Why: The sine approaches 1 while the cosine approaches zero, so their ratio grows without limit. In slope terms, the terminal side becomes nearly vertical and a nearly vertical line has an enormous slope. This is §3.7's vertical asymptote behaviour, arriving in a trigonometric setting.
Worked example
A line can have any slope at all.
\[ \text{What values can } \tan\theta \text{ take?} \]
Recall the slope reading
Why: The tangent is the terminal side's slope.
Ask what slopes exist
Why: Any real number, for a non-vertical line.
Note the exception
Why: Vertical lines have no slope.
State the range
Why: Every real number is attained.
Figure (svg): The solution to Worked example explain the tangent's range shown as a ladder of expressions, one row per legal move
\[ \text{range of } \tan = (-\infty,\infty) \]
Verify: contrast with the sine and cosine
Why: Those two are coordinates on a circle of radius 1, so they are bounded between negative 1 and 1. The tangent is a ratio of them and is unbounded, because the denominator can be arbitrarily small. The slope reading makes this obvious rather than surprising.
OpenStax, Precalculus, §5.3 The Other Trigonometric Functions §5.3, pp. 688-689
Trap
\[ -1\le\tan\theta\le 1 \]
Apply the bound that holds for sine and cosine
Why: The tangent is built from them, so the same bound is assumed to carry over.
The tangent's values are taken to lie between negative 1 and 1.
The tangent is unbounded. It is a ratio, and dividing a small number by an even smaller one gives a large result.
At 89 degrees the tangent is about 57; at 89.9 degrees it is about 573. It grows without bound as the angle approaches 90.
A ratio of bounded quantities need not be bounded, since the denominator can approach zero. Reading the tangent as a slope makes this immediate, because lines can be arbitrarily steep.
Faded example
A terminal side passes through the point at (2, 6).
Fill in the blanks
\tan\theta = \frac23} = ___
Why: The tangent is the vertical coordinate over the horizontal one, which is the slope of the line through the origin and that point. Note this works for any point on the terminal side, not only the one on the unit circle, since scaling both coordinates leaves the ratio unchanged.
Sorting
Coordinates are bounded; ratios of them need not be.
Sort into buckets
Sort each function.
Explain it to yourself
The tangent repeats every half turn, not every full one.
Discussion prompt
Explain why, using the slope reading.
Hint: What happens to the terminal side after a half turn?
Answer:
After a half turn the terminal side points in the opposite direction — but it lies on the same line through the origin, just on the other side.
The tangent is that line's slope, and a line has one slope regardless of which direction along it you travel. So the tangent takes the same value.
The sine and cosine do change, because the point moves to the opposite side of the origin and both coordinates flip sign. But their ratio is unaffected, since both flips cancel — which is exactly why the tangent's period is half of theirs.
Section
Section 4
Concept
Dividing the Pythagorean identity by the cosine squared or by the sine squared produces two further identities involving the derived functions.
\[ 1+\tan^2\theta=\sec^2\theta, \qquad \cot^2\theta+1=\csc^2\theta \]
The domain restriction in the third point is easy to overlook and is genuinely there. The tangent and secant version says nothing at 90 degrees, because both sides are undefined — which is consistent, since the division that produced it was by zero at that angle.
Figure (svg): The two derived Pythagorean identities shown as consequences of dividing the original one by the cosine squared and by the sine squared
OpenStax, Precalculus, §5.3 The Other Trigonometric Functions §5.3, pp. 689-692
Picture it
Each derived version is one division away from the original.
Figure (svg): The two derived Pythagorean identities shown as consequences of dividing the original one by the cosine squared and by the sine squared
The notes record where each is valid. Both inherit a restriction from the division that produced them, which is why neither holds at every angle.
Worked example
Divide every term by the cosine squared.
\[ \text{Derive } 1+\tan^2\theta=\sec^2\theta. \]
Start from the original
Why: The unit circle's equation.
\[ \cos ^{2} + \sin ^{2} = 1 \]
Divide every term by cosine squared
Why: Valid where the cosine is nonzero.
\[ 1 + \sin ^{2} / \cos ^{2} = 1 / \cos ^{2} \]
Recognise the middle term
Why: It is the tangent squared.
\[ 1 + \tan ^{2} \]
Recognise the right term
Why: It is the secant squared.
\[ = \sec ^{2} \]
Figure (svg): The two derived Pythagorean identities shown as consequences of dividing the original one by the cosine squared and by the sine squared
\[ 1+\tan^2\theta=\sec^2\theta \]
Verify: test at a special angle
Why: At 45 degrees the tangent is 1 and the secant is root 2. One plus 1 is 2, and root 2 squared is 2 — agreeing. Testing at one special angle is a cheap check on any identity, and it catches a mis-derived version immediately.
OpenStax, Precalculus, §5.3 The Other Trigonometric Functions §5.3, pp. 690-691
Faded example
Divide the Pythagorean identity through by the sine squared.
Fill in the blanks
\frac1csc + ___ = \frac______ \;\Longrightarrow\; \cot^2 + 1 = ___^2
Why: The middle term becomes sine squared over sine squared, which is 1, and the right becomes the cosecant squared. Every term must be divided, and the middle one becoming 1 rather than staying is the step most often missed.
Worked example
The identity relates two functions directly.
\[ \text{If } \tan\theta=\tfrac{3}{4} \text{ and } \theta \text{ is in quadrant I, find } \sec\theta. \]
Apply the derived identity
Why: One plus the tangent squared.
\[ \sec ^{2} = 1 + \frac{9}{16} \]
Compute
Why: Sixteen sixteenths plus nine.
\[ \frac{25}{16} \]
Take the square root
Why: Both signs possible.
\[ \sec = +- \frac{5}{4} \]
Use the quadrant
Why: Cosine and therefore secant are positive in quadrant I.
\[ \sec = \frac{5}{4} \]
Figure (svg): The solution to Worked example use it to find a value shown as a ladder of expressions, one row per legal move
\[ \sec\theta=\tfrac{5}{4} \]
Verify: check by finding the coordinates
Why: A secant of five fourths means a cosine of four fifths, and a tangent of three quarters then gives a sine of three fifths. Those satisfy the original identity, since sixteen plus nine over twenty-five is 1. This is the 3-4-5 triangle, which is why the numbers are clean.
OpenStax, Precalculus, §5.3 The Other Trigonometric Functions §5.3, pp. 691-692
Error analysis
A student derives the second identity.
Annotate
On: \( \cos^2+\sin^2=1 \;\Longrightarrow\; \cot^2+\sin^2=\csc^2 \)
Every term must be divided, not just the convenient ones. Testing the result at a special angle is a two-line check that catches this kind of slip immediately.
Prediction
The identity relating tangent and secant was obtained by dividing by the cosine squared.
Predict first
Where does it fail to say anything?
Correct: Where the cosine is zero, since both sides are undefined.
Why: The derivation divided by the cosine squared, which is illegitimate where the cosine is zero. And at those angles both the tangent and the secant are undefined anyway, so the identity has nothing to say. The restriction is inherited from the division and is consistent with the functions' own domains.
Sorting
Match the functions involved to the right identity.
Sort into buckets
Sort each pair of functions.
Explain it to yourself
Two extra identities look like two more things to remember.
Discussion prompt
Explain how to recover both in a few seconds.
Hint: What do you start from and what do you do to it?
Answer:
Start from the Pythagorean identity, which is itself just the unit circle's equation. Then divide every term by either the cosine squared or the sine squared.
Dividing by the cosine squared turns the sine squared into a tangent squared and the 1 into a secant squared, giving the first. Dividing by the sine squared gives the second the same way.
So there is one identity and one operation, rather than three facts. That is worth the ten seconds of derivation, because a misremembered identity produces wrong answers that look plausible, while a derived one is right by construction.
Section
Section 5
Concept
Negating an angle reflects its point across the horizontal axis, so the cosine is unchanged and the sine changes sign. The derived functions inherit that behaviour.
The reflection is the whole argument. Negating an angle sweeps the same amount the other way, putting the point at the same horizontal position and the opposite vertical one — so anything built from the horizontal coordinate is unaffected and anything built from the vertical one flips.
Figure (svg): The unit circle with the four quadrantal angles marked, showing which trigonometric functions are undefined at each because a denominator vanishes there
OpenStax, Precalculus, §5.3 The Other Trigonometric Functions §5.3, pp. 682-690
Picture it
The quadrantal angles and the reflection structure they sit in.
Figure (svg): The unit circle with the four quadrantal angles marked, showing which trigonometric functions are undefined at each because a denominator vanishes there
A negated angle reflects across the horizontal axis, which preserves the horizontal coordinate and flips the vertical one — and that single fact determines all six parities.
Worked example
Trace the reflection through the definition.
\[ \text{Is the tangent even, odd, or neither?} \]
Recall the reflection
Why: Negating reflects across the horizontal axis.
Write the tangent at the negated angle
Why: Sine over cosine.
\[ (-\sin) / \cos \]
Factor out the sign
Why: The minus comes to the front.
\[ -(\sin / \cos) \]
Compare with the original
Why: It is the negative.
Figure (svg): The solution to Worked example determine a parity shown as a ladder of expressions, one row per legal move
\[ \tan(-\theta)=-\tan\theta \]
Verify: check at a special angle
Why: The tangent of 45 degrees is 1 and of negative 45 degrees is negative 1 — negated, as odd requires. In slope terms, reflecting a line across the horizontal axis reverses its slope, which is the same fact seen geometrically.
OpenStax, Precalculus, §5.3 The Other Trigonometric Functions §5.3, pp. 683-685
Sorting
The horizontal coordinate survives the reflection; the vertical one flips.
Sort into buckets
Sort each function.
Worked example
Six functions, all determined by the reflection.
\[ \text{Classify all six as even or odd.} \]
Start with the two coordinates
Why: Cosine unchanged, sine flipped.
Take reciprocals
Why: A reciprocal preserves parity.
Form the ratios
Why: One part flips in each.
Count
Why: One even pair and four odd.
Figure (svg): The solution to Worked example build the full table shown as a ladder of expressions, one row per legal move
\[ \cos,\sec \text{ even}; \; \sin,\csc,\tan,\cot \text{ odd} \]
Verify: notice the pattern
Why: Only the two built purely from the horizontal coordinate are even, since that is the coordinate the reflection preserves. Everything involving the vertical coordinate an odd number of times is odd. The tangent involves it once and is odd; a hypothetical function involving it twice would be even.
OpenStax, Precalculus, §5.3 The Other Trigonometric Functions §5.3, pp. 685-690
Trap
\[ \cos(-\theta)=-\cos\theta \]
Apply the sine's behaviour to the cosine
Why: Negating the input is taken to negate the output for both.
The cosine is treated as an odd function.
The cosine is even: negating the angle leaves it unchanged. The reflection across the horizontal axis preserves the horizontal coordinate.
Check at 60 degrees: the cosine is one half, and at negative 60 degrees it is also one half. The sine, by contrast, goes from root 3 over 2 to its negative.
The two coordinates behave differently under the reflection, so the functions built from them do too. Only the ones involving the vertical coordinate change sign.
Faded example
The sine of 40 degrees is about 0.643.
Fill in the blanks
\sin(-40^\circ) = -0.643, \qquad \cos(-40^\circ) = +\cos(40^\circ)
Why: The sine is odd, so negating the angle negates the value. The cosine is even, so it is unchanged. Both follow from the reflection across the horizontal axis, which flips the vertical coordinate and preserves the horizontal one.
Prediction
An angle is replaced by its negative.
Predict first
What happens to its point on the unit circle?
Correct: It reflects across the horizontal axis.
Why: Sweeping the same amount in the opposite direction lands the same distance the other side of the initial side, which is a reflection in the horizontal axis. That preserves the horizontal coordinate and negates the vertical one, determining every function's parity at once.
Explain it
Six parities look like six facts.
Discussion prompt
Explain to a classmate how to work out any of them in five seconds.
Hint: What does negating the angle do to the picture?
Answer:
Negating the angle reflects the point across the horizontal axis. So the horizontal coordinate stays and the vertical one flips sign — that is the only fact needed.
Then ask how many vertical coordinates the function involves. Cosine and secant involve none, so they are unchanged and therefore even. Sine, cosecant, tangent and cotangent each involve one, so each picks up a single sign flip and is odd.
A good explanation stresses that this is one picture rather than six memorised entries, and that it also matches §1.5's definitions: an even function is symmetric in the vertical axis, which the cosine's graph will visibly be in §6.1.
Comparison
Fill the blanks from memory. Everything after the first two rows is arithmetic.
Comparison matrix
| definition | undefined where | parity | |
|---|---|---|---|
| sine | vertical coordinate | never | odd |
| cosine | horizontal coordinate | never | even |
| tangent | sine over cosine | cosine is zero | odd |
| secant | one over cosine | cosine is zero | even |
| cosecant | one over sine | sine is zero | odd |
The third column is entirely determined by the second: a function fails exactly where its denominator vanishes, which is §3.7's rule in a new setting.
Pattern
Five steps, and the first two are §5.2's work.
Step 3 is worth making explicit rather than discovering by dividing. Noticing a zero coordinate before forming the ratio prevents writing down a value where none exists.
OpenStax Algebra and Trigonometry 2e, §7.4 The Other Trigonometric Functions §7.4
Check
The pairing is crosswise.
Check your understanding
What is the secant the reciprocal of?
Answer: A
Why: The secant is one over the cosine, and the cosecant is one over the sine. The naming pairs the co- word with the non-co- one, which is counterintuitive and is the one thing in this section worth memorising.
Check
Ask which denominator vanishes.
Check your understanding
Where is the cotangent undefined?
Answer: A
Why: The cotangent is the cosine over the sine, so it fails where the sine vanishes — at 0 and 180 degrees and every 180 degrees from there. A ratio is undefined exactly where its denominator is zero.
Check
One division from the original.
Check your understanding
Which identity follows from dividing the Pythagorean identity by the cosine squared?
Answer: A
Why: Dividing every term by the cosine squared turns the cosine squared into 1, the sine squared into the tangent squared, and the 1 into the secant squared. Testing at 45 degrees confirms: 1 plus 1 is 2, and root 2 squared is 2.
Real world
The tangent's slope reading is what makes it the function of choice for gradients and angles of elevation.
Discussion prompt
A road sign says a gradient of 12 percent. What angle is that, and which function converts between them?
Hint: What does a 12 percent gradient mean as a rise over a run?
Answer:
A 12 percent gradient means a rise of 12 for every 100 horizontally, so the slope is 0.12. That slope is the tangent of the angle the road makes with the horizontal.
So the angle is the inverse tangent of 0.12, which is about 6.8 degrees. Steeper than it sounds to walk and shallower than most people picture from the number.
The tangent is the natural function here precisely because it is the slope, by this section's reading. Any question relating an angle to a rise over a run — roof pitches, ramp regulations, camera angles — is a tangent question for the same reason.
Commit first
State your confidence along with your answer.
Predict first
Why is the tangent undefined at 90 degrees?
Correct: Because the cosine is zero there, so the ratio has a zero denominator.
Why: The tangent is the sine over the cosine, and at 90 degrees the point is at the top of the circle with a horizontal coordinate of zero. Equivalently, the terminal side is vertical and a vertical line has no slope. The tangent is unbounded rather than bounded, so the last option is doubly wrong.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate why this section introduces almost nothing new despite naming four functions.
Hint: What are the four defined in terms of?
Answer:
All four are defined from the cosine and sine — two ratios and two reciprocals. So every value, every sign and every special-angle entry follows by arithmetic from §5.2, with nothing new to look up.
The only genuinely new behaviour is the undefined points, and even those are familiar: a ratio fails where its denominator is zero, exactly as in Chapter 3's rational functions.
A good explanation adds the one thing that does need memorising: the secant pairs with the cosine, not with the sine. That crosswise naming is the section's only arbitrary fact, and everything else can be derived on the spot.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The first contains the section's only arbitrary fact and is worth fixing firmly. The third is the reading that explains the tangent's asymptotes, its unbounded range and its shorter period all at once, so it repays the time.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Write the two defined functions and the four derived ones with their definitions, marking clearly which reciprocal pairs with which. Beside each derived function, note where it is undefined and why. Then draw a terminal side with its slope triangle, labelling the rise and run as the sine and cosine, and write the three Pythagorean identities with the division that produces each derived one.
If your undefined column is filled in from the denominators rather than from memory, the section's only genuinely new behaviour has been understood rather than learned.
Recap
Five things, and only one of them was new information.
| if you remember one thing | it should be this |
|---|---|
| about the definitions | only two functions are defined; four are arithmetic |
| about the naming | secant pairs with cosine, crosswise, and that is arbitrary |
| about undefined points | a ratio fails where its denominator vanishes, as always |
| about the tangent | it is the slope of the terminal side |
Section 5.4 closes the chapter by connecting all six back to right triangles, where the ratios first came from and where most applications still live.
OpenStax, Precalculus, §5.3 The Other Trigonometric Functions §5.3, pp. 674-692 — everything on these slides traces back here
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