All six circular functions lifted off the unit circle: expressed as ratios of x, y and r for any point on the terminal side, and as ratios of the three sides of a right triangle of any size. Covers reconstructing a point from a single function value plus a quadrant, the angle of inclination and the classic one- and two-sighting height problems, and closes with the domains and ranges of all six functions including the extended interval notation the book uses to write them.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 10 — Foundations of Trigonometry
§10.3 The Six Circular Functions and Fundamental Identities, pp. 752-759
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 752-759 — the pages these objectives are drawn from
Warm-up
You can already find the cosine and sine from a point on the terminal side. Four more functions are waiting, and they need no new machinery.
Discussion prompt
A point on the terminal side is three comma negative four, so the radius is 5, the cosine is three fifths and the sine is negative four fifths. Without looking anything up, write down the tangent and the secant.
Hint: Use the quotient and reciprocal identities, or go straight back to the definitions in terms of x, y and r.
Answer:
\[ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{-4/5}{3/5} = -\frac{4}{3} = \frac{y}{x} \]
\[ \sec(\theta) = \frac{1}{\cos(\theta)} = \frac{5}{3} = \frac{r}{x} \]
Both routes agree, and the second is the point of this lesson: the tangent is just y over x and the secant is just r over x, with no reference to the unit circle at all. The fifths cancelled because they were never needed.
Concept
A point on the terminal side gives you three numbers: its two coordinates and its distance from the origin. Every one of the six circular functions is a ratio of two of those three, and which ratio it is you can read straight off the unit-circle definition.
Theorem 10.9 — If the point with coordinates x and y lies on the terminal side of an angle at distance r from the origin, then each of the six circular functions of that angle is the corresponding ratio of x, y and r.
\[ \cos = \tfrac{x}{r}, \; \sin = \tfrac{y}{r}, \; \sec = \tfrac{r}{x}, \; \csc = \tfrac{r}{y}, \; \tan = \tfrac{y}{x}, \; \cot = \tfrac{x}{y} \]
Because r is a distance it is always positive, so every sign in every one of the six values comes from the signs of the coordinates. That single fact removes all the sign bookkeeping from these problems.
Figure (svg): A point Q with coordinates x and y on the terminal side of an angle, at distance r from the origin, with all six circular functions written as ratios of x, y and r
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 752-753
Section
Section 1
Concept
Given a point on the terminal side, compute the radius once and then every function is a single division. Nothing needs to be looked up and the angle itself is never identified.
Notice that the six values come in three reciprocal pairs, so in practice there are only three divisions to do and three reciprocals to write down.
Figure (svg): A point Q with coordinates x and y on the terminal side of an angle, at distance r from the origin, with all six circular functions written as ratios of x, y and r
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 753-753
Picture it
The three pairs behave completely differently, and the reason is visible in the formulas.
Figure (svg): Two columns pairing each circular function with its range, showing that cosine and sine are bounded, secant and cosecant avoid the middle, and tangent and cotangent are unbounded
Cosine and sine divide by the largest of the three lengths, so they come out small. Secant and cosecant divide by one of the smaller ones, so they come out large. Tangent and cotangent divide one leg by the other, and a leg can be arbitrarily small compared with the other.
Worked example
Example 10.3.4, part 1. One radius, then six divisions.
\[ \text{The terminal side of } \theta \text{ contains } Q(3, -4). \text{ Find all six circular functions.} \]
Compute the radius
Why: Nine plus sixteen is twenty-five, whose root is five.
\[ r = 5 \]
Write the two with r downstairs
Why: Cosine is x over r and sine is y over r.
\[ \cos = \frac{3}{5}, \sin = -\frac{4}{5} \]
Reciprocate for secant and cosecant
Why: Secant is r over x and cosecant is r over y.
\[ \sec = \frac{5}{3}, \csc = -\frac{5}{4} \]
Divide the coordinates for tangent and cotangent
Why: Tangent is y over x and cotangent is x over y.
\[ \tan = -\frac{4}{3}, \cot = -\frac{3}{4} \]
Figure (svg): The solution to Worked example all six from the point 3, -4 shown as a ladder of expressions, one row per legal move
\[ \cos = \tfrac{3}{5}, \; \sin = -\tfrac{4}{5}, \; \sec = \tfrac{5}{3}, \; \csc = -\tfrac{5}{4}, \; \tan = -\tfrac{4}{3}, \; \cot = -\tfrac{3}{4} \]
Verify: check the signs against the quadrant
Why: The point is right and down, so quadrant four: cosine and secant positive, everything else negative. That is exactly the pattern above. And each reciprocal pair multiplies to 1, which is a second free check.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 753-753
Fill the middle
The terminal side contains the point negative 8 comma 15.
Fill in the blanks
r = 17 \;\Longrightarrow\; \tan(\theta) = \frac-15/817/15 = ___, \quad \csc(\theta) = \frac___}___ = ___
Why: Sixty-four plus two hundred twenty-five is two hundred eighty-nine, whose root is 17 — eight, fifteen and seventeen are a Pythagorean triple. The tangent is y over x, which is negative because the coordinates have opposite signs, and the cosecant is r over y, positive because y is positive. The point is in quadrant two, where exactly sine and cosecant are positive.
Worked example
The radius is still positive; only the coordinates change sign.
\[ \text{The terminal side of } \theta \text{ contains } (-5, 12). \text{ Find all six.} \]
Compute the radius
Why: Twenty-five plus one hundred forty-four is one hundred sixty-nine, whose root is thirteen.
\[ r = 13 \]
Cosine and sine
Why: Divide each coordinate by 13.
\[ \cos = -\frac{5}{13}, \sin = \frac{12}{13} \]
Secant and cosecant
Why: Reciprocate each of those.
\[ \sec = -\frac{13}{5}, \csc = \frac{13}{12} \]
Tangent and cotangent
Why: Divide the coordinates by each other, in each order.
\[ \tan = -\frac{12}{5}, \cot = -\frac{5}{12} \]
Figure (svg): The solution to Worked example a point in quadrant two shown as a ladder of expressions, one row per legal move
\[ \cos = -\tfrac{5}{13}, \; \sin = \tfrac{12}{13}, \; \sec = -\tfrac{13}{5}, \; \csc = \tfrac{13}{12}, \; \tan = -\tfrac{12}{5}, \; \cot = -\tfrac{5}{12} \]
Verify: check the quadrant pattern
Why: The point is left and up, so quadrant two: only sine and cosecant are positive, and indeed only those two came out positive. Four negatives and two positives is exactly the quadrant two signature.
Trap
\[ (-6, -8): \quad r = -10 \quad\Longrightarrow\quad \cos(\theta) = \frac{-6}{-10} = \frac{3}{5} \]
Make the radius negative because both coordinates are
Why: It looks tidier for the signs to cancel, and the arithmetic runs smoothly.
The point is in quadrant three, where the cosine must be negative. This answer says it is positive, so the sign information has been destroyed.
\[ r = \sqrt{36 + 64} = 10 \quad\Longrightarrow\quad \cos(\theta) = \frac{-6}{10} = -\frac{3}{5} \]
Take the radius positive always, since it is a distance
Why: The square root symbol returns the non-negative root, so this happens automatically unless it is overridden.
All the sign information lives in the coordinates, and r is there only to set the scale. If r ever comes out negative, a distance has been mismeasured — and the resulting values will have the sign pattern of the wrong quadrant.
Sorting
The point is negative 3 comma negative 3, in quadrant three.
Sort into buckets
Sort the six functions by sign at this angle.
Prediction
A point on the terminal side is doubled, from two comma three to four comma six.
Predict first
What happens to the six function values?
Correct: All six are unchanged.
Why: Doubling the point doubles x, doubles y and doubles r, so every one of the six ratios has both its numerator and its denominator doubled and is therefore unchanged. That is exactly what must happen, since the two points lie on the same terminal side and so belong to the same angle. Any answer in which the values changed would mean the functions were not functions of the angle at all.
Socratic
Cosine and sine can never exceed 1, and secant and cosecant can never be smaller than 1 in size. Tangent has no such limit.
Discussion prompt
Explain the difference by looking at which of x, y and r appear in each formula.
Hint: Which of the three is always the largest?
Answer:
The radius r is the hypotenuse of the little triangle, so it is always at least as large as either coordinate. Cosine and sine divide a coordinate by r, so their quotients are at most 1; secant and cosecant divide r by a coordinate, so their quotients are at least 1.
Tangent and cotangent are the only two that do not involve r at all. They divide one leg by the other, and there is no constraint forcing one leg to be comparable to the other — a nearly vertical terminal side has an enormous y and a tiny x, giving an enormous tangent.
\[ \tan(\theta) = \frac{y}{x} \quad\text{with}\quad x \to 0 \;\Longrightarrow\; |\tan(\theta)| \to \infty \]
Section
Section 2
Concept
If you are given one function value and a quadrant, you can invent a convenient point on the terminal side rather than working through the Pythagorean identity. Any point on the ray will do, so choose the one with the simplest numbers.
The freedom to choose is genuine: a cotangent of negative four could be four over negative one or negative four over one or eight over negative two, and every choice gives the same six answers. Pick the one that keeps the arithmetic smallest.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 753-753
Picture it
Whatever point you choose, you are choosing a triangle, and every such triangle for a given angle is similar to every other.
Figure (svg): A point Q with coordinates x and y on the terminal side of an angle, at distance r from the origin, with all six circular functions written as ratios of x, y and r
That similarity is why the choice does not matter, and it is the same argument that made Theorem 10.3 true in the first place.
Worked example
Example 10.3.4, part 2. The quadrant does the work of choosing the signs.
\[ \theta \text{ is in Quadrant IV with } \cot(\theta) = -4. \text{ Find the other five circular functions.} \]
Read the cotangent as x over y
Why: Cotangent is the x-coordinate over the y-coordinate, so that quotient must be negative four.
\[ \frac{x}{y} = -4 \]
Choose signs to match quadrant four
Why: There x is positive and y is negative, so write negative four as four over negative one.
\[ x = 4, y = -1 \]
Compute the radius
Why: Sixteen plus one is seventeen, which does not simplify.
\[ r = \sqrt{17} \]
Read off the remaining five and rationalise
Why: Each is a ratio of two of the three, with surds rationalised.
Figure (svg): The solution to Worked example five values from a cotangent shown as a ladder of expressions, one row per legal move
\[ \cos = \tfrac{4\sqrt{17}}{17}, \; \sin = -\tfrac{\sqrt{17}}{17}, \; \sec = \tfrac{\sqrt{17}}{4}, \; \csc = -\sqrt{17}, \; \tan = -\tfrac{1}{4} \]
Verify: check the reciprocal of the given value
Why: The tangent came out negative one quarter, and the reciprocal of negative four is indeed negative one quarter. And the sign pattern is quadrant four: cosine and secant positive, the other four negative.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 753-753
Matching
In each case one function value and one quadrant are given.
Match the pairs
Why: Tangent is y over x, so the ratio fixes the sizes and the quadrant fixes the two signs. A positive tangent occurs in quadrants one and three, where the coordinates agree in sign; a negative tangent in quadrants two and four, where they differ. All four points have radius 5, so all four give values of size three fifths and four fifths, differing only in their signs.
Worked example
Same value, different quadrant. Only the sign choice changes.
\[ \theta \text{ is in Quadrant II with } \cot(\theta) = -4. \text{ Find } \cos(\theta) \text{ and } \sin(\theta). \]
Read the cotangent as x over y again
Why: The quotient is still negative four.
\[ \frac{x}{y} = -4 \]
Choose signs to match quadrant two
Why: There x is negative and y is positive, so write negative four as negative four over one.
\[ x = -4, y = 1 \]
Compute the radius
Why: Sixteen plus one is seventeen again, since squaring removed the signs.
\[ r = \sqrt{17} \]
Read off the two values
Why: Divide each coordinate by the radius and rationalise.
\[ \cos = -4 \sqrt{17} / 17, \sin = \sqrt{17} / 17 \]
Figure (svg): The solution to Worked example choosing the signs the other way shown as a ladder of expressions, one row per legal move
\[ \cos(\theta) = -\frac{4\sqrt{17}}{17} \qquad \sin(\theta) = \frac{\sqrt{17}}{17} \]
Verify: compare with the previous example
Why: Both values are exactly the negatives of the quadrant four answers, which is right: the two terminal sides are opposite rays through the origin, so the two points are negatives of each other. The cotangent could not distinguish them, which is precisely why the quadrant had to be given.
Error analysis
A student is told the tangent of theta is negative two thirds with theta in quadrant two, and picks a point.
Annotate
On: \( \tan(\theta) = -\tfrac{2}{3} \;\Longrightarrow\; x = 3, \; y = -2 \)
A negative fraction can always be written with the minus sign on top or on the bottom, and the two choices land in opposite quadrants. The quadrant is the only thing that decides between them, so read it before writing the point down.
Faded example
An angle in quadrant three has secant equal to negative 5 over 3. Find the sine.
Fill in the blanks
\sec = \frac-3-4 = -\frac-4/5___ \;\Longrightarrow\; r = 5, \; x = ___ \;\Longrightarrow\; y = ___ \;\Longrightarrow\; \sin = ___
Why: The secant is r over x, and since r is always positive the minus sign must belong to x, giving x equal to negative three with r equal to five. Then y squared is twenty-five minus nine, which is sixteen, so y is plus or minus four — and quadrant three requires it negative. The sine is y over r, which is negative four fifths.
Two truths and a lie
Rule out the statements that are true. The survivor is the false one.
Eliminate the wrong options
One of these statements about reconstructing a point is wrong.
Survives elimination: B
Why: Statement B is false, and the exception is instructive. If the given value is a tangent or cotangent, the quadrant is genuinely needed and then everything follows. But if the given function is undefined in the stated quadrant, or if the value lies outside that function's range — a secant of one half, say — then no such angle exists and nothing is determined. The claim needs the value to be attainable, which is a real condition rather than a technicality.
Edge cases
You are told the tangent of theta is 0 and asked to reconstruct a point.
Discussion prompt
What does that force about the point, which quadrants are possible, and what happens to the cotangent?
Hint: A fraction is zero exactly when its numerator is.
Answer:
Tangent is y over x, and a fraction is zero exactly when its numerator is zero, so y equals 0. The point lies on the x-axis, at either one comma zero or negative one comma zero on the unit circle.
So the angle is quadrantal and lies in no quadrant at all — the question's premise that a quadrant can be given quietly fails.
The cotangent is x over y, and with y equal to zero it is undefined. This is the general pattern: a function is zero exactly where its reciprocal is undefined, since one over zero cannot exist. Every zero of tangent is a hole in cotangent, and vice versa, which is exactly what the two graphs show in Lesson 10.5c.
Section
Section 3
Concept
Place an acute angle of a right triangle in standard position with its adjacent side along the positive x-axis. The opposite vertex is then a point on the terminal side at distance c from the origin, so the previous theorem applies verbatim with a, b and c in place of x, y and r.
Theorem 10.10 — For an acute angle in a right triangle with adjacent side a, opposite side b and hypotenuse c: the tangent is b over a, the secant is c over a, the cosecant is c over b, and the cotangent is a over b.
\[ \tan = \tfrac{b}{a}, \quad \sec = \tfrac{c}{a}, \quad \csc = \tfrac{c}{b}, \quad \cot = \tfrac{a}{b} \]
Together with the cosine and sine relations from Lesson 10.2c, this gives all six. The familiar mnemonic covers only the first three; the other three are simply their reciprocals.
Figure (svg): A right triangle with acute angle theta, adjacent side a, opposite side b and hypotenuse c, with all six circular functions written as ratios of the three sides
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 753-753
Picture it
Every one of the six is a ratio of two of the three side lengths, and there are exactly six such ordered ratios.
Figure (svg): A right triangle with acute angle theta, adjacent side a, opposite side b and hypotenuse c, with all six circular functions written as ratios of the three sides
That there are exactly six functions is not a coincidence. Three lengths taken two at a time in order give six ratios, and the six circular functions are precisely those six.
Worked example
The skill is choosing which of the six to use, not computing once you have chosen.
\[ \text{A right triangle has an acute angle } \theta \text{ with } \cos(\theta) = \tfrac{7}{25}. \text{ Find all six values.} \]
Read the cosine as adjacent over hypotenuse
Why: So the adjacent side is 7 and the hypotenuse is 25, up to a common scale factor.
\[ a = 7, c = 25 \]
Find the opposite side by the Pythagorean theorem
Why: Six hundred twenty-five minus forty-nine is five hundred seventy-six, whose root is 24.
\[ b = 24 \]
Write the three primary ratios
Why: Cosine, sine and tangent from the three sides.
\[ \cos 7 / 25, \sin 24 / 25, \tan 24 / 7 \]
Reciprocate for the other three
Why: Secant, cosecant and cotangent.
\[ \sec 25 / 7, \csc 25 / 24, \cot 7 / 24 \]
Figure (svg): The solution to Worked example choose the relation that fits shown as a ladder of expressions, one row per legal move
\[ \cos = \tfrac{7}{25}, \; \sin = \tfrac{24}{25}, \; \tan = \tfrac{24}{7}, \; \sec = \tfrac{25}{7}, \; \csc = \tfrac{25}{24}, \; \cot = \tfrac{7}{24} \]
Verify: check the identity
Why: Forty-nine over six hundred twenty-five plus five hundred seventy-six over six hundred twenty-five is 1. All six values are positive, which is right because an acute angle is in quadrant one.
Matching
In every case one side and one acute angle are known.
Match the pairs
Why: Choose the function whose ratio contains exactly the known side and the wanted side. Tangent is opposite over adjacent, cotangent is its reciprocal, secant is hypotenuse over adjacent and cosecant is hypotenuse over opposite. Each can of course be replaced by dividing with cosine or sine instead, which is what most people actually do — but naming the right function first makes the setup unambiguous.
Worked example
When neither the hypotenuse nor the quantity you want touches it, tangent is the relation to reach for.
\[ \text{In a right triangle, } \theta = 40^\circ \text{ and the side adjacent to it is } 18. \text{ Find the opposite side.} \]
Identify what is known and what is wanted
Why: Adjacent is known, opposite is wanted, hypotenuse is neither.
Choose the relation containing exactly those two
Why: Tangent is opposite over adjacent, and it is the only one of the six that avoids the hypotenuse in that order.
\[ \tan \theta = \frac{b}{a} \]
Substitute and solve
Why: Multiply both sides by the adjacent side.
\[ b = 18 \tan(40 ^\circ) \]
Evaluate
Why: The tangent of 40 degrees is about 0.8391.
\[ b = 15.1\text{ approx} \]
Figure (svg): The solution to Worked example a missing side from a tangent shown as a ladder of expressions, one row per legal move
\[ b = 18\tan(40^\circ) \approx 15.1 \]
Verify: check the proportion
Why: Forty degrees is less than 45, so the opposite side should be shorter than the adjacent one — and 15.1 is less than 18. At exactly 45 degrees they would be equal, which is a useful landmark for every problem of this shape.
Trap
A right triangle has legs 5 and 12 with the right angle at the bottom left. A student computes the tangent of the upper-left angle as twelve fifths, because 12 is the vertical leg.
\[ \tan(\theta) = \frac{12}{5} \]
Take the vertical leg as opposite and the horizontal as adjacent
Why: That is how the standard picture is drawn, so the habit transfers even when the angle has moved.
But for the upper angle the vertical leg is the ADJACENT one — it is one of the two sides forming that angle. Opposite and adjacent are defined relative to the angle, not to the page.
\[ \text{for the upper angle: } \tan(\theta) = \frac{5}{12} \]
Identify opposite and adjacent relative to the angle in question
Why: The opposite side is the one not touching the angle; the adjacent leg is the one that does, excluding the hypotenuse.
The reliable test: the opposite side is the one you do not touch when you put a finger on the angle. The two angles of a right triangle swap opposite and adjacent between them, which is exactly why their tangents are reciprocals and why the cofunction identities of Lesson 10.4a exist.
Faded example
A right triangle has an acute angle with cotangent 3 over 4 and hypotenuse 20.
Fill in the blanks
\cot = \frac54 = \frac12___ \;\Longrightarrow\; a = 3k, \; b = 4k, \; c = ___k = 20 \;\Longrightarrow\; k = ___ \;\Longrightarrow\; a = ___
Why: A cotangent of three quarters means the legs are in the ratio three to four, so they are 3k and 4k for some positive k. The hypotenuse is then 5k by the three-four-five triple. Setting 5k equal to 20 gives k equal to 4, so the adjacent side is 12 and the opposite is 16. Checking, 144 plus 256 is 400, which is 20 squared.
Prediction
The two acute angles of a right triangle are theta and its complement.
Predict first
How does the tangent of one compare with the tangent of the other?
Correct: They are reciprocals of each other.
Why: The side opposite one acute angle is adjacent to the other, and vice versa. So if one tangent is b over a, the other is a over b, and those are reciprocals. Equivalently, the tangent of one equals the cotangent of the other, which is the cofunction identity for tangent that Lesson 10.4a proves for every angle rather than just for acute ones.
Explain it to yourself
There are exactly six circular functions, no more and no fewer.
Discussion prompt
Explain why six is the right number, using the triangle picture. What would a seventh function have to be?
Hint: Count the ways of choosing two of three things in order.
Answer:
A right triangle relative to a chosen acute angle has exactly three relevant lengths: adjacent, opposite and hypotenuse. A ratio uses two of them, and the order matters, so there are three times two, which is six ordered pairs — and each one is a circular function.
\[ \tfrac{a}{c}, \; \tfrac{b}{c}, \; \tfrac{c}{a}, \; \tfrac{c}{b}, \; \tfrac{b}{a}, \; \tfrac{a}{b} \]
A seventh function would have to be a ratio of two of three lengths that is not already on the list, and there is no such ratio. The six are exhaustive, which is why no textbook ever introduces a seventh, and why the six fall so naturally into three reciprocal pairs.
Section
Section 4
Concept
The angle of inclination, also called the angle of elevation, is the angle between a horizontal base line and the line of sight to an object. Measuring it and one distance is enough to find a height that cannot be reached.
angle of inclination — The angle whose initial side is a horizontal base line and whose terminal side is the line of sight to an object above that line. Also called the angle of elevation.
The angle of depression is the same idea measured downward from a horizontal line, and it equals the angle of elevation from the other end because they are alternate angles between parallel lines.
Figure (svg): An observer on the ground looking up at the top of a tower, with the angle of inclination marked between the horizontal base line and the line of sight
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 753-754
Picture it
When the base of the object cannot be reached, a second sighting from a known extra distance supplies the missing equation.
Figure (svg): Two observation points 200 feet apart on level ground, sighting the top of a tall tree at angles of elevation of forty-five degrees from the near point and thirty degrees from the far point
The unknown x is the distance from the near point to the base, and it is eliminated between the two equations rather than being wanted for its own sake.
Worked example
Example 10.3.5, part 1. The clocktower at Lakeland Community College.
\[ \text{From a point } 30 \text{ feet away, the angle of inclination to the top of a tower is } 60^\circ. \text{ Find its height.} \]
Sketch and label
Why: The 30 feet is the horizontal distance, adjacent to the angle; the height is opposite it.
\[ \text{adjacent } 30,\text{ opposite } h \]
Choose the relation
Why: Tangent is opposite over adjacent, which is exactly the pair in play.
\[ \tan 60 ^\circ = \frac{h}{30} \]
Solve for the height
Why: Multiply both sides by 30.
\[ h = 30 \tan(60 ^\circ) \]
Evaluate exactly then approximate
Why: The tangent of 60 degrees is root three.
\[ h = 30 \sqrt{3} = 51.96 \]
Figure (svg): The solution to Worked example one sighting shown as a ladder of expressions, one row per legal move
\[ h = 30\tan(60^\circ) = 30\sqrt{3} \approx 51.96 \text{ ft} \]
Verify: check the proportion
Why: Sixty degrees is steeper than 45, so the height should exceed the horizontal distance — and 52 is more than 30. Specifically it should be root three times as large, about 1.73 times, and 30 times 1.73 is about 52.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 754-754
Prediction
You walk directly away from a tower, doubling your distance from its base.
Predict first
What happens to the angle of elevation to its top?
Correct: It decreases, but by less than half.
Why: The tangent of the angle is the height over the distance, so doubling the distance halves the tangent. But the angle is not proportional to its tangent, and halving the tangent reduces the angle by less than half. For example, an elevation of 60 degrees has tangent root three, about 1.732; halving that gives 0.866, whose angle is about 40.9 degrees — a drop of 19 degrees rather than 30. This is exactly why angle and tangent must be kept distinct in these problems.
Worked example
Example 10.3.5, part 2. A California Redwood, sighted from two points 200 feet apart.
\[ \text{Two sightings } 200 \text{ ft apart give angles of inclination } 45^\circ \text{ and } 30^\circ. \text{ Find the height.} \]
Name both unknowns
Why: The height h and the distance x from the near point to the base.
Write one equation per sighting
Why: The far sighting is 200 feet further from the base than the near one.
\[ \tan 45 = \frac{h}{x}, \tan 30 = \frac{h}{x + 200} \]
Use the first to eliminate x
Why: The tangent of 45 degrees is 1, so x equals h.
\[ x = h \]
Substitute and solve the resulting linear equation
Why: Clearing fractions gives three h equals h root three plus 200 root three, then gather the h terms.
\[ h = 200 \sqrt{3} / (3 - \sqrt{3}) \]
Figure (svg): The solution to Worked example two sightings shown as a ladder of expressions, one row per legal move
\[ h = \frac{200\sqrt{3}}{3 - \sqrt{3}} \approx 273.20 \text{ ft} \]
Verify: check both sightings
Why: With h about 273.2, the near distance x is also about 273.2, and the tangent of 45 degrees is indeed 273.2 over 273.2, which is 1. The far distance is about 473.2, and 273.2 over 473.2 is about 0.577, which is root three over three — the tangent of 30 degrees. Both equations hold.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 754-755
Error analysis
In the two-sighting problem, a student writes the second equation.
Annotate
On: \( \tan(30^\circ) = \frac{h}{x - 200} \)
The check that settles it every time: a shallower angle of elevation means you are further away. If your setup pairs the smaller angle with the smaller distance, the sketch has been misread.
Faded example
From the top of a 60 metre cliff the angle of depression to a boat is 25 degrees. Find the boat's distance from the base.
Fill in the blanks
\tan(25^\circ) = \frac60}tan(25 deg) \;\Longrightarrow\; d = \frac129___} \approx ___ \text___
Why: The angle of depression from the cliff top equals the angle of elevation from the boat, since they are alternate angles across the parallel horizontals. So the 60 metre height is opposite the 25 degree angle and the distance is adjacent to it. The tangent of 25 degrees is about 0.4663, so the distance is about 128.7 metres. The shallow angle means a distance much larger than the height, which the answer confirms.
Missing information
A problem says: the angle of elevation to the top of a building is 38 degrees. How tall is the building?
Discussion prompt
Why can this not be answered, and what single extra measurement would suffice? Explain why the angle alone determines a shape but not a size.
Hint: How many buildings are consistent with a 38 degree sighting?
Answer:
Infinitely many buildings fit. An angle of elevation of 38 degrees fixes only the ratio of height to distance, at about 0.781. A 10 metre building sighted from 12.8 metres away and a 100 metre building sighted from 128 metres away give exactly the same angle.
The missing measurement is one length — most naturally the horizontal distance to the base, but the slant distance along the line of sight would do just as well, using the sine instead of the tangent.
This is the same point as similar triangles: angles determine shape, lengths determine size, and no amount of angle measurement will ever produce a length on its own. Every triangle problem in the rest of this course needs at least one side given, and Lesson 11.2 will say so explicitly.
Real world
A surveyor needs the height of a mountain whose base is inaccessible, standing on level ground. Two sightings are taken 1000 metres apart along a straight line towards the peak, giving elevations of 32 degrees and 48 degrees.
Discussion prompt
Set up the two equations and describe how to eliminate the unknown distance. Then say why the two-sighting method is the standard field technique rather than a textbook artifice.
Hint: Let x be the distance from the nearer point to the point directly below the peak.
Answer:
\[ \tan(48^\circ) = \frac{h}{x}, \qquad \tan(32^\circ) = \frac{h}{x + 1000} \]
Solve the first for x, giving h over the tangent of 48 degrees, and substitute into the second. That leaves one linear equation in h alone, which rearranges to h equal to 1000 divided by the difference of the two cotangents — about 1263 metres.
This is the standard field technique precisely because the base is inaccessible. You cannot measure to a point inside a mountain, but you can pace out a baseline on flat ground and measure two angles with a theodolite. The whole of practical surveying, and the parallax method astronomers use for stellar distances, is this same two-sighting idea: an inaccessible distance is reached by measuring an accessible one plus two angles.
Section
Section 5
Concept
Read as functions of a real number, the six differ sharply in both what they accept and what they produce. Every difference traces back to which of x, y and r sits downstairs.
The book writes these unions of infinitely many intervals using extended interval notation, a union indexed over all the integers. The notation is cumbersome but it is saying something simple: between consecutive holes, the function is defined everywhere.
| Function | Domain | Range |
|---|---|---|
| cosine, sine | all real numbers | from -1 to 1 inclusive |
| secant | all except odd multiples of pi over 2 | size at least 1 |
| cosecant | all except multiples of pi | size at least 1 |
| tangent | all except odd multiples of pi over 2 | all real numbers |
| cotangent | all except multiples of pi | all real numbers |
Figure (svg): A number line marked with the excluded points for secant and tangent at odd multiples of pi over two, and separately the excluded points for cosecant and cotangent at multiples of pi
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 755-757
Picture it
Four functions have holes, and there are only two distinct patterns of hole.
Figure (svg): A number line marked with the excluded points for secant and tangent at odd multiples of pi over two, and separately the excluded points for cosecant and cotangent at multiples of pi
Secant and tangent share a denominator so they share their holes, and cosecant and cotangent likewise. The holes are spaced pi apart in both patterns, offset from each other by pi over two.
Worked example
The reasoning matters more than the notation.
\[ \text{Find the domain of } F(t) = \sec(t). \]
Identify the denominator
Why: Secant is one over cosine, so the exclusions are where the cosine vanishes.
\[ \text{exclude } \cos t = 0 \]
Solve that equation
Why: The cosine is zero exactly on the y-axis, at pi over two and three pi over two, and everything coterminal.
\[ t = \frac{\pi}{2} + \pi k \]
Describe the excluded set compactly
Why: The two families combine into one stepping by pi, since the two base angles differ by pi.
\[ \text{odd multiples of } \frac{\pi}{2} \]
State the domain
Why: Everything except that set, which is a union of open intervals between consecutive holes.
Figure (svg): The solution to Worked example write the domain of secant shown as a ladder of expressions, one row per legal move
\[ \text{dom}(\sec) = \left\{ t : t \ne \tfrac{\pi}{2} + \pi k, \; k \in \mathbb{Z} \right\} \]
Verify: check a value in a hole and one outside
Why: At pi over two the cosine is zero, so the secant genuinely does not exist there. At pi over three the cosine is one half, so the secant is 2 — defined, as the domain claims.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 755-756
Comparison
Fill the blanks from the definitions, not from memory.
Comparison matrix
| Function | Excluded from the domain | Range |
|---|---|---|
| cosine | nothing | -1 to 1 |
| secant | odd multiples of pi over 2 | size at least 1 |
| cosecant | multiples of pi | size at least 1 |
| tangent | odd multiples of pi over 2 | all real numbers |
| cotangent | multiples of pi | all real numbers |
The domain column has only three distinct entries and the range column has only three. Six functions, but the structure is genuinely three by three.
Worked example
Knowing the range turns some questions into one-line answers.
\[ \text{Do } \sec(t) = \tfrac{1}{2} \text{ and } \csc(t) = -42 \text{ have solutions?} \]
Check the first against the range of secant
Why: The range excludes everything strictly between negative one and one, and one half is in that gap.
\[ \frac{1}{2}\text{ not in range} \]
Conclude for the first
Why: No angle has a secant of one half.
Check the second against the range of cosecant
Why: Negative forty-two has absolute value well above 1, so it is in the range.
\[ -42\text{ is in range} \]
Say where those solutions are
Why: A negative cosecant means a negative sine, which is quadrants three and four.
Figure (svg): The solution to Worked example use the range to reject an equation shown as a ladder of expressions, one row per legal move
\[ \sec(t) = \tfrac{1}{2}: \text{ none} \qquad \csc(t) = -42: \text{ two families of solutions} \]
Verify: sanity-check the second
Why: A cosecant of negative 42 means a sine of about negative 0.0238, a very small negative number, so the angle is just below the x-axis — just past pi in quadrant three, and just below two pi in quadrant four. Both exist, so two families is right.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 758-758
Trap
A student reasons: cosine and sine are bounded by 1, so all the circular functions are bounded, and an equation like tangent of t equals 1000 has no solution.
\[ \tan(t) = 1000 \quad\text{has no solution} \]
But the tangent is a ratio of two coordinates with no radius involved, and there is nothing forcing one coordinate to be comparable to the other.
\[ \tan(t) = 1000 \quad\Longrightarrow\quad t \approx 1.5698 + \pi k \]
Check the range of the specific function, not of cosine and sine
Why: Tangent and cotangent have range all of the real numbers, so every equation of this form has solutions.
An angle just under pi over two has a tiny cosine and a sine near 1, so their quotient is enormous. The three pairs have three different ranges — bounded, outside a bounded interval, and unbounded — and confusing them is what makes this error possible.
Sorting
Check each required value against that function's range.
Sort into buckets
Sort each equation by whether any real number satisfies it.
Notation
This is the book's extended interval notation for the domain of secant. Read what each piece is doing.
Annotate
On: \( \bigcup_{k = -\infty}^{\infty} \left( \frac{(2k+1)\pi}{2}, \; \frac{(2k+3)\pi}{2} \right) \)
The notation is heavy for what it says, and the book admits as much. The set-builder form, t not equal to pi over two plus pi k, is usually the more readable of the two and means exactly the same thing.
Counterexample
A student proposes: since tangent is undefined at pi over two, and cotangent is the reciprocal of tangent, cotangent must also be undefined at pi over two.
Discussion prompt
Compute the cotangent of pi over two and explain what is wrong with the reasoning.
Hint: Compute it from the definition as x over y rather than as one over tangent.
Answer:
\[ \cot\left(\tfrac{\pi}{2}\right) = \frac{\cos(\pi/2)}{\sin(\pi/2)} = \frac{0}{1} = 0 \]
The cotangent is perfectly well defined at pi over two and equals zero. The reasoning fails because the reciprocal relationship only holds where both functions are defined.
The correct statement of the pattern is the reverse of what the student said: a function is zero exactly where its reciprocal is undefined. Tangent blows up at pi over two, so cotangent is zero there; cotangent blows up at zero, so tangent is zero there. Reading the reciprocal identity as though it were valid everywhere is the error, and the identity's own restriction is what forbids it.
Comparison
Fill the blanks from memory. Every column says the same thing about the same six functions.
Comparison matrix
| Function | Unit circle | Any point | Right triangle |
|---|---|---|---|
| cosine | x | x over r | adjacent over hypotenuse |
| sine | y | y over r | opposite over hypotenuse |
| tangent | y over x | y over x | opposite over adjacent |
| secant | 1 over x | r over x | hypotenuse over adjacent |
| cosecant | 1 over y | r over y | hypotenuse over opposite |
| cotangent | x over y | x over y | adjacent over opposite |
Tangent and cotangent are identical in the first two columns, because they never involved r in the first place. That is the same fact that makes their range unbounded.
Pattern
Whether the problem gives you a point, a value, a triangle or a sighting, the same five moves cover it.
Before starting an equation, check the value against that function's range. A secant of one half or a cosine of two has no solutions, and noticing that first saves the whole problem.
OpenStax Algebra and Trigonometry 2e, §7.2 Right Triangle Trigonometry §7.2
Check
All six from a point. Compute the radius first.
Check your understanding
The terminal side of theta contains the point negative 12 comma 5. What is the secant of theta?
Answer: B
Why: The radius is the square root of 144 plus 25, which is the square root of 169, namely 13. The secant is r over x, which is 13 over negative 12, giving negative thirteen twelfths. The negative sign comes from the x-coordinate, since the radius is always positive.
Check
A right triangle. Choose the relation containing both sides.
Check your understanding
A ladder leans against a wall making a 65 degree angle with the ground, with its foot 2.5 metres from the wall. How long is the ladder?
Answer: D
Why: The 2.5 metres is adjacent to the 65 degree angle and the ladder is the hypotenuse, so the relation containing both is the cosine: cosine of 65 degrees equals 2.5 over the length. The cosine of 65 degrees is about 0.4226, so the length is 2.5 divided by that, about 5.92 metres.
Check
Domain and range. Check the value against the function's range.
Check your understanding
Which of these equations has no solutions?
Answer: C
Why: The secant is one over the cosine, and since the cosine never exceeds 1 in absolute value, the secant never falls below 1 in absolute value. The value 0.8 lies strictly between negative one and one, which is exactly the gap the range of secant excludes, so no angle has a secant of 0.8.
Real world
An aircraft at a constant altitude is tracked by a ground radar. At one moment the elevation angle is 20 degrees and thirty seconds later, with the aircraft having flown directly overhead-ward along a straight line, it is 55 degrees. The aircraft's ground speed is 240 metres per second.
Discussion prompt
Find the aircraft's altitude. Explain which unknown gets eliminated and why the radar cannot determine the altitude from a single sighting no matter how accurate it is.
Hint: Let x be the horizontal distance at the second sighting. The first sighting is further out by the distance flown.
Answer:
\[ \text{distance flown} = 240 \cdot 30 = 7200 \text{ m} \]
\[ \tan(55^\circ) = \frac{h}{x}, \qquad \tan(20^\circ) = \frac{h}{x + 7200} \]
Solving the first for x and substituting eliminates x, leaving a linear equation in h. Working it through gives an altitude of about 3390 metres.
A single sighting cannot do it because one angle fixes only the ratio of altitude to horizontal distance, and both are unknown. No improvement in angular accuracy helps, because the missing information is not precision but a second equation. Real radar solves this differently, by measuring the slant range directly with timing — which supplies the hypotenuse and turns one sighting into enough. When you cannot get a second angle, get a length.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Two students pick different points on the same terminal side and compute the tangent. What happens?
Correct: They get the same answer.
\[ \frac{ky}{kx} = \frac{y}{x} \quad\text{for any } k > 0 \]
Why: Both points lie on the same ray from the origin, so one is a positive multiple of the other. The tangent is y over x, and multiplying both coordinates by the same positive factor leaves that quotient unchanged. The fourth option is a trap: two points on the same terminal side are necessarily in the same quadrant, since a ray from the origin cannot cross into another quadrant, so the stated condition is automatically satisfied and adds nothing.
Explain it
They know SOH CAH TOA and are being asked to find a height using two sightings. They have never seen a problem with two unknowns.
Discussion prompt
In no more than five sentences, explain why one sighting is not enough and how the second one rescues the situation.
Hint: Count the unknowns and count the equations.
Answer:
A usable answer: with one sighting you know an angle but you have two unknowns — how tall the thing is and how far away its base is. One equation cannot pin down two numbers, and that is why the answer feels out of reach no matter how you rearrange it.
Walking a measured distance and sighting again gives you a second equation, because the new distance is the old one plus the amount you walked. Two equations and two unknowns can be solved, and the distance you never wanted gets eliminated along the way. The height is what survives.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: The six ratios are fixed by remembering that r is always downstairs for cosine and sine, upstairs for secant and cosecant, and absent from tangent and cotangent. Sign choice is fixed by writing the quadrant's sign pattern down before choosing the point. The two-sighting setup is fixed by naming both unknowns explicitly and writing one equation per sighting before touching any algebra. Domain and range are fixed by asking what is downstairs and whether the reciprocal of something bounded can be small. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
Draw a coordinate plane with a terminal side in quadrant two and a point on it labelled with x, y and r. Beside it, write all six functions as ratios of those three, grouping them into the three reciprocal pairs. Underneath, draw a right triangle labelled adjacent, opposite and hypotenuse, and write the same six ratios in those words, drawing a line linking each to its partner above. On the right, draw an angle of elevation problem with a two-sighting setup, label both unknowns, and write the two equations without solving them. Along the bottom, draw two number lines: on the first mark the holes in the domain of secant and tangent, and on the second the holes for cosecant and cotangent. Beside each, write the range of the functions concerned. Finally, circle the two functions on your page whose range is all real numbers, and write one sentence explaining what they have in common that causes it.
The circled pair should be tangent and cotangent, and what they have in common is that neither involves r. Every other function has the hypotenuse in one position or the other, and the hypotenuse is the largest of the three lengths, which is what bounds them.
Recap
Five things, and the last one turns a whole class of equations into one-line answers.
| If the question says | Your first move is |
|---|---|
| The terminal side contains this point | Compute r, take it positive, then form six ratios |
| Given one value and a quadrant | Invent a point whose signs match that quadrant |
| Find this side of the right triangle | Name the relation containing the known and wanted sides |
| The angle of elevation is | Sketch it; the height is opposite and the ground distance adjacent |
| Does this equation have a solution | Check the value against that function's range first |
That completes Section 10.3. Every one of the six functions is now defined, computable from any starting information, and bounded or not according to its shape. Section 10.4 turns to the relationships between them at different angles — what happens to a value when the angle is negated, added to another, or doubled.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.3 The Six Circular Functions and Fundamental Identities §10.3, pp. 752-759 — everything on these slides traces back here
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