The Reference Angle Theorem and what it buys: the acute angle a terminal side makes with the x-axis determines the size of both coordinates, and the quadrant supplies the two signs, so five memorised values generate all sixteen special points on the unit circle. Covers the four subtraction rules, the denominator shortcut in radian measure, angles built symmetrically from a given one, and the first trigonometric equations — where a single value produces two families of infinitely many solutions.
Subject: Trigonometry · 65 slides · symbolic lesson
Open the interactive version of this deck
Title
Trigonometry · Chapter 10 — Foundations of Trigonometry
§10.2 The Unit Circle: Cosine and Sine, pp. 722-730
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 — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 722-730 — the pages these objectives are drawn from
Warm-up
You can find the cosine and sine of three acute angles exactly. You are about to discover that this is already enough for twelve more.
Discussion prompt
The point for pi over six is root three over two comma one half. Where is the point for five pi over six, and what are its coordinates? Do not compute anything — argue from the picture.
Hint: How far is each terminal side from the x-axis? Which axis is each one near?
Answer:
Both terminal sides make an angle of pi over six with the x-axis — one with the positive half, one with the negative half. So the two points are at the same height and the same distance from the vertical axis, just on opposite sides of it.
\[ \left(\tfrac{\sqrt{3}}{2}, \tfrac{1}{2}\right) \quad\longrightarrow\quad \left(-\tfrac{\sqrt{3}}{2}, \tfrac{1}{2}\right) \]
The sine is unchanged and the cosine has flipped sign, exactly as reflecting across the y-axis demands. That observation, made precise, is the whole of this lesson.
Concept
For a non-quadrantal angle, the reference angle is the acute angle between its terminal side and the x-axis. Because the unit circle is symmetric about both axes and the origin, the point for the angle and the point for its reference angle are always mirror images — so their coordinates agree up to sign.
reference angle — For a non-quadrantal angle in standard position, the acute angle its terminal side makes with the x-axis. It is measured to the positive x-axis in quadrants one and four, and to the negative x-axis in quadrants two and three.
\[ \cos(\theta) = \pm\cos(\alpha), \qquad \sin(\theta) = \pm\sin(\alpha) \]
The theorem deliberately leaves the signs as plus-or-minus, because they are not its business. The reference angle knows the size of each coordinate; only the quadrant knows the sign. Keeping those two jobs separate is what makes the method reliable.
Figure (svg): The angle five pi over six and the angle pi over six drawn on the same unit circle, showing that the two points are reflections of each other across the y-axis so their coordinates differ only in the sign of the first
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 721-722
Section
Section 1
Concept
The reference angle is always measured to the x-axis, never to the y-axis, and always comes out acute. Which subtraction achieves that depends on the quadrant, but the question being asked is the same in all four: how far is this terminal side from horizontal?
In degrees the same four rules read theta, 180 minus theta, theta minus 180, and 360 minus theta. If the angle is negative or larger than one revolution, reduce it into the interval from zero to two pi first and then apply the table.
| Quadrant of theta | Reference angle alpha | Measured to |
|---|---|---|
| I | alpha equals theta | the positive x-axis |
| II | alpha equals pi minus theta | the negative x-axis |
| III | alpha equals theta minus pi | the negative x-axis |
| IV | alpha equals two pi minus theta | the positive x-axis |
Figure (svg): The reference angle drawn for an angle in each of the four quadrants, measured to the positive x-axis in quadrants one and four and to the negative x-axis in quadrants two and three
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 721-722
Picture it
Rather than memorising four formulas, look at the four pictures until the single idea behind them is obvious.
Figure (svg): The reference angle drawn for an angle in each of the four quadrants, measured to the positive x-axis in quadrants one and four and to the negative x-axis in quadrants two and three
In every case the dashed wedge lies between the terminal side and the horizontal, and it is always acute. If your reference angle came out obtuse, you measured to the wrong axis.
Worked example
Example 10.2.3, part 1. Find the reference angle, then attach the signs.
\[ \text{Find } \cos(225^\circ) \text{ and } \sin(225^\circ). \]
Place the angle and name the quadrant
Why: Two hundred twenty-five overshoots the negative x-axis at 180, landing in quadrant three.
Find the reference angle
Why: In quadrant three, subtract 180 from the angle to reach the negative x-axis.
\[ \alpha = 225 - 180 = 45 ^\circ \]
Look up the size of each coordinate
Why: The reference angle is 45 degrees, whose cosine and sine are both root two over two.
\[ \text{both sizes are } \sqrt{2} / 2 \]
Attach the signs from the quadrant
Why: In quadrant three both coordinates are negative.
Figure (svg): The solution to Worked example a quadrant three angle shown as a ladder of expressions, one row per legal move
\[ \cos(225^\circ) = -\frac{\sqrt{2}}{2} \qquad \sin(225^\circ) = -\frac{\sqrt{2}}{2} \]
Verify: check the identity and the picture
Why: One half plus one half is 1, so the pair is legitimate. And 225 degrees points down and to the left, so both coordinates should indeed be negative and equal in size, since the angle is diagonal.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 722-723
Matching
Reduce into one revolution first where you need to.
Match the pairs
Why: For 150 degrees, in quadrant two, subtract from 180 to get 30. For 210 degrees, in quadrant three, subtract 180 to get 30 again — the same reference angle in a different quadrant, which is why the signs will differ. For 315 degrees, in quadrant four, subtract from 360 to get 45. Negative 60 degrees is coterminal with 300 degrees in quadrant four, and 360 minus 300 is 60.
Worked example
Example 10.2.3, part 4. Reduce first, then proceed as usual.
\[ \text{Find } \cos\left(\frac{7\pi}{3}\right) \text{ and } \sin\left(\frac{7\pi}{3}\right). \]
Compare with one revolution
Why: Two pi written with denominator three is six pi over three, and seven thirds exceeds six thirds.
\[ 7 \pi / 3 > 2 \pi \]
Subtract one revolution
Why: Removing a whole turn does not move the terminal side.
\[ 7 \pi / 3 - 6 \pi / 3 = \frac{\pi}{3} \]
Recognise the reduced angle
Why: Pi over three is already acute and in quadrant one, so it is its own reference angle.
\[ \alpha = \frac{\pi}{3}, QI \]
Read off both values with quadrant one signs
Why: Both coordinates are positive in quadrant one, so no sign change is needed.
\[ \cos = \frac{1}{2}, \sin = \sqrt{3} / 2 \]
Figure (svg): The solution to Worked example an angle past one revolution shown as a ladder of expressions, one row per legal move
\[ \cos\left(\frac{7\pi}{3}\right) = \frac{1}{2} \qquad \sin\left(\frac{7\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
Verify: use coterminality directly
Why: Seven pi over three and pi over three differ by exactly two pi, so they are coterminal and must share a point. Any coterminal pair agrees on both cosine and sine, so quoting the pi over three values is not just a shortcut, it is forced.
Trap
\[ \theta = 120^\circ \quad\Longrightarrow\quad \alpha = 120^\circ - 90^\circ = 30^\circ \]
Subtract 90 because the terminal side is past the vertical
Why: The nearest axis really is the y-axis here, so measuring to it feels natural.
This gives a reference angle of 30 degrees, which would make the cosine root three over two in size. But the cosine of 120 degrees has size one half, so the answer is wrong.
\[ \theta = 120^\circ \quad\Longrightarrow\quad \alpha = 180^\circ - 120^\circ = 60^\circ \]
Measure to the x-axis, always
Why: The theorem is stated for the angle with the x-axis, because the reflections that justify it are reflections across the axes of the coordinate system as it is set up.
\[ \cos(120^\circ) = -\cos(60^\circ) = -\tfrac{1}{2}, \qquad \sin(120^\circ) = \sin(60^\circ) = \tfrac{\sqrt{3}}{2} \]
A quick check that catches this: the two reference angles 30 and 60 give different-sized coordinates, so sketch the point and ask whether it is high or wide. At 120 degrees the point is high and close to the vertical, which means a small cosine — one half, not root three over two.
Sorting
The denominator shortcut makes this fast in radians.
Sort into buckets
Sort each angle by its reference angle.
Prediction
Two angles have the same reference angle but lie in different quadrants.
Predict first
What must be true of their cosines and sines?
Correct: The values agree in size but may differ in sign.
Why: That is precisely what the Reference Angle Theorem states. The reference angle fixes the distance of the point from each axis, so the magnitudes of both coordinates are determined; only the quadrant decides the signs. The fourth option is too strong: two angles with the same reference angle in quadrants one and two share a sine exactly, with no sign change at all, so it is not true that both values flip.
Edge cases
The reference angle is defined only for a non-quadrantal angle.
Discussion prompt
What goes wrong if you try to find the reference angle of pi over two? And how does the book still manage to use pi over two when solving cosine of theta equals zero?
Hint: Try each of the four subtraction rules on it and see what you get.
Answer:
The terminal side of pi over two lies along the y-axis, so the acute angle it makes with the x-axis is a right angle — which is not acute. The definition asks for something that does not exist, and the four quadrant rules disagree with each other about which one applies, since the angle is in no quadrant.
The book is explicit that pi over two is not technically a reference angle, but uses it anyway when solving cosine of theta equals zero, because the symmetry argument still works: the two solutions on the y-axis really are mirror images across the x-axis. This is a case where the machinery is being borrowed slightly outside its stated domain, and saying so is more honest than quietly widening the definition.
Section
Section 2
Concept
The Reference Angle Theorem produces plus-or-minus deliberately. Resolving it is a separate question with a separate answer, and running the two steps separately is what keeps them both correct.
There is a well-known mnemonic for this. It is not needed and it is worth resisting: reciting a mnemonic is slower and less reliable than picturing the point and asking whether it is left of centre and below centre.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 722-722
Picture it
This is the figure the book says should be committed to memory. It is generated entirely by the previous two sections.
Figure (svg): The complete unit circle with all sixteen special angles marked, each labelled with its exact coordinates, the figure the book says should be committed to memory
Do not memorise it as sixteen facts. Memorise three first-quadrant points, four quadrantal points, and the two rules — reference angle for the size, quadrant for the sign — and the figure regenerates itself in seconds.
Worked example
Example 10.2.3, part 2. Here the two signs differ, which is where care is needed.
\[ \text{Find } \cos\left(\frac{11\pi}{6}\right) \text{ and } \sin\left(\frac{11\pi}{6}\right). \]
Place the angle
Why: Eleven twelfths of a revolution lands just short of the positive x-axis, in quadrant four.
Find the reference angle
Why: In quadrant four, subtract from two pi.
\[ \alpha = 2 \pi - 11 \pi / 6 = \frac{\pi}{6} \]
Take the sizes from the reference angle
Why: At pi over six the coordinates are root three over two and one half.
\[ \text{sizes } \sqrt{3} / 2\text{ and } \frac{1}{2} \]
Attach quadrant four signs
Why: Cosine positive, sine negative.
\[ \cos = \sqrt{3} / 2, \sin = -\frac{1}{2} \]
Figure (svg): The solution to Worked example quadrant four, mixed signs shown as a ladder of expressions, one row per legal move
\[ \cos\left(\frac{11\pi}{6}\right) = \frac{\sqrt{3}}{2} \qquad \sin\left(\frac{11\pi}{6}\right) = -\frac{1}{2} \]
Verify: check against the picture
Why: Just below the positive x-axis means far right and slightly down: a large positive cosine and a small negative sine. Root three over two is about 0.87 and negative one half is small and negative, matching.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 723-723
Comparison
Fill the blanks from the picture, not from a mnemonic.
Comparison matrix
| Quadrant | Cosine | Sine |
|---|---|---|
| I | positive | positive |
| II | negative | positive |
| III | negative | negative |
| IV | positive | negative |
Read the cosine column: positive, negative, negative, positive — it is negative exactly in the two left-hand quadrants. Read the sine column: positive, positive, negative, negative — negative exactly in the two lower quadrants. Each column is answering one question about direction.
Worked example
Example 10.2.3, part 3. The sign of the angle and the signs of the values are different questions.
\[ \text{Find } \cos\left(-\frac{5\pi}{4}\right) \text{ and } \sin\left(-\frac{5\pi}{4}\right). \]
Sweep the angle clockwise
Why: Five pi over four is five eighths of a revolution, clockwise.
\[ \text{past } 180 ^\circ\text{ clockwise by } 45 ^\circ \]
Identify the quadrant
Why: Five eighths of a turn clockwise lands above the negative x-axis, in quadrant two.
Find the reference angle
Why: The terminal side makes an angle of pi over four with the negative x-axis.
\[ \alpha = 5 \pi / 4 - \pi = \frac{\pi}{4} \]
Attach quadrant two signs
Why: Cosine negative, sine positive, with both sizes root two over two.
\[ \cos = -\sqrt{2} / 2, \sin = \sqrt{2} / 2 \]
Figure (svg): The solution to Worked example a negative angle shown as a ladder of expressions, one row per legal move
\[ \cos\left(-\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2} \qquad \sin\left(-\frac{5\pi}{4}\right) = \frac{\sqrt{2}}{2} \]
Verify: convert to a positive coterminal angle
Why: Adding two pi gives three pi over four, which is in quadrant two with reference angle pi over four — the same conclusion. The negative sign on the angle did not make either value negative; only the quadrant did that, and only to one of them.
Error analysis
A student finds the cosine and sine of two pi over three.
Annotate
On: \( \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2}, \qquad \sin\left(\frac{2\pi}{3}\right) = -\frac{\sqrt{3}}{2} \)
The failure mode is treating a quadrant as though it flipped everything. Ask the two questions separately every time: is the point left of the vertical axis, and is it below the horizontal axis? In quadrant two the answers are yes and no.
Faded example
Find the cosine and sine of seven pi over four.
Fill in the blanks
\alpha = 2\pi - \fracpi/4sqrt2/2 = -sqrt2/2 \quad\text___\quad\Longrightarrow\quad \cos = ___, \;\sin = ___
Why: Two pi written with denominator four is eight pi over four, and subtracting seven pi over four leaves pi over four. That is the reference angle, so both coordinates have size root two over two. Quadrant four makes the cosine positive and the sine negative. The denominator shortcut confirms the reference angle instantly: a reduced denominator of four always means pi over four.
Discrimination
Two angles share a sine when their points are at the same height.
Sort into buckets
Sort each pair by whether the two angles have the same sine.
Explain it to yourself
Twelve of the sixteen special points use only three distinct pairs of numbers.
Discussion prompt
Explain why that is, and say exactly how many facts you actually have to remember to reproduce the whole circle.
Hint: Count the reference angles and count the quadrants.
Answer:
There are only three non-quadrantal reference angles among the special angles — pi over six, pi over four and pi over three — and each occurs once in each of the four quadrants. Three reference angles times four quadrants is the twelve non-quadrantal points.
So the facts to remember are: three coordinate pairs, the four quadrantal points which are read off the axes and need no memory at all, and the two rules — reference angle for size, quadrant for sign. That is three things memorised rather than sixteen, and the three are the ones you derived from triangles in the previous lesson.
Section
Section 3
Concept
For a special angle written as a reduced fraction times pi, the denominator alone names the reference angle. This is the one genuinely convenient thing about radian measure that has nothing to do with theory, and the book says so.
The fraction must be in lowest terms for this to work. Two pi over four is really pi over two, a quadrantal angle, and reading its denominator as four would give the wrong answer.
Figure (svg): A table showing that a radian measure whose reduced denominator is six has reference angle pi over six, denominator four has pi over four, and denominator three has pi over three
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 724-724
Picture it
Combine this with the quadrant and every special angle is a two-second evaluation.
Figure (svg): A table showing that a radian measure whose reduced denominator is six has reference angle pi over six, denominator four has pi over four, and denominator three has pi over three
The numerator tells you the quadrant and the denominator tells you the reference angle. Between them they determine both coordinates completely.
Worked example
Use the shortcut for the size and the numerator for the quadrant.
\[ \text{Find the cosine and sine of } \frac{5\pi}{3}, \; \frac{3\pi}{4}, \; \frac{7\pi}{6}. \]
Read the reference angles off the denominators
Why: Denominators three, four and six give pi over three, pi over four and pi over six.
\[ \alpha = \frac{\pi}{3}, \frac{\pi}{4}, \frac{\pi}{6} \]
Locate each quadrant from the numerator
Why: Five sixths of a turn is QIV; three eighths is QII; seven twelfths is QIII.
Attach the signs
Why: QIV is plus then minus, QII is minus then plus, QIII is minus then minus.
Write out all six values
Why: Sizes from the reference angle, signs from the quadrant.
Figure (svg): The solution to Worked example evaluate three angles at speed shown as a ladder of expressions, one row per legal move
\[ \frac{5\pi}{3}: \left(\tfrac{1}{2}, -\tfrac{\sqrt{3}}{2}\right) \quad \frac{3\pi}{4}: \left(-\tfrac{\sqrt{2}}{2}, \tfrac{\sqrt{2}}{2}\right) \quad \frac{7\pi}{6}: \left(-\tfrac{\sqrt{3}}{2}, -\tfrac{1}{2}\right) \]
Verify: check each against the identity
Why: One quarter plus three quarters is 1; one half plus one half is 1; three quarters plus one quarter is 1. All three points lie on the unit circle, and each set of signs matches the quadrant named.
Sorting
A special angle is one whose exact values come from the two triangles or from an axis.
Sort into buckets
Sort each angle by whether it has exact cosine and sine values you can write in surds.
Worked example
The shortcut fails silently on an unreduced fraction, so reducing is not optional.
\[ \text{Find the cosine and sine of } \frac{6\pi}{4}. \]
Notice the fraction is not in lowest terms
Why: Six and four share a factor of two.
\[ 6 \pi / 4\text{ is not reduced} \]
Reduce it
Why: Dividing top and bottom by two gives three pi over two.
\[ 6 \pi / 4 = 3 \pi / 2 \]
Recognise the reduced angle
Why: A denominator of two means a quadrantal angle, not a reference angle of pi over two used as a size.
Read the point off the axis
Why: Three pi over two is three quarters of a turn, pointing straight down.
\[ P = (0, -1) \]
Figure (svg): The solution to Worked example a fraction that must be reduced first shown as a ladder of expressions, one row per legal move
\[ \frac{6\pi}{4} = \frac{3\pi}{2} \quad\Longrightarrow\quad \cos = 0, \; \sin = -1 \]
Verify: see what the shortcut would have said
Why: Read carelessly, the denominator four would suggest a reference angle of pi over four and coordinates of size root two over two — completely wrong. Reducing first is what prevents that, and it costs one glance.
Trap
\[ \theta = \frac{2\pi}{5} \quad\Longrightarrow\quad \text{denominator } 5 \quad\Longrightarrow\quad \alpha = \frac{\pi}{5} \]
Read the denominator as the reference angle, as with 3, 4 and 6
Why: The pattern seems general, and nothing in the notation flags that five is different.
The reference angle of two pi over five is actually two pi over five itself, since the angle is acute and in quadrant one. And in any case pi over five is not an angle whose cosine and sine have nice exact values.
\[ \theta = \frac{2\pi}{5} \in \text{QI} \quad\Longrightarrow\quad \alpha = \theta = \frac{2\pi}{5} \approx 72^\circ \]
Use the quadrant rules, which always work, rather than the shortcut, which works only for 3, 4 and 6
Why: The shortcut is a coincidence of the special angles, not a theorem.
The deeper point: the shortcut identifies which special angle you have, not how to find a reference angle in general. Denominators of three, four and six are exactly the ones arising from the two special triangles. Any other denominator means the angle is not special, its cosine and sine have no exact form in surds, and a calculator is required.
Fill the middle
Evaluate the cosine of four pi over three.
Fill in the blanks
\fracIIIpi/3 \in \text-1/2___, \quad \alpha = \frac______ - \pi = ___ \quad\Longrightarrow\quad \cos\left(\frac______\right) = ___
Why: Four pi over three is two thirds of a revolution, which lands between pi and three pi over two — quadrant three. Subtracting pi gives the reference angle pi over three, matching what the denominator predicted. The cosine of pi over three is one half, and quadrant three makes it negative.
Ranking
Evaluate each with the shortcut, then compare the signed values.
Put in order
Why: The values are negative one, negative root three over two which is about negative 0.87, negative one half, zero, and positive one half. They are already in increasing order as listed, which is no accident: the five angles are in decreasing order from pi down to pi over three, and cosine increases as the angle decreases across the upper half of the circle, because the point is sliding rightward.
Explain it
A classmate is drawing the whole triangle every single time, even for angles like seven pi over six.
Discussion prompt
Teach them the denominator shortcut in four sentences or fewer, and give them the one warning that comes with it.
Hint: Say what the denominator does and what the numerator does.
Answer:
A usable answer: for an angle written as a reduced multiple of pi, the denominator names the reference angle — six means pi over six, four means pi over four, three means pi over three — so the sizes of both coordinates are known immediately. The numerator tells you which quadrant you are in, which fixes the two signs. Between them you have both values without drawing anything.
The warning: reduce the fraction first, and remember the shortcut only works for denominators of three, four and six. Six pi over four looks like a denominator of four but is really three pi over two, a quadrantal angle, and an angle like two pi over five has no exact value at all.
Section
Section 4
Concept
A problem may give you the cosine and sine of some unnamed acute angle alpha and ask for the values at pi plus alpha, or two pi minus alpha. These are handled by exactly the same method: locate the terminal side, find that alpha is the reference angle, and read off the signs.
That last case is the one worth dwelling on. Its reference angle is the complement of alpha, and the coordinates come out swapped rather than merely re-signed — which is a preview of the cofunction identities in Lesson 10.4a.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 725-727
Picture it
Every one of these constructions is a walk around the same figure.
Figure (svg): The complete unit circle with all sixteen special angles marked, each labelled with its exact coordinates, the figure the book says should be committed to memory
If you can place pi plus alpha on this picture for an alpha of your choosing, you can place any of them.
Worked example
Example 10.2.4. Alpha is acute with cosine five thirteenths, so its sine is twelve thirteenths.
\[ \text{Given acute } \alpha \text{ with } \cos(\alpha) = \tfrac{5}{13}, \; \sin(\alpha) = \tfrac{12}{13}, \text{ find the values at } \pi + \alpha \text{ and } 2\pi - \alpha. \]
Locate pi plus alpha
Why: Rotate half a revolution, then a further alpha. Since alpha is acute the result is just past the negative x-axis.
Identify its reference angle
Why: The terminal side makes an angle of alpha with the negative x-axis.
Attach quadrant three signs
Why: Both coordinates negative.
\[ \cos = -\frac{5}{13}, \sin = -\frac{12}{13} \]
Repeat for two pi minus alpha
Why: A full revolution then backing up alpha lands just below the positive x-axis, in quadrant four, again with reference angle alpha.
\[ \cos = \frac{5}{13}, \sin = -\frac{12}{13} \]
Figure (svg): The solution to Worked example pi plus alpha and two pi minus alpha shown as a ladder of expressions, one row per legal move
\[ \cos(\pi + \alpha) = -\tfrac{5}{13}, \; \sin(\pi + \alpha) = -\tfrac{12}{13} \qquad \cos(2\pi - \alpha) = \tfrac{5}{13}, \; \sin(2\pi - \alpha) = -\tfrac{12}{13} \]
Verify: check the identity for both
Why: Twenty-five plus one hundred forty-four over one hundred sixty-nine is 1 in each case, since squaring removes every sign. And each pair of signs matches its named quadrant.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 725-726
Matching
Take alpha to be acute throughout.
Match the pairs
Why: Pi minus alpha stops just short of the negative x-axis, in quadrant two. Pi plus alpha goes just past it, into quadrant three. Two pi minus alpha backs up from a full turn into quadrant four. Three pi minus alpha is one and a half revolutions then backing up alpha, which lands in quadrant two — the same place as pi minus alpha, since three pi and pi are coterminal.
Worked example
Example 10.2.4, part 2d. Read this one slowly — the book asks you to see why it is different before continuing.
\[ \text{With the same } \alpha, \text{ find } \cos\left(\frac{\pi}{2} + \alpha\right) \text{ and } \sin\left(\frac{\pi}{2} + \alpha\right). \]
Locate the angle
Why: A quarter turn, then a further alpha, puts the terminal side in quadrant two.
Check whether alpha is the reference angle
Why: The terminal side makes an angle of alpha with the y-axis, not with the x-axis, so alpha is not the reference angle.
Use symmetry instead
Why: Equal angles subtend equal chords, so the new point is the old point rotated a quarter turn, which swaps the coordinates and negates one.
Read off the result
Why: The x-coordinate becomes minus the old y, and the y-coordinate becomes the old x.
\[ \cos = -\frac{12}{13}, \sin = \frac{5}{13} \]
Figure (svg): The solution to Worked example the case where the theorem does not apply shown as a ladder of expressions, one row per legal move
\[ \cos\left(\frac{\pi}{2} + \alpha\right) = -\frac{12}{13} \qquad \sin\left(\frac{\pi}{2} + \alpha\right) = \frac{5}{13} \]
Verify: check the identity and the quadrant
Why: One hundred forty-four plus twenty-five over one hundred sixty-nine is 1. And quadrant two demands a negative cosine and a positive sine, which is what came out. Note that the values swapped rather than merely changing sign — the signature of a quarter-turn rotation rather than a reflection.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 727-727
Error analysis
A student computes the cosine of pi over two plus alpha for an acute alpha.
Annotate
On: \( \cos\left(\frac{\pi}{2} + \alpha\right) = -\cos(\alpha) \)
The test is quick and it is worth running every time: is the terminal side measured from the x-axis or the y-axis? Adding a multiple of pi keeps alpha as the reference angle; adding an odd multiple of pi over two does not.
Faded example
Alpha is acute with sine three fifths and cosine four fifths. Find the values at pi minus alpha.
Fill in the blanks
\pi - \alpha \in \textII-4/5, \text3/5 \alpha \;\Longrightarrow\; \cos = ___, \;\sin = ___
Why: Subtracting an acute angle from pi stops just short of the negative x-axis, so the angle is in quadrant two with reference angle alpha. Quadrant two negates the cosine and leaves the sine positive. This is a result worth recognising on sight: the sine of pi minus alpha always equals the sine of alpha, which is exactly why the Law of Sines will have an ambiguous case in Lesson 11.2.
Prediction
Alpha is acute. Consider the two angles pi minus alpha and pi plus alpha.
Predict first
How do their sines compare?
Correct: They are negatives of each other.
Why: Both angles have reference angle alpha, so both sines have size equal to the sine of alpha. But pi minus alpha is in quadrant two where the sine is positive, and pi plus alpha is in quadrant three where it is negative. Geometrically the two points are reflections of each other across the x-axis, which preserves the horizontal coordinate and flips the vertical one — so in fact their cosines are equal while their sines are opposite.
Socratic
Adding pi to an angle keeps alpha as the reference angle. Adding pi over two does not.
Discussion prompt
Explain the difference geometrically. What does adding pi do to the point, and what does adding pi over two do?
Hint: One is a reflection through a point; the other is a rotation.
Answer:
Adding pi sends the point to the diametrically opposite point, which is the reflection through the origin. That negates both coordinates but keeps them in their places, so the sizes are unchanged and alpha remains the reference angle.
Adding pi over two rotates the point a quarter turn. A quarter-turn rotation sends the point with coordinates x and y to the point with coordinates negative y and x — it exchanges the roles of the two coordinates. So the new cosine is built from the old sine, and the reference angle is the complement of alpha rather than alpha.
Any multiple of pi preserves the reference angle; any odd multiple of pi over two swaps the two functions. That distinction is the seed of the cofunction identities and of the phase-shift relationship between the sine and cosine graphs in Lesson 10.5a.
Section
Section 5
Concept
Asking which angles have a given cosine is asking where a vertical line cuts the unit circle. It cuts in two places, so there are two reference-angle positions, and each of them repeats every revolution.
\[ \cos(\theta) = \tfrac{1}{2} \;\Longrightarrow\; \theta = \tfrac{\pi}{3} + 2\pi k \;\text{ or }\; \theta = \tfrac{5\pi}{3} + 2\pi k, \quad k \in \mathbb{Z} \]
The answer to such a question is never a single angle. It is a description of infinitely many, and leaving out the arbitrary integer k leaves out almost all of the answer.
Figure (svg): The horizontal line x equals one half cutting the unit circle at two points, one in quadrant one and one in quadrant four, showing the two families of solutions to the equation cosine theta equals one half
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 728-729
Picture it
The picture is the method. The line's height is the value; the crossings are the solutions.
Figure (svg): The horizontal line y equals negative one half cutting the unit circle at two points in quadrants three and four, giving the two families of solutions to the equation sine theta equals negative one half
Because the value here is negative the line sits below the axis, so the crossings are in quadrants three and four — the two quadrants where the sine is negative.
Worked example
Example 10.2.5, part 1. Find all angles, not just the obvious one.
\[ \text{Solve } \cos(\theta) = \frac{1}{2} \text{ for all } \theta. \]
Draw the vertical line at one half
Why: Cosine is the x-coordinate, so the solutions are where the circle has x equal to one half.
\[ \text{line } x = \frac{1}{2}\text{ cuts twice} \]
Identify the two quadrants
Why: A positive cosine means the crossings are to the right of the y-axis, in quadrants one and four.
Find the reference angle
Why: A cosine of size one half belongs to the angle pi over three.
\[ \alpha = \frac{\pi}{3} \]
Write one solution per quadrant and add the families
Why: Quadrant one gives pi over three; quadrant four gives two pi minus pi over three, which is five pi over three.
\[ \frac{\pi}{3} + 2 \pi k\text{ and } 5 \pi / 3 + 2 \pi k \]
Figure (svg): The solution to Worked example solve a cosine equation shown as a ladder of expressions, one row per legal move
\[ \theta = \frac{\pi}{3} + 2\pi k \quad\text{or}\quad \theta = \frac{5\pi}{3} + 2\pi k, \quad k \in \mathbb{Z} \]
Verify: test one member of each family
Why: Taking k equal to 1 in the first family gives seven pi over three, which is coterminal with pi over three and so has cosine one half. Taking k equal to negative 1 in the second gives negative pi over three, which is in quadrant four with reference angle pi over three, so its cosine is also one half.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 728-728
Prediction
Consider the equation cosine of theta equals 1.
Predict first
How many families of solutions does it have?
Correct: One family of solutions, not the usual two.
Why: The vertical line at x equals 1 is tangent to the unit circle, touching it at exactly one point, one comma zero. So there is a single position of the terminal side, and the solution is theta equals zero plus two pi k, a single family. The value 1 and the value negative 1 are the two exceptional cases where the usual two crossings collapse into one, which is exactly what happens at a maximum or minimum of the cosine graph in Lesson 10.5a.
Worked example
Example 10.2.5, part 2. The sign of the value chooses the quadrants.
\[ \text{Solve } \sin(\theta) = -\frac{1}{2} \text{ for all } \theta. \]
Draw the horizontal line at negative one half
Why: Sine is the y-coordinate, so this time the line is horizontal and below the axis.
\[ \text{line } y = -\frac{1}{2}\text{ cuts twice} \]
Identify the two quadrants
Why: A negative sine means the crossings are below the x-axis, in quadrants three and four.
Find the reference angle from the size
Why: Ignore the sign: a sine of size one half belongs to pi over six.
\[ \alpha = \frac{\pi}{6} \]
Write both families
Why: Quadrant three gives pi plus pi over six; quadrant four gives two pi minus pi over six.
\[ 7 \pi / 6 + 2 \pi k\text{ and } 11 \pi / 6 + 2 \pi k \]
Figure (svg): The solution to Worked example solve a sine equation with a negative value shown as a ladder of expressions, one row per legal move
\[ \theta = \frac{7\pi}{6} + 2\pi k \quad\text{or}\quad \theta = \frac{11\pi}{6} + 2\pi k, \quad k \in \mathbb{Z} \]
Verify: check an alternative form of the second family
Why: The book notes that negative pi over six is also a quadrant four solution, so that family could be written as negative pi over six plus two pi k. Taking k equal to 1 there gives eleven pi over six, so the two descriptions list the same angles — a reminder that a family can be written correctly in more than one way.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 728-729
Trap
\[ \cos(\theta) = \tfrac{1}{2} \quad\Longrightarrow\quad \theta = \frac{\pi}{3} \]
Find the reference angle and stop
Why: Pi over three genuinely is a solution, and it is the one that comes to mind first, so it feels like the answer.
But the question asked for all angles. This answer omits the entire quadrant four family and every coterminal angle in both families — it names one angle out of infinitely many.
\[ \theta = \frac{\pi}{3} + 2\pi k \quad\text{or}\quad \theta = \frac{5\pi}{3} + 2\pi k, \quad k \in \mathbb{Z} \]
Give one solution per crossing, then add the coterminal family to each
Why: A line cuts the circle twice, so there are two positions, and each repeats every revolution.
Two checks catch the omission. Did I use both crossings? A horizontal or vertical line meets the circle twice unless it is tangent at plus or minus one. Did I write plus two pi k? Without it the answer is a finite list where an infinite family was required.
Fill the middle
Solve the equation sine of theta equals root two over two.
Fill in the blanks
\alpha = pi/4, \quad \textI3 pi/4 \text___ \;\Longrightarrow\; \theta = \frac______ + 2\pi k \;\text___\; \theta = ___ + 2\pi k
Why: A sine of root two over two is positive, so the horizontal line sits above the axis and cuts in quadrants one and two. The reference angle is pi over four. Quadrant one gives pi over four directly, and quadrant two gives pi minus pi over four, which is three pi over four. Each family then carries plus two pi k.
Elimination
The equation is cosine of theta equals negative root three over two.
Eliminate the wrong options
Which of these gives all the solutions?
Survives elimination: B
Why: A cosine of negative root three over two puts the vertical line to the left of the y-axis, cutting in quadrants two and three, and the size root three over two gives reference angle pi over six. Quadrant two gives pi minus pi over six, which is five pi over six; quadrant three gives pi plus pi over six, which is seven pi over six. Each carries plus two pi k. Note that the step in a solution family is two pi and not pi, because it is coterminality that generates the repetition.
Missing information
A problem says: solve sine of theta equals one half for theta in the interval from zero to two pi.
Discussion prompt
How does the added interval change the shape of the answer compared with solving for all theta? Give both answers and say which is more useful and when.
Hint: How many members of each family fall inside an interval of width two pi?
Answer:
\[ \text{all } \theta: \quad \theta = \tfrac{\pi}{6} + 2\pi k \;\text{ or }\; \theta = \tfrac{5\pi}{6} + 2\pi k \]
\[ \theta \in [0, 2\pi): \quad \theta = \tfrac{\pi}{6} \;\text{ or }\; \theta = \tfrac{5\pi}{6} \]
The interval has width exactly two pi, so each infinite family contributes exactly one member. The answer collapses from two families to two numbers.
Which is more useful depends on the question behind it. A physical problem asking when a wheel is at a certain position wants the whole family, because it recurs. A problem asking for the angles of a triangle wants the restricted answer, because a triangle's angle cannot be 13 pi over 6. Lesson 10.7a is entirely about managing this distinction.
Comparison
These are inverse tasks, and confusing which one you are doing is a real source of error. Fill the blanks from memory.
Comparison matrix
| Evaluating | Solving | |
|---|---|---|
| You are given | an angle | a value |
| You want | a value | every angle |
| How many answers | exactly one | infinitely many, in two families |
| The picture | a point on the circle | a line cutting the circle |
| Role of the quadrant | supplies the signs | the value's sign selects the quadrants |
Notice that the quadrant does opposite jobs in the two columns. When evaluating you know the quadrant and use it to fix signs; when solving you know the sign and use it to find the quadrants.
Pattern
Whether you are evaluating a special angle or solving a basic equation, the same five moves cover it.
When solving rather than evaluating, run this in reverse: the sign of the given value names the two candidate quadrants, its size names the reference angle, and each quadrant contributes one solution plus a coterminal family.
Check
A reference angle. Sketch it before choosing.
Check your understanding
What is the reference angle for 240 degrees?
Answer: B
Why: Two hundred forty degrees lies between 180 and 270, so it is in quadrant three, and there the reference angle is the angle minus 180. That gives 60 degrees, which is acute as a reference angle must be.
Check
A special value. Reference angle for the size, quadrant for the sign.
Check your understanding
What is the sine of five pi over four?
Answer: B
Why: Five pi over four is five eighths of a revolution, which lands in quadrant three. The reduced denominator of four gives a reference angle of pi over four, whose sine has size root two over two. Quadrant three makes the sine negative, so the answer is negative root two over two.
Check
Solving. Remember that the answer is a family, not a number.
Check your understanding
Which describes all solutions of sine of theta equals 1?
Answer: B
Why: The horizontal line at y equals 1 is tangent to the unit circle, touching only at the point zero comma one, so there is a single terminal side rather than two. That position is pi over two, and it repeats every full revolution, giving one family with step two pi.
Real world
A Ferris wheel of radius 20 metres has its centre 22 metres above the ground. A rider boards at the lowest point and the wheel turns counterclockwise. The rider's height above the ground is 22 plus 20 times the sine of the angle turned from the horizontal.
Discussion prompt
At which angles is the rider exactly 32 metres up? Give all of them, and explain why a single answer would be the wrong shape of answer for this question.
Hint: Set up the equation, solve for the sine, and then use the two-families method.
Answer:
\[ 22 + 20\sin(\theta) = 32 \;\Longrightarrow\; \sin(\theta) = \tfrac{1}{2} \]
\[ \theta = \tfrac{\pi}{6} + 2\pi k \quad\text{or}\quad \theta = \tfrac{5\pi}{6} + 2\pi k \]
There are two positions per revolution at that height — once climbing on the right of the wheel and once descending on the left — and the wheel keeps turning, so each recurs every revolution.
A single answer would be the wrong shape because the physical situation is genuinely periodic. Asking when the rider is 32 metres up has infinitely many correct answers, and the two families are exactly the statement 'twice per turn, every turn'. Discarding the family would answer a question nobody asked.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Two angles have the same reference angle. Must they have the same cosine?
Correct: No, but the cosines always have the same absolute value.
\[ \cos\left(\tfrac{\pi}{3}\right) = \tfrac{1}{2} \qquad \cos\left(\tfrac{2\pi}{3}\right) = -\tfrac{1}{2} \]
Why: The reference angle fixes how far the point is from the y-axis, so it fixes the size of the cosine completely. What it cannot fix is the side: the point may be to the left or to the right, so the cosine may be positive or negative. Both the second and fourth options say something true, but the second is the full statement, since it describes what is guaranteed in every case rather than naming a condition under which equality happens to hold. Concretely, pi over three and two pi over three share the reference angle pi over three and have cosines of one half and negative one half.
Explain it
They have memorised the sixteen-point unit circle as sixteen separate facts and keep forgetting half of them the night before a test.
Discussion prompt
In no more than five sentences, teach them how to regenerate the whole circle from three facts. Then give them the one check that catches a sign error every time.
Hint: Separate the size question from the sign question.
Answer:
A usable answer: you only need three points, the ones at thirty, forty-five and sixty degrees in the first quadrant. Every other non-quadrantal point is one of those three with signs attached. Find the reference angle — how far the arm is from the horizontal axis — and that tells you the two numbers; then look at which quadrant you are in, and that tells you the two signs. The four points on the axes need no memory at all, since you can just read them off.
The check: square both values and add them. You must get exactly 1, because the point is on a circle of radius one. That catches a wrong pair instantly, though not a sign error — for signs, ask separately whether the point is left of centre and whether it is below centre.
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: Reference angles are fixed by always measuring to the x-axis and checking the result is acute. Signs are fixed by asking two separate questions about left and below rather than reciting a mnemonic. The described-angle cases are fixed by asking whether the terminal side is measured from the x-axis or the y-axis, since a multiple of pi preserves the reference angle and an odd multiple of pi over two does not. Solution families are fixed by two habits: use both crossings, and always write plus two pi k. 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 large unit circle with axes and reproduce the complete sixteen-point figure from memory, labelling every point with its exact coordinates and every angle in radians. Do not consult the slides while you draw it; consult them afterwards and mark your errors in a different colour. Around the outside, write in each quadrant the reference angle rule for that quadrant and the two signs. In the top corner, write the three coordinate pairs that generate all twelve non-quadrantal points, and draw an arrow from each to the four points on the circle it generates. In the bottom corner, solve completely and write out both families for the equation cosine of theta equals negative one half, and then draw the vertical line on your circle that produces those two solutions. Finally, circle the two points on your figure where a vertical line would cut the circle only once, and say what is special about the cosine there.
The circled points should be one comma zero and negative one comma zero, where the cosine is 1 and negative 1. Those are the only cosine values giving a single family rather than two, because the vertical line is tangent there — and they are exactly the maximum and minimum of the cosine graph you will meet in Lesson 10.5a.
Recap
Five things, and the third one replaces sixteen memorised facts with three.
| If the question says | Your first move is |
|---|---|
| Find the exact value at this angle | Reduce into one revolution, then name the quadrant |
| The angle is negative or over two pi | Add or subtract whole revolutions first |
| Given cos alpha, find cos of pi plus alpha | Locate the terminal side and check the reference angle |
| Find all angles with this cosine | Draw the vertical line at that value |
| Find all angles with this sine | Draw the horizontal line at that value |
Everything so far has lived on a circle of radius one. The next lesson lifts that restriction: for a point on a circle of any radius, the coordinates turn out to be the radius times the cosine and the radius times the sine — which is what finally connects this material back to right triangles of any size.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 722-730 — everything on these slides traces back here
Want this taught 1-on-1? Alexander tutors Trigonometry — $55/session, free consultation.