Cosine and sine defined as the two coordinates of the point where an angle's terminal side crosses the unit circle, rather than as ratios in a triangle. Covers reading the quadrantal values straight off the axes, deriving the values at pi over six, pi over four and pi over three from the two special right triangles together with the circle equation, the Pythagorean identity as the circle equation with its coordinates renamed, and recovering a missing coordinate from the identity plus quadrant information.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 10 — Foundations of Trigonometry
§10.2 The Unit Circle: Cosine and Sine, pp. 717-722
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. 717-722 — the pages these objectives are drawn from
Warm-up
You already know the equation of a circle of radius one centred at the origin. That equation is about to become the single most important identity in this course.
Discussion prompt
Write down the equation of the circle of radius 1 centred at the origin. Now: if a point on that circle has x-coordinate three fifths, what are the possible y-coordinates, and what extra information would pin down which one it is?
Hint: Substitute and solve. The plus-or-minus is the whole point of the question.
Answer:
\[ x^2 + y^2 = 1 \qquad \left(\tfrac{3}{5}\right)^2 + y^2 = 1 \;\Longrightarrow\; y^2 = \tfrac{16}{25} \;\Longrightarrow\; y = \pm\tfrac{4}{5} \]
Two points on the circle share that x-coordinate, one above the axis and one below. The algebra genuinely cannot choose between them — only knowing which half of the circle the point is in can.
That is the exact structure of every problem in the last section of this lesson. The identity narrows the answer to two values; the quadrant picks one.
Concept
Put an angle in standard position. Its terminal side crosses the unit circle at exactly one point. The first coordinate of that point is called the cosine of the angle and the second is called the sine.
cosine and sine — For an angle in standard position, let P be the point where its terminal side meets the unit circle. The x-coordinate of P is the cosine of the angle and the y-coordinate of P is the sine of the angle.
\[ P\big(\cos(\theta), \; \sin(\theta)\big) \text{ on } x^2 + y^2 = 1 \]
Each angle determines exactly one such point, so each determines exactly one cosine and exactly one sine. That is what makes these rules functions rather than merely correspondences, and it is worth checking rather than assuming.
Figure (svg): An angle theta in standard position with its terminal side crossing the unit circle at a point P, whose x-coordinate is labelled cosine of theta and whose y-coordinate is labelled sine of theta
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 717-717
Section
Section 1
Concept
If you have met sine and cosine before, it was probably as ratios of sides in a right triangle. That definition is true but crippled: a triangle's angles are all acute, so it can say nothing about an angle of 200 degrees or a negative one.
So nothing you learned before is wrong. It is being extended, and the extension is what makes the periodic functions of Section 10.5 possible.
Figure (svg): An angle theta in standard position with its terminal side crossing the unit circle at a point P, whose x-coordinate is labelled cosine of theta and whose y-coordinate is labelled sine of theta
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 717-717
Picture it
Because cosine and sine are coordinates, their signs are the signs of coordinates and nothing more.
Figure (svg): A grid of the four quadrants showing the sign of the x-coordinate and the y-coordinate in each, and therefore the sign of cosine and sine there
Many students memorise a mnemonic for this. It is faster and more reliable to picture the point and ask whether it is left of centre and whether it is below centre.
Worked example
Example 10.2.1, part 1. This one needs no computation whatsoever.
\[ \text{Find } \cos(270^\circ) \text{ and } \sin(270^\circ). \]
Place the angle in standard position
Why: Two hundred seventy degrees is three quarters of a counterclockwise revolution.
Find where the terminal side meets the unit circle
Why: Straight down, one unit from the origin.
\[ P = (0, -1) \]
Read the first coordinate as the cosine
Why: The definition says cosine is the x-coordinate, and here that is zero.
\[ \cos 270 ^\circ = 0 \]
Read the second coordinate as the sine
Why: Sine is the y-coordinate, which is negative one.
\[ \sin 270 ^\circ = -1 \]
Figure (svg): The solution to Worked example a quadrantal angle shown as a ladder of expressions, one row per legal move
\[ \cos(270^\circ) = 0 \qquad \sin(270^\circ) = -1 \]
Verify: check against the circle equation
Why: Zero squared plus negative one squared is 1, so the point really is on the unit circle. Any pair of values failing that check cannot be a cosine and sine of the same angle.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 717-717
Matching
All four points are read straight off the axes.
Match the pairs
Why: Sweep each angle and see where the terminal side crosses the circle. Zero has not moved from the positive x-axis, so the point is one comma zero. A quarter turn is straight up, half a turn is straight left, and three quarters of a turn is straight down. Every one of these has one coordinate zero and the other plus or minus one, which is the signature of a quadrantal angle.
Worked example
Example 10.2.1, part 2. The negative sign changes the direction of the sweep and nothing else.
\[ \text{Find } \cos(-\pi) \text{ and } \sin(-\pi). \]
Interpret the sign
Why: Negative means clockwise, and pi is half a revolution.
Place the terminal side
Why: Half a turn clockwise ends on the negative x-axis, the same ray positive pi reaches.
Find the point on the unit circle
Why: One unit to the left of the origin.
\[ P = (-1, 0) \]
Read off both coordinates
Why: Cosine is the first, sine is the second.
\[ \cos = -1, \sin = 0 \]
Figure (svg): The solution to Worked example a negative quadrantal angle shown as a ladder of expressions, one row per legal move
\[ \cos(-\pi) = -1 \qquad \sin(-\pi) = 0 \]
Verify: compare with positive pi
Why: Positive pi is half a turn counterclockwise and lands on the same ray, so it must give the same point and the same values. Coterminal angles always agree on cosine and sine, because they share the point.
Trap
\[ \theta = \tfrac{\pi}{2} \quad\Longrightarrow\quad P = (0, 1) \quad\Longrightarrow\quad \cos\left(\tfrac{\pi}{2}\right) = 1 \]
Take the nonzero coordinate as the cosine
Why: Cosine is named first and comes first alphabetically, so the first thing that looks substantial gets assigned to it.
But the point is zero comma one, and its FIRST coordinate is zero. Cosine is the x-coordinate whether or not that coordinate happens to be interesting.
\[ \theta = \tfrac{\pi}{2} \quad\Longrightarrow\quad P = (0, 1) \quad\Longrightarrow\quad \cos\left(\tfrac{\pi}{2}\right) = 0, \;\sin\left(\tfrac{\pi}{2}\right) = 1 \]
Match the coordinates to the names by position, not by size
Why: Cosine is always the horizontal coordinate and sine is always the vertical one.
A memory hook that survives: sine rhymes with climb, and climbing is vertical. Or simply notice that the letters c and x come before s and y in their respective alphabets, and the pairing follows that order.
Sorting
Cosine is the x-coordinate, so this is a question about left and right.
Sort into buckets
Sort each angle by whether its cosine is positive, negative, or zero.
Prediction
An angle has cosine equal to negative 0.8.
Predict first
What can you conclude about where its terminal side is?
Correct: It is in quadrant two or quadrant three.
Why: A negative cosine means a negative x-coordinate, which puts the point to the left of the y-axis. That is quadrants two and three together, and cosine alone cannot distinguish them because both contain points with x equal to negative 0.8 — one above the axis and one below. Deciding between them requires the sign of the sine, which is exactly the extra information the last section of this lesson always supplies.
Socratic
The book asks the reader to verify that these rules really do define functions.
Discussion prompt
What would have to go wrong for cosine to fail to be a function of the angle, and why can that not happen here?
Hint: A function gives exactly one output per input. What is the input, and what is the output?
Answer:
It would fail if some angle produced two different cosines. The input is an angle; the output is the x-coordinate of a point.
It cannot happen because an angle in standard position has exactly one terminal side, and a ray from the origin crosses the unit circle exactly once. One angle, one ray, one point, one x-coordinate.
Notice that the reverse is emphatically not a function: one cosine value corresponds to infinitely many angles, since every coterminal angle shares the point and there is usually a second angle in the same revolution too. That failure is why Lesson 10.6a has to restrict the domain before an inverse can exist at all.
Section
Section 2
Concept
For an angle that does not land on an axis there is nothing to read off, so the coordinates must be computed. The method is always the same: drop a perpendicular from the point to the x-axis and use the resulting right triangle together with the circle equation.
For 45 degrees the triangle has two equal angles, so it is isosceles and y equals x. Substituting into the circle equation gives two x squared equals one, so x is root two over two.
Figure (svg): The angle pi over four in standard position, with a perpendicular dropped from the point on the unit circle to the x-axis forming a forty-five forty-five ninety triangle whose two legs are equal
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 718-718
Picture it
The picture is the method. Every non-quadrantal value in this course comes from a version of it.
Figure (svg): The angle pi over four in standard position, with a perpendicular dropped from the point on the unit circle to the x-axis forming a forty-five forty-five ninety triangle whose two legs are equal
The plus-or-minus that comes out of the square root is resolved by the quadrant: P is in quadrant one here, so x is positive.
Worked example
Example 10.2.1, part 3. Follow the two-equation method exactly.
\[ \text{Find } \cos(45^\circ) \text{ and } \sin(45^\circ). \]
Drop a perpendicular from P to the x-axis
Why: The legs of the resulting triangle have lengths x and y, the coordinates being sought.
Use the geometry to relate the legs
Why: The triangle has angles 45, 45 and 90, so it is isosceles and the legs are equal.
\[ y = x \]
Substitute into the equation of the unit circle
Why: P lies on the circle, so its coordinates satisfy that equation.
\[ x ^{2} + x ^{2} = 1 \]
Solve and choose the sign from the quadrant
Why: Two x squared equals one gives x equal to plus or minus root one half; P is in quadrant one so x is positive.
\[ x = \sqrt{2} / 2 \]
Figure (svg): The solution to Worked example cosine and sine of 45 degrees shown as a ladder of expressions, one row per legal move
\[ \cos(45^\circ) = \frac{\sqrt{2}}{2} \qquad \sin(45^\circ) = \frac{\sqrt{2}}{2} \]
Verify: check the identity
Why: Root two over two squared is one half, and one half plus one half is 1, so the point is on the unit circle. It is also reassuring that the two values are equal, which they must be for an angle exactly halfway between the axes.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 718-718
Fill the middle
The derivation at 45 degrees, with the middle removed.
Fill in the blanks
y = x \quad\text2x^2\quad x^2 + y^2 = 1 \;\Longrightarrow\; 1/2 = 1 \;\Longrightarrow\; x^2 = sqrt2/2 \;\Longrightarrow\; x = ___
Why: Substituting y equal to x turns the circle equation into two x squared equals one, so x squared is one half and x is plus or minus one over root two. Rationalising gives root two over two, and the positive root is taken because the point is in quadrant one. Every step is forced once the two equations are written down.
Worked example
The identical angle in radians. Worth doing once so the two notations stop feeling like different problems.
\[ \text{Find } \cos\left(\frac{\pi}{4}\right) \text{ and } \sin\left(\frac{\pi}{4}\right). \]
Recognise the angle
Why: Pi over four divided by two pi is one eighth of a revolution, which is 45 degrees.
\[ \frac{\pi}{4} = 45 ^\circ \]
Note that the point is unchanged
Why: Cosine and sine depend on the terminal side, not on the units used to name the angle.
Quote the values
Why: They were just derived, and no new work is needed.
\[ \text{both } \sqrt{2} / 2 \]
Rationalise if the answer came out as one over root two
Why: One over root two and root two over two are the same number; the second is the conventional form.
\[ 1 / \sqrt{2} = \sqrt{2} / 2 \]
Figure (svg): The solution to Worked example the same values, at pi over four shown as a ladder of expressions, one row per legal move
\[ \cos\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \approx 0.707 \]
Verify: check the size
Why: Both coordinates are about 0.707, which is less than 1 as every coordinate on the unit circle must be, and the point sits diagonally up and to the right — exactly where an eighth of a turn should land it.
Error analysis
A student tries to find the coordinates at 45 degrees using the triangle alone.
Annotate
On: \( y = x \quad\Longrightarrow\quad x = 1, \; y = 1 \quad\Longrightarrow\quad P = (1, 1) \)
Every coordinate on the unit circle has absolute value at most 1. Any answer with a coordinate bigger than 1 is wrong before you check anything else, and that single test catches a great many slips in this chapter.
Estimation
Root two over two is the exact value of both coordinates at 45 degrees.
Predict first
Roughly what decimal is that, and does the size make sense?
Correct: About 0.707.
Why: Root two is about 1.414, and half of that is about 0.707. The size is exactly right: both coordinates must be under 1 because the point is on the unit circle, and both must be equal and reasonably large because the point sits diagonally, halfway between the axes. The distractor 1.41 is root two itself, which would be the distance from the origin to the point one comma one — outside the circle.
Prediction
Consider the angle 135 degrees, which is 45 degrees past the vertical.
Predict first
Without computing, what will its coordinates be?
Correct: Root two over two with the x-coordinate made negative.
Why: The point at 135 degrees is the reflection across the y-axis of the point at 45 degrees, since the two angles are equally far from the vertical on either side. Reflecting across the y-axis negates the x-coordinate and leaves the y-coordinate alone, so the point is negative root two over two comma root two over two. That matches the sign rule: quadrant two has negative cosine and positive sine. This reflection idea is exactly what the Reference Angle Theorem formalises in the next lesson.
Counterexample
A student proposes: the cosine of an angle is always bigger than its sine for angles in quadrant one.
Discussion prompt
Find an angle in quadrant one that makes this false, and then state exactly where the changeover happens.
Hint: Try both ends of quadrant one, and then the middle.
Answer:
At 60 degrees the cosine is one half and the sine is root three over two, about 0.866. The sine is larger, so the claim fails.
\[ \cos(60^\circ) = \tfrac{1}{2} < \tfrac{\sqrt{3}}{2} = \sin(60^\circ) \]
The changeover is at exactly 45 degrees, where the two are equal at root two over two. Below 45 degrees the point is low and far right, so the cosine wins; above 45 degrees the point is high and close in, so the sine wins. That single crossing is visible on the graphs in Lesson 10.5a as the point where the two curves meet.
Section
Section 3
Concept
The same two-equation method handles 30 and 60 degrees, and the geometry supplies a different first relation: in a 30-60-90 triangle the side opposite the 30 degree angle is half the hypotenuse.
That swap is not a coincidence. Thirty and sixty degrees are complementary, and Lesson 10.4a will show that complementary angles always exchange their cosine and sine — that is what the co in cosine means.
Figure (svg): The angle pi over six in standard position with a perpendicular dropped to the x-axis forming a thirty sixty ninety triangle, beside the angle pi over three whose triangle is the same shape rotated
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 719-719
Picture it
Look at the two pictures side by side and notice that the triangles are congruent, just oriented differently.
Figure (svg): The angle pi over six in standard position with a perpendicular dropped to the x-axis forming a thirty sixty ninety triangle, beside the angle pi over three whose triangle is the same shape rotated
If you remember only that the pair of values is one half and root three over two, the only remaining question is which goes with which — and the picture answers it: the smaller angle has the smaller height.
Worked example
Example 10.2.1, part 4. Same method, different first relation.
\[ \text{Find } \cos\left(\frac{\pi}{6}\right) \text{ and } \sin\left(\frac{\pi}{6}\right). \]
Drop the perpendicular and identify the triangle
Why: The angle at the origin is 30 degrees, so the triangle is 30-60-90 with hypotenuse 1.
\[ 30 - 60 - 90,\text{ hypotenuse } 1 \]
Use the geometry to find y
Why: The leg opposite the 30 degree angle is half the hypotenuse.
\[ y = \frac{1}{2} \]
Substitute into the circle equation
Why: Solve for the remaining coordinate.
\[ x ^{2} + \frac{1}{4} = 1,\text{ so } x ^{2} = \frac{3}{4} \]
Take the root and pick the sign
Why: The square root of three quarters is root three over two, and P is in quadrant one so x is positive.
\[ x = \sqrt{3} / 2 \]
Figure (svg): The solution to Worked example cosine and sine of pi over six shown as a ladder of expressions, one row per legal move
\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \qquad \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]
Verify: check the identity
Why: Three quarters plus one quarter is 1, so the point is on the circle. And the sine being the smaller of the two is right for a shallow angle: at 30 degrees the point has barely risen but has moved a long way across.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 719-719
Comparison
This is the table the book says must be memorised. Fill it from the pictures rather than from memory, and it will stick.
Comparison matrix
| Angle in degrees | Angle in radians | Cosine | Sine |
|---|---|---|---|
| 0 | 0 | 1 | 0 |
| 30 | pi over 6 | sqrt3 over 2 | 1 over 2 |
| 45 | pi over 4 | sqrt2 over 2 | sqrt2 over 2 |
| 60 | pi over 3 | 1 over 2 | sqrt3 over 2 |
| 90 | pi over 2 | 0 | 1 |
Read the cosine column downward: 1, root three over two, root two over two, one half, 0. It decreases steadily as the angle grows, because the point is sliding leftward. The sine column is the same list read upward.
Worked example
Example 10.2.1, part 5. Predict the answer from the previous one before working it.
\[ \text{Find } \cos(60^\circ) \text{ and } \sin(60^\circ). \]
Drop the perpendicular
Why: The angle at the origin is now 60 degrees, so the 30 degree angle sits at the top of the triangle.
\[ 30 - 60 - 90,\text{ rotated} \]
Use the geometry to find x
Why: The leg opposite the 30 degree angle is half the hypotenuse, and that leg is now the horizontal one.
\[ x = \frac{1}{2} \]
Substitute into the circle equation
Why: Same substitution, other unknown.
\[ \frac{1}{4} + y ^{2} = 1,\text{ so } y ^{2} = \frac{3}{4} \]
Take the root and pick the sign
Why: P is in quadrant one, so y is positive.
\[ y = \sqrt{3} / 2 \]
Figure (svg): The solution to Worked example cosine and sine of 60 degrees shown as a ladder of expressions, one row per legal move
\[ \cos(60^\circ) = \frac{1}{2} \qquad \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
Verify: compare with 30 degrees
Why: At 30 degrees the values were root three over two and one half; here they are one half and root three over two — the same pair swapped, exactly as predicted for complementary angles. And the sine being the larger is right for a steep angle.
Trap
\[ \cos\left(\tfrac{\pi}{6}\right) = \tfrac{1}{2} \qquad \sin\left(\tfrac{\pi}{6}\right) = \tfrac{\sqrt{3}}{2} \]
Recall that the values at 30 degrees are one half and root three over two, then guess the order
Why: Both angles use the same two numbers, so the pairing is the only thing that can go wrong, and it is precisely the thing not written down.
This says that at 30 degrees the point is high and close in, which would mean a shallow angle produces a tall point. Sketching it makes the contradiction obvious.
\[ \cos\left(\tfrac{\pi}{6}\right) = \tfrac{\sqrt{3}}{2} \qquad \sin\left(\tfrac{\pi}{6}\right) = \tfrac{1}{2} \]
Sketch the angle and ask which coordinate is bigger
Why: A small angle keeps the point near the positive x-axis: far across, barely up. So the x-coordinate must be the large one.
The rule that never fails: for an acute angle, the smaller the angle the larger the cosine. Thirty degrees is the shallower of the two, so it gets the larger value, root three over two, as its cosine.
Ranking
All five are cosines of acute angles.
Put in order
Why: The values are 0, one half, about 0.707, about 0.866, and 1. Cosine decreases across quadrant one because as the angle opens the point slides leftward, shrinking its x-coordinate from 1 all the way to 0. Knowing that cosine decreases and sine increases across quadrant one lets you sanity-check any value in the table without recalling it exactly.
Faded example
The 30-60-90 triangle for the angle pi over three.
Fill in the blanks
x = 1/2 \quad\text3/4\quad x^2 + y^2 = 1 \;\Longrightarrow\; y^2 = sqrt3/2 \;\Longrightarrow\; y = ___
Why: At 60 degrees the horizontal leg is the one opposite the 30 degree angle, so it is half the hypotenuse, giving x equal to one half. Then y squared is 1 minus one quarter, which is three quarters, so y is root three over two, taken positive because the point is in quadrant one. About 0.866, which is appropriately close to the top of the circle for a steep angle.
Explain it to yourself
The values at 30 and 60 degrees are the same two numbers in opposite orders.
Discussion prompt
Explain why that happens, using the picture rather than a formula. What is the relationship between the two angles that causes it?
Hint: What do 30 and 60 add up to, and what does reflecting a point across the diagonal line y equals x do to its coordinates?
Answer:
The two angles are complementary — they add to 90 degrees — so they sit symmetrically on either side of the 45 degree diagonal. Reflecting a point across the line y equals x swaps its coordinates, and that reflection is exactly what carries the 30 degree point to the 60 degree point.
So the swap is forced by the symmetry, not by a coincidence of the numbers. In Lesson 10.4a this becomes the cofunction identity: the cosine of an angle equals the sine of its complement, for every angle and not just these two. The prefix co in cosine is short for complement, and this is what it is recording.
Section
Section 4
Concept
The point corresponding to any angle lies on the unit circle, so its coordinates satisfy x squared plus y squared equals one. But the coordinates are the cosine and the sine. Substituting gives the most important identity in trigonometry.
identity — An equation that is true for every value of the variable, as opposed to an equation that is true only for particular values. The Pythagorean identity holds for every angle without exception.
\[ \cos^2(\theta) + \sin^2(\theta) = 1 \]
The notation is a genuine nuisance and the book apologises for it: cosine squared of theta means the square of the cosine, not the cosine of theta squared. The convention is universal and you have to live with it.
Figure (svg): The unit circle equation x squared plus y squared equals one, with x replaced by cosine of theta and y replaced by sine of theta to give the Pythagorean identity
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 720-720
Picture it
There is no derivation to memorise here, only a substitution to notice.
Figure (svg): The unit circle equation x squared plus y squared equals one, with x replaced by cosine of theta and y replaced by sine of theta to give the Pythagorean identity
This will be generalised in Lesson 10.3a into three Pythagorean identities, all obtained from this one by dividing through by a square.
Worked example
The identity is a test as well as a tool. Use it to reject impossible answers.
\[ \text{Could an angle have } \cos(\theta) = \tfrac{2}{3} \text{ and } \sin(\theta) = \tfrac{2}{3}\text{?} \]
State what the identity requires
Why: The two squares must add to exactly 1 for any genuine pair.
\[ \cos ^{2} + \sin ^{2} = 1 \]
Substitute the proposed values
Why: Two thirds squared is four ninths, twice.
\[ \frac{4}{9} + \frac{4}{9} = \frac{8}{9} \]
Compare with 1
Why: Eight ninths is less than 1, so the identity fails.
\[ \frac{8}{9}\text{ is not } 1 \]
Interpret the failure geometrically
Why: The point two thirds comma two thirds is inside the unit circle, not on it.
Figure (svg): The solution to Worked example check a proposed pair of values shown as a ladder of expressions, one row per legal move
\[ \tfrac{4}{9} + \tfrac{4}{9} = \tfrac{8}{9} \ne 1 \quad\Longrightarrow\quad \text{no such angle} \]
Verify: find what would work
Why: For two equal coordinates the identity needs two x squared equal to 1, giving x equal to root two over two, about 0.707. Two thirds is about 0.667, slightly too small, which is why the point falls just inside the circle.
Sorting
The identity decides every one of these, and so does asking whether the point is on the circle.
Sort into buckets
Sort each pair by whether it can be the cosine and sine of some angle.
Worked example
Example 10.2.2, part 3. A case where the identity leaves no ambiguity at all.
\[ \text{If } \sin(\theta) = 1, \text{ find } \cos(\theta). \]
Substitute into the identity
Why: Sine squared is one squared, which is 1.
\[ \cos ^{2} \theta + 1 = 1 \]
Solve for the cosine squared
Why: Subtracting 1 from both sides leaves zero.
\[ \cos ^{2} \theta = 0 \]
Take the square root
Why: Zero has only one square root, so no plus-or-minus appears.
\[ \cos \theta = 0 \]
Note why no quadrant was needed
Why: The ambiguity that usually requires quadrant information vanishes when the square is zero.
Figure (svg): The solution to Worked example an angle with sine equal to 1 shown as a ladder of expressions, one row per legal move
\[ \cos^2(\theta) = 1 - 1 = 0 \quad\Longrightarrow\quad \cos(\theta) = 0 \]
Verify: identify the angle
Why: A sine of 1 means the y-coordinate is 1, and the only point on the unit circle with that y-coordinate is zero comma one, at the top. Its x-coordinate is indeed 0. The angle is pi over two, or anything coterminal with it.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 720-720
Error analysis
A student evaluates the identity at pi over three.
Annotate
On: \( \cos^2\left(\frac{\pi}{3}\right) = \cos\left(\frac{\pi^2}{9}\right) \)
When in doubt, write the bracket you mean: the square of the cosine is unambiguous where cosine squared is not. This is one of the very few places in mathematics where the standard notation is genuinely bad, and the authors say so.
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 the Pythagorean identity is wrong.
Survives elimination: B
Why: Statement B is the false one. The identity gives the sine squared, and taking a square root introduces a plus-or-minus that the identity itself cannot resolve. Both the point above the axis and the point below it share a cosine, and something else — the quadrant, or an inequality restricting the angle — must choose between them. That is precisely the structure of every problem in the next section.
Fill the middle
An angle has cosine equal to negative three fifths.
Fill in the blanks
\left(-\tfrac16/25+- 4/5\right)^2 + \sin^2(\theta) = 1 \;\Longrightarrow\; \sin^2(\theta) = ___ \;\Longrightarrow\; \sin(\theta) = ___
Why: Negative three fifths squared is positive nine twenty-fifths, and subtracting from 1 leaves sixteen twenty-fifths. The square root is plus or minus four fifths, and the sign genuinely cannot be determined from what was given: a cosine of negative three fifths occurs in both quadrant two, where the sine is positive, and quadrant three, where it is negative.
Edge cases
The identity says the two squares add to 1.
Discussion prompt
What does that force about the possible values of cosine and sine individually? State the range of each, and explain what would be geometrically true of an angle whose cosine was 1.2.
Hint: A square is never negative.
Answer:
\[ \cos^2(\theta) = 1 - \sin^2(\theta) \le 1 \quad\Longrightarrow\quad -1 \le \cos(\theta) \le 1 \]
Since sine squared is never negative, cosine squared can never exceed 1, so the cosine always lies between negative one and one inclusive. The identical argument bounds the sine the same way.
An angle with cosine 1.2 would need a point on the unit circle with x-coordinate 1.2, which would be at distance at least 1.2 from the origin — outside a circle of radius 1. No such point exists. This bound is the reason the graphs in Lesson 10.5a never leave the strip between negative one and one, and the reason arccosine in Lesson 10.6a has domain only that interval.
Section
Section 5
Concept
The standard problem of this section gives you one of the two values plus some information about where the angle is, and asks for the other. The identity does the arithmetic and the quadrant does the deciding.
An inequality such as pi less than theta less than three pi over two is naming a quadrant in disguise. Translate it before doing anything else.
Figure (svg): A grid of the four quadrants showing the sign of the x-coordinate and the y-coordinate in each, and therefore the sign of cosine and sine there
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 720-721
Picture it
There is no need for a mnemonic. Ask where the point is and what that does to its coordinates.
Figure (svg): A grid of the four quadrants showing the sign of the x-coordinate and the y-coordinate in each, and therefore the sign of cosine and sine there
Quadrant three is the one worth double-checking, since both coordinates are negative there and it is easy to negate only one of them.
Worked example
Example 10.2.2, part 1. The quadrant is handed to you, so only the sign choice needs care.
\[ \text{If } \theta \text{ is a Quadrant II angle with } \sin(\theta) = \tfrac{3}{5}, \text{ find } \cos(\theta). \]
Substitute the known sine into the identity
Why: Three fifths squared is nine twenty-fifths.
\[ \cos ^{2} \theta + \frac{9}{25} = 1 \]
Solve for the cosine squared
Why: One minus nine twenty-fifths is sixteen twenty-fifths.
\[ \cos ^{2} \theta = \frac{16}{25} \]
Take the square root with both signs
Why: Do not choose yet; the algebra genuinely permits both.
\[ \cos \theta = +- \frac{4}{5} \]
Choose the sign from the quadrant
Why: In quadrant two the x-coordinates are negative, so the cosine is negative.
\[ \cos \theta = -\frac{4}{5} \]
Figure (svg): The solution to Worked example quadrant named directly shown as a ladder of expressions, one row per legal move
\[ \cos(\theta) = -\frac{4}{5} \]
Verify: check the identity and the quadrant
Why: Sixteen twenty-fifths plus nine twenty-fifths is 1, so the pair is legitimate. And the point negative four fifths comma three fifths is left of the y-axis and above the x-axis, which is quadrant two as required.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 720-720
Translation
Problems disguise the quadrant in several ways. Recognising them costs nothing and saves the whole problem.
Match the pairs
Why: An inequality between two quadrantal angles names the quadrant between them: pi over two to pi is the second quarter of the revolution, and three pi over two to two pi is the fourth. A pair of signs names it directly, since the quadrant is determined by the signs of the two coordinates. Note that two different conditions here both describe quadrant four, which is the point: the same quadrant arrives dressed several different ways.
Worked example
Example 10.2.2, part 2. Translate the inequality into a quadrant first.
\[ \text{If } \pi < \theta < \frac{3\pi}{2} \text{ with } \cos(\theta) = -\frac{\sqrt{5}}{5}, \text{ find } \sin(\theta). \]
Translate the inequality into a quadrant
Why: Between pi and three pi over two is the third quarter of a revolution.
Substitute into the identity
Why: Negative root five over five squared is five over twenty-five, which is one fifth.
\[ \frac{1}{5} + \sin ^{2} \theta = 1 \]
Solve and take both roots
Why: One minus one fifth is four fifths, whose square root is two over root five.
\[ \sin \theta = +- 2 \sqrt{5} / 5 \]
Choose the sign from the quadrant
Why: In quadrant three the y-coordinates are negative.
\[ \sin \theta = -2 \sqrt{5} / 5 \]
Figure (svg): The solution to Worked example quadrant given as an inequality shown as a ladder of expressions, one row per legal move
\[ \sin(\theta) = -\frac{2\sqrt{5}}{5} \]
Verify: check the identity
Why: One fifth plus four fifths is 1. And both coordinates came out negative, which is exactly what quadrant three requires — a useful cross-check, since getting only one of them negative is the usual slip here.
Trap
\[ \theta \text{ in QIII}, \; \sin(\theta) = -\tfrac{1}{2} \quad\Longrightarrow\quad \cos^2(\theta) = \tfrac{3}{4} \quad\Longrightarrow\quad \cos(\theta) = \tfrac{\sqrt{3}}{2} \]
Take the square root and write the positive value
Why: The square root symbol denotes the positive root, so writing the positive value feels like following a rule.
But we are solving cosine squared equals three quarters, and that equation has two solutions. The positive root is the value of the radical; it is not automatically the value of the cosine.
\[ \cos^2(\theta) = \tfrac{3}{4} \quad\Longrightarrow\quad \cos(\theta) = \pm\tfrac{\sqrt{3}}{2} \quad\Longrightarrow\quad \cos(\theta) = -\tfrac{\sqrt{3}}{2} \]
Keep both roots, then let the quadrant decide
Why: Quadrant three has negative x-coordinates, so the cosine is the negative one.
The distinction matters: the radical sign means the positive root, but solving an equation of the form u squared equals c produces both. Writing the plus-or-minus explicitly and then striking one out is the habit that keeps them apart.
Faded example
An angle in quadrant four has cosine equal to five thirteenths. Find its sine.
Fill in the blanks
\left(\tfrac144/169+- 12/13\right)^2 + \sin^2(\theta) = 1 \;\Longrightarrow\; \sin^2(\theta) = -12/13 \;\Longrightarrow\; \sin(\theta) = ___ \;\Longrightarrow\; \sin(\theta) = ___
Why: Twenty-five over one hundred sixty-nine subtracted from 1 leaves one hundred forty-four over one hundred sixty-nine, whose square root is twelve thirteenths. In quadrant four the y-coordinates are negative, so the sine is negative twelve thirteenths. Five, twelve and thirteen form a Pythagorean triple, which is why the arithmetic came out exact — textbook problems in this section almost always use one.
Elimination
An angle in quadrant three has sine equal to negative four fifths. Rule out the impossible answers for its cosine.
Eliminate the wrong options
Which value is the cosine of this angle?
Survives elimination: B
Why: The identity gives cosine squared equal to 1 minus sixteen twenty-fifths, which is nine twenty-fifths, so the cosine is plus or minus three fifths. Quadrant three requires a negative x-coordinate, so it is negative three fifths. Notice that the three wrong answers fail for three different reasons — wrong sign, fails the identity, and outside the possible range — and each of those is worth checking independently.
Missing information
A problem states: an angle has cosine equal to seven twenty-fifths. Find its sine.
Discussion prompt
Why can this not be answered as asked, and what is the smallest thing you could add to make the answer unique?
Hint: Work the identity through and look at what you are left holding.
Answer:
\[ \sin^2(\theta) = 1 - \tfrac{49}{625} = \tfrac{576}{625} \;\Longrightarrow\; \sin(\theta) = \pm\tfrac{24}{25} \]
The identity narrows the answer to exactly two values, and nothing in the question chooses between them. A positive cosine puts the angle in quadrant one or quadrant four, and the sine is positive in the first and negative in the second.
The smallest sufficient addition is the sign of the sine, or equivalently naming one of the two quadrants, or any inequality pinning the angle to one of them. Note that giving the cosine's sign again would add nothing — it is already known, and it is precisely what fails to distinguish the two cases.
Comparison
Fill the blanks from memory. Which method applies is decided entirely by where the terminal side lands.
Comparison matrix
| Situation | Method | Work required |
|---|---|---|
| Terminal side on an axis | Read the point off the axis | none |
| Terminal side at 45 degrees to an axis | 45-45-90 triangle plus the circle equation | two equations, two unknowns |
| Terminal side at 30 or 60 degrees | 30-60-90 triangle plus the circle equation | two equations, two unknowns |
| One value known, quadrant known | Pythagorean identity | solve, then pick the sign from the quadrant |
| One value known, quadrant unknown | identity gives two answers | the problem is underdetermined |
Every row except the last produces a definite answer. The last row is not a harder problem, it is an incomplete one, and saying so is the correct response.
Pattern
Whether the question gives you an angle or gives you a value, the same five moves cover it.
When the problem gives you a value rather than an angle, steps three and four collapse into a single use of the identity, but the sign choice at the end is unchanged and is still where the marks are lost.
Check
A quadrantal angle. Read it off the picture.
Check your understanding
What is the cosine of pi?
Answer: C
Why: The angle pi is half a revolution, so its terminal side lies along the negative x-axis and meets the unit circle at the point negative one comma zero. Cosine is the x-coordinate, which is negative one.
Check
A special angle. Sketch it before choosing.
Check your understanding
What is the sine of pi over six?
Answer: A
Why: Pi over six is 30 degrees, a shallow angle, so the point on the unit circle has barely risen above the x-axis. Its y-coordinate, which is the sine, is the smaller of the two values, namely one half. The 30-60-90 triangle confirms it: the leg opposite the 30 degree angle is half the hypotenuse, and the hypotenuse here is 1.
Check
The identity plus a quadrant. Keep the plus-or-minus until the last step.
Check your understanding
An angle in quadrant two has sine equal to eight seventeenths. What is its cosine?
Answer: B
Why: Sixty-four over two hundred eighty-nine subtracted from 1 leaves two hundred twenty-five over two hundred eighty-nine, whose square root is fifteen seventeenths. In quadrant two the x-coordinates are negative, so the cosine is negative fifteen seventeenths. Eight, fifteen and seventeen are a Pythagorean triple, which is why the numbers came out whole.
Real world
A robot arm of length 1 metre is anchored at the origin and its shoulder joint reports its angle from horizontal. A controller needs the gripper's x and y position in the workspace.
Discussion prompt
Write the two coordinates in terms of the reported angle. Then explain what changes if the arm is 1.8 metres long instead, and why the controller software would be written for the unit circle first regardless.
Hint: The definition is literally a statement about coordinates. Then think about scaling.
Answer:
\[ x = \cos(\theta) \qquad y = \sin(\theta) \]
\[ \text{arm of length } r: \quad x = r\cos(\theta), \qquad y = r\sin(\theta) \]
For a unit-length arm the coordinates are the cosine and sine, with nothing else to do — that is the definition, not an application of it. For an arm of length r every coordinate scales by r, because the point sits on a circle of radius r rather than radius 1.
Controllers are written against the unit circle because it separates the two things that can vary: the direction, carried entirely by cosine and sine, and the reach, carried entirely by r. Change the arm and only r changes; the trigonometry is untouched. That separation is exactly what Lesson 10.2c formalises when it moves beyond the unit circle.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Is there an angle whose cosine is 1 and whose sine is also 1?
Correct: No, because the identity would give 2 rather than 1.
\[ 1^2 + 1^2 = 2 \ne 1 \]
\[ \cos(45^\circ) = \sin(45^\circ) = \tfrac{\sqrt{2}}{2} \quad \checkmark \]
Why: One squared plus one squared is 2, which violates the Pythagorean identity, so no such angle exists. Geometrically, the point one comma one is at distance root two from the origin and lies outside the unit circle entirely. The fourth option is a tempting but false reason: cosine and sine are equal at 45 degrees and again at 225 degrees, where both are root two over two and both are negative root two over two respectively. They can be equal; they simply cannot both be 1.
Explain it
They learned sine as opposite over hypotenuse and are confused about how an angle of 200 degrees could possibly have a sine, since there is no triangle with a 200 degree angle in it.
Discussion prompt
In no more than five sentences, explain the coordinate definition to them and say why it agrees with the triangle definition wherever the triangle definition applies.
Hint: Draw the unit circle and put a point on it.
Answer:
A usable answer: forget triangles for a moment. Draw a circle of radius one, sweep the angle out from the positive x-axis, and mark where the arm crosses the circle. The across coordinate of that point is the cosine and the up coordinate is the sine — and every angle has such a point, including 200 degrees, where it lands down and to the left so both coordinates are negative.
It agrees with the old definition because for an acute angle, dropping a line from that point to the x-axis makes a right triangle whose hypotenuse is 1. Opposite over hypotenuse is then the height over 1, which is just the height — the y-coordinate. The old definition was the new one all along, restricted to the corner of the circle where a triangle happens to fit.
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: Which coordinate is fixed by remembering that cosine goes with x, both being first. The pairing at 30 and 60 is fixed by sketching and asking whether the point is high or wide. Sign choice is fixed by drawing the point and asking whether it is left of centre and below centre, rather than reciting a mnemonic. Derivation is fixed by always writing both equations, the triangle relation and the circle equation, before solving anything. 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 through the centre. Mark the four quadrantal points and label each with its coordinates. Then mark the three first-quadrant points at thirty, forty-five and sixty degrees, and label each with its exact coordinates. Beside the forty-five degree point, draw the small right triangle and write the two equations that determine it, the triangle relation and the circle equation. In the top right corner, write the Pythagorean identity and, underneath it, the equation of the unit circle, with an arrow showing that one becomes the other by substitution. Around the outside of the circle, write in each quadrant the signs of the cosine and the sine there. Finally, in the bottom corner, work this problem completely: an angle in quadrant three has cosine negative five thirteenths, find its sine, showing the plus-or-minus step explicitly before you resolve it.
The final answer should be negative twelve thirteenths. If you got positive twelve thirteenths, you resolved the sign from the algebra instead of from the quadrant, which is the single most common error in this lesson.
Recap
Five things, and the first one is the definition everything else in this course is built on.
| If the question says | Your first move is |
|---|---|
| Find cosine and sine of this angle | Draw the angle and locate the point on the circle |
| The angle is quadrantal | Read both coordinates off the axis; no work needed |
| The angle is 30, 45 or 60 degrees | Drop the perpendicular and write both equations |
| One value is given plus a quadrant | Substitute into the identity, keep the plus-or-minus |
| An inequality bounds the angle | Translate it into a quadrant before anything else |
The next lesson takes the reflection idea that appeared when 135 degrees was compared with 45 degrees and makes it a theorem. Once the Reference Angle Theorem is available, the five values in the table above generate the cosine and sine of every special angle on the whole circle.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 717-722 — everything on these slides traces back here
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