Defines cosine and sine as the coordinates of the point where an angle's terminal side meets the unit circle, which makes them defined for every angle. Derives the Pythagorean identity as the circle's own equation, establishes the exact values at the special angles, and reads the sign of each function in each quadrant directly off the coordinate axes.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 5 — Trigonometric Functions
§5.2 Unit Circle: Sine and Cosine Functions, pp. 651-673
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §5.2 Unit Circle: Sine and Cosine Functions §5.2, pp. 651-673 — the pages these objectives are drawn from
Warm-up
You have probably met sine and cosine as ratios in a right triangle. That definition has a limitation.
Discussion prompt
In a right triangle, the sine of an angle is the opposite over the hypotenuse. What is the sine of 120 degrees?
Hint: Can a right triangle contain a 120 degree angle?
Answer:
It cannot be answered from that definition. A right triangle's other two angles must sum to 90 degrees, so neither can be 120 — there is no triangle to read the ratio from.
Yet the sine of 120 degrees is a perfectly good number, and the graphs in Chapter 6 will need it. So the triangle definition is too narrow: it works only for acute angles.
This section gives a definition that works for every angle, by putting the angle in standard position and reading coordinates off a circle. The triangle definition survives as a special case, which §5.4 will show.
Concept
Put the angle in standard position and find where its terminal side crosses the unit circle. The horizontal coordinate of that point is the cosine and the vertical coordinate is the sine.
unit circle — The circle of radius 1 centred at the origin. For an angle in standard position, the point where the terminal side meets it has coordinates given by the cosine and the sine of that angle.
\[ (\cos\theta,\;\sin\theta) \text{ is the point on the unit circle at angle } \theta \]
Defining them as coordinates rather than as ratios is what makes them functions of any angle. There is a point on the circle for every angle, positive, negative or larger than a full turn, so there is a cosine and a sine for every one of them.
Figure (svg): The unit circle with a terminal side meeting it at a point, whose horizontal coordinate is labelled cosine and whose vertical coordinate is labelled sine
OpenStax, Precalculus, §5.2 Unit Circle: Sine and Cosine Functions §5.2, pp. 651-656
Section
Section 1
Concept
For an angle in standard position, the terminal side meets the unit circle at exactly one point. Its coordinates are the cosine and the sine.
The last point is the origin of everything in Chapter 6. Because a full turn returns to the same point, both functions repeat with period 2 pi — and periodicity is what makes them the right tools for describing anything that oscillates.
Figure (svg): The unit circle with a terminal side meeting it at a point, whose horizontal coordinate is labelled cosine and whose vertical coordinate is labelled sine
OpenStax, Precalculus, §5.2 Unit Circle: Sine and Cosine Functions §5.2, pp. 651-658
Picture it
One angle, one point, two numbers.
Figure (svg): The unit circle with a terminal side meeting it at a point, whose horizontal coordinate is labelled cosine and whose vertical coordinate is labelled sine
Nothing about this construction requires the angle to be acute, which is exactly the limitation of the triangle definition that it removes.
Worked example
The point sits on an axis, so the coordinates are obvious.
\[ \text{Find } \cos(90^\circ) \text{ and } \sin(90^\circ). \]
Locate the terminal side
Why: A quarter turn counterclockwise.
Find where it meets the unit circle
Why: One unit up from the origin.
\[ \text{the point } (0, 1) \]
Read the horizontal coordinate
Why: It is the cosine.
\[ \cos = 0 \]
Read the vertical coordinate
Why: It is the sine.
\[ \sin = 1 \]
Figure (svg): The unit circle with a terminal side meeting it at a point, whose horizontal coordinate is labelled cosine and whose vertical coordinate is labelled sine
\[ \cos 90^\circ=0, \qquad \sin 90^\circ=1 \]
Verify: check against the identity
Why: Zero squared plus 1 squared is 1, satisfying the Pythagorean identity. The quadrantal angles are worth knowing without computation, since their points sit exactly on the axes and their coordinates are 0 and plus or minus 1 in some order.
OpenStax, Precalculus, §5.2 Unit Circle: Sine and Cosine Functions §5.2, pp. 652-654
Sorting
Sketch the terminal side and see whether the point is high or far right.
Sort into buckets
Sort each angle by which of cosine and sine is larger in size.
Worked example
Reduce to a coterminal angle first.
\[ \text{Find } \cos(750^\circ). \]
Subtract full turns
Why: Seven hundred fifty minus 720.
\[ \text{coterminal with } 30 \]
Note why that is valid
Why: The same terminal side, so the same point.
Recall the value at 30 degrees
Why: From the special values.
\[ \sqrt{3}\text{ over } 2 \]
State the answer
Why: Unchanged by the turns.
\[ \sqrt{3}\text{ over } 2 \]
Figure (svg): The solution to Worked example an angle beyond a full turn shown as a ladder of expressions, one row per legal move
\[ \cos 750^\circ=\cos 30^\circ=\tfrac{\sqrt{3}}{2} \]
Verify: confirm the reduction
Why: Seven hundred fifty minus 30 is 720, which is exactly two full turns — so the two angles share a terminal side and therefore share a point on the circle. This is why the functions are periodic, and it is the first place that periodicity does real work.
OpenStax, Precalculus, §5.2 Unit Circle: Sine and Cosine Functions §5.2, pp. 654-658
Trap
\[ \text{at } 30^\circ \text{ the point is } (\tfrac{1}{2},\tfrac{\sqrt3}{2}), \text{ so } \cos 30^\circ=\tfrac{1}{2} \]
Read a coordinate as the cosine
Why: One of the two coordinates is selected and named.
The vertical coordinate has been taken as the cosine.
The point at 30 degrees is the other way round: the cosine is the larger value, root 3 over 2, and the sine is one half.
A picture settles it. Thirty degrees is close to the horizontal axis, so its point is far to the right and only slightly up — a large horizontal coordinate and a small vertical one.
Sketch the terminal side before reading values. Whether the point is high or far right is visible immediately, and it decides which coordinate is the larger without any recall.
Prediction
Cosine and sine are coordinates of a point on a circle of radius 1.
Predict first
What values can they take?
Correct: Anything from -1 to 1 inclusive.
Why: A point on the unit circle is never more than 1 unit from the origin in any direction, so neither coordinate can exceed 1 in size. Both endpoints are attained at the quadrantal angles. This is why an equation setting a sine equal to 2 has no solutions.
Faded example
Find the sine and cosine of 180 degrees.
Fill in the blanks
\text-1 (0, ___), \text___ \cos 180^\circ = ___ \text___ \sin 180^\circ = ___
Why: A half turn puts the terminal side on the negative horizontal axis, meeting the circle at the point one unit to the left of the origin. Its horizontal coordinate is negative 1 and its vertical coordinate is 0, giving the cosine and sine respectively.
Socratic
The triangle definition came first historically.
Discussion prompt
What does the coordinate definition buy that the triangle one cannot?
Hint: Which angles can appear in a right triangle?
Answer:
The triangle definition works only for acute angles, since a right triangle's other angles are both under 90 degrees. It cannot assign a sine to 120 degrees or to a negative angle at all.
The coordinate definition works for every angle, because every terminal side meets the circle somewhere. So the functions have all real numbers as their domain, which is what makes them functions in the sense of Chapter 1.
It also makes the Pythagorean identity obvious and the periodicity immediate, since a full turn returns to the same point. Two facts that would need separate proofs from the triangle definition fall out of this one for free.
Section
Section 2
Concept
A point on the unit circle satisfies x squared plus y squared equals 1. Since the coordinates are the cosine and the sine, that equation is the Pythagorean identity.
\[ \cos^2\theta+\sin^2\theta=1 \]
The third and fourth points are how it is used. Knowing the sine, the identity gives the cosine's size but not its sign, because both a positive and a negative value square to the same thing. The quadrant is what settles it, which is why the two ideas are always used together.
Figure (svg): The unit circle with a right triangle formed by the coordinates of a point on it, showing that the Pythagorean identity is the circle's equation
OpenStax, Precalculus, §5.2 Unit Circle: Sine and Cosine Functions §5.2, pp. 658-662
Picture it
The two coordinates and the radius form a right triangle with hypotenuse 1.
Figure (svg): The unit circle with a right triangle formed by the coordinates of a point on it, showing that the Pythagorean identity is the circle's equation
The Pythagorean theorem on that triangle gives exactly the identity. It is one theorem applied to one triangle, not a separate result about trigonometric functions.
Worked example
The identity gives the size; the quadrant gives the sign.
\[ \text{If } \sin\theta=\tfrac{3}{5} \text{ and } \theta \text{ is in quadrant II, find } \cos\theta. \]
Substitute into the identity
Why: Cosine squared plus sine squared is 1.
\[ \cos ^{2} + \frac{9}{25} = 1 \]
Isolate the cosine squared
Why: Subtract nine twenty-fifths.
\[ \cos ^{2} = \frac{16}{25} \]
Take the square root
Why: Both signs are possible so far.
\[ \cos = +- \frac{4}{5} \]
Use the quadrant to choose
Why: Cosine is negative in the second quadrant.
\[ \cos = -\frac{4}{5} \]
Figure (svg): The unit circle with a right triangle formed by the coordinates of a point on it, showing that the Pythagorean identity is the circle's equation
\[ \cos\theta=-\tfrac{4}{5} \]
Verify: check the identity holds
Why: Sixteen twenty-fifths plus nine twenty-fifths is 1 — correct. And the second quadrant is up and to the left, so the horizontal coordinate must be negative while the vertical one is positive, which matches both signs. The identity alone could not have chosen between plus and minus four fifths.
OpenStax, Precalculus, §5.2 Unit Circle: Sine and Cosine Functions §5.2, pp. 659-661
Faded example
Given that the cosine is 5 over 13, find the size of the sine.
Fill in the blanks
\sin^2 = 1 - \frac14412 = \frac___}___ \;\Longrightarrow\; |\sin| = \frac___}___
Why: One hundred sixty nine minus 25 is 144, whose square root is 12. So the sine is plus or minus twelve thirteenths, and the quadrant decides which. This is the 5-12-13 right triangle, which is why the numbers come out whole.
Worked example
One substitution.
\[ \text{Derive the identity from the unit circle's equation.} \]
Write the unit circle's equation
Why: Radius 1 centred at the origin.
\[ x ^{2} + y ^{2} = 1 \]
Recall what the coordinates are
Why: For a point at angle theta.
\[ x = \cos, y = \sin \]
Substitute
Why: Replacing the coordinates by their names.
\[ \cos ^{2} + \sin ^{2} = 1 \]
Note it holds for every angle
Why: Every terminal side meets the circle.
Figure (svg): The solution to Worked example why the identity is the circle's equation shown as a ladder of expressions, one row per legal move
\[ x^2+y^2=1 \;\Longrightarrow\; \cos^2\theta+\sin^2\theta=1 \]
Verify: notice what was assumed
Why: Only the definition of the unit circle and the definition of the two functions. No triangle, no special angles, and no restriction on theta — which is why the identity holds for every angle including the quadrantal ones, where a triangle-based argument would break down.
OpenStax, Precalculus, §5.2 Unit Circle: Sine and Cosine Functions §5.2, pp. 661-662
Error analysis
A student finds a cosine from a sine.
Annotate
On: \( \sin\theta=\tfrac{3}{5}, \; \theta \text{ in QIII} \;\Longrightarrow\; \cos\theta=\tfrac{4}{5} \)
The identity gives sizes and never signs, because squaring destroys them. Reading the quadrant is a separate step and it is not optional.
Prediction
The Pythagorean identity is used to find one function from the other.
Predict first
What does it leave undetermined?
Correct: The sign, which the quadrant must supply.
Why: The identity involves squares, and squaring destroys sign information, so taking a square root leaves two possibilities. Only knowing which quadrant the angle terminates in decides between them. This is the same structure as the plus-or-minus in §3.8's inverse problems.
Sorting
They must satisfy the identity.
Sort into buckets
Sort each proposed pair of cosine and sine values.
Explain it to yourself
The Pythagorean identity is often presented as something to memorise.
Discussion prompt
Explain why it needs no memorising at all.
Hint: What is the equation of a circle of radius 1?
Answer:
A circle of radius 1 centred at the origin has equation x squared plus y squared equals 1. That is the distance formula, and it is not a trigonometric fact.
The cosine and sine are the coordinates x and y of a point on that circle. So substituting their names into the equation gives the identity directly.
There is nothing extra to remember. If you can write the circle's equation, you have the identity, and the name honours Pythagoras because the circle's equation is his theorem applied to the radius and the two coordinates.
Section
Section 3
Concept
The angles of 30, 45 and 60 degrees have exact cosine and sine values, and the pattern in them makes the table easier to hold than a list of six numbers.
The root sequence is the memorable form. Written as root 0 over 2, root 1 over 2, root 2 over 2, root 3 over 2, root 4 over 2, the sines are a single pattern, and the cosines are the same pattern read the other way — which is a great deal less to hold than six unrelated numbers.
Figure (svg): The unit circle with the special angles of the first quadrant marked and their exact coordinates labelled
OpenStax, Precalculus, §5.2 Unit Circle: Sine and Cosine Functions §5.2, pp. 662-668
Picture it
Five angles, and their coordinates follow one pattern.
Figure (svg): The unit circle with the special angles of the first quadrant marked and their exact coordinates labelled
Reading up the circle, the vertical coordinates increase and the horizontal ones decrease, which is the two sequences running in opposite directions.
Worked example
The diagonal makes both coordinates equal.
\[ \text{Find } \cos 45^\circ \text{ and } \sin 45^\circ. \]
Locate the terminal side
Why: Exactly halfway between the axes.
Note the coordinates are equal
Why: Symmetric about the diagonal.
\[ \cos = \sin \]
Apply the identity
Why: Twice the square is 1.
\[ 2 \cos ^{2} = 1 \]
Solve
Why: Take the positive root.
\[ \sqrt{2}\text{ over } 2 \]
Figure (svg): The unit circle with the special angles of the first quadrant marked and their exact coordinates labelled
\[ \cos 45^\circ=\sin 45^\circ=\tfrac{\sqrt{2}}{2} \]
Verify: check the identity
Why: Two times one half is 1 — correct, since root 2 over 2 squared is 2 over 4, which is one half. Deriving this from the symmetry and the identity is quicker than recalling it, and it works even when the memory of the table has faded.
OpenStax, Precalculus, §5.2 Unit Circle: Sine and Cosine Functions §5.2, pp. 663-665
Matching
The sines increase as the angle rises towards 90 degrees.
Match the pairs
Why: Written as roots of 0, 1, 2 and 3 over 2, the four values form a single increasing pattern. The cosines run the same sequence backwards, which is why the whole first-quadrant table reduces to one pattern read in two directions.
Worked example
Thirty and sixty degrees swap their values.
\[ \text{Compare the values at } 30^\circ \text{ and } 60^\circ. \]
Recall the values at 30
Why: The point is far right and slightly up.
\[ \cos \sqrt{3} / 2, \sin 1 / 2 \]
Recall the values at 60
Why: The point is high and slightly right.
\[ \cos 1 / 2, \sin \sqrt{3} / 2 \]
Note the pattern
Why: The two values have swapped.
Explain why
Why: The angles sum to 90, so they are complementary.
Figure (svg): The solution to Worked example the complementary pair shown as a ladder of expressions, one row per legal move
\[ \cos 30^\circ=\sin 60^\circ, \qquad \sin 30^\circ=\cos 60^\circ \]
Verify: state the general rule
Why: The cosine of an angle equals the sine of its complement, for any acute angle. That is where the name cosine comes from — it is the sine of the complement. This halves the table: knowing the first quadrant's sines gives all its cosines for free.
OpenStax, Precalculus, §5.2 Unit Circle: Sine and Cosine Functions §5.2, pp. 665-668
Trap
\[ \sin 60^\circ=\tfrac{1}{2} \]
Recall a value from the table
Why: One of the two small-angle values is selected.
The value belonging to 30 degrees is assigned to 60.
Sixty degrees is close to vertical, so its point is high up — a large vertical coordinate. The sine is root 3 over 2, about 0.87.
One half is the sine of 30 degrees, whose point is low. Sketching the terminal side settles it in two seconds.
Judge by the picture rather than by recall. A large angle in the first quadrant has a large sine and a small cosine, and that alone distinguishes the two values without any memory of which is which.
Faded example
Write the sines at 0, 30, 45, 60 and 90 as roots over 2.
Fill in the blanks
\frac2}4, \frac___}___, \frac___}}}___, \frac___}___, \frac___}}}___
Why: The radicands run 0, 1, 2, 3, 4 in order, so the sines are 0, one half, root 2 over 2, root 3 over 2 and 1. The last two simplify to familiar forms, and the pattern makes the whole row recoverable from a single rule.
Prediction
The cosine of an angle equals the sine of its complement.
Predict first
What is the cosine of 20 degrees equal to?
Correct: The sine of 70 degrees.
Why: The complement of 20 is 70, since they sum to 90. This is where the name cosine comes from — it is the sine of the complementary angle — and it is why the first-quadrant table's two rows are reverses of each other.
Socratic
Most angles have sines and cosines that are irrational and unmemorable.
Discussion prompt
What is special about 30, 45 and 60 degrees?
Hint: What triangles have these angles?
Answer:
They come from two special triangles. The 45-45-90 triangle is half a square, and the 30-60-90 is half an equilateral triangle — both constructible with exact side ratios by the Pythagorean theorem.
So their coordinates can be computed exactly using square roots, rather than only approximated. Most angles have no such construction and their values are irrational numbers with no closed form.
That is the whole reason these three recur everywhere in the course. Exercises are built around them because their values can be written down, and the reference-angle technique in the next section extends them to every quadrant.
Section
Section 4
Concept
Cosine is a horizontal coordinate and sine is a vertical one, so each is positive exactly where its own coordinate is positive.
The mnemonics taught for this are unnecessary and occasionally misremembered. Asking whether the point is to the right and whether it is above answers both questions directly, and it cannot be recalled wrongly because it is not a fact being recalled.
Figure (svg): The four quadrants labelled with which of cosine and sine are positive in each, reading directly off the signs of the coordinates
OpenStax, Precalculus, §5.2 Unit Circle: Sine and Cosine Functions §5.2, pp. 665-670
Picture it
Each function follows its own coordinate.
Figure (svg): The four quadrants labelled with which of cosine and sine are positive in each, reading directly off the signs of the coordinates
Reading off the picture is faster and more reliable than a mnemonic, because the picture is the definition rather than a summary of consequences.
Worked example
Locate the quadrant and read off each coordinate's sign.
\[ \text{An angle terminates in quadrant III. What are the signs of its cosine and sine?} \]
Locate quadrant III
Why: Down and to the left.
Read the horizontal coordinate
Why: To the left of the vertical axis.
Read the vertical coordinate
Why: Below the horizontal axis.
State both
Why: Both negative.
\[ \cos < 0, \sin < 0 \]
Figure (svg): The four quadrants labelled with which of cosine and sine are positive in each, reading directly off the signs of the coordinates
\[ \cos\theta<0 \text{ and } \sin\theta<0 \]
Verify: check with a specific angle
Why: Take 210 degrees, in the third quadrant. Its reference angle is 30, so the sizes are root 3 over 2 and one half — and both carry minus signs, giving negative root 3 over 2 and negative one half. The signs and the sizes are found independently and combined.
OpenStax, Precalculus, §5.2 Unit Circle: Sine and Cosine Functions §5.2, pp. 666-668
Sorting
Sine follows the vertical coordinate.
Sort into buckets
Sort each quadrant.
Worked example
Run the reasoning backwards.
\[ \text{If } \cos\theta<0 \text{ and } \sin\theta>0, \text{ which quadrant?} \]
Interpret the cosine's sign
Why: The point is left of the vertical axis.
Interpret the sine's sign
Why: The point is above the horizontal axis.
Combine
Why: Upper and left.
Check
Why: Second quadrant is indeed up and to the left.
Figure (svg): The solution to Worked example identify the quadrant from the signs shown as a ladder of expressions, one row per legal move
\[ \text{quadrant II} \]
Verify: confirm with an angle
Why: Take 120 degrees. Its point is up and to the left, so its cosine is negative one half and its sine is root 3 over 2 — matching both stated signs. Each sign restricts the point to a half plane, and two half planes intersect in one quadrant.
OpenStax, Precalculus, §5.2 Unit Circle: Sine and Cosine Functions §5.2, pp. 668-670
Trap
\[ \text{in quadrant IV, sine is positive} \]
Recall the mnemonic for which functions are positive where
Why: A memorised phrase is applied to determine the sign.
The sine is declared positive in the fourth quadrant.
The fourth quadrant is below the horizontal axis, so the vertical coordinate is negative and the sine is negative. The cosine is positive there, since the point is to the right.
Sketching the quadrant settles it instantly, without any recall.
Ask whether the point is above the axis and whether it is to the right. Those two questions answer both signs directly, and they are the definition rather than a consequence of it — so there is nothing to misremember.
Faded example
An angle of 300 degrees terminates in the fourth quadrant.
Fill in the blanks
\cos 300^\circ \textpositivenegative, \qquad \sin 300^\circ \text______
Why: The fourth quadrant is to the right of the vertical axis and below the horizontal one, so the horizontal coordinate is positive and the vertical one is negative. The values are one half and negative root 3 over 2, whose sizes come from the reference angle of 60 degrees.
Prediction
An angle has a positive cosine and a negative sine.
Predict first
Where does it terminate?
Correct: Quadrant IV.
Why: A positive cosine puts the point to the right of the vertical axis and a negative sine puts it below the horizontal one — the lower right region, which is the fourth quadrant. Each sign halves the plane and the two halves intersect in exactly one quadrant.
Counterexample
A classmate says sine and cosine always have the same sign.
Discussion prompt
Give a quadrant where they differ, and say how often that happens.
Hint: How many quadrants have one coordinate positive and one negative?
Answer:
Quadrant II has a positive sine and a negative cosine, and quadrant IV has the reverse. Either one disproves the claim.
In fact they differ in half the quadrants — two of the four. They agree only in quadrants I, where both are positive, and III, where both are negative.
The reason is that the two functions track different coordinates, and the two coordinates change sign at different places: the horizontal one at the vertical axis and the vertical one at the horizontal axis. Those two boundaries cut the plane into four regions with all four sign combinations.
Section
Section 5
Concept
The reference angle is the acute angle between the terminal side and the horizontal axis. Its trigonometric values give the sizes, and the quadrant gives the signs.
The separation of size and sign is what makes the technique work. Two independent facts are combined: the reference angle handles the magnitude and the quadrant handles the direction, and neither can supply the other.
Figure (svg): The four quadrants labelled with which of cosine and sine are positive in each, reading directly off the signs of the coordinates
OpenStax, Precalculus, §5.2 Unit Circle: Sine and Cosine Functions §5.2, pp. 668-673
Picture it
The reference angle gives the size and this picture gives the sign.
Figure (svg): The four quadrants labelled with which of cosine and sine are positive in each, reading directly off the signs of the coordinates
Every exact value in the whole chapter is one first-quadrant number with one of these signs attached, which is a much smaller table than it first appears.
Worked example
Size from the reference angle, sign from the quadrant.
\[ \text{Find } \sin 240^\circ. \]
Identify the quadrant
Why: Between 180 and 270.
Find the reference angle
Why: Two hundred forty minus 180.
\[ 60 ^\circ \]
Recall the size
Why: The sine of 60.
\[ \sqrt{3}\text{ over } 2 \]
Apply the sign
Why: Sine is negative in quadrant III.
\[ -\sqrt{3} / 2 \]
Figure (svg): The four quadrants labelled with which of cosine and sine are positive in each, reading directly off the signs of the coordinates
\[ \sin 240^\circ=-\tfrac{\sqrt{3}}{2} \]
Verify: check the picture
Why: Two hundred forty degrees points down and to the left, so its vertical coordinate is negative — matching the sign. And it is closer to the vertical axis than to the horizontal one, so the vertical coordinate is the larger in size, matching root 3 over 2 rather than one half.
OpenStax, Precalculus, §5.2 Unit Circle: Sine and Cosine Functions §5.2, pp. 669-671
Faded example
For an angle of 200 degrees.
Fill in the blanks
200 - 180 = 20 \text___
Why: Two hundred is 20 past 180, so the reference angle is 20 degrees. The terminal side is just past the negative horizontal axis, which the small reference angle reflects. Both coordinates are negative there, since the angle is in the third quadrant.
Worked example
Measure the reference angle back from a full turn.
\[ \text{Find } \cos 315^\circ. \]
Identify the quadrant
Why: Between 270 and 360.
Find the reference angle
Why: Three sixty minus 315.
\[ 45 ^\circ \]
Recall the size
Why: The cosine of 45.
\[ \sqrt{2}\text{ over } 2 \]
Apply the sign
Why: Cosine is positive in quadrant IV.
\[ +\sqrt{2} / 2 \]
Figure (svg): The solution to Worked example a fourth quadrant angle shown as a ladder of expressions, one row per legal move
\[ \cos 315^\circ=\tfrac{\sqrt{2}}{2} \]
Verify: check against the sine
Why: At the same angle the sine is negative root 2 over 2, since the point is below the axis but equally far right. Their squares sum to one half plus one half, which is 1 — satisfying the Pythagorean identity, which is a check on both values at once.
OpenStax, Precalculus, §5.2 Unit Circle: Sine and Cosine Functions §5.2, pp. 671-673
Error analysis
A student finds the reference angle for 150 degrees.
Annotate
On: \( \text{reference angle} = 150-90 = 60^\circ \)
The reference angle is always to the horizontal axis. Sketching the terminal side shows immediately whether it is close to that axis or far from it, which settles the size before any subtraction.
Sorting
It depends on the quadrant.
Sort into buckets
Sort each angle by how its reference angle is computed.
Prediction
An angle's reference angle is 30 degrees.
Predict first
What does that determine?
Correct: The sizes of its trigonometric values, but not their signs.
Why: The reference angle determines how far the point is from each axis, which fixes the magnitudes. Which side of each axis it lies on is a separate question answered by the quadrant. The two pieces of information are independent and both are needed.
Explain it
Every exact value comes from a size and a sign.
Discussion prompt
Explain to a classmate why finding a trigonometric value of a non-acute angle takes two separate steps.
Hint: What does each step determine?
Answer:
The reference angle determines how far the point sits from each axis, which is the size of each coordinate. It is always acute, so its values come from the small first-quadrant table.
The quadrant determines which side of each axis the point is on, which is the sign of each coordinate. That is a completely separate fact and the reference angle cannot supply it.
So a value is assembled from one number and one sign, found independently. A good explanation stresses that this is why only the first quadrant's table has to be known: everything else is that table plus a sign.
Comparison
Fill the blanks from memory. Everything follows from which coordinate each one is.
Comparison matrix
| cosine | sine | |
|---|---|---|
| which coordinate | horizontal | vertical |
| positive in quadrants | I and IV | I and II |
| value at 0 degrees | 1 | 0 |
| value at 90 degrees | 0 | 1 |
| at 45 degrees | root 2 over 2 | root 2 over 2, the same |
The first row generates every other. Knowing which coordinate each function is makes the signs, the quadrantal values and the special angles all readable off a sketch.
Pattern
Five steps, and the last two are independent of each other.
Steps 4 and 5 answer different questions and neither can substitute for the other. Combining a correct size with a wrong sign is the commonest failure, and the final identity check catches it.
Check
Which coordinate is which.
Check your understanding
The terminal side of an angle meets the unit circle at the point with coordinates -0.6 and 0.8. What is the cosine?
Answer: A
Why: The cosine is the horizontal coordinate, which is negative 0.6. The sine is the vertical one, 0.8. Checking the identity: 0.36 plus 0.64 is 1, so the point really is on the unit circle.
Check
Size from the reference angle, sign from the quadrant.
Check your understanding
What is the cosine of 150 degrees?
Answer: A
Why: The reference angle is 30 degrees, whose cosine is root 3 over 2, and the second quadrant has a negative cosine. So the value is negative root 3 over 2.
Check
The identity gives size, not sign.
Check your understanding
If the sine of an angle is 0.6 and the angle is in quadrant II, what is the cosine?
Answer: A
Why: The identity gives a cosine squared of 1 minus 0.36, which is 0.64, so the size is 0.8. The second quadrant has a negative cosine, giving negative 0.8.
Real world
The unit circle definition is what makes rotation computable, which is the basis of computer graphics.
Discussion prompt
How does a graphics program rotate a shape on screen?
Hint: What are the coordinates of a point after it rotates about the origin?
Answer:
Rotating a point about the origin sends it to a new point whose coordinates are combinations of the cosine and sine of the rotation angle. That formula comes directly from this section's definition.
The unit circle definition is essential because rotations are by any angle, not only acute ones. A triangle-based definition could not handle a rotation of 200 degrees at all.
Every frame of every video game and every animated interface performs millions of these calculations. The coordinates on a circle are the computation, which is why cosine and sine are among the most-executed functions in any graphics system.
Commit first
State your confidence along with your answer.
Predict first
Why does the Pythagorean identity hold for every angle?
Correct: Because it is the unit circle's equation with the coordinates renamed.
Why: Every terminal side meets the unit circle, and every point on that circle satisfies the equation that its coordinates' squares sum to 1. Since the coordinates are the cosine and sine, the identity holds for every angle including the quadrantal ones and the negative ones, where no triangle exists at all.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate why the unit circle definition is better than the right triangle one.
Hint: Which angles can each definition handle?
Answer:
The triangle definition works only for acute angles, because a right triangle cannot contain an angle of 120 degrees or a negative one. So it assigns no value to most angles.
The circle definition works for every angle, since every terminal side meets the circle somewhere. That makes cosine and sine genuine functions with all real numbers as their domain.
It also makes two important facts free: the Pythagorean identity is just the circle's equation, and the periodicity follows from a full turn returning to the same point. A good explanation notes that the triangle definition is not wrong — it is the special case for acute angles, which §5.4 will confirm.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The fourth is where most computational errors occur, since it requires combining two independent facts and either can be got wrong. The third is worth drilling until the values are automatic, because the rest of the chapter uses them constantly.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Draw the unit circle and mark the special angles of the first quadrant with their exact coordinates. Beside it, write the sines as roots over 2 to show the pattern. Then mark the four quadrants with the signs of cosine and sine in each, and work one example of a third or fourth quadrant angle end to end: reduce, find the quadrant, find the reference angle, look up the size, attach the sign.
If your worked example separates the size step from the sign step, you have the method that makes the whole circle available from a five-entry table.
Recap
Five things, and the first is the definition everything else rests on.
| if you remember one thing | it should be this |
|---|---|
| about the definition | cosine is horizontal, sine is vertical - everything follows |
| about the identity | it is the circle's equation, so there is nothing to memorise |
| about signs | each function takes the sign of its own coordinate |
| about reference angles | they give the size, and the quadrant gives the sign |
Section 5.3 defines the remaining four functions as ratios of these two, which is where the tangent and its undefined points arrive.
OpenStax, Precalculus, §5.2 Unit Circle: Sine and Cosine Functions §5.2, pp. 651-673 — everything on these slides traces back here
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