Lifting the coordinate definition off the unit circle by similar triangles: on a circle of radius r the point at angle theta has coordinates r cosine theta and r sine theta, and conversely the cosine and sine can be read off any point on the terminal side as x over r and y over r. Includes the equations of circular motion as explicit functions of time, right triangle trigonometry recovered as the first-quadrant special case, and the domain and range of cosine and sine once they are read as functions of a real number.
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. 730-736
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. 730-736 — the pages these objectives are drawn from
Warm-up
Every value you have computed so far has come from a circle of radius one. Real problems almost never involve a circle of radius one.
Discussion prompt
A point sits on a circle of radius 5 centred at the origin, at an angle of 60 degrees from the positive x-axis. On the unit circle that angle gives the point one half comma root three over two. What do you think the coordinates are on the radius-5 circle, and why?
Hint: The two points are on the same ray from the origin. What is the relationship between two points on the same ray at different distances?
Answer:
\[ \left(\tfrac{1}{2}, \tfrac{\sqrt{3}}{2}\right) \quad\longrightarrow\quad \left(\tfrac{5}{2}, \tfrac{5\sqrt{3}}{2}\right) \]
Both coordinates are multiplied by 5, because the second point is five times as far out along the same ray. That is the whole content of this lesson, and the only work is proving that the guess is right rather than merely plausible.
Concept
Take an angle and follow its terminal side out to a circle of radius r instead of to the unit circle. Dropping perpendiculars from both crossing points produces two similar right triangles, and similarity forces each coordinate to be exactly r times its unit-circle counterpart.
Theorem 10.3 — If the point with coordinates x and y lies on the terminal side of an angle and on the circle of radius r about the origin, then x equals r times the cosine of the angle and y equals r times the sine. Equivalently the cosine is x over r and the sine is y over r.
\[ x = r\cos(\theta), \qquad y = r\sin(\theta) \]
Read the second way round, this says the cosine and sine can be recovered from any point on the terminal side, since r is just the distance from that point to the origin. That is far more useful than the first reading and it is what most of the exercises use.
Figure (svg): Two nested circles of radius one and radius r sharing a centre, with one terminal side crossing both, and the two right triangles formed by dropping perpendiculars shown to be similar
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 730-730
Section
Section 1
Concept
The proof is one line of similarity. The angle at the origin is shared, both triangles have a right angle, so they are similar and corresponding sides are in proportion.
The argument as stated covers an acute angle. Reflecting through the axes and the origin extends it to every non-quadrantal angle, and the quadrantal cases can be checked directly.
Figure (svg): Two nested circles of radius one and radius r sharing a centre, with one terminal side crossing both, and the two right triangles formed by dropping perpendiculars shown to be similar
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 730-730
Picture it
Changing the radius changes some quantities and leaves others alone, and knowing which is which prevents most of the errors in this lesson.
Figure (svg): Two columns separating what changes and what does not when the circle's radius is changed from one to r
The cosine and sine are on the right-hand list. They are ratios, and a ratio cannot see the size of the circle it was measured in — which is exactly what makes them functions of the angle alone.
Worked example
Straight application of the theorem, in the direction it is stated.
\[ \text{Find the coordinates of the point at angle } \frac{5\pi}{6} \text{ on the circle } x^2 + y^2 = 36. \]
Identify the radius
Why: The circle equation has r squared equal to 36, so the radius is 6.
\[ r = 6 \]
Find the unit-circle values
Why: Five pi over six is in quadrant two with reference angle pi over six.
\[ \cos = -\sqrt{3} / 2, \sin = \frac{1}{2} \]
Multiply each by the radius
Why: The theorem says both coordinates scale by r.
\[ x = 6(-\sqrt{3} / 2), y = 6(\frac{1}{2}) \]
Simplify
Why: Six times root three over two is three root three.
\[ x = -3 \sqrt{3}, y = 3 \]
Figure (svg): The solution to Worked example coordinates on a circle of radius r shown as a ladder of expressions, one row per legal move
\[ \big(-3\sqrt{3}, \; 3\big) \]
Verify: check the point is on the circle
Why: Negative three root three squared is 27, and 3 squared is 9. Their sum is 36, which is r squared. And the point is left of the y-axis and above the x-axis, which is quadrant two as required.
Prediction
An angle's terminal side passes through the point three comma four.
Predict first
What is the radius of the circle that point lies on?
Correct: 5.
Why: The radius is the distance from the origin, which is the square root of nine plus sixteen, the square root of 25, which is 5. Three, four and five are the smallest Pythagorean triple and it appears constantly in these exercises. The distractor 7 comes from adding the coordinates rather than using the distance formula, which is the most common slip.
Worked example
Example 10.2.6, part 2. This is the approximation that was simply asserted back in Lesson 10.1c.
\[ \text{The Earth's radius at the Equator is about } 3960 \text{ miles. Find the radius of the circle traced by a point at } 41.628^\circ \text{ north.} \]
Set up the cross-section
Why: A slice through the poles is a circle of radius 3960 with the Equator as the x-axis.
\[ \text{circle of radius } 3960 \]
Identify what is being asked for
Why: The latitude circle's radius is the horizontal distance from the axis, which is the x-coordinate of the point.
Apply the theorem
Why: The x-coordinate is the radius times the cosine of the angle.
\[ x = 3960 \cos(41.628 ^\circ) \]
Evaluate with a calculator in degree mode
Why: This angle is not special, so an exact value is not available.
\[ x = 2960\text{ approx} \]
Figure (svg): The solution to Worked example the radius of the Earth at a latitude shown as a ladder of expressions, one row per legal move
\[ x = 3960\cos(41.628^\circ) \approx 2960 \text{ miles} \]
Verify: sanity-check the direction
Why: The latitude circle must be smaller than the equatorial one, and 2960 is indeed less than 3960. It should also shrink to zero at the pole, and the cosine of 90 degrees is zero, which it does.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 731-731
Trap
\[ Q(6, 8) \text{ on the terminal side} \quad\Longrightarrow\quad \cos(\theta) = 6, \; \sin(\theta) = 8 \]
Read the coordinates as the cosine and sine directly
Why: That is exactly what the definition said - on the unit circle. Here the circle is not the unit circle.
Both answers exceed 1, and no cosine or sine can. The point six comma eight is at distance 10 from the origin, so the values must be divided by 10.
\[ r = \sqrt{6^2 + 8^2} = 10 \quad\Longrightarrow\quad \cos(\theta) = \tfrac{6}{10} = \tfrac{3}{5}, \; \sin(\theta) = \tfrac{8}{10} = \tfrac{4}{5} \]
Compute the radius first, then divide each coordinate by it
Why: The theorem read backwards says the cosine is x over r, not x.
The check is free and it never fails: a cosine or sine larger than 1 in absolute value is impossible. Any time an answer breaks that bound, the radius has been forgotten or mis-divided.
Fill the middle
The terminal side of an angle contains the point negative five comma twelve.
Fill in the blanks
r = \sqrt13 = -5/13 \quad\Longrightarrow\quad \cos(\theta) = 12/13, \quad \sin(\theta) = ___
Why: Twenty-five plus one hundred forty-four is one hundred sixty-nine, whose square root is 13. Dividing each coordinate by 13 gives the cosine and the sine. Notice that the radius is always taken positive, being a distance, so the sign of each value comes entirely from the sign of the coordinate — here negative for the cosine and positive for the sine, which is quadrant two.
Sorting
One bound rules out most wrong answers instantly.
Sort into buckets
Sort each number by whether it can be the cosine of some angle.
Socratic
Any point on the terminal side, other than the origin, can be used to compute the cosine and sine.
Discussion prompt
Two students pick different points on the same terminal side and both compute x over r. Explain why they must get the same answer.
Hint: What is the relationship between their two points?
Answer:
Their two points lie on the same ray from the origin, so one is a positive multiple of the other: if the second student's point is k times the first student's, then both x and r are multiplied by k, and the quotient is unchanged.
\[ \frac{kx}{kr} = \frac{x}{r} \]
This is the same invariance that made radian measure well defined in Lesson 10.1b, and it is the reason cosine and sine are functions of the angle alone. If the answer depended on which point you picked, they would not be functions of the angle at all.
Section
Section 2
Concept
The version of the theorem that actually gets used says the cosine is x over r and the sine is y over r. Given any point on the terminal side you can produce both values without ever identifying the angle.
\[ \cos(\theta) = \frac{x}{r}, \qquad \sin(\theta) = \frac{y}{r}, \qquad r = \sqrt{x^2 + y^2} \]
The radius is a distance and is therefore always taken positive. All the sign information lives in the coordinates, which is why a point in quadrant three automatically produces two negative values with no extra reasoning.
Figure (svg): A coordinate plane with the point four comma negative two marked, a terminal side drawn from the origin through it, and the radius computed as the square root of twenty
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 730-731
Picture it
The angle is never named in this calculation, and it does not need to be.
Figure (svg): A coordinate plane with the point four comma negative two marked, a terminal side drawn from the origin through it, and the radius computed as the square root of twenty
This is how the cosine and sine of a non-special angle are handled exactly rather than approximately: describe the angle by a point rather than by a measure.
Worked example
Example 10.2.6, part 1. Watch the rationalising at the end.
\[ \text{The terminal side of } \theta \text{ contains } Q(4, -2). \text{ Find } \cos(\theta) \text{ and } \sin(\theta). \]
Compute the radius
Why: Sixteen plus four is twenty, and root twenty simplifies to two root five.
\[ r = \sqrt{20} = 2 \sqrt{5} \]
Divide the x-coordinate by the radius
Why: Four over two root five reduces to two over root five.
\[ \cos = \frac{4}{2 \sqrt{5}} = 2 / \sqrt{5} \]
Rationalise
Why: Multiply top and bottom by root five.
\[ \cos = 2 \sqrt{5} / 5 \]
Do the same for the y-coordinate
Why: Negative two over two root five is negative one over root five, which rationalises to negative root five over five.
\[ \sin = -\sqrt{5} / 5 \]
Figure (svg): The solution to Worked example values from the point 4, -2 shown as a ladder of expressions, one row per legal move
\[ \cos(\theta) = \frac{2\sqrt{5}}{5} \qquad \sin(\theta) = -\frac{\sqrt{5}}{5} \]
Verify: check the identity
Why: Four fifths plus one fifth is 1, since two root five over five squared is twenty over twenty-five and root five over five squared is five over twenty-five. And the point is right and down, so a positive cosine with a negative sine is exactly quadrant four.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 731-731
Matching
The point determines both without any computation.
Match the pairs
Why: Each has radius 13, since five, twelve and thirteen form a Pythagorean triple, so all four give values of size five thirteenths and twelve thirteenths. Only the signs differ, and each sign is simply inherited from the corresponding coordinate because the radius is positive. This is the cleanest illustration that the reference angle fixes the sizes and the quadrant fixes the signs.
Worked example
The signs take care of themselves if you let the coordinates carry them.
\[ \text{The terminal side of } \theta \text{ contains } (-8, -6). \text{ Find } \cos(\theta) \text{ and } \sin(\theta). \]
Compute the radius
Why: Sixty-four plus thirty-six is one hundred, whose root is ten. The radius is positive even though both coordinates are negative.
\[ r = 10 \]
Divide each coordinate by the radius
Why: Negative eight over ten and negative six over ten.
\[ \cos = -\frac{8}{10}, \sin = -\frac{6}{10} \]
Reduce both fractions
Why: Both divide by two.
\[ \cos = -\frac{4}{5}, \sin = -\frac{3}{5} \]
Confirm the quadrant
Why: Both values negative means quadrant three, and the point is indeed left and down.
Figure (svg): The solution to Worked example a point in quadrant three shown as a ladder of expressions, one row per legal move
\[ \cos(\theta) = -\frac{4}{5} \qquad \sin(\theta) = -\frac{3}{5} \]
Verify: check the identity
Why: Sixteen twenty-fifths plus nine twenty-fifths is 1. Note that the negative signs vanish under squaring, which is why the identity can never detect a sign error — the quadrant check is a genuinely separate test and both are worth doing.
Error analysis
A student computes the cosine from the point negative three comma four.
Annotate
On: \( r = -5 \quad\Longrightarrow\quad \cos(\theta) = \frac{-3}{-5} = \frac{3}{5} \)
The square root symbol already returns the non-negative root, so r comes out positive automatically unless it is interfered with. The signs are the coordinates' job and only the coordinates' job.
Faded example
The terminal side contains the point one comma negative one.
Fill in the blanks
r = \sqrtsqrt2 = sqrt2/2 \;\Longrightarrow\; \cos(\theta) = \frac-sqrt2/2___} = ___, \quad \sin(\theta) = ___
Why: The radius is root two. Dividing gives one over root two for the cosine, which rationalises to root two over two, and negative one over root two for the sine, which is negative root two over two. The point is on the diagonal in quadrant four, so this is the angle seven pi over four — and the values agree with the unit circle, as they must.
Discrimination
Two points give the same angle when they lie on the same ray from the origin.
Sort into buckets
Sort each point by whether it lies on the same terminal side as the point three comma four.
Edge cases
The method needs a point on the terminal side, and any such point will do.
Discussion prompt
What goes wrong if the point chosen is the origin itself? And what does that tell you about the one thing the origin fails to do?
Hint: Compute r and then try to divide.
Answer:
\[ r = \sqrt{0^2 + 0^2} = 0 \quad\Longrightarrow\quad \cos(\theta) = \frac{0}{0} \]
The radius comes out zero and both quotients are undefined. But the deeper problem is geometric rather than algebraic: the origin lies on every terminal side, so it identifies no direction at all.
Every other point on a ray determines that ray uniquely. The origin is the one point shared by all of them, which is exactly why the theorem excludes it and why a terminal side is described by a point other than its endpoint.
Section
Section 3
Concept
Section 10.1 opened by asking where an object on a circular path is at time t and then admitted it could not yet answer. It can now. If the angular velocity is constant then the angle swept by time t is omega times t, and the theorem converts that angle into coordinates.
\[ x = r\cos(\omega t), \qquad y = r\sin(\omega t) \]
The convention is that the object starts at the point r comma zero when t is zero, and a positive omega means counterclockwise motion while a negative one means clockwise.
Figure (svg): An object on a circle of radius r whose angle at time t is omega times t, with its coordinates written as r cosine of omega t and r sine of omega t
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 732-732
Picture it
One angle, growing steadily with time, and two coordinates reading off its cosine and sine.
Figure (svg): An object on a circle of radius r whose angle at time t is omega times t, with its coordinates written as r cosine of omega t and r sine of omega t
Watching the horizontal coordinate alone as the point goes round is watching a cosine curve being drawn. That is precisely the animation Lesson 10.5a formalises.
Worked example
Example 10.2.7. Every number here was computed back in Lesson 10.1c.
\[ \text{With } r = 2960 \text{ miles and } \omega = \frac{\pi}{12} \text{ per hour, write the equations of motion.} \]
Recall the general form
Why: The theorem applied at angle omega t.
\[ x = r \cos(\omega t), y = r \sin(\omega t) \]
Substitute the radius
Why: Two thousand nine hundred sixty miles, the radius of the latitude circle.
\[ r = 2960 \]
Substitute the angular velocity
Why: One revolution in 24 hours is two pi over 24, which is pi over 12 per hour.
\[ \omega = \frac{\pi}{12} \]
State the units
Why: Coordinates in miles, time in hours.
Figure (svg): The solution to Worked example the equations of motion of a point on the Earth shown as a ladder of expressions, one row per legal move
\[ x = 2960\cos\left(\frac{\pi}{12}t\right), \qquad y = 2960\sin\left(\frac{\pi}{12}t\right) \]
Verify: check at two times
Why: At t equal to 0 the equations give 2960 and 0, the starting point, as required. At t equal to 12, half a day, the angle is pi and the equations give negative 2960 and 0 — diametrically opposite, which is exactly where a point is half a rotation later.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 732-732
Prediction
An object moves on a circle of radius 5 with angular velocity pi over 4 radians per second, starting at five comma zero.
Predict first
At what time does it first return to its starting point?
Correct: 8 seconds.
Why: The object returns when omega t equals two pi, so t equals two pi divided by pi over four, which is 8 seconds. That time is the period of the motion, and notice that the radius played no part: how long a full revolution takes depends only on the angular velocity, which is exactly the point made in Lesson 10.1c about omega being shared while v is not.
Worked example
Once the equations are written, a position is a substitution.
\[ \text{A wheel of radius } 3 \text{ m turns at } \frac{\pi}{6} \text{ rad/s from } (3, 0). \text{ Where is the marked point at } t = 4 \text{ s?} \]
Write the equations
Why: Substitute the radius and the angular velocity into the general form.
\[ x = 3 \cos(\pi t / 6), y = 3 \sin(\pi t / 6) \]
Find the angle at the given time
Why: Omega times t is pi over six times four.
\[ \theta = 2 \pi / 3 \]
Evaluate the cosine and sine
Why: Two pi over three is in quadrant two with reference angle pi over three.
\[ \cos = -\frac{1}{2}, \sin = \sqrt{3} / 2 \]
Multiply each by the radius
Why: Three times negative one half, and three times root three over two.
\[ x = -\frac{3}{2}, y = 3 \sqrt{3} / 2 \]
Figure (svg): The solution to Worked example find a position at a given time shown as a ladder of expressions, one row per legal move
\[ \left(-\frac{3}{2}, \; \frac{3\sqrt{3}}{2}\right) \]
Verify: check the distance from the centre
Why: Nine quarters plus twenty-seven quarters is thirty-six quarters, which is 9, and the square root of 9 is 3 — the radius, as it must be at every time. And after four seconds at pi over six per second the point should be a third of the way round, which is quadrant two.
Trap
\[ \omega = 30^\circ\text{/s}, \; t = 3 \quad\Longrightarrow\quad x = r\cos(90) \]
Multiply the angular velocity by the time and feed the result straight in
Why: The arithmetic is right, but 90 here means 90 degrees while the surrounding formulas assume radians.
A calculator in radian mode will read 90 as 90 radians, which is more than fourteen revolutions, and return a value with no relation to the intended quarter turn.
\[ \omega = \tfrac{\pi}{6}\text{ rad/s}, \; t = 3 \quad\Longrightarrow\quad x = r\cos\left(\tfrac{\pi}{2}\right) = 0 \]
Convert the angular velocity to radians per unit time before writing the equations
Why: Thirty degrees per second is pi over six radians per second.
This is the same unit discipline as Lesson 10.1c and it fails the same silent way. Convert omega to radians per unit time once, when you write the equations, and every substitution afterwards is safe.
Fill the middle
A point starts at two comma zero and turns at pi over three radians per second. Find its position at t equal to 2 seconds.
Fill in the blanks
\theta = \frac2 pi/3-1(2) = sqrt3 \;\Longrightarrow\; x = 2\cos\left(___\right) = ___, \quad y = 2\sin\left(___\right) = ___
Why: The angle after two seconds is two pi over three, which is in quadrant two with reference angle pi over three. Its cosine is negative one half and its sine is root three over two, and multiplying each by the radius 2 gives negative one and root three. The distance from the centre is the square root of one plus three, which is 2 — the radius, as required.
Invariant
Follow the object on a circle of radius 4 with angular velocity pi over 2 radians per second.
Step through it
Both coordinates have changed at every step. What quantity has not changed at any of them?
The distance from the centre is 4 at every single time, because the squares of the coordinates always sum to r squared — that is the Pythagorean identity multiplied by r squared. The object stays on its circle by construction, which is the guarantee these equations are built to provide.
Real world
A pedal on a bicycle turns on a circle of radius 17 centimetres about the crank axis, at a steady 80 revolutions per minute.
Discussion prompt
Write the equations of motion of the pedal relative to the crank axis, and then say what the vertical coordinate alone is describing as time passes.
Hint: Convert the rate first, then read the second equation as a function of time on its own.
Answer:
\[ \omega = 80 \cdot \frac{2\pi}{60} = \frac{8\pi}{3} \text{ rad/s} \]
\[ x = 17\cos\left(\tfrac{8\pi}{3}t\right), \qquad y = 17\sin\left(\tfrac{8\pi}{3}t\right) \]
The vertical coordinate on its own is the pedal's height above the axis as a function of time, and it oscillates smoothly between negative 17 and 17, returning to each height twice per revolution.
That function — a constant times the sine of a constant times t — is a sinusoid, and it is the shape of almost every oscillation in nature. Section 10.5 graphs it and Section 11.1 uses it to model tides, temperature and alternating current. The bicycle pedal is the simplest honest example of where it comes from.
Section
Section 4
Concept
Take a right triangle with an acute angle, and place it in the first quadrant with that angle in standard position and its adjacent side along the positive x-axis. The opposite vertex is then a point on a circle whose radius is the hypotenuse, and the theorem applies unchanged.
Theorem 10.4 — If an acute angle sits in a right triangle with adjacent side a, opposite side b and hypotenuse c, then its cosine is a over c and its sine is b over c.
\[ \cos(\theta) = \frac{a}{c} = \frac{\text{adjacent}}{\text{hypotenuse}}, \qquad \sin(\theta) = \frac{b}{c} = \frac{\text{opposite}}{\text{hypotenuse}} \]
So the definition you may have met first is a corollary of the definition used here, restricted to acute angles. Nothing has been thrown away, and the coordinate definition is strictly more general.
Figure (svg): A right triangle with an acute angle theta, its adjacent side a, opposite side b and hypotenuse c, drawn again in the first quadrant with the angle in standard position so the point a comma b lies on a circle of radius c
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 733-733
Picture it
Nothing about the triangle changes when it is placed in standard position. Only its description does.
Figure (svg): A right triangle with an acute angle theta, its adjacent side a, opposite side b and hypotenuse c, drawn again in the first quadrant with the angle in standard position so the point a comma b lies on a circle of radius c
Because the Pythagorean theorem gives a squared plus b squared equals c squared, the vertex really does lie on a circle of radius c, which is the only hypothesis the theorem needed.
Worked example
Example 10.2.8. A right triangle with a 30 degree angle whose adjacent side is 7.
\[ \text{A right triangle has an acute angle of } 30^\circ \text{ with adjacent side } 7. \text{ Find the other angle and both missing sides.} \]
Find the missing angle
Why: The three angles sum to 180 and one of them is the right angle.
\[ 180 - 30 - 90 = 60 ^\circ \]
Set up the cosine relation for the hypotenuse
Why: Cosine is adjacent over hypotenuse, and the adjacent side is the known one.
\[ \cos 30 ^\circ = \frac{7}{c} \]
Solve for the hypotenuse
Why: Rearranged, c is 7 divided by the cosine, and the cosine of 30 degrees is root three over two.
\[ c = \frac{7}{\sqrt{3} / 2} = 14 \sqrt{3} / 3 \]
Find the opposite side with the sine relation
Why: Sine is opposite over hypotenuse, so the opposite side is the hypotenuse times the sine.
\[ b = c \sin 30 ^\circ = 7 \sqrt{3} / 3 \]
Figure (svg): The solution to Worked example solve the triangle shown as a ladder of expressions, one row per legal move
\[ 60^\circ, \qquad b = \frac{7\sqrt{3}}{3} \approx 4.04, \qquad c = \frac{14\sqrt{3}}{3} \approx 8.08 \]
Verify: check with the Pythagorean theorem
Why: Seven squared is 49, and 7 root three over three squared is 147 over 9, which is about 16.3. Their sum is about 65.3. And 14 root three over three squared is 588 over 9, about 65.3. The two agree, so the sides are consistent.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 733-734
Matching
In each case one side and one acute angle are known.
Match the pairs
Why: All four come from the same two equations rearranged. Cosine is adjacent over hypotenuse, so multiplying gives the adjacent side and dividing gives the hypotenuse; sine is opposite over hypotenuse and behaves the same way. Rather than memorising four formulas, write the one relation that contains both the known and the wanted side, then rearrange it.
Worked example
The standard applied shape. Draw the triangle before writing anything.
\[ \text{From } 40 \text{ m away, a tower's top has an angle of elevation of } 35^\circ. \text{ How tall is the tower?} \]
Draw the triangle and label the sides
Why: The horizontal distance is adjacent to the angle; the height is opposite it.
\[ \text{adjacent } 40,\text{ opposite } h \]
Choose the relation containing both
Why: Cosine and sine both involve the hypotenuse, which is neither known nor wanted. Their quotient does not.
Express that quotient in terms of the two known relations
Why: The sine over the cosine is the opposite over the adjacent, since the hypotenuse cancels.
\[ \frac{h}{40} = \sin 35 / \cos 35 \]
Solve and evaluate
Why: The quotient of sine and cosine at 35 degrees is about 0.7002.
\[ h = 40(0.7002) = 28.0 m \]
Figure (svg): The solution to Worked example find a side from an angle of elevation shown as a ladder of expressions, one row per legal move
\[ h = 40 \cdot \frac{\sin(35^\circ)}{\cos(35^\circ)} \approx 28.0 \text{ m} \]
Verify: sanity-check the proportion
Why: An elevation of 35 degrees is less than 45, so the height should be less than the horizontal distance — and 28 is less than 40. At exactly 45 degrees the two would be equal, which is a useful landmark for checking this kind of answer.
Error analysis
A ladder 10 metres long leans against a wall at 70 degrees to the ground. A student finds how high it reaches.
Annotate
On: \( \text{height} = 10\cos(70^\circ) \approx 3.42 \text{ m} \)
The check that catches this instantly: a steep ladder should reach nearly its own length up the wall, and 3.42 out of 10 is not nearly. Ask whether the answer should be close to the hypotenuse or close to zero before computing it.
Faded example
A right triangle has an acute angle of 45 degrees and hypotenuse 12.
Fill in the blanks
a = 12\cos(45^\circ) = 12 \cdot sqrt2/2 = 6 sqrt2, \qquad b = 12\sin(45^\circ) = ___
Why: At 45 degrees the cosine and sine are both root two over two, so both legs come out equal at six root two, about 8.49. That the two legs are equal is exactly right for a 45-45-90 triangle, and it is a free check on the arithmetic. Confirming with Pythagoras: seventy-two plus seventy-two is one hundred forty-four, which is twelve squared.
Estimation
A right triangle has hypotenuse 20 and an acute angle of 80 degrees.
Predict first
Roughly how long is the side opposite that angle?
Correct: About 19.7.
Why: The sine of 80 degrees is about 0.985, very close to 1, so the opposite side is close to the whole hypotenuse. A steep angle means the opposite side does almost all the work. The distractor 3.5 is 20 times the cosine, the adjacent side, which is short precisely because the angle is steep; and 20.3 can be rejected immediately since no leg can exceed the hypotenuse.
Counterexample
A student proposes: the sine of an angle is the opposite side divided by the hypotenuse, so sine is only defined for angles in a triangle, which means angles between 0 and 90 degrees.
Discussion prompt
Give a specific counterexample and explain what the student has confused.
Hint: Which definition came first in this course, and which is the special case?
Answer:
\[ \sin(210^\circ) = -\tfrac{1}{2} \]
The sine of 210 degrees is perfectly well defined and equals negative one half, yet no triangle contains a 210 degree angle. The student has taken a special case for the definition.
The order matters. Cosine and sine are defined as coordinates on the unit circle, for every angle. Theorem 10.4 then observes that when the angle happens to be acute, those coordinates can be read off a right triangle as ratios. The triangle description is a consequence with a restricted domain, and this is why the book develops the circle first even though the triangle is more familiar.
Section
Section 5
Concept
Because each real number t was identified with an angle of t radians back in Lesson 10.1b, the cosine and sine can be regarded as ordinary functions of a real variable. Nothing changes computationally; what changes is that they can now be graphed and studied like any other function.
Theorem 10.5 — The functions cosine and sine each have domain all real numbers and range the closed interval from negative one to one.
\[ f(t) = \cos(t), \qquad g(t) = \sin(t) \]
Whether you read the input as an angle in radians or as a plain real number is entirely your choice, and the book says so. Every property proved for one reading holds for the other.
Figure (svg): Two number lines showing that the domain of cosine and sine is all real numbers while the range of each is the closed interval from negative one to one
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 734-735
Picture it
Two very different pictures: an unrestricted input line and a tightly bounded output interval.
Figure (svg): Two number lines showing that the domain of cosine and sine is all real numbers while the range of each is the closed interval from negative one to one
That asymmetry — everything in, only a narrow band out — is what forces the graphs of Section 10.5 to oscillate rather than grow, and what forces the inverses of Section 10.6 to have restricted domains.
Worked example
The book's point is that this is not a new kind of problem. Only the letter changed.
\[ \text{Solve } \sin(t) = -\frac{1}{2} \text{ for all real } t. \]
Read t as an angle in radians
Why: The identification of real numbers with angles makes this legitimate and costs nothing.
Find the reference angle from the size
Why: A sine of size one half belongs to pi over six.
\[ \text{reference } \frac{\pi}{6} \]
Find the two quadrants from the sign
Why: A negative sine puts the crossings below the x-axis, in quadrants three and four.
Write both families
Why: One solution per quadrant, each with its coterminal family.
\[ 7 \pi / 6 + 2 \pi k, 11 \pi / 6 + 2 \pi k \]
Figure (svg): The solution to Worked example solve an equation in t shown as a ladder of expressions, one row per legal move
\[ t = \frac{7\pi}{6} + 2\pi k \quad\text{or}\quad t = \frac{11\pi}{6} + 2\pi k, \quad k \in \mathbb{Z} \]
Verify: compare with the earlier solution
Why: This is identical to Example 10.2.5 part 2 from the previous lesson, where the variable was called theta. Only the name of the variable differs, which is exactly the book's point: reading the input as a number rather than an angle changes nothing.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 735-735
Sorting
The range settles each of these without any solving.
Sort into buckets
Sort each equation by whether any real number satisfies it.
Worked example
Knowing the range turns some questions into one-line answers.
\[ \text{Does } \cos(t) = \frac{5}{4} \text{ have any solutions? Does } 3\sin(t) = 2\text{?} \]
Check the first against the range
Why: Five quarters is 1.25, which is outside the interval from negative one to one.
\[ \frac{5}{4} > 1 \]
Conclude for the first
Why: No point on the unit circle has an x-coordinate of 1.25, so no angle works.
Isolate the sine in the second
Why: Divide both sides by 3.
\[ \sin t = \frac{2}{3} \]
Check that against the range
Why: Two thirds is about 0.667, comfortably inside the interval.
Figure (svg): The solution to Worked example decide whether an equation has solutions shown as a ladder of expressions, one row per legal move
\[ \cos(t) = \tfrac{5}{4}: \text{ no solutions} \qquad \sin(t) = \tfrac{2}{3}: \text{ two families of solutions} \]
Verify: say what the second's solutions look like
Why: Two thirds is not a special value, so the reference angle has no exact form and would need an inverse sine from Lesson 10.6a. But the range argument already settles that solutions exist, and that is a genuinely separate question from finding them.
Trap
A student reasons: since the cosine only takes values between negative one and one, the input must also be restricted to that interval.
\[ \cos(5) \quad\text{is undefined, because } 5 > 1 \]
This confuses the output restriction with an input restriction. They are different questions about different sets.
\[ \cos(5) \approx 0.284 \]
Check the domain, not the range, before evaluating
Why: The domain of cosine is every real number, so 5 is a perfectly legitimate input.
Five radians is an angle a bit past three pi over two, in quadrant four, where the cosine is positive and modest. The output is 0.284, safely inside the range.
The confusion is worth naming because it reverses in Lesson 10.6a: arccosine has domain the interval from negative one to one and unbounded range, because an inverse swaps exactly these two roles.
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 domain and range is wrong.
Survives elimination: B
Why: Statement B is false because the endpoints are attained rather than approached. The cosine equals exactly 1 at t equal to zero and exactly negative 1 at t equal to pi, since the points one comma zero and negative one comma zero are genuinely on the circle. The range is the closed interval. This matters directly in Lesson 10.6a, where arccosine is defined on that closed interval including both ends.
Notation
This line summarises the domain and range result. Read what each part is claiming.
Annotate
On: \( f(t) = \cos(t): \quad \text{dom} = (-\infty, \infty), \quad \text{ran} = [-1, 1] \)
Contrast this with tangent in Lesson 10.3a, whose domain genuinely does have infinitely many exclusions and whose range is all real numbers. Cosine and sine are the well-behaved pair; the other four functions are where the exceptions live.
Explain it to yourself
The book says the distinction between t as a real number and as an angle in radians is often blurred.
Discussion prompt
Explain what makes that blurring legitimate here, and name one place where a similar blurring would not be legitimate.
Hint: What does the wrapping function from Lesson 10.1b actually establish?
Answer:
It is legitimate because the wrapping function sets up a genuine correspondence between real numbers and angles: each real t names the arc of length t from the point one comma zero, which names an angle of t radians. The two readings pick out exactly the same point, so any statement true of one is true of the other.
It would not be legitimate for degrees. The number 60 does not correspond to an angle of 60 degrees under any natural identification — the correspondence that exists is with 60 radians. That is why cos(60) in a programming language means the cosine of 60 radians and returns about negative 0.95, which surprises people who expected one half. The blurring is a privilege radians have and degrees do not.
Comparison
Fill the blanks from memory. The right-hand column reduces to the left when r is 1.
Comparison matrix
| Unit circle | Circle of radius r | |
|---|---|---|
| Equation | x squared plus y squared equals 1 | x squared plus y squared equals r squared |
| Coordinates at angle theta | cosine theta, sine theta | r cosine theta, r sine theta |
| Cosine from a point | the x-coordinate itself | x divided by r |
| Arc for angle theta | theta | r theta |
| Position at time t | cosine omega t, sine omega t | r cosine omega t, r sine omega t |
Every entry in the right column is the left column multiplied by r. The unit circle is not a different case, it is the case where the multiplier happens to be invisible.
Pattern
Whether the problem hands you a point, a triangle or a moving object, the same five moves cover it.
For motion problems, insert one step before all of these: convert the angular velocity into radians per unit time. Everything downstream assumes it.
OpenStax Algebra and Trigonometry 2e, §7.2 Right Triangle Trigonometry §7.2
Check
Values from a point. Compute the radius first.
Check your understanding
The terminal side of an angle contains the point negative 7 comma 24. What is the sine of that angle?
Answer: A
Why: The radius is the square root of forty-nine plus five hundred seventy-six, which is the square root of 625, namely 25. The sine is the y-coordinate over the radius, which is 24 over 25. Seven, twenty-four and twenty-five are a Pythagorean triple, which is why it came out whole.
Check
A right triangle. Draw it before choosing.
Check your understanding
A right triangle has hypotenuse 26 and an acute angle whose cosine is five thirteenths. How long is the side adjacent to that angle?
Answer: B
Why: The cosine is the adjacent side over the hypotenuse, so the adjacent side is the hypotenuse times the cosine: 26 times five thirteenths is 10. Checking, the opposite side would be 26 times twelve thirteenths, which is 24, and 100 plus 576 is 676, which is 26 squared.
Check
Circular motion. Convert the rate first.
Check your understanding
An object moves on a circle of radius 6 with angular velocity pi over 2 radians per second, starting at six comma zero. Where is it at time t equals 3 seconds?
Answer: C
Why: The angle after three seconds is pi over two times three, which is three pi over two — three quarters of a revolution. That points straight down, so the coordinates are the radius times cosine and sine of three pi over two, giving 6 times 0 and 6 times negative 1, which is zero comma negative six.
Real world
A camera is mounted on a rotating platform of radius 0.6 metres, turning once every 30 seconds, and its position is logged as x and y in metres from the platform centre. A colleague asks you to convert a log entry back into the angle the platform had turned through.
Discussion prompt
Given a logged position of negative 0.3 comma negative 0.52, find the cosine and sine of the angle, name the quadrant, and identify the angle. Then say why logging the coordinates rather than the angle is arguably the better engineering choice.
Hint: Compute the radius from the point and check it against the known platform radius.
Answer:
\[ r = \sqrt{0.3^2 + 0.52^2} \approx 0.60 \quad\checkmark \]
\[ \cos(\theta) = \frac{-0.3}{0.6} = -\tfrac{1}{2}, \qquad \sin(\theta) = \frac{-0.52}{0.6} \approx -\tfrac{\sqrt{3}}{2} \]
Both values negative puts the angle in quadrant three, and the sizes one half and root three over two give reference angle pi over three, so the angle is four pi over three, or 240 degrees.
Logging coordinates is arguably better because the computed radius is a free consistency check: it came out 0.60, matching the known platform radius, which confirms the log entry is not corrupted. An angle logged alone carries no such redundancy. This is the same reason the identity is worth checking on every answer in this lesson — a quantity that must equal a known value is a free test.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Two points lie on the same terminal side, one at distance 3 from the origin and one at distance 12. How do the cosines of their angles compare?
Correct: The two cosines are equal.
\[ \frac{4x}{4r} = \frac{x}{r} \]
Why: They lie on the same terminal side, so they belong to the same angle, and cosine is a function of the angle alone. Concretely, the far point's coordinates are four times the near point's, but its radius is also four times as large, so the quotient x over r is unchanged. The fourth option is a distractor with a grain of truth in it: the quadrant does determine the sign of the shared value, but it cannot make the two differ from each other, since both points are in the same quadrant.
Explain it
They learned SOH CAH TOA and are annoyed that this course seems to have replaced it with circles for no reason.
Discussion prompt
In no more than five sentences, explain what the circle definition buys that the triangle definition cannot, and reassure them that their triangle knowledge is still correct.
Hint: Ask what a triangle cannot contain.
Answer:
A usable answer: everything you learned is still true, and this course proves it as a theorem rather than assuming it. The problem with the triangle definition is that a triangle's angles are all under 90 degrees, so it simply has nothing to say about an angle of 200 degrees, or a negative one, or one that has gone round twice.
The circle handles all of them, because every angle has a terminal side and every terminal side crosses the circle somewhere. When the angle does happen to be acute, drop a line from that crossing point to the horizontal axis and you have your triangle back, with adjacent over hypotenuse sitting exactly where the cosine was. The circle is the general case and the triangle is the corner of it you already knew.
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: Values from a point are fixed by always computing r first and always taking it positive. Motion equations are fixed by converting omega to radians per unit time before anything else. The triangle choice is fixed by writing the one relation containing both the known and the wanted side, rather than recalling four rearrangements. Domain and range are fixed by remembering that the input is an angle and every angle exists, while the output is a coordinate on a circle of radius one. 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 two concentric circles, a small one of radius one and a larger one of radius r, with a single terminal side crossing both. Drop perpendiculars from both crossing points and shade the two similar triangles, then write beside them the proportion that gives the theorem, and the theorem itself in both directions. Below that, draw a coordinate plane with the point negative six comma eight marked and work out the cosine and sine completely, showing the radius calculation. To the right, draw a circle with an object on it and write the two equations of motion, labelling what r, omega and t each mean. In the bottom left, draw a right triangle and label adjacent, opposite and hypotenuse, and write the two ratios. In the bottom right, write the domain and range of cosine, and beside each write one sentence saying where that restriction comes from. Finally, circle the one quantity on your whole page that is a ratio rather than a length, and say why that is the reason it does not change when r changes.
You should have circled the cosine and sine themselves, along with the side ratios in the triangle, which are the same thing. Everything else on the page — coordinates, arc lengths, side lengths, positions — is a length and scales with r.
Recap
Five things, and the first one is what makes the other four possible.
| If the question says | Your first move is |
|---|---|
| The terminal side contains this point | Compute the radius from the distance formula |
| Find the coordinates on a circle of radius r | Get the unit-circle values, then multiply by r |
| An object moves with angular velocity omega | Convert omega to radians per unit time |
| Solve this right triangle | Label adjacent, opposite and hypotenuse relative to the angle |
| Does this equation have a solution | Check the required value against the interval from -1 to 1 |
That completes Section 10.2. Cosine and sine are now fully defined, computable and bounded. The next section defines the other four circular functions as combinations of these two, which immediately raises a question cosine and sine never had to face: what happens where a denominator is zero?
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.2 The Unit Circle: Cosine and Sine §10.2, pp. 730-736 — everything on these slides traces back here
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