Taking new projectsCanvas course design

Canvas course pages, built from your course materials

I build Canvas unit hubs and before-class prep pages from your existing book, slides, and handouts. I do this instructional design work from Boston, remotely, and I taught high school math for more than twelve years.

Message me on LinkedIn Email contact@megan-warren.com

Fixed price, quoted by scope before any work starts. I reply within two business days.

What the unit hub and prep pages look like

The Trigonometry Unit 2 hub shown in a plain browser frame and a phone frame: unit goals with typeset math, under a purple Built for Canvas header. — Built for Canvas: unit hub, shown outside Canvas

The Trigonometry Section 2.6 prep page in a browser frame and a phone frame, the phone showing a Check your answer box opened to a worked answer. — Built for Canvas: Trigonometry prep page, shown outside Canvas

The Discrete Math and Linear Algebra Section 2.4 prep page on probability, in a browser frame with one answer open and a phone frame showing the page title. — Built for Canvas: Discrete Math prep page, shown outside Canvas

Three phone frames side by side: the hub's sections table, the Trigonometry prep page with an answer open, and the top of the Discrete Math prep page. — Built for Canvas: hub and prep pages on a phone, shown outside Canvas

Try the Trigonometry Unit 2 hub

Below is the hub as students see it in Canvas, shown here outside it. The Prep links open each prep page right here, and the other links open the public book, slides, and games.

Open the unit hub (Trigonometry, Unit 2, with all six prep pages)

Unit 2 hub: The Unit Circle

By the end of Unit 2 you will be able to:

Before solving a two-sides-and-an-angle (SSA) setup, count its triangles by comparing side $a$ with the height $h = b\sin A$ . Then solve each possible triangle, catching the second triangle's angle as the supplement $B' = 180^\circ - B$ .
Measure rotation the circle's own way. Set an angle in standard position, name its coterminal angles, and define the radian. Convert between degrees and radians, and find the arc length and sector area a central angle cuts off.
Read the sine and cosine of any angle, not merely an acute one, as the coordinates where its terminal side meets the unit circle. Take tangent as that side's slope. Off the axes, take every sign from the quadrant the angle lies in. Mnemonic: ASTC, All Star Trig Class. Take every absolute value from a reference angle $\theta'$ .
Build the 45-45-90 and 30-60-90 triangles from a square and an equilateral triangle. Use the exact values they give, with reference angles and signs, at the twelve unit-circle angles off the axes. State the cofunction identities that pair each function with its cofunction.
Define cosecant, secant, and cotangent as the reciprocals of sine, cosine, and tangent. Evaluate them at the 16 unit-circle angles, and locate where each one is undefined. Keep the reciprocal $\frac{1}{\sin\theta}$ apart from the inverse sine $\sin^{-1}$ .
Read the quotient identities $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and $\cot\theta = \frac{\cos\theta}{\sin\theta}$ straight off the coordinates. Rebuild all six function values from a single one, and rewrite expressions in sine and cosine.

Everything for Unit 2 is on this page: the before-class prep pages, the slides from class, the book, the games, and the assignments.

Open Unit 2 in the book

Sections

Section	Open	Practice	Games and activities
2.1 The Law of Sines: The Ambiguous Case	Prep · Slides · Book	Exercises	Game: Which law? Try it: The ambiguous case
2.2 Angle Terms and Radian Measurement	Prep · Slides · Book	Exercises	Try it: What a radian is
2.3 Sine, Cosine, and Tangent in Quadrants II, III, and IV	Prep · Slides · Book	Exercises	Try it: The unit circle
2.4 Special Right Triangles and Cofunctions	Prep · Slides · Book	Exercises	Game: The unit circle challenge
2.5 The Reciprocal Trig Functions	Prep · Slides · Book	Exercises
2.6 The Quotient Identities	Prep · Slides · Book	Exercises	Game: The six-function challenge

Review and assignments

Section 2.R Recap: Check Yourself (one check per goal)
Section 2.P Mixed Practice (every Unit 2 skill, shuffled)
Unit 2 Problem Set (a Canvas assignment)
Unit 2 Assessment (in class, on paper)

Back to the unit hub

Section 2.1 Prep: The Law of Sines: The Ambiguous Case

Lesson goal: Count and solve every triangle that a two-sides-and-a-non-included-angle (SSA) setup allows.

Before Part 1

Try these from memory first. Then use your notes to finish any you could not. Open Check your answer when you are done.

1. In triangle $ABC$ , $A = 40^\circ$ , $B = 75^\circ$ , and side $a = 10$ cm. Use the Law of Sines to find side $b$ , to three decimal places.

Check your answer

Pair each side with its opposite angle: $\dfrac{b}{\sin 75^\circ} = \dfrac{10}{\sin 40^\circ}$ .

Multiply both sides by $\sin 75^\circ$ : $b = \dfrac{10\sin 75^\circ}{\sin 40^\circ}$ .

$b \approx 15.027$ cm.

2. In a right triangle, the hypotenuse is $8$ and one acute angle is $35^\circ$ . How long is the side opposite that $35^\circ$ angle, to three decimal places?

Check your answer

Sine is opposite over hypotenuse: $\sin 35^\circ = \dfrac{\text{opposite}}{8}$ .

Opposite side $= 8\sin 35^\circ \approx 4.589$ .

3. A triangle has two angles measuring $52^\circ$ and $34^\circ$ . Find the third angle.

Check your answer

The three angles of a triangle add to $180^\circ$ .

Third angle $= 180^\circ - 52^\circ - 34^\circ = 94^\circ$ .

Before Part 2

1. SSA count (Part 1): $A = 30^\circ$ , $b = 12$ , $a = 5$ . How many triangles?

Check your answer

$a = 5 < b = 12$ , so find the height.

$h = 12\sin 30^\circ = 12 \cdot \dfrac{1}{2} = 6$ .

$a = 5 < h = 6$ : side $a$ cannot reach the base line. 0 triangles.

2. Law of Sines: in triangle $ABC$ , $A = 45^\circ$ , $B = 70^\circ$ , $a = 12$ . Find side $b$ , to three decimal places.

Check your answer

$\dfrac{b}{\sin 70^\circ} = \dfrac{12}{\sin 45^\circ}$ .

$b = \dfrac{12\sin 70^\circ}{\sin 45^\circ} \approx 15.947$ .

3. An angle measures $52^\circ$ . What is its supplement (the angle that adds with it to $180^\circ$ )?

Check your answer

Supplementary angles add to $180^\circ$ .

$180^\circ - 52^\circ = 128^\circ$ .

One more from last time

State the Law of Cosines for side c.

Check your answer

$c^2 = a^2 + b^2 - 2ab\cos C$ .

Open this section in the book

Back to the unit hub

Section 2.2 Prep: Angle Terms and Radian Measurement

Lesson goal: Measure any rotation in standard position, name its coterminal angles, convert between degrees and radians, and use the radian to compute arc length and sector area.

Before Part 1

Try these from memory first. Then use your notes to finish any you could not. Open Check your answer when you are done.

1. An angle in a triangle is $B = 65^\circ$ . Find its supplement $B' = 180^\circ - B$ .

Check your answer

The supplement is $B' = 180^\circ - B$ .

$B' = 180^\circ - 65^\circ = 115^\circ$ .

2. A right triangle has an acute angle $\theta$ with opposite side $8$ and hypotenuse $17$ . (a) Write $\sin\theta$ as a fraction. (b) Find $\theta$ with $\sin^{-1}$ , to the nearest degree.

Check your answer

(a) Sine is opposite over hypotenuse: $\sin\theta = \frac{8}{17}$ .

(b) With the calculator in degree mode, $\theta = \sin^{-1}\left(\frac{8}{17}\right) \approx 28.072^\circ$ .

To the nearest degree, $\theta \approx 28^\circ$ .

3. Name the quadrant of the point $(-3, 5)$ on the coordinate plane.

Check your answer

The $x$ -coordinate $-3$ is negative, so the point is left of the $y$ -axis.

The $y$ -coordinate $5$ is positive, so the point is above the $x$ -axis.

Left of the $y$ -axis and above the $x$ -axis is Quadrant II.

Before Part 2

1. Find the angle in $[0^\circ, 360^\circ)$ coterminal with $430^\circ$ .

Check your answer

Subtract a full turn: $430^\circ - 360^\circ = 70^\circ$ .

$70^\circ$ is in $[0^\circ, 360^\circ)$ , so the answer is $70^\circ$ .

2. Take a $3$ - $4$ - $5$ right triangle and a $6$ - $8$ - $10$ right triangle. In each, find the sine of the angle opposite the shortest side. Why are the two values equal?

Check your answer

In the $3$ - $4$ - $5$ triangle, the opposite side is $3$ and the hypotenuse is $5$ , so the sine is $\frac{3}{5} = 0.6$ .

In the $6$ - $8$ - $10$ triangle, the opposite side is $6$ and the hypotenuse is $10$ , so the sine is $\frac{6}{10} = 0.6$ .

The second triangle is the first scaled by $2$ . The opposite side and the hypotenuse both double, so their ratio does not change.

3. Write the circumference and the area of a circle of radius $r$ .

Check your answer

Circumference: $C = 2\pi r$ , the distance around the circle.

Area: $A = \pi r^2$ .

One more from last time

How do you count SSA triangles by comparing a with the height h = b sin A?

Check your answer

Compute $h = b\sin A$ (A the given angle, b its adjacent side), then compare a with h and b.

Open this section in the book

Back to the unit hub

Section 2.3 Prep: Sine, Cosine, and Tangent in Quadrants II, III, and IV

Lesson goal: Read sine and cosine as coordinates on the unit circle, name the sign of any angle from its quadrant (mnemonic: ASTC, All Star Trig Class), and evaluate anywhere with the reference angle for the absolute value and the quadrant for the sign.

Before Part 1

Try these from memory first. Then use your notes to finish any you could not. Open Check your answer when you are done.

1. Find the angle in $[0^\circ, 360^\circ)$ coterminal with $495^\circ$ . Then name the quadrant its terminal side lands in.

Check your answer

Subtract one full turn: $495^\circ - 360^\circ = 135^\circ$ .

$135^\circ$ is between $90^\circ$ and $180^\circ$ .

Answer: $135^\circ$ , in Quadrant II.

2. For the point $P = (-4, -3)$ : (a) name its quadrant, (b) find its distance from the origin.

Check your answer

(a) Both coordinates are negative, so $P$ is in Quadrant III.

(b) Distance $= \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ .

Answer: (a) Quadrant III; (b) $5$ .

3. A right triangle has an acute angle $\theta$ with opposite side $3$ , adjacent side $4$ , and hypotenuse $5$ . Find $\sin\theta$ , $\cos\theta$ , and $\tan\theta$ .

Check your answer

$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5}$ .

$\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{4}{5}$ .

$\tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4}$ .

Before Part 2

1. From ASTC: (a) In which quadrant is sine positive but cosine negative? (b) Which single function is positive in Quadrant III?

Check your answer

(a) Sine positive means $y > 0$ , and cosine negative means $x < 0$ . That is Quadrant II.

(b) In Quadrant III, $x < 0$ and $y < 0$ . Only $\tan\theta = \frac{y}{x}$ is positive there.

2. Find the angle in $[0^\circ, 360^\circ)$ coterminal with $-70^\circ$ . Then name its quadrant.

Check your answer

Add one full turn: $-70^\circ + 360^\circ = 290^\circ$ .

$290^\circ$ is between $270^\circ$ and $360^\circ$ .

Answer: $290^\circ$ , in Quadrant IV.

3. Compute $180^\circ - 118^\circ$ and $360^\circ - 295^\circ$ .

Check your answer

$180^\circ - 118^\circ = 62^\circ$ .

$360^\circ - 295^\circ = 65^\circ$ .

Subtractions like these give reference angles, coming up next.

One more from last time

Find the angle in $[0^\circ, 360^\circ)$ coterminal with $430^\circ$ .

Check your answer

$70^\circ$

Open this section in the book

Back to the unit hub

Section 2.4 Prep: Special Right Triangles and Cofunctions

Lesson goal: Build the two special right triangles from a square and an equilateral triangle, read the exact values of sine, cosine, and tangent at 30, 45, and 60 degrees, rebuild the first-quadrant table with the first-quadrant sine pattern, extend those values to all 16 unit-circle angles with reference angles and the ASTC mnemonic, and use the cofunction identities.

Before Part 1

Try these from memory first. Then use your notes to finish any you could not. Open Check your answer when you are done.

1. For $300^\circ$ : (a) name the quadrant, (b) find the reference angle $\theta'$ , (c) say which of $\sin$ , $\cos$ , $\tan$ is positive there.

Check your answer

(a) $300^\circ$ is between $270^\circ$ and $360^\circ$ , so it is in Quadrant IV.

(b) The reference angle is $\theta' = 360^\circ - 300^\circ = 60^\circ$ .

2. (a) A right triangle has two legs equal to $1$ . Find the exact hypotenuse. (b) A second right triangle has a leg $2$ and hypotenuse $4$ . Find the exact other leg.

Check your answer

(a) Legs $1$ and $1$ : the hypotenuse is $\sqrt{1^2 + 1^2} = \sqrt{2}$ .

(b) Leg $2$ and hypotenuse $4$ : the other leg is $\sqrt{4^2 - 2^2} = \sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}$ .

3. Rationalize the denominator: (a) $\dfrac{1}{\sqrt{2}}$ , (b) $\dfrac{1}{\sqrt{3}}$ .

Check your answer

Multiply the numerator and the denominator by the radical in the denominator.

(a) $\dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$ .

(b) $\dfrac{1}{\sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}$ .

Before Part 2

1. From the special triangles, give the exact $\sin$ , $\cos$ , and $\tan$ of $45^\circ$ .

Check your answer

The $45$ - $45$ - $90$ triangle has sides $1$ , $1$ , $\sqrt{2}$ .

$\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ , and $\tan 45^\circ = \frac{1}{1} = 1$ .

2. For $\frac{5\pi}{3}$ : (a) name the quadrant, (b) find the reference angle, (c) say which function is positive there.

Check your answer

(a) $\frac{5\pi}{3}$ is between $\frac{3\pi}{2}$ and $2\pi$ , so it is in Quadrant IV.

(b) $\theta' = 2\pi - \frac{5\pi}{3} = \frac{6\pi}{3} - \frac{5\pi}{3} = \frac{\pi}{3}$ .

3. Subtract the fractions: $\dfrac{\pi}{2} - \dfrac{\pi}{5}$ .

Check your answer

Use the common denominator $10$ : $\frac{\pi}{2} = \frac{5\pi}{10}$ and $\frac{\pi}{5} = \frac{2\pi}{10}$ .

$\frac{5\pi}{10} - \frac{2\pi}{10} = \frac{3\pi}{10}$ .

One more from last time

On the unit circle, what are $\cos\theta$ and $\sin\theta$ ?

Check your answer

$\cos\theta = x$ and $\sin\theta = y$ , the coordinates of the crossing point

Open this section in the book

Back to the unit hub

Section 2.5 Prep: The Reciprocal Trig Functions

Lesson goal: Evaluate cosecant, secant, and cotangent anywhere on the circle by taking the reciprocal of sine, cosine, or tangent. Keep the reciprocal (a number) apart from the output of an inverse trig function (an angle).

Before Part 1

Try these from memory first. Then use your notes to finish any you could not. Open Check your answer when you are done.

1. Evaluate exactly from the special-angle table: (a) $\sin\frac{\pi}{6}$ , (b) $\cos\frac{\pi}{4}$ , (c) $\tan\frac{\pi}{3}$ .

Check your answer

(a) From the special-angle table, $\sin\frac{\pi}{6} = \frac{1}{2}$ .

(b) From the table, $\cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$ .

2. Rationalize each denominator: (a) $\frac{2}{\sqrt{3}}$ , (b) $\frac{1}{\sqrt{2}}$ .

Check your answer

(a) Multiply by $\frac{\sqrt{3}}{\sqrt{3}}$ , which equals $1$ : $\frac{2}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$ .

(b) Multiply by $\frac{\sqrt{2}}{\sqrt{2}}$ : $\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ .

3. Write the reciprocal of each number: (a) $\frac{4}{5}$ , (b) $-6$ , (c) $2$ .

Check your answer

(a) Swap the numerator and the denominator: the reciprocal of $\frac{4}{5}$ is $\frac{5}{4}$ .

(b) $-6 = \frac{-6}{1}$ , so its reciprocal is $-\frac{1}{6}$ . The sign stays the same.

Before Part 2

1. Evaluate exactly: (a) $\sec\frac{\pi}{6}$ , (b) $\cot\frac{\pi}{6}$ .

Check your answer

(a) Secant is the reciprocal of cosine: $\sec\frac{\pi}{6} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$ .

(b) Cotangent is the reciprocal of tangent: $\cot\frac{\pi}{6} = \frac{1}{\frac{\sqrt{3}}{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}$ .

2. For $\theta = \frac{5\pi}{6}$ : (a) which quadrant? (b) what is the reference angle $\theta'$ ? (c) what are the signs of $\sin\theta$ and $\cos\theta$ ?

Check your answer

(a) $\frac{5\pi}{6}$ is between $\frac{\pi}{2}$ and $\pi$ , so it is in Quadrant II.

(b) $\theta' = \pi - \frac{5\pi}{6} = \frac{\pi}{6}$ .

3. Evaluate exactly: $\sin\frac{5\pi}{6}$ and $\cos\frac{5\pi}{6}$ .

Check your answer

The reference angle $\frac{\pi}{6}$ gives the absolute values $\frac{1}{2}$ and $\frac{\sqrt{3}}{2}$ .

The Quadrant II signs give $\sin\frac{5\pi}{6} = \frac{1}{2}$ and $\cos\frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$ .

One more from last time

What are the side ratios of the $45$ - $45$ - $90$ and $30$ - $60$ - $90$ triangles?

Check your answer

$1 : 1 : \sqrt{2}$ and $1 : \sqrt{3} : 2$

Open this section in the book

Back to the unit hub

Section 2.6 Prep: The Quotient Identities

Lesson goal: State the quotient identities $\tan\theta = \frac{\sin\theta}{\cos\theta}$ and $\cot\theta = \frac{\cos\theta}{\sin\theta}$ by reading tangent as the slope of the terminal side, use them to evaluate and to build all six functions from one value and a quadrant, rewrite expressions in sine and cosine and simplify, and locate where each quotient is undefined.

Before Part 1

Try these from memory first. Then use your notes to finish any you could not. Open Check your answer when you are done.

1. $\cos\frac{\pi}{3} = \frac{1}{2}$ . Use a reciprocal identity to find $\sec\frac{\pi}{3}$ .

Check your answer

Secant is the reciprocal of cosine: $\sec\frac{\pi}{3} = \frac{1}{\cos\frac{\pi}{3}}$ .

$\dfrac{1}{\;\frac{1}{2}\;} = 1 \cdot \frac{2}{1} = 2$ , so $\sec\frac{\pi}{3} = 2$ .

2. Name the quadrant of $\frac{2\pi}{3}$ . Then give $\sin\frac{2\pi}{3}$ and $\cos\frac{2\pi}{3}$ exactly.

Check your answer

$\frac{2\pi}{3}$ is between $\frac{\pi}{2}$ and $\pi$ , so it is in Quadrant II.

Its reference angle is $\pi - \frac{2\pi}{3} = \frac{\pi}{3}$ .

In Quadrant II, sine is positive and cosine is negative: $\sin\frac{2\pi}{3} = \frac{\sqrt{3}}{2}$ and $\cos\frac{2\pi}{3} = -\frac{1}{2}$ .

3. Divide, then rationalize the denominator (bottom): $\dfrac{\;\frac{1}{2}\;}{\;\frac{\sqrt{2}}{2}\;}$ .

Check your answer

Dividing by a fraction is multiplying by its reciprocal: $\frac{1}{2} \cdot \frac{2}{\sqrt{2}} = \frac{1}{\sqrt{2}}$ .

Rationalize: $\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ .

Before Part 2

1. Given $\sin\theta = \frac{3}{5}$ with $\theta$ in Quadrant I, find $\cos\theta$ and $\tan\theta$ exactly.

Check your answer

Sizes: opposite $3$ and hypotenuse $5$ , so the adjacent side is $\sqrt{5^2 - 3^2} = \sqrt{16} = 4$ .

Quadrant I makes every function positive: $\cos\theta = \frac{4}{5}$ .

$\tan\theta = \dfrac{\;\frac{3}{5}\;}{\;\frac{4}{5}\;} = \frac{3}{5} \cdot \frac{5}{4} = \frac{3}{4}$ .

2. On $[0, 2\pi)$ : (a) where is $\sec\theta$ undefined? (b) Why?

Check your answer

(a) At $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ .

(b) $\sec\theta = \frac{1}{\cos\theta}$ , and $\cos\theta = 0$ at those two angles. A fraction with denominator $0$ has no value.

3. Give $\sin\pi$ and $\cos\frac{\pi}{2}$ .

Check your answer

At $\pi$ the point is $(-1, 0)$ , so $\sin\pi = 0$ .

At $\frac{\pi}{2}$ the point is $(0, 1)$ , so $\cos\frac{\pi}{2} = 0$ .

One more from last time

What are cosecant, secant, and cotangent?

Check your answer

The reciprocals of sine, cosine, tangent: $\csc\theta = \frac{1}{\sin\theta}$ , $\sec\theta = \frac{1}{\cos\theta}$ , $\cot\theta = \frac{1}{\tan\theta}$ .

Open this section in the book

What each Canvas page includes

Unit goalsWhat students will be able to do by the end of the unit, with the math typeset.
A table of sectionsEach section linked to its prep page, the book, the slides, and the games.
A prep page for each sectionA short read before class, with “Check your answer” boxes that open to a worked answer.

How a Canvas project runs

Tell me which course and which units.Send what you already have: the book, the slides, the handouts. An outline of the units is plenty.
Get a fixed price and a date.Quoted by scope, before any work starts.
Research comes first.I study your course, your students, and how your Canvas is set up, and I look up anything I’m not sure of.
Get the finished pages.Ready to paste into Canvas, and you own the files and the source.
Keep it current.I can show your team how to update them, or handle the updates myself whenever you need them.

Want Canvas pages built from your course materials?

Message me on LinkedIn Email contact@megan-warren.com

I reply within two business days, with a quote or a question.

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