 55 lessons
 1 quizzes
 10 week duration

Numerical Computation
Here you'll be introduced to the bare basics of mathematics. Topics include commonly used words and phrases, symbols, and how to follow the order of operations.

Measurements
An introduction to numerical computation. Emphasis is placed on scientific and engineering notation, the rule of significant figures, and converting between SI and Imperial units.

Trigonometry with Right Triangles
Here we focus on right angle triangles within quadrant I of an xy plane. None of the angles we evaluate here are greater than 90°. A unit on trigonometry with oblique triangles is covered later.

Trigonometry with Oblique Triangles
This unit is a continuation of trigonometry with right triangles except we'll extend our understanding to deal with angles *greater* than 90°. Resolving and combining vectors will be covered at the end of this unit.

Vector Analysis

Introduction to Algebra

Factoring

Solving Equations

Functions and Graphs

Geometry
This unit focuses on analyzing and understand the characteristics of various shapes, both 2D and 3D.
 Identify, measure, and calculate different types of straight lines and angles
 Calculate the interior angles of polygons
 Solve problems involving a variety of different types of triangles
 Calculate the area of a variety of different types of quadrilaterals
 Solve problems involving circles
 Calculate the areas and volumes of different solids

Introduction to Statistics
CAST Rule and Reference Angles
Given the right triangle shown below, a summary of all the trigonometric functions we’ve learned so far are listed:
Trigonometric Functions:
$\mathbf{sine}\mathbf{}\mathit{\theta}=\mathrm{sin}\theta =\frac{opposite}{hypotenuse}=\frac{y}{r}\phantom{\rule{0ex}{0ex}}\mathbf{cosine}\mathbf{}\mathit{\theta}=\mathrm{cos}\theta =\frac{adjacent}{hypotenuse}=\frac{x}{r}\phantom{\rule{0ex}{0ex}}\mathbf{tangent}\mathbf{}\mathbf{\theta}=\mathrm{tan}\theta =\frac{opposite}{adjacent}=\frac{y}{x}\phantom{\rule{0ex}{0ex}}$
Reciprocal Trigonometric Functions:
$\mathbf{cosecant}\mathbf{}\mathbf{\theta}=\mathrm{csc}\theta =\frac{hypotenuse}{opposite}=\frac{r}{y}\phantom{\rule{0ex}{0ex}}\mathbf{secant}\mathbf{}\mathbf{\theta}=\mathrm{sec}\theta =\frac{hypotenuse}{adjacent}=\frac{r}{x}\phantom{\rule{0ex}{0ex}}\mathbf{cotangent}\mathbf{}\mathbf{\theta}=\mathrm{cot}\theta =\frac{adjacent}{opposite}=\frac{x}{y}\phantom{\rule{0ex}{0ex}}$
Inverse Trigonometric Functions:
${\mathrm{sin}}^{\u20131}\left(\frac{\mathit{y}}{\mathit{r}}\right)=\mathrm{arcsin}\left(\frac{\mathit{y}}{\mathit{r}}\right)=\theta \phantom{\rule{0ex}{0ex}}{\mathrm{cos}}^{\u20131}\left(\frac{\mathit{x}}{\mathit{r}}\right)=\mathrm{arccos}\left(\frac{\mathit{y}}{\mathit{r}}\right)=\theta \phantom{\rule{0ex}{0ex}}{\mathrm{tan}}^{\u20131}\left(\frac{\mathit{y}}{\mathit{x}}\right)=\mathrm{arctan}\left(\frac{\mathit{y}}{\mathit{r}}\right)=\theta \phantom{\rule{0ex}{0ex}}$
Inverse Reciprocal Trigonometric Functions:
${\mathrm{csc}}^{\u20131}\left(\frac{r}{y}\right)=\mathrm{arccsc}\left(\frac{r}{y}\right)=\theta \phantom{\rule{0ex}{0ex}}{\mathrm{sec}}^{\u20131}\left(\frac{r}{x}\right)=\mathrm{arcsec}\left(\frac{r}{x}\right)=\theta \phantom{\rule{0ex}{0ex}}{\mathrm{cot}}^{\u20131}\left(\frac{x}{y}\right)=\mathrm{arccot}\left(\frac{x}{y}\right)=\theta \phantom{\rule{0ex}{0ex}}$
In the last unit, we explored all six of these trigonometric functions using angles found within a right triangle. You discovered that no matter what trigonometric function you used, the ratio always came out positive. The reason is that all the angles we ever used were acute (less than 90°). Now, using the same techniques as before, we can find ratios of angles greater than 90°, but various precautions need to be taken.
Take, for example, a line segment drawn on a Cartesian plane (xy plane) positioned in the 2nd quadrant (shown below in blue).
 Remember that a Cartesian plane is used to map coordinates on a twodimensional graph. Splitting a plane into two axes creates four quadrants: I, II, III, and IV.
The blue line segment, which we’ll call the terminal side extends beyond 90° (let’s assume that the blue line is at an angle of 135°), and the tip of that line segment has the coordinates (–6, 5). Extending a vertical line from the tip to the xaxis generates a perfect right triangle (shown below). Given our previous assumption of 135°, the inside angle of this right triangle is calculated by taking 180° minus 135°. 45° will now serve as the acute reference angle for this terminal side.
Now using this right triangle whose side lengths are –6 and 5, this is how the trigonometric ratios would look. Keep in mind that the hypotenuse (r) can be found using the Pythagorean theorem, and will always be positive. In our case here, the hypotenuse is ≈ 7.8.
$\mathrm{sin}\theta =\frac{5}{7.8}\to {\mathbf{+}}{\mathbf{positive}}{\mathbf{}}{\mathbf{output}}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\theta =\frac{\u20136}{7.8}\to {\mathbf{\u2013}}{\mathbf{negative}}{\mathbf{}}{\mathbf{output}}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\theta =\frac{\u20136}{5}\to {\mathbf{\u2013}}{\mathbf{negative}}{\mathbf{}}{\mathbf{output}}\phantom{\rule{0ex}{0ex}}$Therefore, the trigonometric ratio for 135° is the same as the ratio for 45°! Try it yourself, type sin(135°) and sin(45°) into your calculator and you’ll get approximately 0.7071 for both. This means you now have to report both 135° and 45° if you’re asked to find theta for sin θ = 0.7071. Previously, we would have only reported 45°.
Takehome message:
If you ever want to find out the ratio of an angle that’s between 90° and 180°, first subtract your angle from 180°. This will give you an acute reference angle. Next, evaluate that angle using any trigonometric function on your calculator to get the ratio. If you use sine, the output will always be positive, while the output of cosine and tangent in this quadrant will always be negative.
Without investigating what happens in each quadrant, a summary of what to expect is illustrated below.
Instead of trying to remember which trigonometric functions are negative in which quadrants, just remember C A S T:
 Cosine is always positive in the 4th quadrant
 ALL are positive in the 1st quadrant
 Sine is always positive in the 2nd quadrant
 Tangent is always positive in the 3rd quadrant
Another thing to keep in mind is that depending where the terminal side lies, how to find the reference angle differs. In quadrant three, you subtract 180° or π rad from the given angle. In quadrant four, you subtract the angle from 360° or 2π rad. This is summarized in the video below:
Summary on how to find the reference angle (denoted θ’):
$\theta =\theta \u2018\mathbf{[}\mathbf{Quadrant}\mathbf{}\mathbf{I}\mathbf{]}\phantom{\rule{0ex}{0ex}}180\xb0\u2013\theta =\theta \u2018\mathbf{[}\mathbf{Quadrant}\mathbf{}\mathbf{II}\mathbf{]}\phantom{\rule{0ex}{0ex}}\theta \u2013180\xb0=\theta \u2018\mathbf{[}\mathbf{Quadrant}\mathbf{}\mathbf{III}\mathbf{]}\phantom{\rule{0ex}{0ex}}360\xb0\u2013\theta =\theta \u2018\mathbf{[}\mathbf{Quadrant}\mathbf{}\mathbf{IV}\mathbf{]}$