The Timpson Lyceum · College Algebra

Lecture 11 (College Algebra)

Overview

This lecture is a crash course in the trigonometry and coordinate-geometry tools that algebra leans on. It begins with the Cartesian (xy) plane and the one-to-one correspondence between points and ordered pairs, then proves the Pythagorean theorem from scratch. From there it builds radian measure, arc length, and the unit-circle definitions of sine and cosine, deriving the exact values at the special angles. It rounds out the six trig functions, then returns to algebra with the distance formula, slope, and the slope criteria for parallel and perpendicular lines. The payoff is a toolkit you reuse constantly: measuring distances, computing arc lengths, and reasoning about lines through their slopes.

Key Concepts

The Cartesian (xy) plane

Two perpendicular directed lines — the x-axis (horizontal) and y-axis (vertical) — meeting at the origin.
A point is a geometric object; an ordered pair $(x, y)$ is a pair of numbers. The coordinate plane lets us identify each point with exactly one ordered pair and vice versa (a one-to-one correspondence).
$x$ is the directed distance of the point from the y-axis (positive to the right); $y$ is the directed distance from the x-axis (positive upward).
Older texts call $x$ the abscissa and $y$ the ordinate; modern usage just says x-coordinate and y-coordinate.

Pythagorean theorem

For a right triangle with legs $a$ and $b$ and hypotenuse $c$ : $a^2 + b^2 = c^2$ .
The square-within-a-square proof shows that the two legs' squares sum to the square on the hypotenuse — read it as "the squares on the shorter sides add to the square on the longest side," not as a memorized letter formula.

Radian measure

For a circle of radius $r$ : diameter $d = 2r$ , and $\pi = \dfrac{C}{d}$ (circumference over diameter), so the circumference is $C = 2\pi r$ .
A radian is the angle whose arc length equals one radius. Going all the way around is $2\pi$ radians ( $= 360^\circ$ ); half-way is $\pi$ ( $= 180^\circ$ ); a quarter is $\dfrac{\pi}{2}$ ( $= 90^\circ$ ).
Convert with $180^\circ = \pi$ rad. Special angles: $30^\circ = \dfrac{\pi}{6}$ , $45^\circ = \dfrac{\pi}{4}$ , $60^\circ = \dfrac{\pi}{3}$ .

Arc length

On a circle of radius $r$ , an angle $\theta$ (in radians) subtends arc length $s = r\theta$ . This clean formula only works in radians — it is the main reason radians are preferred.

Sine and cosine on the unit circle

On the unit circle (radius $1$ ), the point at angle $\theta$ has coordinates $(\cos\theta, \sin\theta)$ : cosine is the x-coordinate, sine is the y-coordinate.
Exact values worth knowing:

\begin{aligned} \cos 0 &= 1, & \sin 0 &= 0 \\ \cos\tfrac{\pi}{6} &= \tfrac{\sqrt{3}}{2}, & \sin\tfrac{\pi}{6} &= \tfrac{1}{2} \\ \cos\tfrac{\pi}{4} &= \tfrac{\sqrt{2}}{2}, & \sin\tfrac{\pi}{4} &= \tfrac{\sqrt{2}}{2} \\ \cos\tfrac{\pi}{3} &= \tfrac{1}{2}, & \sin\tfrac{\pi}{3} &= \tfrac{\sqrt{3}}{2} \\ \cos\tfrac{\pi}{2} &= 0, & \sin\tfrac{\pi}{2} &= 1 \end{aligned}

Values in other quadrants come from reflecting these points, picking up sign changes (e.g. $\cos\tfrac{3\pi}{4} = -\tfrac{\sqrt{2}}{2}$ ).
For a circle of radius $r$ , the point at angle $\theta$ is $(r\cos\theta,\; r\sin\theta)$ . So in a right triangle with hypotenuse $r$ and angle $\theta$ , the adjacent leg is $r\cos\theta$ and the opposite leg is $r\sin\theta$ .

The six trigonometric functions

With a point $(x, y)$ at distance $r$ from the origin, the six ratios are:

\begin{aligned} \cos\theta &= \tfrac{x}{r}, & \sin\theta &= \tfrac{y}{r}, & \tan\theta &= \tfrac{y}{x} \\ \sec\theta &= \tfrac{r}{x}, & \csc\theta &= \tfrac{r}{y}, & \cot\theta &= \tfrac{x}{y} \end{aligned}

Reciprocal/quotient relations: $\sec\theta = \dfrac{1}{\cos\theta}$ , $\csc\theta = \dfrac{1}{\sin\theta}$ , $\cot\theta = \dfrac{1}{\tan\theta}$ , and $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$ .
Reading the geometry: cosine gives the x-coordinate, sine gives the y-coordinate, and tangent gives the slope ( $\tfrac{y}{x} = \tfrac{\text{rise}}{\text{run}}$ ).

Distance formula

The distance between $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ is the hypotenuse of a right triangle with legs $|x_2 - x_1|$ and $|y_2 - y_1|$ :

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

Distance is always taken as the non-negative root.

Slope and lines

The slope of the non-vertical line through $P_1$ and $P_2$ is $m = \dfrac{y_2 - y_1}{x_2 - x_1} = \tan\theta$ , where $\theta$ is the line's inclination. In $y = mx + b$ , $m$ is this slope. Vertical lines have no slope.
Parallel lines: $l_1 \parallel l_2 \iff m_1 = m_2$ (equal inclinations, equal slopes).
Collinear points: $A$ , $B$ , $C$ are collinear $\iff$ slope of $AB$ = slope of $AC$ .
Perpendicular lines: $l_1 \perp l_2 \iff m_1 = -\dfrac{1}{m_2}$ (slopes are negative reciprocals); equivalently $m_1 m_2 = -1$ .

Worked Example

Find the point on the x-axis equidistant from $A(5, 4)$ and $B(-2, 3)$ .

A point on the x-axis has the form $P(x, 0)$ . Equidistant means $d(P, A) = d(P, B)$ ; squaring removes the radicals, so set $d_1^{\,2} = d_2^{\,2}$ using the distance formula:

\begin{aligned} (x - x_A)^2 + (0 - y_A)^2 &= (x - x_B)^2 + (0 - y_B)^2 \\ (x - 5)^2 + (0 - 4)^2 &= (x + 2)^2 + (0 - 3)^2 \\ x^2 - 10x + 25 + 16 &= x^2 + 4x + 4 + 9 \\ -10x + 41 &= 4x + 13 \\ 28 &= 14x \\ x &= 2 \end{aligned}

So $P = (2, 0)$ is the point on the x-axis equidistant from $A$ and $B$ .

Summary

The coordinate plane pairs every point with a unique ordered pair $(x, y)$ , where $x$ and $y$ are directed distances from the y- and x-axes.
The Pythagorean theorem $a^2 + b^2 = c^2$ underlies the trigonometry, distance, and slope results that follow.
Radian measure ( $180^\circ = \pi$ ) makes arc length clean: $s = r\theta$ , and circumference is $C = 2\pi r$ .
On the unit circle, $\cos\theta$ and $\sin\theta$ are the x- and y-coordinates; memorize the exact values at $0, \tfrac{\pi}{6}, \tfrac{\pi}{4}, \tfrac{\pi}{3}, \tfrac{\pi}{2}$ and reflect for other quadrants.
The six trig functions are ratios of $x$ , $y$ , $r$ ; tangent doubles as the slope $\tfrac{\text{rise}}{\text{run}}$ .
Distance is $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ ; slope is $m = \tfrac{y_2 - y_1}{x_2 - x_1}$ , with $m_1 = m_2$ for parallel lines and $m_1 = -\tfrac{1}{m_2}$ for perpendicular ones.