Defining Ratios in the Cartesian Plane Revision Notes for Grade 10 NSC Matric Mathematics

Defining Ratios in the Cartesian Plane

Introduction to extended trigonometric definitions

Trigonometric ratios can be extended beyond right-angled triangles to work with any angle in the Cartesian plane. This extension allows us to define trigonometric functions for angles of any size, not just those between 0° and 90°.

infoNote

The key insight is that trigonometric ratios depend only on the size of the angle, not on the lengths of the triangle's sides. When we plot any point on the Cartesian plane and draw a line from the origin to that point, we can create a right-angled triangle and calculate trigonometric ratios using the coordinates.

Finding angles using coordinate points

To find an angle when given specific coordinate points, follow these steps:

Plot the point on the Cartesian plane
Draw a line from the origin (0,0) to the point
Create a right-angled triangle by drawing a perpendicular line to the x-axis
Use trigonometric ratios to calculate the angle

For a point with coordinates $(x, y)$ , we can calculate the angle using:

$\tan θ = \frac{\text{opposite}}{\text{adjacent}} = \frac{y}{x}$

lightbulbExample

Worked Example: Finding an angle from coordinates

For point P(2,3):

Opposite side = 3, Adjacent side = 2
$\tan θ = \frac{3}{2}$
Therefore $θ = 56.3°$

General definitions of trigonometric ratios

When working with any point P(x,y) on the Cartesian plane, we can define all trigonometric ratios in terms of the coordinates and the radius r.

infoNote

The radius r is the distance from the origin to point P, calculated using Pythagoras' theorem:

$r^2 = x^2 + y^2$

The six trigonometric ratios

Primary ratios:

$\sin θ = \frac{y}{r}$ (opposite/hypotenuse)
$\cos θ = \frac{x}{r}$ (adjacent/hypotenuse)
$\tan θ = \frac{y}{x}$ (opposite/adjacent)

Reciprocal ratios:

$\text{cosec } θ = \frac{r}{y}$ (1/sin θ)
$\sec θ = \frac{r}{x}$ (1/cos θ)
$\cot θ = \frac{x}{y}$ (1/tan θ)

chatImportant

The radius r is always positive since it represents a distance.

The CAST diagram

The CAST diagram shows which trigonometric ratios are positive in each quadrant of the Cartesian plane.

Quadrant signs explained

infoNote

Quadrant I (0° to 90°):

All ratios are positive
Both x and y coordinates are positive

Quadrant II (90° to 180°):

Sin and cosec are positive
y is positive, x is negative

Quadrant III (180° to 270°):

Tan and cot are positive
Both x and y coordinates are negative

Quadrant IV (270° to 360°):

Cos and sec are positive
x is positive, y is negative

Memory aid: "All Students Take Chemistry"

Special angles in the Cartesian plane

Two important special angles have exact values that are essential to memorise.

chatImportant

Angle of 0° When θ = 0°, the point lies on the positive x-axis:

$\sin 0° = \frac{0}{r} = 0$
$\cos 0° = \frac{r}{r} = 1$
$\tan 0° = \frac{0}{r} = 0$

Angle of 90° When θ = 90°, the point lies on the positive y-axis:

$\sin 90° = \frac{r}{r} = 1$
$\cos 90° = \frac{0}{r} = 0$
$\tan 90° = \frac{r}{0} = \text{undefined}$

Worked example 1: Ratios in the Cartesian plane

lightbulbExample

Worked Example: Calculating trigonometric ratios

Given: Point P(-3,4) with angle θ between OP and the positive x-axis.

Find: cos θ, 3 tan θ, and ½ cosec θ

Step 1: Calculate the radius using Pythagoras' theorem

$r^2 = x^2 + y^2$
$r^2 = (-3)^2 + (4)^2$
$r^2 = 9 + 16 = 25$
$r = 5$

Step 2: Substitute values into the required ratios

x = -3, y = 4, r = 5

Step 3: Calculate each ratio

$\cos θ = \frac{x}{r} = \frac{-3}{5}$
$3 \tan θ = 3 \cdot \frac{y}{x} = 3 \cdot \frac{4}{-3} = -4$
$\frac{1}{2} \text{cosec } θ = \frac{1}{2} \cdot \frac{r}{y} = \frac{1}{2} \cdot \frac{5}{4} = \frac{5}{8}$

Worked example 2: Finding unknown coordinates

lightbulbExample

Worked Example: Finding coordinates and proving identities

Given: Angle XOK = θ in the third quadrant, where X is on the positive x-axis and K(-5,y). OK = 13 units.

Find: The value of y and prove that $\tan^2θ + 1 = \sec^2θ$

Step 1: Use Pythagoras' theorem to find y

$r^2 = x^2 + y^2$
$13^2 = (-5)^2 + y^2$
$169 = 25 + y^2$
$y^2 = 144$
$y = ±12$

Step 2: Determine the correct sign Since θ is in the third quadrant, y must be negative. Therefore: $y = -12$

Step 3: Prove the identity With x = -5, y = -12, r = 13:

LHS: $\tan^2θ + 1 = \left(\frac{y}{x}\right)^2 + 1 = \left(\frac{-12}{-5}\right)^2 + 1 = \frac{144}{25} + 1 = \frac{169}{25}$

RHS: $\sec^2θ = \left(\frac{r}{x}\right)^2 = \left(\frac{13}{-5}\right)^2 = \frac{169}{25}$

Since LHS = RHS, the identity is proven.

Exam tips

infoNote

Always sketch the point on the coordinate plane when solving problems
Remember that r is always positive as it represents distance
Use the CAST diagram to determine the sign of trigonometric ratios
Check which quadrant the angle lies in before determining signs
When solving without a calculator, exact values and fractions are expected

bookmarkSummary

Key Points to Remember:

Trigonometric ratios can be extended to any angle using coordinates in the Cartesian plane
The general definitions are: $\sin θ = \frac{y}{r}$ , $\cos θ = \frac{x}{r}$ , $\tan θ = \frac{y}{x}$ , with $r^2 = x^2 + y^2$
The CAST diagram shows which ratios are positive: All (I), Sin (II), Tan (III), Cos (IV)
Special angles have exact values: $\sin 0° = 0$ , $\cos 0° = 1$ , $\sin 90° = 1$ , $\cos 90° = 0$ , $\tan 90° = \text{undefined}$
Always sketch the problem to visualise the quadrant and determine the correct signs

Defining Ratios in the Cartesian Plane (Grade 10 NSC Matric Mathematics): Revision Notes