Trigonometric Ratios

Introduction

The word trigonometry comes from the Greek words "trigonon" which means triangle and "metron" which means to measure. Trigonometry deals with determination of measures of sides and angles of triangles by means of relevant trigonometric functions. The trigonometric functions are sine, cosine, tangent, secant, cosecant, and cotangent of an angle. In this chapter you will learn about trigonometric ratios, trigonometric identities, compound angle formulae, double angle formulae, trigonometric equations, factor formulae, radians and small angles, trigonometric functions, and inverse of trigonometric functions. The competencies developed can be applied in various real life situations such as in solving problems related to astronomy, navigation, architecture, oceanography, and in creation of maps. Generally, trigonometry has great practical importance to builders, architects, surveyors, engineers, and users in many other fields.

The ratios of sides of a right-angled triangle with respect to any of its acute angles are known as the trigonometric ratios of that particular angle. The three sides of the right-angled triangle are hypotenuse (the longest side), opposite (side across from the angle), and adjacent (base of the angle). Consider the right-angled triangle OAB in Figure below

The hypotenuse of the right-angled triangle is OB, OA is the adjacent side to angle θ, and AB is the opposite side to angle θ. The three basic trigonometric ratios with their abbreviations are defined as follows:

$\sin \theta = \frac{\text{Length of opposite Side AB}}{\text{Length of hypotenuse Side OB}}$

$\cos \theta = \frac{\text{Length of adjacent Side OA}}{\text{Length of hypotenuse Side OB}}$

$\tan \theta = \frac{\text{Length of opposite Side AB}}{\text{Length of adjacent Side OA}}$

Now, using Pythagoras' theorem, the following relation is obtained:

$OA^2 + AB^2 = OB^2$

The reciprocal of the trigonometric functions of sine, cosine, and tangent are cosecant, secant, and cotangent, respectively. These are defined from Figure 7.1 as follows:

$\csc \theta = \frac{1}{\sin \theta} = \frac{\text{Length of hypotenuse Side OB}}{\text{Length of opposite Side AB}}$

$\sec \theta = \frac{1}{\cos \theta} = \frac{\text{Length of hypotenuse Side OB}}{\text{Length of adjacent Side OA}}$

$\cot \theta = \frac{1}{\tan \theta} = \frac{\text{Length of adjacent Side OA}}{\text{Length of opposite Side AB}}$

Example 1

From the figure, $\tan 39^\circ = \frac{11}{AH}$

$AH = \frac{11}{\tan 39^\circ}$

$HC = 19 - AH = 19 - \frac{11}{\tan 39^\circ}$

Using Pythagoras' theorem, for the right-angled triangle HBC:

$BH^2 + HC^2 = BC^2$

$11^2 + HC^2 = x^2$

Substitute HC to find the value of x:

$x^2 = 11^2 + \left(19 - \frac{11}{\tan 39^\circ}\right)^2$

$x \approx 12.3$

Therefore, $x = 12.3$ cm.

Example 2

Since A is a right angle, both triangles ABC and ABD in the figure, are right-angled triangles.

Using Pythagoras' theorem:

From triangle BAD:

$AD^2 = 14^2 - 10^2$

From triangle BAC:

$AC^2 = 16^2 - 10^2$

Thus, $x = AC - AD$

$x = \sqrt{16^2 - 10^2} - \sqrt{14^2 - 10^2}$

$x = \sqrt{156} - \sqrt{96}$

$x \approx 2.69$

Therefore, $x = 2.69$ cm.

Example 3

If $\tan \theta = 3$, where θ is an acute angle, find the value of each of the following leaving your answer in surd form.

1. sec θ

Given $\tan \theta = 3 = \frac{3}{1}$

Using the Pythagoras' theorem,

$AB^2 + BC^2 = AC^2$

$1^2 + 3^2 = AC^2$

$AC^2 = 4$

$AC = \sqrt{10}$ units.

$\sec \theta = \frac{1}{\cos \theta}$

$\cos \theta = \frac{1}{\sqrt{10}}$

$\sec \theta = \sqrt{10}$

2. cot θ

$\cot \theta = \frac{1}{\tan \theta}$

$\cot \theta = \frac{1}{3}$

3. cosec θ

$\csc \theta = \frac{1}{\sin \theta}$

$\sin \theta = \frac{3}{\sqrt{10}}$

$\csc \theta = \frac{\sqrt{10}}{3}$