Overview of Sin, Cos, and Tan Functions

Introduction

Trigonometry is the study of triangles. In this article, we’ll focus on sin, cos, and tan which are the three trigonometric functions that specifically concern right-angled triangles. They are mathematically referred to as sine, cosine, and tangent, but shortened to sin, cos, and tan.

SOHCAHTOA

First, we need to be able to label each side of a right-angled triangle:

The hypotenuse is always the longest side; it is the one opposite the right-angle.

The opposite side is the side that is opposite to the angle.

The adjacent side is the side that is adjacent (next to) the angle.

If we let O be the opposite, A be the adjacent and H be the hypotenuse, then these are shortened to:

Exact Trigonometric Ratios for 0°, 30°, 45°, 60° and 90°

The trigonometric ratios for the angles 30°, 45° and 60° can be calculated using two special triangles. Using an equilateral triangle split into two right-angled triangles. we can find exact values for the trigonometric ratios of 30° and 60°.

A square with side lengths of 1 cm can be used to calculate accurate values for the trigonometric ratios of 45°.

The accurate trigonometric ratios for 0°, 30°, 45°, 60° and 90° are:

Examples

1. Find the length of side (x) in the diagram below:

The angle is (60^o). We are given the hypotenuse and need to find the adjacent side. This formula which connects these three is:

[cos(theta)=frac{adj}{hyp}~~~~implies~~~~cos(60)=frac{x}{13}]

Using the exact value for (cos(60)) and rearranging the equation gives

[x=13cos(60)~~~~implies~~~~x=13timesfrac{1}{2}=6.5cm.]

2. Calculate the size of the angle (theta) in the diagram below:

We are given the opposite and hypotenuse so we need the ratio for sine.

[sin(theta)=frac{opp}{hyp}~~~~implies~~~~sin(theta)=frac{8}{10.5}]

Rearranging the equation and using a calculator gives

[sin^{-1}left(frac{8}{10.5}right)=49.63^o.]

3. The diagram shows two right-angled triangles. Find the value of (x).

First, let’s work out the opposite side of the upper triangle since we know the length of one of the sides and an angle. We can use:

[tan(theta)=frac{opp}{adj}~~~~implies~~~~tan(50)=frac{opp}{5}.]

This means the opposite side of the upper triangle is

[5timestan(50)approx5.958767…]

Now we know the adjacent side of the lower triangle and we want to find the opposite side. Using

[tan(theta)=frac{opp}{adj}~~~~implies~~~~tan(38)=frac{opp}{5.958767…}.]

Therefore, the missing side (x) is

[5.958767…timestan(38)approx 4.6554997…=4.66cm.]