Trigonometric rations

Trigonometric Ratios

Trigonometry is all about triangles. In this chapter we are going to deal with right-angled triangles. Consider the right-angled triangle below:

The sides are given names according to their properties relating to the angle.

Adjacent side is adjacent (next to) to the angle.

Opposite side is opposite the angle.

Hypotenuse side is the longest side.

Sine, Cosine and Tangent of an Angle using a Right Angled Triangle

Trigonometry is good at finding the missing side or angle of a right-angled triangle. The special functions, sine, cosine and tangent help us. They are simply one side of a triangle divided by another. See similar triangles below:

$$\frac{CB}{AB} = \frac{ED}{AD} = \frac{GF}{AF} = t$$ $$\frac{AB}{AC} = \frac{AD}{AE} = \frac{AF}{AG} = c$$ $$\frac{CB}{AC} = \frac{ED}{AE} = \frac{FG}{AG} = s$$

The ratios of the corresponding sides are:

Where by t, c and s are constant ratios called tangent (t), cosine (c) and sine (s) of the angle respectively.

The right-angled triangle can be used to define trigonometrical ratios as follows:

• Tangent = $\frac{\text{opposite side}}{\text{adjacent side}}$

• Sine = $\frac{\text{opposite side}}{\text{hypotenuse}}$

• Cosine = $\frac{\text{adjacent side}}{\text{hypotenuse}}$

The short form of tangent is tan, that of sine is sin and that of cosine is cos.

The simple way to remember the definition of sine, cosine and tangent is the word SOHCAHTOA. This means sine is Opposite (O) over Hypotenuse (H); cosine is Adjacent (A) over Hypotenuse (H); and tangent is Opposite (O) over Adjacent (A).

SO	TO	CA
H	A	H

Example 1

Given a triangle below, find sine, cosine and tangent of the angle indicated.

Solution

Case 1:

$$\sin \alpha = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{9}{15} = 0.6$$

Case 2:

$$\cos \alpha = \frac{\text{adjacent}}{\text{hypotenuse}}$$

To get the value of adjacent side, use Pythagoras theorem:

$$\text{Adjacent side} = (\text{hypotenuse})^2 - (\text{opposite})^2$$ $$\text{Adjacent side} = 15^2 - 9^2$$ $$\text{Adjacent side} = 12$$

Thus, $\cos \alpha = \frac{12}{15} = 0.8$

Case 3:

$$\tan \alpha = \frac{\text{opposite}}{\text{adjacent}} = \frac{9}{12} = 0.75$$

Example 2

Given that $\cos 40^\circ = \frac{40}{41}$, find the value of $\tan 40^\circ$ and $\sin 40^\circ$.

Solution

Cosine of an angle = $\frac{\text{adjacent}}{\text{hypotenuse}}$; thus, adjacent side = 40 and hypotenuse side = 41.

Opposite side = $(\text{hypotenuse})^2 - (\text{adjacent})^2$

Opposite side = $41^2 - 40^2$

Opposite side = 9

Therefore, $\tan 40^\circ = \frac{\text{opposite}}{\text{adjacent}} = \frac{9}{40}$ and $\sin 40^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{9}{41}$