# Trigonometric Functions

##### Jun 11, 2015 - Trigonometry Trigonometric Functions

At the heart of trigonometry lies the study of the relationship between the angles of a triangle and the length of its sides. There exists 6 such relationships, known as the trigonometric functions. This article in the series on trigonometry aims to define the trigonometric functions as a set of ratios and show how they can be derived from certain common angles.

To begin consider an angle $\alpha$ in standard position and a terminal side forming an angle between 0° and 90°, with a point $\text{(x,y)}$ anywhere on the terminal side that is not the origin. Such an angle is shown in Figure 1.

Recall from the original article that rotating a terminal side away from the initial side by 360° forms a circle. This makes the distance $r$, shown in Figure 1, the radius of a circle that is centered at $\text{(0,0)}$. The relationship between each of these elements, $r$, $x$, and $y$, can be expressed in the distance formula $r = \sqrt{x^{2} + y^{2}}$. From this formula, six ratios can be formed, otherwise known as the trigonometric ratios.

#### Trigonometric Ratios

$\text{sin}\ \alpha = \frac{y}{r}$ $\text{cos}\ \alpha = \frac{x}{r}$ $\text{tan}\ \alpha = \frac{y}{x}$
$\text{csc}\ \alpha = \frac{r}{y}$ $\text{sec}\ \alpha = \frac{r}{x}$ $\text{cot}\ \alpha = \frac{x}{y}$

Notice that the cosecant is the reciprocal of the sine, the secant the reciprical of the cosine, and the cotangent the reciprical of the tangent. These relationships are important enough to be given names, the reciprocal identies. An identity is a relation where a variable, in this case angle $\alpha$, appears on both sides of the equation and no matter the angle, the relationship remains true.

#### Reciprocal Identies

$\text{csc}\ \alpha = \frac{1}{\text{sin} \alpha}$ $\text{sec}\ \alpha = \frac{1}{\text{cos} \alpha}$ $\text{cot}\ \alpha = \frac{1}{\text{tan} \alpha}$

### Trigonometric Functions in Use

Armed with the above knowledge about the trigonometric ratios, it is now possibe to find the value of each of the trigonometric functions for an angle in standard position. In this case, consider the point $\text{(1,-3)}$ on the terminal side of an angle $\alpha$ in standard position. The playing pieces in the trigonometric ratio game are $x = 1$, and $y = -3$, from the point on the terminal side, and $r$, which can be found with $r = \sqrt{1^{2} + -3^{2}} = \sqrt{10}$.

Now, finding the trigonometric ratios is as simple as filling in the variables and simplifying.

$\text{sin}\ \alpha = \frac{y}{r} = \frac{-3}{\sqrt{10}} = \frac{-3}{\sqrt{10}} * \frac{\sqrt{10}}{\sqrt{10}} = \frac{-3\sqrt{10}}{10}$ $\text{csc}\ \alpha = \frac{\sqrt{10}}{-3}$
$\text{cos}\ \alpha = \frac{x}{r} = \frac{1}{\sqrt{10}} = \frac{1}{\sqrt{10}} * \frac{\sqrt{10}}{\sqrt{10}} = \frac{\sqrt{10}}{10}$ $\text{sec}\ \alpha = \frac{r}{x} = \frac{\sqrt{10}}{1} = \sqrt{10}$
$\text{tan}\ \alpha = \frac{y}{x} = \frac{-3}{1} = -3$ $\text{cot}\ \alpha = \frac{x}{y} = \frac{1}{-3}$

An interesting aspect of the trigonometric functions is that it applies to any point on the terminal side of the ange that is not the origin. This aspect is higlighted in Figure 2, which shows $r$ and the $\text{(x,y)}$ point similar to what was just used for determining the values of the trigonometric functions. It also shows the proportional relationship between this and any other $\text{(x,y)}$ coordinate on the terminal side.

Certain types of angles have a terminal side that lies on either the $x$ or the $y$ axis. These are known as quadrantal angles. Solutions for the trigonometric functions for these angles can be found just like any other angle, however, thinking about them in terms of a unit circle highlights a simpler way at solving them.

Since the radius for a unit circle is always 1, so $r = 1$. Then, the appropriate $\text{(x,y)}$ coordinate is selected based on axis upon which the terminal side lies. For example the $\text{sin}\ 90°$ has $r = 1$, $x = 0$, and $y = 1$. Since the $\text{sin}\ \alpha = \frac{y}{r}$ then:

$\text{sin}\ 90° = \frac{y}{r} = \frac{1}{1} = 1$

Therefore, the $\text{sin}\ 90°$ is $1$.

## 45° Angles and 45-45-90 Triangles

In the previous articles, it has been mentioned that there are certain special angles. One of these types of special angles defines angles that are multiples of 45°, but not quadrantal. A 45° angle is also part of a special kind of triangle, known as a 45-45-90 triangle, which has 2 sides of the same length. Using the Pythagorean theorem, it is a simple matter to find the length of the hypotenuse.

How can this be useful for finding the values of the trigonometric functions? Simply put, any angle in standard position that is a multiple of 45° can be turned into a 45-45-90 triangle as shown in Figure 5.

Here, the Pythagorean theorem is applied, yielding $\sqrt{2}$ for the value of $r$. Now, finding the values for the trigonometric functions are as easy as filling in the blanks of the trigonometric ratios.

$\text{sin}\ 45° = \frac{y}{r} = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ $\text{cos}\ 45° = \frac{x}{r} = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ $\text{tan}\ 45° = \frac{y}{x} = \frac{1}{1} = 1$

Again, this applies to any angle that is a multiple of 45° that is not also quadrantal. To highlight this, see Figure 6 which depicts a 135° angle.

The 135° angle produces the same results then as the 45° angle for the trigonometric functions:

$\text{sin}\ 135° = \frac{y}{r} = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ $\text{cos}\ 135° = \frac{x}{r} = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ $\text{tan}\ 135° = \frac{y}{x} = \frac{1}{1} = 1$

## 30° Angles and 30-60-90 Triangles

Another special type of triangle is the 30-60-90 triangle which can be used to produced the special angles which are multiples of 30°. This type of triangle can be created by bisecting an equalateral triangle, or a triangle with all sides of the same length. This results in a 30-60-90 triangle like the one shown in Figure 7.

Similar to how the 45-45-90 triangle could be created from a 45° angle in standard position, so too can a 30-60-90 triangle be created from 30° and 60° triangles. as shown in Figure 8 and Figure 9.

Now, finding the values for the trigonometric functions works just like it did for the 45° angles. Using the Pythagorean theorem to find the values of $r$, it is a simple matter to, for example, find the $\text{sin}\ 30°$ and the $\text{sin}\ 60°$

$\text{sin}\ 30° = \frac{y}{r} = \frac{1}{2}$ $\text{sin}\ 60° = \frac{y}{r} = \frac{\sqrt{3}}{2}$

## Sine and Cosine of Common Angles

A common theme for finding the values for the trigonometric functions via the trigonometric ratios should be apparent by this point: finding them requires for the angle in question to be in standard position. If an angle is not in standard position, it must first be translated into the standard position, a topic for another time. However, the sines and cosines of angles that are multiples of 30° and 45° are found so often in the real world that having them memorized can be useful.

$\text{sin}\ 0° = \frac{\sqrt{0}}{2} = 0$ $\text{cos}\ 0° = \frac{\sqrt{4}}{2} = 1$
$\text{sin}\ 30° = \frac{\sqrt{1}}{2} = \frac{1}{2}$ $\text{cos}\ 30° = \frac{\sqrt{3}}{2}$
$\text{sin}\ 45° = \frac{\sqrt{2}}{2}$ $\text{cos}\ 45° = \frac{\sqrt{2}}{2}$
$\text{sin}\ 60° = \frac{\sqrt{3}}{2}$ $\text{cos}\ 60° = \frac{\sqrt{1}}{2} = \frac{1}{2}$
$\text{sin}\ 90° = \frac{\sqrt{4}}{2} = 1$ $\text{cos}\ 90° = \frac{\sqrt{0}}{2} = 0$

## Conclusion

This wraps up the first discussion on the trigonometric functions. These are so important that they will be seen time and time again through this series on trigonometry. The next article will dive deeper into the trigonometric functions as they apply to right angles. Until then!