# Sine, Cosine and Tangent and Their Inverse Functions

This entry is only valid for right triangles/right-angled triangles

The sine, cosine and tangent of an angle are more than just buttons on your calculator! The three functions describe the relationships between two sides in a right triangle. There are three different functions as there are three possible relationships. When you study the formulas, it is useful to compare them to the figure.

The relationship between the sides of a right triangle and the angles in the triangle is as follows:

Formula

### Sine,Cosine,TangentandTheirInverseFunctions

$\begin{array}{llll}\hfill \mathrm{sin}B& =\frac{b}{c}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}B={\mathrm{sin}}^{-1}\phantom{\rule{-0.17em}{0ex}}\left(\frac{b}{c}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \mathrm{cos}B& =\frac{a}{c}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}B={\mathrm{cos}}^{-1}\phantom{\rule{-0.17em}{0ex}}\left(\frac{a}{c}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \mathrm{tan}B& =\frac{b}{a}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}B={\mathrm{tan}}^{-1}\phantom{\rule{-0.17em}{0ex}}\left(\frac{b}{a}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Rule

### TheTwoUses

• Finding the angles of a right triangle if you know two of the sides.

• Finding the sides in a right triangle if you know one side and one angle.

Example 1

In a right triangle $△ABC$ you know that the hypotenuse is $c=10$ and one leg is $b=8$. You want to find the angle $\angle B$, which is the angle opposite to $b$.

Then you calculate

 $B={\mathrm{sin}}^{-1}\phantom{\rule{-0.17em}{0ex}}\left(\frac{8}{10}\right)\approx 53.13\text{°}.$

Example 2

A right triange $△ABC$ has a leg $b=3$ and angle $B=\text{}30\text{}\text{°}$. Find the adjacent leg $a$ to the angle $\angle B$.

As you know the angle $\angle B$ and the opposite leg to the angle $\angle B$, it is natural to select the tangent function. You enter it into the formula and get $\begin{array}{llllllll}\hfill \mathrm{tan}30\text{°}& =\frac{3}{a}\phantom{\rule{2em}{0ex}}& \hfill & |\cdot a\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill a\cdot \mathrm{tan}30\text{°}& =3\phantom{\rule{2em}{0ex}}& \hfill & |:\mathrm{tan}30\text{°}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill a& =\frac{3}{\mathrm{tan}30\text{°}}\approx 5.2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Example 3

Find the hypotenuse $c$ in the right triangle $△ABC$ when you know that $a=3$ and that angle $\angle B=\text{}60\text{}\text{°}$.

As you need to find the hypotenuse and have been given the adjacent leg to the angle $\angle B$, it is cosine that is the best fit. You enter it into the formula and get: $\begin{array}{llllllll}\hfill \mathrm{cos}60\text{°}& =\frac{3}{c}\phantom{\rule{2em}{0ex}}& \hfill & |\cdot c\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill c\cdot \mathrm{cos}60\text{°}& =3\phantom{\rule{2em}{0ex}}& \hfill & |:\mathrm{cos}60\text{°}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill c& =\frac{3}{\mathrm{cos}60\text{°}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill c& =6\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$