# Basic trigonometric functions

Word trigonometry comes from Greek words trigonon and metron, in translation triangle and measurement. Trigonometry is a branch of mathematics which studies relations among lengths of sides and angles in a triangle. Trigonometry was, through history,  mostly used in astronomy to track the movements of stars and measure their distance and Egyptians used it to determine when the Nile will be flooded.

The triangles which are mostly are in the focus of trigonometry are right triangles, they have two legs and a hypotenuse, the side opposite  to the right angle.

Let take a triangle $\bigtriangleup ABC$ with the hypotenuse $AB$:

$sin(\alpha)=\frac{a}{c}$

$Sin(\alpha)$ is a ratio of the leg $a$, the leg opposite to the angle $\alpha$ and the hypotenuse, side denoted by $c$.

$cos(\alpha)=\frac{b}{c}$

$Cos(\alpha)$ is a ratio of the leg $b$, the leg adjacent to the angle $\alpha$ and the hypotenuse.

$tan(\alpha)=\frac{a}{b}$

$Tan(\alpha)$ is a ratio of the leg $a$, the leg opposite to the angle $\alpha$ and the leg $b$, the leg adjacent to the angle $\alpha$.

$cot(\alpha)=\frac{b}{a}$

$Cot(\alpha)$ is a ratio of the leg $b$, the leg adjacent to the angle $\alpha$ and the leg $a$, the leg opposite to the angle $\alpha$.

$cot(\alpha)=\frac{1}{tan(\alpha)}$

What are the opposite and adjacent legs? Take a look at the angle $\beta$. It is enclosed by the hypotenuse and the leg $a$. The leg $a$ is its adjacent side and the remaining side $b$ is its opposite side.

Example 1. Calculate all angles and sides if the hypotenuse in a right triangle is equal to 5 and angle to the point $A$ is equal to $30^{\circ}$.

When solving problems such as this it is advisable to draw a sketch. In this example either sine or cosine can be used. The order in which the problem is solved doesn’t effect the final result. Angle to the point $B$ is  equal to $30^{\circ}$ and the opposite side of this angle is $AC$.

$sin(30^{\circ}) = \frac{AC}{AB}$

$sin(30^{\circ}) = \frac{AC}{5}$

$AC = 5 \cdot sin(30^{\circ})=2.5$

Adjacent side is $BC$ so cosine is used.

$cos(30^{\circ}) = \frac{BC}{AB}$

$cos(30^{\circ}) = \frac{BC}{5}$

$BC = 5 \cdot cos(30^{\circ})=4.33$

Now that all three sides are known, the angles can be calculated easily.

$\alpha= 180^{\circ} – 30^{\circ} – 90^{\circ} = 60^{\circ}$

Example 2. Calculate all angles and sides in a right triangle if the length of one leg is equal to 3 cm and $\alpha = 50^{\circ}$.

It isn’t specified which leg is 3 cm long, so let say $a = 3$. In this example we’ll use tangent and cotangent.

Side $a$ is the opposite and side $b$ is the adjacent side for angle $\alpha$.

$tan(\alpha) = \frac{a}{b}$

$b = \frac{a}{tan(\alpha)}$

$b = \frac{3}{tan(50^{\circ})}=2.52$

$sin(\alpha) = \frac{a}{c}$

$sin(50^{\circ}) = \frac{3}{c}$

$c = \frac{3}{sin(50^{\circ})} = 3.92$

## Trigonometric functions and triangle similarity

Two triangles are similar if all of their corresponding angles are of equal measure and all sides corresponding are proportional.

What does this imply when observed in right triangle?

Let’s say we have two similar right triangles $\bigtriangleup ABC$ and $\bigtriangleup A’B’C’$.

From the premise that the triangles $\\bigtriangleup ABC$ and $\\bigtriangleup A’B’C’$ are similar follows:

$\frac{a}{c} = \frac{a’}{c’}$

Side $a$ is the adjacent side of  the angle $\beta$ and $b’$ is the adjacent side of $\beta’$, and also $c$ is the hypotenuse of the triangle $ABC$ and $c’$ is the hypotenuse of triangle $A’B’C’$, from this follows to:

$sin(\beta) = sin(\beta’)$

In conclusion, all trigonometric values of any two similar triangles are the same.

## Trigonometric functions – application problems

Through the following few examples we’ll show some basic application of trigonometry.

Example 1. Imagine you are 190 centimeters tall and you decide to go for a walk. Your shadow falls under the angle of $60^{\circ}$. How long is your shadow?

First, draw a sketch. You and your shadow enclose the right angle, the sun rays and your shadow enclose an angle of $60^{\circ}$. From this information, the lenght of the one side, and the value of the opposite angle, you can conclude that the best option is to use the tangent function.

$tan(60^{\circ}) =$ (your height) $/$ shadow

shadow $=$ your height $\cdot tan(60^{\circ})$

shadow $= 190 \cdot tan(60^{\circ})$

shadow $= 329$

Example 2. You have to climb on a roof of a cabin that is 2 meters high. You put your ladder on the ground under the $30^{\circ}$. What height are your ladders?

$sin(30^{\circ}) =$ (height of the cabin) $/$ (height of the ladders)

height of the ladders $= \frac{2}{sin(30^{\circ})}$

height of the ladders $= 4$ meters

## Reciprocal trigonometric functions

Other than our basic trigonometric functions – sine, cosine, tangent and cotangent there are a lot more. Some of them are called reciprocal trigonometric functions cosecant and secant.

Cosecant is the reciprocal of the sine function. This means that if

Secant is the reciprocal of the cosine function. This means that if

Example 1. Find values of sine, cosine, tangent, cotangent, secant and cosecant for the given triangle in angle $\alpha$.

$sin( \alpha) = \frac{opposite}{hypotenuse} = \frac{7}{8.6} = 0.813$

$cos( \alpha) = \frac{adjacent}{hypotenuse} = \frac{5}{cos(8.6)} = 0.58$

$tan( \alpha) = \frac{sin( \alpha)}{cos( \alpha)} = \frac{0.813}{0.58} = 1.4$

$cot( \alpha) = \frac{cos( \alpha)}{sin( \alpha)} = \frac{1}{tan( \alpha)} = \frac{1}{1.4} = 0.71$

$csc( \alpha) = \frac{1}{sin( \alpha)} = \frac{1}{0.813} = 1.23$

$sec( \alpha) = \frac{1}{cos( \alpha)} = \frac{1}{cos(0.58)} = 1.72$

## Inverse trigonometric functions

Each trigonometric function has its own inverse. Inverse functions “undo” trigonometric functions, or :

$sin^{-1}(sin(x)) = x$.

$sin^{-1}(x)$ = arcsinx $\rightarrow$ The arc that has a sine of x.

$cos^{-1}(x)$ = arccosx $\rightarrow$ The arc that has a cosine of x.

$tan^{-1}(x)$ = arctanx $\rightarrow$ The arc that has a tangent of x.

$cot^{-1}(x)$ = arccotx $\rightarrow$ The arc that has a cotangent of x.

Inverse trigonometric functions have various applications in real life situations, below are a few examples.

Example 1. Bob went fishing. He traveled in his boat for 5 meters to the west, but didn’t find any fishes there, so he decided to travel north. When he was 7 meters away from the last point, he noticed that a storm was coming, so he had to go back to the coast. The fastest way back is directly towards the coast. At which angle must Bob sail back?

Bob’s trip started at the point $A$, which represents the coast. He first traveled 5 meters to the west, to the point $C$, and then changed the direction for $90^{\circ}$ and went north for 7 meters to the point $B$. His path looks like a right triangle. The lenghts of the legs are known so the angle in the point $B$ can easily be calculated using he function tangent.

$tan(\alpha) = \frac{5}{7} \rightarrow \alpha = 35,54^{\circ}$

Example 2. Angela is walking her dog Polo. Leash is 5 meters long, and Polo is 3 meters away from Angela. Under which angle is Angela holding a leash?

First, let’s draw a sketch:

Angela, the leash and the ground are making a right triangle with hypotenuse whose length is equal to 5, and one leg 3. Since the known leg is opposite from the angle we are looking for, we can use sine function.

$sin \alpha = \frac{3}{5} \rightarrow \alpha = 36,87^{\circ}$

## Basic trigonometric functions worksheets

Trigonometric ratios in a right triangles (171.3 KiB, 945 hits)

Trigonometric ratios of given angles (156.2 KiB, 608 hits)

Trigonometric ratios of inverse function in a right triangle (193.9 KiB, 701 hits)

Using trigonometric functions to find a missing side of a right triangle (256.1 KiB, 555 hits)