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Overview of Sin, Cos, and Tan Functions

Learn more about sin, cos, and tan trigonometry functions in solving triangular equations.
The image shows an architectural structure in black and white made of triangles.
© University of Nottingham

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.

A right-angled triangle, the angle of 90 degrees is labelled with a square, another angle is labelled as x. The longest side is labelled as “Hypotenuse”, the side that is opposite to angle x is labelled “Opposite”. Finally, the side that has both the right angle and angle x is labelled “Adjacent”.

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°.

In the left of this image, there is a right-angled triangle. The angles in the triangle are 30 degrees and 60 degrees. There is also a square for one of the angles to represent the right angle. The side opposite the angle of 60 degrees is labelled as 2 cm, whilst the side opposite the angle of 30 degrees is labelled as 1 cm. The side opposite the right angle is unlabelled. In the right of this image, there is a square with a width of 1cm and a height of 1cm. The square is made from two right-angled triangles, such that the diagonal of the square that goes from the top left of the square to the bottom right of the square is the hypotenuses of the two right-angled triangles. One of the right angles has angles labelled 45 degrees, 45 degrees and a square to represent the right angle.

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

Examples

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

In this image there is a right-angled triangle. The right-angle is labelled with a square, there is another angle of 60 degrees. The side opposite to the right-angle is labelled as 13cm. The side that is opposite the unlabelled angle is labelled x cm.

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:

In this image there is a right-angled triangle. The right-angle is labelled with a square, there is another angle which is labelled as theta degrees. The side opposite to the right-angle is labelled as 10.5 cm. The side that is opposite the angle of theta degrees is labelled as 8 cm.

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).

The image shows two right-angled triangled which are joined together. The first right-angled triangle has one angle labelled with a square and another angle labelled 50 degrees, the other angle is unlabelled. There is a side of 5cm which is the side opposite the unlabelled angle. The second right-angled triangle has an angle labelled with a square and another angle labelled 38 degrees. The third angle is unlabelled. The side opposite the angle of 38 degrees is labelled x cm. The side that is opposite the angle of 50 degrees in the first triangle, and the side that is opposite the unlabelled angle in the second triangle are joined together and appear to be the same length.

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.]

 

© University of Nottingham
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