What are the six trigonometry functions?
Last Updated :
27 Mar, 2024
Trigonometry can be defined as the branch of mathematics that determines and studies the relationships between the sides of a triangle and angles subtended by them. Trigonometry is basically used in the case of right-angled triangles. Trigonometric functions define the relationships between the 3 sides and the angles of a triangle. There are 6 trigonometric functions mainly. Before going into the study of the trigonometric functions we will learn about the 3 sides of a right-angled triangle.
The three sides of a right-angled triangle are as follows,
- Base The side on which the angle θ lies is known as the base.
- Perpendicular It is the side opposite to the angle θ  in consideration.
- Hypotenuse It is the longest side in a right-angled triangle and opposite to the 90° angle.
Trigonometric Functions
Trigonometry has 6 basic trigonometric functions, they are sine, cosine, tangent, cosecant, secant, and cotangent. Now let’s look into the trigonometric functions. The six trigonometric functions are as follows,
- sine It is represented as sin θ and is defined as the ratio of perpendicular and hypotenuse.
- cosine It is represented as cos θ and is defined as the ratio of base and hypotenuse.
- tangent It is represented as tan θ and is defined as the ratio of sine and cosine of an angle. Thus the definition of tangent comes out to be the ratio of perpendicular and base.
- cosecant It is the reciprocal of sin θ and is represented as cosec θ.
- secant It is the reciprocal of cos θ and is represented as sec θ.
- cotangent It is the reciprocal of tan θ and is represented as cot θ.
What are the six trigonometry functions?
The six trigonometric functions have formulae for the right-angled triangles, the formulae help in identifying the lengths of the sides of a right-angled triangle, lets take a look at all those formulae,
Trigonometric Functions | Formulae |
sin θ | [Tex]\frac{P}{H}[/Tex] |
cos θ | [Tex]\frac{B}{H}[/Tex] |
tan θ | [Tex]\frac{P}{B}[/Tex] |
cosec θ | [Tex]\frac{H}{P}[/Tex] |
sec θ | [Tex]\frac{H}{B}[/Tex] |
cot θ | [Tex]\frac{B}{P}[/Tex] |
The below table shows the values of these functions at some standard angles,
Function | 0° | 30° | 45° | 60° | 90° |
[Tex]sin\theta = \frac{P}{H}[/Tex] | [Tex]0[/Tex] | [Tex]\frac{1}{2}[/Tex] | [Tex]\frac{1}{\sqrt2}[/Tex] | [Tex]\frac{\sqrt3}{2}[/Tex] | [Tex]1[/Tex] |
[Tex]cos\theta = \frac{B}{H}[/Tex] | [Tex]1[/Tex] | [Tex]\frac{\sqrt3}{2}[/Tex] | [Tex]\frac{1}{\sqrt2}[/Tex] | [Tex]\frac{1}{2}[/Tex] | [Tex]0[/Tex] |
[Tex]tan\theta = \frac{sin\theta}{cos\theta}=\frac{P}{B}[/Tex] | [Tex]0[/Tex] | [Tex]\frac{1}{\sqrt3}[/Tex] | [Tex]1[/Tex] | [Tex]\sqrt3[/Tex] | ∞ |
[Tex]cosec\theta = \frac{H}{P}[/Tex] | ∞ | [Tex]2[/Tex] | [Tex]\sqrt2[/Tex] | [Tex]\frac{2}{\sqrt3}[/Tex] | [Tex]1[/Tex] |
[Tex]sec\theta = \frac{H}{B}[/Tex] | [Tex]1[/Tex] | [Tex]\frac{2}{\sqrt3}[/Tex] | [Tex]\sqrt2[/Tex] | [Tex]2[/Tex] | ∞ |
[Tex]cot\theta = \frac{B}{P}[/Tex] | ∞ | [Tex]\sqrt3[/Tex] | [Tex]1[/Tex] | [Tex]\frac{1}{\sqrt3}[/Tex] | [Tex]0[/Tex] |
Note: It is advised to remember the first 3 trigonometric functions and their values at these standard angles for ease of calculations.
Sample Problems
Question 1: Evaluate sine, cosine, and tangent in the following figure.
Solution:Â
Given [Tex]P=3, B=4, H=5[/Tex]
Using the trigonometric formulas for sine, cosine and tangent,
[Tex]sin\theta=\frac{P}{H}=\frac{3}{5}[/Tex]
[Tex]cos\theta=\frac{B}{H}=\frac{4}{5}[/Tex]
[Tex]tan\theta=\frac{P}{B}=\frac{3}{4}[/Tex]
Question 2: In the same triangle evaluate secant, cosecant, and cotangent.Â
Solution:Â
As it is known the values of sine, cosine and tangent, we can easily calculate the required ratios.
[Tex]cosec\theta=\frac{1}{sin\theta}=\frac{5}{3}[/Tex]
[Tex]sec\theta=\frac{1}{cos\theta}=\frac{5}{4}[/Tex]
[Tex]cot\theta=\frac{1}{tan\theta}=\frac{4}{3}[/Tex]
Question 3: Given [Tex]tan\theta=\frac{6}{8}[/Tex], evaluate sin θ.cos θ.
Solution:Â
[Tex]tan\theta=\frac{P}{B}[/Tex]
Thus P=6, B=8
Using Pythagoras theorem,
H2=P2+B2
H2=36+64=100
Therefore, H =10
Now, [Tex]sin\theta= \frac{6}{10}[/Tex]
[Tex]cos\theta=\frac{8}{10}[/Tex]
Question 4: If [Tex]cot\theta = \frac{12}{13}[/Tex], evaluate tan2θ.
Solution:Â
Given [Tex]cot\theta=\frac{12}{13}[/Tex]
Thus [Tex]tan\theta=\frac{1}{cot\theta}=\frac{13}{12}[/Tex]
[Tex]\therefore tan^2\theta=\frac{169}{144}[/Tex]
Question 5: In the given triangle, verify sin2θ+cos2θ = 1
Solution:Â
Given P=12, B=5, H=13
Thus [Tex]sin\theta=\frac{12}{13}[/Tex]
[Tex]cos\theta=\frac{5}{13}[/Tex]
[Tex]sin^2\theta=144/169[/Tex]
[Tex]cos^2\theta=25/169[/Tex]
[Tex]sin^2\theta+cos^2\theta=\frac{169}{169}=1[/Tex]
Hence verified.
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