What are the six trigonometry functions?

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:

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. 

Solution: 

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: 

[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: 

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: 

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.