Six Trigonometric Functions (original) (raw)

Last Updated : 23 Jul, 2025

Trigonometry can be defined as the branch of mathematics that determines and studies the relationships between the sides of a triangle and the angles subtended by them. Trigonometry is 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,

Right-Triangle

Right Triangle

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,

What are 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
cos θ
tan θ
cosec θ
sec θ
cot θ

The below table shows the values of these functions at some standard angles,

**Function ** **30° **45° **60° **90°
sin\theta = \frac{P}{H} 0 \frac{1}{2} \frac{1}{\sqrt2} \frac{\sqrt3}{2} 1
cos\theta = \frac{B}{H} 1 \frac{\sqrt3}{2} \frac{1}{\sqrt2} \frac{1}{2} 0
tan\theta = \frac{sin\theta}{cos\theta}=\frac{P}{B} 0 \frac{1}{\sqrt3} 1 \sqrt3
cosec\theta = \frac{H}{P} 2 \sqrt2 \frac{2}{\sqrt3} 1
sec\theta = \frac{H}{B} 1 \frac{2}{\sqrt3} \sqrt2 2
cot\theta = \frac{B}{P} \sqrt3 1 \frac{1}{\sqrt3} 0

**Note: It is advised to remember the first 3 trigonometric functions and their values at these standard angles for ease of calculations.

Sample Problems on Six Trigonometric Functions

**Problem 1: Evaluate sine, cosine, and tangent in the following figure.

Right-Triangle(3-4-5)

**Solution:

Given,

Using the trigonometric formulas for sine, cosine and tangent,

sin\theta=\frac{P}{H}=\frac{3}{5}

cos\theta=\frac{B}{H}=\frac{4}{5}

tan\theta=\frac{P}{B}=\frac{3}{4}

**Problem 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.

cosec\theta=\frac{1}{sin\theta}=\frac{5}{3}

sec\theta=\frac{1}{cos\theta}=\frac{5}{4}

cot\theta=\frac{1}{tan\theta}=\frac{4}{3}

Problem 3: Given tan\theta=\frac{6}{8}, evaluate sin θ.cos θ.

**Solution:

tan\theta=\frac{P}{B}

Thus P = 6, B = 8

Using Pythagoras theorem,

H2 = P2 + B2

H2= 36 + 64 = 100

Therefore, H =10

Now, sin\theta= \frac{6}{10}

cos\theta=\frac{8}{10}

Problem 4: If cot\theta = \frac{12}{13}, evaluate tan 2 θ.

**Solution:

Given cot\theta=\frac{12}{13}

Thus tan\theta=\frac{1}{cot\theta}=\frac{13}{12}

\therefore tan^2\theta=\frac{169}{144}

**Problem 5: In the given triangle, verify **sin 2 θ + cos 2 θ = 1

Right-Triangle(51213)

**Solution:

Given,

Thus sin\theta=\frac{12}{13}

cos\theta=\frac{5}{13}

sin^2\theta=144/169

cos^2\theta=25/169

sin^2\theta+cos^2\theta=\frac{169}{169}=1

Hence verified.