Six Trigonometric Functions

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,

• Base: The side(RQ) on which the angle θ lies is known as the base.

• Perpendicular: It is the side(PQ) opposite to the angle θ  in consideration.

• Hypotenuse: It is the longest side(PR) 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 Function: It is represented as sin θ and is defined as the ratio of perpendicular and hypotenuse.

• Cosine Function: It is represented as cos θ and is defined as the ratio of base and hypotenuse.

• Tangent Function: 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 Function: It is the reciprocal of sin θ and is represented as cosec θ.

• Secant Function: It is the reciprocal of cos θ and is represented as sec θ.

• Cotangent Function: It is the reciprocal of tan θ and is represented as cot θ.

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	0°	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.

Solution: 

Given,

• P = 3
• B = 4
• H = 5

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 tan²θ.

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 sin²θ + cos²θ = 1

Solution: 

Given,

• P = 12
• B = 5
• H = 13

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.