Trigonometry

Trigonometry

Introduction to Trigonometry

Trigonometry is one of the important branches of mathematics and this concept is given by a Greek mathematician Hipparchus. Basically, it is the study of triangles where we deal with the angles and sides of the triangle. To be more specific, its all about a right-angled triangle. It is one of those divisions in mathematics that helps in finding the angles and missing sides of a triangle. The angles are either measured in radians or degrees.

This branch divides into two sub-branches called plane trigonometry and spherical geometry. Here in this theory, you will learn about the trigonometric formulas, functions, and ratios, Right-Angled Triangles, etc.

Trigonometric Functions and Ratios

The trigonometric ratios of a triangle are also called the trigonometric functions. Sine, cosine, and tangent are 3 important trigonometric functions and are abbreviated as sin, cos, and tan. Let us see how are these ratios or functions, evaluated in the case of a right-angled triangle.

Consider a right-angled triangle, where the longest side is called the hypotenuse, and the sides opposite to the hypotenuse is referred to as the adjacent and opposite.

Trigonometry Basics

The trigonometric ratios are calculated by the below formulas using above figure.

Functions

Abbreviation

Relationship to sides of a right triangle

Sine Function sin Opposite / Hypotenuse
Tangent Function tan Opposite / Adjacent
Cosine Function cos Adjacent / Hypotenuse
Cosecant Function cosec Hypotenuse / Opposite
Secant Function sec Hypotenuse / Adjacent
Cotangent Function cot Adjacent / Opposite

Trigonometry Table

Trigonometry Ratios table

Unit Circle

The concept of unit circle helps us to measure the angles of cos, sin and tan directly since the centre of the circle is located at the origin and radius is 1. Consider theta be an angle then,

Trigonometry angles

Trigonometry Formula

The Trigonometric formulas or Identities are the equations which are true in the case of Right Angled Triangles. Some of the special trigonometric identities are as given below –

  1. Pythagorean Identities
  • sin ² θ + cos ² θ = 1
  • tan 2 θ + 1 = sec2 θ
  • cot2 θ + 1 = cosec2 θ
  • sin 2θ = 2 sin θ cos θ
  • cos 2θ = cos² θ – sin² θ
  • tan 2θ = 2 tan θ / (1 – tan² θ)
  • cot 2θ = (cot² θ – 1) / 2 cot θ
  1. Sum and Difference identities-

For angles u and v, we have the following relationships:

  • sin(u + v) = sin(u)cos(v) + cos(u)sin(v)
  • cos(u + v) = cos(u)cos(v) – sin(u)sin(v)
  • tan(u+v) = \(\frac{tan(u)\ +\ tan(v)}{1-tan(u)\ tan(v)}\)
  • sin(u – v) = sin(u)cos(v) – cos(u)sin(v)
  • cos(u – v) = cos(u)cos(v) + sin(u)sin(v)
  • tan(u-v) = \(\frac{tan(u)\ -\ tan(v)}{1+tan(u)\ tan(v)}\)
  1. If A, B and C are angles and a, b and c are the sides of a triangle, then,

         Sine Laws

  • a/sinA = b/sinB = c/sinC

        Cosine Laws

  • c= a+ b– 2ab cos C
  • a= b+ c– 2bc cos A
  • b= a+ c– 2ac cos B

Applications

  • Its applications is in various fields like oceanography, seismology, meteorology, physical sciences, astronomy, acoustics, navigation, electronics, etc.
  • It is also helpful to measure the height of the mountain, find the distance of long rivers, etc.

Word Questions or Problems

Example 1: Two friends, Rakesh and Vishal started climbing a pyramid-shaped hill. Rakesh climbs 315 mtr and finds that the angle of depression is 72.3 degrees from his starting point. How high he is from the ground.

Solution: Let m is the height above the ground.

To find: Value of m

Trigonometry Questions

To solve m, use sine ratio.

Sin 72.30 = m/315

0.953 = m/315

m= 315 x 0.953

m=300.195 mtr

The man is 300.195 mtr  above the ground.

Example 2:  A man is observing a pole of height 55 foot. According to his measurement, pole cast a 23 feet long shadow. Can you help him to know the angle of elevation of the sun from the tip of shadow?

Solution:

Trigonometry Problems

Let x be the angle of elevation of the sun, then

tan x  = 55/23 = 2.391

x = tan-1(2.391)

or x = 67.30 degrees

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Practise This Question

cos θcosec θ+1+cos θcosec θ1= ___