JEE MAIN 2020 NEWS

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# JEE Main Study Notes for Trigonometry: Basic Formulas, Solving Methods and Previous Year Questions One of the keys to solving questions from trigonometry in JEE Main is familiarity with its formulas. Carrying a weightage of around 7% in JEE Main, basic formulas of trigonometry play a vital role while solving questions. Every year, around 4-5 questions are asked directly from the topic in JEE Main. Also, there are some more questions from other topics in which trigonometry is imbibed. Toppers recommend to learn all formulas and have conceptual clarity of their application to become a pioneer in this topic. Click here to read JEE Main Mathematics Syllabus

• Trigonometric Ratios and concepts are used in Algebra and Calculus (as a direct function or as a way to simplify functions).
• Questions can be asked from trigonometric identities, equations, heights and distances, and inverse trigonometric functions.

Read the article to refer to important trigonometric formulas and methods to solve questions.

### Important Topics in Trigonometry (Chapter-wise Weightage)

Below given is the weightage of important topics in trigonometry in JEE Main.

ChapterWeightage (out of 120 marks)
Trigonometric Identities5.67
Solutions of Triangles and Heights & Distances1.33

Tip: Being an essential unit, it should be covered first at the time of preparation as it will help in other units. Read JEE Main Mathematics Preparation Tips

### Trigonometric Ratios (T-Ratios)

To understand Trigonometric Ratios, we take a right angled triangle ABC, where ∠BAC = θ. Below mentioned are the trigonometric ratios for the angle θ.

### Signs of Trigonometric Functions

Two perpendicular lines intersecting at a point O divide a plane in 4 right angles, each is called a quadrant. If θ be the angle that a line OA subtends with the initial line OX, anticlockwise then.

In I Quadrant, 0° < θ < 90°

In II Quadrant, 90° < θ < 180°

In III Quadrant, 180° < θ < 270°

In IV Quadrant, 270° < θ < 360°

The following table illustrates the sign of various trigonometric functions in all the four quadrants

sin θ++--
cos θ+--+
tan θ+-+-
cot θ+-+-
sec θ+--+
cosec θ++--

More precisely, we can discuss the way of increase and decrease of trigonometric functions as described in the following tables:

sin θincreases from 0 to 1decreases from 1 to 0decreases from 0 to -1increases from -1 to 0
cos θdecreases from 1 to 0decreases from 0 to -1increases from -1 to 0increases from 0 to 1
tan θincreases from 0 to increases from -∞ to 0increases from 0 to ∞increases from -∞ to 0
cot θdecreases from ∞ to 0decreases from 0 to -∞decreases from ∞ to 0decreases from 0 to -∞
sec θincreases from 1 to ∞increases from -∞ to -1decreases from -1 to -∞decreases from ∞ to 1
cosec θdecreases from ∞ to 1increases from 1 to ∞increases from -∞ to -1decreases from -1 to -∞

### Basic Formulae of Trigonometric Functions

1. sin θ . cosec θ = 1 or
2. cosθ . sec θ = 1 or
3. tan θ . cot θ = 1 or

### Graphs of Trigonometric Functions

1. Graph of y = sin x

1. Graph of y = cos x

1. Graph of y = tan x

1. Graph of y = cot x

1. Graph of y = sec x and y = cosec x

1. Graph of y = 3 cos 2x

Graph of similar trigonometric functions can be obtained in a similar way.

### Trigonometric Ratios of Compound Angles

• Conversion Table for T-Ratios of –θ in terms of T-Ratios of θ.

sin (– θ) = – sin θ, cos (– θ) = cos θ,

tan (– θ) = – tan θ cot (– θ) = – cot θ,

sec (– θ) = sec θ, cosec (– θ) = – cosec θ

• Conversion Table for T-Ratios of in terms of T-Ratios of θ

• Conversion Table for T-Ratios π ± θ in terms of T-Ratios of θ

sin (π − θ) = sin θ sin (π + θ) = − sin θ

cos (π − θ) = – cos θ cos (π + θ) = − cos θ

tan (π − θ) = – tan θ tan (π + θ) =  tan θ

cot (π − θ) = – cot θ cot (π + θ) = cot θ

sec (π − θ) = – sec θ sec (π + θ) = − sec θ

cosec (π − θ) = cosec θ cosec (π + θ) = − cosec θ

• Conversion Table for T-Ratios of in terms of T-Ratios of θ.

• Conversion Table for T-Ratios of 2π ± θ in terms of T-Ratios of θ.

1. ,
2. ,

1. ;

1.  ;
2. ;
3. ; ;

### Methods of Solving Trigonometric Equations

#### Factor Method

Example : Solve : 2 cos x cos 2x = cos x.

Solution : The given equation is equivalent to the equation cos x (2 cos 2x –1) = 0. This equation is equivalent to the collection of equations

Example : Solve the equation 3 cos2x – 10 cos x + 3 = 0.

Solution :  Assume cos x = y. The given equation assumes the form 3y2 – 10 y +3 = 0. Solving it, we find that

The value y2 = 3 does not satisfy the condition since | cos x | ≤ 1. Consequently,

#### Solving Equations by Introducing an Auxiliary Argument

Equations of the form a sin x + b cos x = c can be solved by dividing the two sides of equation by  and substituting

Example : Solve the equation sin x + cos x =

Solution :

or

#### Use of Half Angle Formulas

Example : Solve :

Solution : cos x – (1 – cos x) = 0

### Solved Examples of Trigonometry

Question: If k = sin (π/18) sin (5π/18) sin (7π/18), then what is the numerical value of k?

Solution: The value of k is given to be k = sin (π/18) sin (5π/18) sin (7π/18).

Hence, k = sin 10° sin 50° sin 70°

= sin 10° sin (60° - 10°). sin (60° + 10°)

= sin 10° [sin 260° - sin 210°]

= sin 10° [(√3/2)2 - sin 210°]

= sin 10° [3/4 - sin 210°]

= 1/4 [3sin 10° - 4sin 310°]

= 1/4 x sin (3 x 10)   (since, sin 3θ = 3sin θ- 4sin 3 θ)

= 1/4 sin 30° = 1/8

Question: Evaluate the value of (1 + cos π/8) (1 + cos 3π/8) (1 + cos 5π/8) (1 + cos 7π/8).

Solution: The given expression is

(1 + cos π/8) (1 + cos 3π/8) (1 + cos 5π/8) (1 + cos 7π/8)

= (1 + cos π/8) (1 + cos 3π/8) (1 - cos 3π/8) (1 - cos π/8)

= (1 – cos 2π/8) (1 - cos 23π/8)

= ¼ [2 sin π/8 sin 3π/8]2

= ¼ [2 sin π/8 cos π/8]2

= ¼ [sin π/4]2

= 1/4. 1/2 = 1/8

### Tips to Study Trigonometry for JEE Main

Trigonometry is a chapter in JEE Main 2020 Mathematics where application of formula is really important. Thus it is important to study the topic in a manner that leads to quick solution of problems. Some of the tips have been given below.

1. Read the theory ONCE, but give formulas a reading of two to three times. Mugging up theorems is not at all suggested.
2. Solve problems and focus more on the equations part by applying the formulas that you have read. Constant application will help you getting well-versed with formulas.
3. Studying this topic from 2 good sources will be enough as the kind of questions that will come from this topic are almost similar.
4. You can make a formula sheet and have a look at it initially when you get stuck, and things will slowly get etched into your brain.
5. Don’t forget to go through the previous year JEE Main question papers.

With the application of above tips, students opting for JEE Main or any other paper for that matter will sure be able to crack questions from Trigonometry in the least time possible. No matter how tough the topic seems to be, constant practice will lead you to ace the paper.

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