NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions (Ex 3.4)

Mathematics is a vitally important system of knowledge in human life and society. It has a distinctly scientific methodology, which implies the necessity of calculations and numerical verifications in order to prove postulates and reach logical conclusions. Mathematics has a long history which can be traced back to early human civilisations like the Babylonian and Indus Valley civilisations. This underlines the fact that Mathematics is a very important applied science and, by extension, a vital academic discipline. For the students of the CBSE Board who are pursuing their senior secondary education in the PCM stream, the NCERT textbook of Mathematics is a mandatory resource for learning.

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NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions (Ex 3.4) Exercise 3.4

The Class 11 Maths NCERT Solutions Chapter 3 Exercise 3.4 have been prepared after extensive and earnestly committed research of the past years’ question papers and sample question papers.This was done to ensure that students could effectively use the Class 11 Maths NCERT Solutions Chapter 3 Exercise 3.4 to familiarise themselves with the pattern of mark distribution in board examinations.The Class 11 Maths NCERT Solutions Chapter 3 Exercise 3.4 have been fashioned in a framework that is easy to understand. Access NCERT Solutions for Class 11 Maths Chapter 3 – Trigonometric Functions

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Trigonometry refers to that branch of the academic discipline of Mathematics that deals with particular functions of angles and their applications to calculations. In layman’s language, it is that stream of Mathematics that is concerned with the relationship between the sides and angles of triangles.

The chronology of Trigonometry as a mathematical sub-discipline could be tracedback to the third century BC when astronomical studies were being conducted through the application of Geometry. The knowledge covered within the sub-discipline of Trigonometry could be practically applied in many ventures in professional and academic spheres such as architecture, surveying, astronomy, physics, engineering, and even crime scene investigations.Morris Kline explained in his renowned work, Mathematical Thought From Ancient to Modern Times, that while trigonometry first appeared in connection with astronomy, it was also used in navigation and the creation of calendars.The foundations of Trigonometry lie in Geometry.The Class 11 Maths NCERT Solutions, Chapter 3, Exercise 3.4, are online learning resources that are highly relevant and can address a variety of common student concerns.Students can use the Class 11 Maths NCERT Solutions Chapter 3 Exercise 3.4 to take personal notes, which can greatly aid in the retention of important information such as formulas, theorems, definitions, and so on.

Terms Used For Trigonometric Functions:

According to common usage, there are six functions of an angle which are utilised in Trigonometry. Their names and abbreviated forms are as follows: sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec) and cosecant (csc). Trigonometric functions are indispensable for discovering unknown angles and distances from known or measured angles in geometrical illustrations.

The Class 11 Maths NCERT Solutions Chapter 3 Exercise 3.4 contain all of the solutions to the exercises presented in the third chapter of the prescribed textbook for Mathematics for students in Class 11.Extramarks has been particularly attentive to providing step-by-step calculations as a part of the Class 11 Maths NCERT Solutions Chapter 3 Exercise 3.4. This would make it easier and more feasible for students to follow a logical sequence of problem solving and verifying obtained conclusions.

Trigonometric Equations

Those equations which include the application and usage of trigonometric functions like sin, cos, tan, cot, etc., are defined as trigonometric equations.

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Theorems Discussed For Trigonometric Functions Exercise 3.4:

The Class 11 Maths NCERT Solutions Chapter 3 Exercise 3.4 have been solved in a detailed and comprehensive way. The formulae applied for solving the exercises provided as part of the Class 11 Maths NCERT Solutions Chapter 3 Exercise 3.4 have been highlighted separately. This has been done so that students would be able to follow the appropriate procedure necessary for reaching an adequate solution through a logically processed series of calculations.

Trigonometric Functions Exercise 3.4:

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Tips to Solve NCERT Solutions of Exercise 3.4 Class 11 Maths Chapter 3

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NCERT Solutions Class 11Maths of Chapter 3 All Exercises

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Q.1

$\begin{array}{l} Find the principal and general solutions of the following \\ equations : \tan x = \sqrt{3} \end{array}$

Ans.

$\begin{array}{l} \begin{array}{l} Since, tan \frac{π}{3} = \sqrt{3} and tan \frac{4 π}{3} = tan (π + \frac{π}{3}) = \tan \frac{π}{3} = \sqrt{3} \\ Therefore, principal solutions are x = \frac{π}{3} and \frac{4 π}{3} . \\ General solution of tanx = \sqrt{3} \\ tanx = \sqrt{3} \\ = \tan \frac{π}{3} \end{array} \\ \begin{array}{l} tanx = \tan (nπ + \frac{π}{3}) \\ \Rightarrow x = nπ + \frac{π}{3}, where n \in Z . \end{array} \end{array}$

Q.2

$\begin{array}{l} Find the principal and general solutions of the following \\ equation : \sec x = 2 \end{array}$

Ans.

$\begin{array}{l} Since, secx = 2 \Rightarrow \sec \frac{π}{3} = 2 and sec \frac{5 π}{3} = \sec (2 π - \frac{π}{3}) = \sec \frac{π}{3} = 2 \\ Therefore, principal solutions are x = \frac{π}{3} and \frac{5 π}{3} . \\ G eneral solution: \\ secx = 2 \\ = \sec \frac{π}{3} \\ = \sec (2 nπ \pm \frac{π}{3}) \\ ∴ x = 2 nπ \pm \frac{π}{3}, where n \in Z . \end{array}$

Q.3

$\begin{array}{l} Find the principal and general solutions of the following \\ equation : \cot x = - \sqrt{3} \end{array}$

Ans.

$\begin{array}{l} Since, cot x = - \sqrt{3} \\ \Rightarrow cot x = - \cot \frac{π}{6} \\ = \cot (π - \frac{π}{6}) \\ = \cot (\frac{5 π}{6}) \\ and \\ cot x = \cot (2 π - \frac{π}{6}) \\ = \cot (\frac{11 π}{6}) \\ Therefore, the principal solution are x = \frac{5 π}{6} and \frac{11 π}{6} . \\ General solution: \\ cotx = - \sqrt{3} \\ = \cot (\frac{5 π}{6}) \\ = \cot (nπ + \frac{5 π}{6}) \\ \Rightarrow x = nπ + \frac{5 π}{6}, where n \in Z . \end{array}$

Q.4

$\begin{array}{l} Find the principal and general solutions of the following \\ equation : cosec x = - 2 \end{array}$

Ans.

$Since, cosec x = cosec (\frac{π}{6})$

$\begin{array}{l} ∴ cosec x = - 2 \\ = - cosec (\frac{π}{6}) \\ = cosec (π + \frac{π}{6}) \\ = cosec (\frac{7 π}{6}) \\ and \\ cosec x = cosec (2 π - \frac{π}{6}) \\ = cosec \frac{11 π}{6} \\ Therefore, the principal solutions are \frac{7 π}{6}, \frac{11 π}{6} . \\ General solution: \\ cosec x = cosec (\frac{7 π}{6}) \\ = cosec (nπ + {(- 1)}^{n} \frac{7 π}{6}) \\ \Rightarrow x = nπ + {(- 1)}^{n} \frac{7 π}{6}, n \in Z . \end{array}$

Q.5

$\begin{array}{l} Find the general solution for each of the following \\ equation : \cos 4 x = \cos 2 x \end{array}$

Ans.

$\begin{array}{l} We have, \\ cos 4x = cos 2x \end{array}$

$\begin{array}{l} = \cos (2 nπ \pm 2 x) [\begin{array}{l} The general solution of cos θ = \cos α \\ \Rightarrow θ = 2 nπ \pm α \end{array}] \\ \Rightarrow 4 x = 2 nπ \pm 2 x \\ For (+) ve sign: \\ 4 x = 2 nπ + 2 x \\ 2 x = 2 nπ \Rightarrow x = nπ \\ For (-) ve sign: \\ 4 x = 2 nπ - 2 x \\ 6 x = 2 nπ \Rightarrow x = \frac{1}{3} nπ \\ The general solutions are x = \frac{1}{3} nπ and nπ, where n \in Z . \end{array}$

Q.6

$\begin{array}{l} Find the general solution for each of the following \\ equation : \cos 3 x + \cos x - \cos 2 x = 0 \end{array}$

Ans.

$\begin{array}{l} We have, \cos 3 x + \cos x - \cos 2 x = 0 \\ 2 \cos (\frac{3 x + x}{2}) \cos (\frac{3 x - x}{2}) - \cos 2 x = 0 \\ 2 \cos 2 x \cos x - \cos 2 x = 0 \\ \cos 2 x (2 \cos x - 1) = 0 \\ \Rightarrow \cos 2 x = 0 \Rightarrow 2 x = (2 n + 1) \frac{π}{2} \\ \Rightarrow x = (2 n + 1) \frac{π}{4}, n \in Z \end{array}$

$\begin{array}{l} or 2 cos x - 1 = 0 \Rightarrow \cos x = \frac{1}{2} = \cos \frac{π}{3} \\ \Rightarrow x = 2 nπ \pm \frac{π}{3}, n \in Z . \end{array}$

Q.7

$\begin{array}{l} Find the general solution for each of the following \\ equation : \sin 2 x + \cos x = 0 \end{array}$

Ans.

$\begin{array}{l} We have, \sin 2 x + \cos x = 0 \\ \Rightarrow 2 \sin x \cos x + \cos x = 0 \\ \Rightarrow \cos x (2 \sin x + 1) = 0 \\ Either \cos x = 0 \Rightarrow x = (2 n + 1) \frac{π}{2}, n \in Z . \\ or (2 \sin x + 1) = 0 \Rightarrow \sin x = - \frac{1}{2} \\ \Rightarrow \sin x = \sin \frac{7 π}{2} \\ \Rightarrow x = nπ + {(- 1)}^{n} \frac{7 π}{2}, n \in Z . \\ Thus, the general solution of given equation is: \\ nπ + {(- 1)}^{n} \frac{7 π}{2} or (2 n + 1) \frac{π}{2}, n \in Z . \end{array}$

Q.8

$\begin{array}{l} Find the general solution for each of the following \\ equation : \sec^{2} 2 x = 1 - \tan 2 x \end{array}$

Ans.

$\begin{array}{l} We have, \sec^{2} 2 x = 1 - \tan 2 x \\ \Rightarrow 1 + \tan^{2} 2 x = 1 - \tan 2 x \end{array}$

$\begin{array}{l} \Rightarrow \tan^{2} 2 x + \tan 2 x = 0 \\ \Rightarrow \tan 2 x (\tan 2 x + 1) = 0 \\ Either tan 2x = 0 \Rightarrow 2 x = nπ \\ \Rightarrow x = \frac{nπ}{2}, where n \in Z . \\ or \tan 2 x + 1 = 0 \Rightarrow \tan 2 x = - 1 \\ \Rightarrow \tan 2 x = - \tan \frac{π}{4} \\ = \tan (π - \frac{π}{4}) \\ \Rightarrow \tan 2 x = \tan \frac{3 π}{4} \\ \Rightarrow 2 x = nπ + \frac{3 π}{4} \\ \Rightarrow x = \frac{nπ}{2} + \frac{3 π}{8}, where n \in Z . \\ The general solution of the given equation is: \\ x = \frac{nπ}{2}, \frac{nπ}{2} + \frac{3 π}{8}, where n \in Z . \end{array}$

Q.9

$\begin{array}{l} Find the general solution for each of the following \\ equation : \sin x + \sin 3 x + \sin 5 x = 0 \end{array}$

Ans.

$\begin{array}{l} We have, \sin x + \sin 3 x + \sin 5 x = 0 \\ \Rightarrow \sin 5 x + \sin x + \sin 3 x = 0 \\ \Rightarrow 2 \sin (\frac{5 x + x}{2}) \cos (\frac{5 x - x}{2}) + \sin 3 x = 0 \end{array}$

$\begin{array}{l} \Rightarrow 2 \sin 3 x \cos 2 x + \sin 3 x = 0 \\ \Rightarrow \sin 3 x (2 \cos 2 x + 1) = 0 \\ Either sin3x = 0 \Rightarrow 3 x = nπ \\ \Rightarrow x = \frac{nπ}{3}, n \in Z . \\ or 2 cos 2x + 1 = 0 \Rightarrow \cos 2 x = - \frac{1}{2} \\ \Rightarrow 2 x = 2 nπ \pm \frac{2 π}{3} \\ \Rightarrow x = nπ \pm \frac{π}{3}, n \in Z . \\ Therefore, the general solution of the given equations is: \\ x = \frac{nπ}{3} or nπ \pm \frac{π}{3}, n \in Z . \end{array}$

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NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions (Ex 3.4)