NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Exercise 3.3

NCERT Solutions for Class 11 Maths Chapter 3 Exercise 3.3 focus on applying trigonometric identities to simplify expressions and solve trigonometric equations. This exercise helps students work with trigonometric ratios and identities in more complex scenarios, helping to reinforce their understanding of the relationships between different functions.

In this exercise, students will learn to simplify trigonometric expressions by using appropriate identities and solve trigonometric equations to find the values of the angles. It also covers applications of identities in real-life situations and deeper trigonometric manipulations.

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Exercise 3.3

Trigonometric Functions

Medium

Q.

Find the values of other five trigonometric functions in Exercises 1 to 5.

$1.\cos \text{x}=-\frac{1}{2}, \text{x} \mathrm{lies} \mathrm{in} \mathrm{third} \mathrm{quadrant}.$

$2.\sin \text{x}=\frac{3}{5}, \text{x} \mathrm{lies} \mathrm{in} \mathrm{second} \mathrm{quadrant}.$

$3.\cot \text{x}=\frac{3}{4}, \text{x} \mathrm{lies} \mathrm{in} \mathrm{third} \mathrm{quadrant}.$

$4.\sec \text{x}=\frac{13}{5}, \text{x} \mathrm{lies} \mathrm{in} \mathrm{fourth} \mathrm{quadrant}.$

$5.\tan \text{x}=-\frac{5}{12}, \text{x} \mathrm{lies} \mathrm{in} \mathrm{second} \mathrm{quadrant}.$

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Trigonometric Functions

Medium

Q.

Prove that:
cos 6x = 32 cos⁶ x – 48cos⁴ x + 18 cos² x – 1

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Trigonometric Functions

Medium

Q.

Prove that:
${(\cos \mathrm{x}-\cos \mathrm{y})}^2+{(\sin \mathrm{x}-\sin \mathrm{y})}^2=4 {\sin }^2 \frac{\mathrm{x}-\mathrm{y}}{2}$

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Trigonometric Functions

Medium

Q.

$\begin{array}{l}\mathrm{Two} \mathrm{trees}, \mathrm{A} \mathrm{and} \mathrm{B} \mathrm{are} \mathrm{on} \mathrm{the} \mathrm{same} \mathrm{side} \mathrm{of} \mathrm{a} \mathrm{river}. \mathrm{From} \mathrm{a}\\ \mathrm{point} \mathrm{C} \mathrm{in} \mathrm{the} \mathrm{river} \mathrm{the} \mathrm{distance} \mathrm{of} \mathrm{the} \mathrm{trees} \mathrm{A} \mathrm{and} \mathrm{B} \mathrm{is} 250 \mathrm{m}\\ \mathrm{and} 300 \mathrm{m}, \mathrm{respectively}. \mathrm{If} \mathrm{the} \mathrm{angle} \mathrm{C} \mathrm{is} 45\mathrm{°}, \mathrm{find} \mathrm{the}\\ \mathrm{distance} \mathrm{between} \mathrm{the} \mathrm{trees} (\mathrm{use} \sqrt{2}=1.44).\end{array}$

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Trigonometric Functions

Difficult

Q.

Two ships leave a port at the same time. One goes 24 km per hour in the direction N45°E and other travels 32 km per hour in the direction S75°E. Find the distance between the ships at the end of 3 hours.

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Trigonometric Functions

Difficult

Q.

A tree stands vertically on a hill side which makes an angle of 15° with the horizontal. From a point on the ground 35 m down the hill from the base of the tree, the angle of elevation of the top of the tree is 60°. Find the height of the tree.

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Trigonometric Functions

Medium

Q.

$\begin{array}{l}\text{For any triangle ABC},\text{ prove that}\\ \frac{\cos \mathrm{A}}{\mathrm{a}}+\frac{\cos \mathrm{B}}{\mathrm{b}}+\frac{\cos \mathrm{C}}{\mathrm{c}}=\frac{{\mathrm{a}}^2 + {\mathrm{b}}^2 + {\mathrm{c}}^2}{2\mathrm{abc}}\end{array}$

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Trigonometric Functions

Medium

Q.

$\begin{array}{l}\text{For any triangle ABC},\text{ prove that}\\ \frac{\mathrm{a} - \mathrm{b}}{\mathrm{c}} = \frac{\sin \left(\frac{\mathrm{A} - \mathrm{B}}{2}\right)}{\cos \frac{\mathrm{C}}{2}}\end{array}$

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Trigonometric Functions

Medium

Q.

$\begin{array}{l}\text{For any triangle ABC},\text{ prove that}\\ \frac{\mathrm{a} + \mathrm{b}}{\mathrm{c}} = \frac{\cos \left(\frac{\mathrm{A} - \mathrm{B}}{2}\right)}{\sin \frac{\mathrm{C}}{2}}\end{array}$

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Trigonometric Functions

Medium

Q.

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

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Trigonometric Functions

Medium

Q.

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

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Trigonometric Functions

Medium

Q.

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

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Trigonometric Functions

Medium

Q.

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

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Trigonometric Functions

Medium

Q.

Prove that:
cos 4x = 1 – 8sin² x cos² x

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Trigonometric Functions

Medium

Q.

$\cos (\frac{\mathrm{\pi }}{4}-\mathrm{x})\cos (\frac{\mathrm{\pi }}{4}-\mathrm{y})-\sin (\frac{\mathrm{\pi }}{4}-\mathrm{x})\sin (\frac{\mathrm{\pi }}{4}-\mathrm{y})=\sin (\mathrm{x}+\mathrm{y})$

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Trigonometric Functions

Medium

Q.

$\begin{array}{l}\mathrm{Prove} \mathrm{that}:\\ \tan 4\mathrm{x} = \frac{4 \tan \mathrm{x} (1 - {\tan }^2\mathrm{x})}{1 - 6 {\tan }^2\mathrm{x} + {\tan }^4\mathrm{x}}\end{array}$

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Trigonometric Functions

Medium

Q.

Prove that:
cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

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Trigonometric Functions

Medium

Q.

$\begin{array}{l}\mathbf{Prove}\mathbf{\hspace{0.17em}}\mathbf{that}\mathbf{:}\\ \frac{\mathbf{cos}\mathbf{ }\mathbf{4}\mathbf{x}\mathbf{\hspace{0.17em}}\mathbf{+}\mathbf{\hspace{0.17em}}\mathbf{cos}\mathbf{\hspace{0.17em}}\mathbf{3}\mathbf{x}\mathbf{\hspace{0.17em}}\mathbf{+}\mathbf{\hspace{0.17em}}\mathbf{cos}\mathbf{ }\mathbf{2}\mathbf{x}}{\mathbf{sin}\mathbf{\hspace{0.17em}}\mathbf{4}\mathbf{x}\mathbf{ }\mathbf{+}\mathbf{\hspace{0.17em}}\mathbf{sin}\mathbf{\hspace{0.17em}}\mathbf{3}\mathbf{x}\mathbf{\hspace{0.17em}}\mathbf{+}\mathbf{ }\mathbf{sin}\mathbf{\hspace{0.17em}}\mathbf{2}\mathbf{x}}\mathbf{\hspace{0.17em}}\mathbf{=}\mathbf{\hspace{0.17em}}\mathbf{cot}\mathbf{ }\mathbf{3}\mathbf{x}\end{array}$

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Trigonometric Functions

Medium

Q.

$\begin{array}{l}\mathbf{Prove}\mathbf{\hspace{0.17em}}\mathbf{that}\mathbf{:}\\ \frac{\mathbf{sin}\mathbf{\hspace{0.17em}}\mathbf{x}\mathbf{ }\mathbf{–}\mathbf{\hspace{0.17em}}\mathbf{sin}\mathbf{\hspace{0.17em}}\mathbf{3}\mathbf{x}}{{\mathbf{sin}}^{\mathbf{2}}\mathbf{x}\mathbf{\hspace{0.17em}}\mathbf{–}\mathbf{ }{\mathbf{cos}}^{\mathbf{2}}\mathbf{x}}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{2}\mathbf{\hspace{0.17em}}\mathbf{sin}\mathbf{\hspace{0.17em}}\mathbf{x}\end{array}$

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Trigonometric Functions

Medium

Q.

Prove that:
cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

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Trigonometric Functions

Medium

Q.

Prove that:
sin 2x + 2 sin 4x + sin 6x = 4 cos² x sin 4x

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Trigonometric Functions

Medium

Q.

Prove that:
cos² 2x – cos² 6x = sin 4x sin 8x

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Trigonometric Functions

Medium

Q.

Prove that:
sin² 6x – sin² 4x = sin 2x sin 10x

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Trigonometric Functions

Medium

Q.

$\cos \hspace{0.17em}(\frac{3\mathrm{\pi }}{4} + \mathrm{x}) - \cos \hspace{0.17em}(\frac{3\mathrm{\pi }}{4}\hspace{0.17em}- \mathrm{x}) =\hspace{0.17em}-\sqrt{2} \sin \mathrm{x}$

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Trigonometric Functions

Medium

Q.

sin (n + 1) x sin (n + 2) x + cos (n + 1) x cos (n + 2) x = cos x

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Trigonometric Functions

Medium

Q.

$\cos \hspace{0.17em}(\frac{3\mathrm{\pi }}{2} +\hspace{0.17em}\mathrm{x})\hspace{0.17em}\cos (2\mathrm{\pi }\hspace{0.17em}+\hspace{0.17em}\mathrm{x})\hspace{0.17em}[\cot \hspace{0.17em}(\frac{3\mathrm{\pi }}{2}\hspace{0.17em}-\hspace{0.17em}\mathrm{x})\hspace{0.17em}+\hspace{0.17em}\cot \hspace{0.17em}(2\mathrm{\pi } + \mathrm{x})] = 1$

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Trigonometric Functions

Medium

Q.

$\begin{array}{l}\mathrm{Prove} \mathrm{the} \mathrm{following}:\\ \frac{\tan (\frac{\mathrm{\pi }}{4} + \mathrm{x})}{\tan (\frac{\mathrm{\pi }}{4} - \mathrm{x})} ={ \left(\frac{1 + \tan \mathrm{x}}{1 - \tan \mathrm{x}}\right)}^2\end{array}$

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Trigonometric Functions

Medium

Q.

$\begin{array}{l}\mathrm{Prove} \mathrm{that}:\\ \sin 3\mathrm{x}+\sin 2\mathrm{x}-\mathrm{sinx}=4\mathrm{sinxcos}\frac{\mathrm{x}}{2}\cos \frac{3\mathrm{x}}{2}\end{array}$

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Question 1: Prove that sin^2(pi/6) + cos^2(pi/3) - tan^2(pi/4) = -1/2

• Ans: Substituting the values of sin(pi/6), cos(pi/3), and tan(pi/4) on the left hand side.
• L.H.S. = (1/2)^2 + (1/2)^2 - (1)^2.
• = 1/4 + 1/4 - 1 = -1/2.
• = R.H.S. Hence proved.

Question 2: Prove that 2 sin^2(pi/6) + cosec^2(7pi/6) cos^2(pi/3) = 3/2

• Ans: L.H.S. = 2 sin^2(pi/6) + cosec^2(7pi/6) cos^2(pi/3).
• Since cosec(7pi/6) = -cosec(pi/6).
• = 2(1/2)^2 + (-cosec(pi/6))^2 (1/2)^2.
• = 2(1/4) + (-2)^2 (1/4) = 1/2 + 4/4 = 3/2.
• = R.H.S. Hence proved.

Question 3: Prove that cot^2(pi/6) + cosec(5pi/6) + 3 tan^2(pi/6) = 6

• Ans: L.H.S. = cot^2(pi/6) + cosec(5pi/6) + 3 tan^2(pi/6).
• Since cosec(5pi/6) = cosec(pi - pi/6) = cosec(pi/6).
• = (sqrt(3))^2 + cosec(pi/6) + 3(1/sqrt(3))^2.
• = 3 + 2 + 3(1/3) = 6.
• = R.H.S. Hence proved.

Question 4: Prove that 2 sin^2(3pi/4) + 2 cos^2(pi/4) + 2 sec^2(pi/3) = 10

• Ans: L.H.S. = 2 sin^2(3pi/4) + 2 cos^2(pi/4) + 2 sec^2(pi/3).
• Since sin(3pi/4) = sin(pi - pi/4) = sin(pi/4).
• = 2(sin(pi/4))^2 + 2(1/sqrt(2))^2 + 2(2)^2.
• = 2(1/2) + 1 + 8 = 10.
• = R.H.S. Hence proved.

Question 5: Find the value of:

• (i) sin 75 degree
  • Ans: sin 75 = sin(45 + 30).
  • Using sin(x+y) = sin x cos y + cos x sin y.
  • = sin 45 cos 30 + cos 45 sin 30 = (1/sqrt(2) * sqrt(3)/2) + (1/sqrt(2) * 1/2).
  • = (sqrt(3) + 1) / 2sqrt(2).
• (ii) tan 15 degree
  • Ans: tan 15 = tan(45 - 30).
  • Using tan(x-y) = (tan x - tan y) / (1 + tan x tan y).
  • Result simplifies to 2 - sqrt(3).

Question 6: Prove cos(pi/4 - x) cos(pi/4 - y) - sin(pi/4 - x) sin(pi/4 - y) = sin(x + y)

• Ans: Using cos(x+y) = cos x cos y - sin x sin y.
• L.H.S. = cos[(pi/4 - x) + (pi/4 - y)] = cos[pi/2 - (x+y)].
• = sin(x+y) = R.H.S. Hence proved.

Question 7: Prove [tan(pi/4 + x) / tan(pi/4 - x)] = [(1 + tan x) / (1 - tan x)]^2

• Ans: Using tan(A+B) and tan(A-B) formulas.
• L.H.S. = [(1 + tan x) / (1 - tan x)] / [(1 - tan x) / (1 + tan x)].
• = [(1 + tan x) / (1 - tan x)]^2 = R.H.S..

Question 8: Prove [cos(pi + x) cos(-x)] / [sin(pi - x) cos(pi/2 + x)] = cot^2 x

• Ans: L.H.S. = [(-cos x)(cos x)] / [(sin x)(-sin x)].
• = -cos^2 x / -sin^2 x = cot^2 x. Hence proved.

Question 9: Prove cos(3pi/2 + x) cos(2pi + x) [cot(3pi/2 - x) + cot(2pi + x)] = 1

• Ans: L.H.S. = sin x cos x [tan x + cot x].
• = sin x cos x [sin x / cos x + cos x / sin x].
• = (sin x cos x) [(sin^2 x + cos^2 x) / (sin x cos x)] = 1 = R.H.S..

Question 10: Prove sin(n+1)x sin(n+2)x + cos(n+1)x cos(n+2)x = cos x

• Ans: Using cos(x-y) = cos x cos y + sin x sin y.
• L.H.S. = cos[(n+2)x - (n+1)x] = cos x = R.H.S..

Question 11: Prove cos(3pi/4 + x) - cos(3pi/4 - x) = -sqrt(2) sin x

• Ans: Using cos A - cos B = -2 sin((A+B)/2) sin((A-B)/2).
• L.H.S. = -2 sin(3pi/4) sin x.
• = -2 (1/sqrt(2)) sin x = -sqrt(2) sin x = R.H.S..

Question 12: Prove sin^2 6x - sin^2 4x = sin 2x sin 10x

• Ans: L.H.S. = (sin 6x + sin 4x)(sin 6x - sin 4x).
• = [2 sin 5x cos x] [2 cos 5x sin x].
• = (2 sin 5x cos 5x) (2 sin x cos x) = sin 10x sin 2x = R.H.S..

Question 13: Prove cos^2 2x - cos^2 6x = sin 4x sin 8x

• Ans: L.H.S. = (cos 2x + cos 6x)(cos 2x - cos 6x).
• = [2 cos 4x cos 2x] [-2 sin 4x sin(-2x)].
• = (2 sin 4x cos 4x) (2 sin 2x cos 2x) = sin 8x sin 4x = R.H.S..

Question 14: Prove sin 2x + 2 sin 4x + sin 6x = 4 cos^2 x sin 4x

• Ans: L.H.S. = (sin 6x + sin 2x) + 2 sin 4x.
• = 2 sin 4x cos 2x + 2 sin 4x = 2 sin 4x (cos 2x + 1).
• = 2 sin 4x (2 cos^2 x) = 4 cos^2 x sin 4x = R.H.S..

Question 15: Prove cot 4x(sin 5x + sin 3x) = cot x(sin 5x - sin 3x)

• Ans: L.H.S. = (cos 4x / sin 4x) [2 sin 4x cos x] = 2 cos 4x cos x.
• R.H.S. = (cos x / sin x) [2 cos 4x sin x] = 2 cos 4x cos x.
• Therefore, L.H.S. = R.H.S..

Questions 16-21 Summary

• 16: (cos 9x - cos 5x) / (sin 17x - sin 3x) = -sin 2x / cos 10x.
• 17: (sin 5x + sin 3x) / (cos 5x + cos 3x) = tan 4x.
• 18: (sin x - sin y) / (cos x + cos y) = tan((x-y)/2).
• 19: (sin x + sin 3x) / (cos x + cos 3x) = tan 2x.
• 20: (sin x - sin 3x) / (sin^2 x - cos^2 x) = 2 sin x.
• 21: (cos 4x + cos 3x + cos 2x) / (sin 4x + sin 3x + sin 2x) = cot 3x.

Question 22: Prove cot x cot 2x - cot 2x cot 3x - cot 3x cot x = 1

• Ans: L.H.S. = cot x cot 2x - cot 3x(cot 2x + cot x).
• = cot x cot 2x - [(cot 2x cot x - 1) / (cot x + cot 2x)] (cot 2x + cot x).
• = cot x cot 2x - (cot 2x cot x - 1) = 1 = R.H.S..

Question 23: Prove tan 4x = [4 tan x (1 - tan^2 x)] / [1 - 6 tan^2 x + tan^4 x]

• Ans: L.H.S. = tan 2(2x) = (2 tan 2x) / (1 - tan^2 2x).
• Substituting tan 2x = (2 tan x) / (1 - tan^2 x).
• Result simplifies to [4 tan x (1 - tan^2 x)] / [1 - 6 tan^2 x + tan^4 x] = R.H.S..

Question 24: Prove cos 4x = 1 - 8 sin^2 x cos^2 x

• Ans: L.H.S. = cos 2(2x) = 1 - 2 sin^2 2x.
• = 1 - 2 (2 sin x cos x)^2 = 1 - 8 sin^2 x cos^2 x = R.H.S..

Question 25: Prove cos 6x = 32 cos^6 x - 48 cos^4 x + 18 cos^2 x - 1

• Ans: L.H.S. = cos 3(2x) = 4 cos^3 2x - 3 cos 2x.
• Substituting cos 2x = 2 cos^2 x - 1.
• = 4(2 cos^2 x - 1)^3 - 3(2 cos^2 x - 1).
• Expanding gives 32 cos^6 x - 48 cos^4 x + 18 cos^2 x - 1 = R.H.S..

FAQs – Class 11 Maths Chapter 3 Exercise 3.3 Trigonometric Functions

Q1. What is the focus of Exercise 3.3?
Exercise 3.3 focuses on simplifying trigonometric expressions using identities and solving trigonometric equations. It challenges students to manipulate and combine different trigonometric identities for a variety of complex problems.

Q2. How are trigonometric equations solved in this exercise?
To solve trigonometric equations:

• Use trigonometric identities to simplify the equation.

• Express the equation in terms of one trigonometric function if possible.

• Solve for the unknown angle using known values of trigonometric functions.

Q3. What are some common trigonometric equations solved in this exercise?

• Basic trigonometric equations like

  $\sin x = 0.5$sinx=0.5 or

  $\cos x = -1$cosx=−1.

• Equations involving multiple trigonometric functions.

• Solving equations for specific angles using identities like the Pythagorean identities or sum and difference formulas.

Q4. Why is Exercise 3.3 important for board exams?
This exercise is essential for mastering the process of simplifying trigonometric expressions and solving trigonometric equations, which are commonly tested in Class 11 board exams. These concepts are foundational for topics in calculus and advanced trigonometry.

Q5. How can students prepare effectively for Exercise 3.3?
Students should:

• Review and memorize all trigonometric identities.

• Practice simplifying complex expressions using identities.

• Solve various trigonometric equations to become comfortable with different types of problems.

Q6. What are the types of problems students should focus on?
Students should focus on problems that involve:

• Simplification of trigonometric expressions using multiple identities.

• Solving trigonometric equations involving different angles.

• Applying inverse trigonometric functions to solve for angles.