NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 3 – Trigonometric Functions

NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 3 – Trigonometric Functions provide a comprehensive set of problems designed to test and reinforce the concepts related to trigonometric ratios, identities, equations, and their applications. This exercise combines various aspects of trigonometry, helping students gain a deeper understanding of the subject and develop problem-solving skills.

The Miscellaneous Exercise covers topics like simplifying trigonometric expressions, solving trigonometric equations, and applying trigonometric identities in real-life scenarios. It also provides practice on angle transformations, sum and difference identities, and solving for unknown angles using trigonometric functions.

NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 3 – Trigonometric Functions

Trigonometric Functions

Medium

Q.

Find the radian measures corresponding to the following degree measures:
(i) 25°
(ii) – 47°30′
(iii) 240°
(iv) 520°

View Solution

Trigonometric Functions

Difficult

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

In any triangle ABC, if a = 18, b = 24, c = 30, find

1. cos A, cos B, cos C
2. sin A, sin B, sin C

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

For any triangle ABC, prove that a (b cos C – c cos B) = b² – c²

View Solution

Trigonometric Functions

Difficult

Q.

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

View Solution

Trigonometric Functions

Difficult

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Difficult

Q.

For any triangle ABC, prove that a cos A + b cos B + c cos C = 2 a sin B sin C

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Difficult

Q.

For any triangle ABC, prove that
(b² – c²) cot A + (c² – a²) cot B + (a² – b²) cot C = 0

View Solution

Trigonometric Functions

Difficult

Q.

$\begin{array}{l}\text{For any triangle ABC},\text{ prove that}\\ \frac{{\mathrm{b}}^2-{\mathrm{c}}^2}{{\mathrm{a}}^2}\sin 2\mathrm{A}+\frac{{\mathrm{c}}^2-{\mathrm{a}}^2}{{\mathrm{b}}^2}\sin 2\mathrm{B}+\frac{{\mathrm{a}}^2-{\mathrm{b}}^2}{{\mathrm{c}}^2}\sin 2\mathrm{C}=0\end{array}$

View Solution

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.

View Solution

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.

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

 $\begin{array}{l}{Find\hspace{0.33em}the\hspace{0.33em}degree\hspace{0.33em}measures\hspace{0.33em}corresponding\hspace{0.33em}to\hspace{0.33em}the\hspace{0.33em}following}\\ {radian\hspace{0.33em}measures\hspace{0.33em}}\quad \left({Use\hspace{0.33em}π}=\frac{22}{7}\right).\\ \left({i}\right){\hspace{0.33em}}\frac{11}{16}\\ \left({ii}\right){\hspace{0.33em}}-4\\ \left({iii}\right){\hspace{0.33em}}\frac{5{π}}{3}\\ \left({iv}\right){\hspace{0.33em}}\frac{7{π}}{6}\end{array}$ 

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

View Solution

Trigonometric Functions

Medium

Q.

 $\begin{array}{l}\mathrm{Find}\quad \mathrm{the}\quad \mathrm{degree}\quad \mathrm{measure}\quad \mathrm{of}\quad \mathrm{the}\quad \mathrm{angle}\quad \mathrm{subtended}\quad \mathrm{at}\quad \mathrm{the}\\ \mathrm{centre}\quad \mathrm{of}\quad \mathrm{a}\quad \mathrm{circle}\quad \mathrm{of}\quad \mathrm{radius}\quad 100\quad \mathrm{cm}\quad \mathrm{by}\quad \mathrm{an}\quad \mathrm{arc}\quad \mathrm{of}\quad \mathrm{length}\quad 22\quad \mathrm{cm}\\ (\mathrm{Use}\quad \mathrm{\pi }=\frac{22}{7}).\end{array}$ 

View Solution

Trigonometric Functions

Medium

Q.

In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.

View Solution

Trigonometric Functions

Medium

Q.

If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.

View Solution

Trigonometric Functions

Medium

Q.

Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length
(i) 10 cm
(ii) 15 cm
(iii) 21 cm

View Solution

Trigonometric Functions

Medium

Q.

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

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

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

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

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

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

View Solution

Trigonometric Functions

Medium

Q.

Find the values of the trigonometric functions in Exercises 6 to 10.

6. sin 765°
7. cosec (– 1410°)
$8. \tan \frac{19\mathrm{\pi }}{3}$

$9. \sin (-\frac{11\mathrm{\pi }}{3})$

$10. \cot (-\frac{15\mathrm{\pi }}{4})$

View Solution

Trigonometric Functions

Medium

Q.

Find the value of: (i) sin75°    (ii) tan15°

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

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

View Solution

Trigonometric Functions

Medium

Q.

$\mathrm{Find}\hspace{0.17em}\sin \frac{\mathrm{x}}{2}, \cos \frac{\mathrm{x}}{2} \mathrm{and} \tan \frac{\mathrm{x}}{2} \mathrm{in} \mathrm{each} \mathrm{of} \mathrm{the} \mathrm{following}:$

$8. \mathrm{t}\mathrm{an} \mathrm{x}=-\frac{4}{3}, \mathrm{x} \mathrm{in} \mathrm{quadrant} \mathrm{II}$

$\mathrm{9. }\mathrm{c}\mathrm{os} \mathrm{x}=-\frac{1}{3},\mathrm{x} \mathrm{in} \mathrm{quadrant} \mathrm{III}$

$10.\mathrm{sinx}=\frac{1}{4},\mathrm{x}\text{\hspace{0.33em}in\hspace{0.33em}quadrant\hspace{0.33em}II }$

View Solution

1. Prove that: 2 cos(pi/13) cos(9pi/13) + cos(3pi/13) + cos(5pi/13) = 0

• Formula used: cos x + cos y = 2 cos((x+y)/2) cos((x-y)/2).
• Step 1: Group the last two terms: 2 cos(pi/13) cos(9pi/13) + [2 cos(4pi/13) cos(pi/13)].
• Step 2: Take 2 cos(pi/13) common: 2 cos(pi/13) [cos(9pi/13) + cos(4pi/13)].
• Step 3: Apply formula again: 2 cos(pi/13) [2 cos(pi/2) cos(5pi/26)].
• Final Step: Since cos(pi/2) = 0, the entire expression becomes 0.

2. Prove that: (sin 3x + sin x) sin x + (cos 3x - cos x) cos x = 0

• Step 1: Expand the brackets: sin 3x sin x + sin^2 x + cos 3x cos x - cos^2 x.
• Step 2: Rearrange: (cos 3x cos x + sin 3x sin x) - (cos^2 x - sin^2 x).
• Step 3: Use identity cos(A-B) and cos 2x: cos(3x - x) - cos 2x.
• Final Step: cos 2x - cos 2x = 0.

3. Prove that: (cos x + cos y)^2 + (sin x - sin y)^2 = 4 cos^2((x+y)/2)

• Step 1: Expand L.H.S: cos^2 x + cos^2 y + 2 cos x cos y + sin^2 x + sin^2 y - 2 sin x sin y.
• Step 2: Group terms: (cos^2 x + sin^2 x) + (cos^2 y + sin^2 y) + 2(cos x cos y - sin x sin y).
• Step 3: Simplify: 1 + 1 + 2 cos(x+y) = 2[1 + cos(x+y)].
• Final Step: Use 1 + cos A = 2 cos^2(A/2): 2[2 cos^2((x+y)/2)] = 4 cos^2((x+y)/2).

4. Prove that: (cos x - cos y)^2 + (sin x - sin y)^2 = 4 sin^2((x-y)/2)

• Step 1: Expand L.H.S: cos^2 x + cos^2 y - 2 cos x cos y + sin^2 x + sin^2 y - 2 sin x sin y.
• Step 2: Group terms: 1 + 1 - 2[cos x cos y + sin x sin y] = 2[1 - cos(x-y)].
• Final Step: Use 1 - cos A = 2 sin^2(A/2): 2[2 sin^2((x-y)/2)] = 4 sin^2((x-y)/2).

5. Prove that: sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

• Step 1: Pair terms: (sin x + sin 5x) + (sin 3x + sin 7x).
• Step 2: Apply sin A + sin B formula: 2 sin 3x cos 2x + 2 sin 5x cos 2x.
• Step 3: Take 2 cos 2x common: 2 cos 2x [sin 3x + sin 5x].
• Final Step: Apply formula again: 2 cos 2x [2 sin 4x cos x] = 4 cos x cos 2x sin 4x.

6. Prove that: [(sin 7x + sin 5x) + (sin 9x + sin 3x)] / [(cos 7x + cos 5x) + (cos 9x + cos 3x)] = tan 6x

• Step 1: Use sum to product formulas in numerator and denominator: [2 sin 6x cos x + 2 sin 6x cos 3x] / [2 cos 6x cos x + 2 cos 6x cos 3x].
• Step 2: Factor out common terms: 2 sin 6x [cos x + cos 3x] / 2 cos 6x [cos x + cos 3x].
• Final Step: sin 6x / cos 6x = tan 6x.

7. Prove that: sin 3x + sin 2x - sin x = 4 sin x cos(x/2) cos(3x/2)

• Step 1: Group first and last terms: (sin 3x - sin x) + sin 2x.
• Step 2: Apply formulas: [2 cos 2x sin x] + 2 sin x cos x.
• Step 3: Take 2 sin x common: 2 sin x [cos 2x + cos x].
• Final Step: Apply cos A + cos B: 2 sin x [2 cos(3x/2) cos(x/2)] = 4 sin x cos(x/2) cos(3x/2).

Finding sin(x/2), cos(x/2), and tan(x/2)

8. If tan x = -4/3, x lies in 2nd quadrant

• Quadrant analysis: x is in 2nd quadrant, so x/2 lies in 1st quadrant (positive values for all).
• Find cos x: sec^2 x = 1 + (-4/3)^2 = 25/9. Since x is in 2nd quadrant, sec x = -5/3 and cos x = -3/5.
• Results:
  • cos(x/2) = sqrt((cos x + 1)/2) = 1/sqrt(5) or sqrt(5)/5.
  • sin(x/2) = 2/sqrt(5) or 2sqrt(5)/5.
  • tan(x/2) = 2.

9. If cos x = -1/3, x lies in 3rd quadrant

• Quadrant analysis: x is in 3rd quadrant, so x/2 lies in 2nd quadrant (sin is positive, cos and tan are negative).
• Results:
  • cos(x/2) = -1/sqrt(3).
  • sin(x/2) = sqrt(2/3).
  • tan(x/2) = -sqrt(2).

10. If sin x = 1/4, x lies in 2nd quadrant

• Find cos x: cos^2 x = 1 - (1/4)^2 = 15/16. In 2nd quadrant, cos x = -sqrt(15)/4.
• Quadrant analysis: x is in 2nd quadrant, so x/2 is in 1st quadrant (all positive).
• Results:
  • sin(x/2) = sqrt((4 + sqrt(15))/8).
  • cos(x/2) = sqrt((4 - sqrt(15))/8).
  • tan(x/2) = 4 + sqrt(15).

FAQs – Class 11 Maths Miscellaneous Exercise Chapter 3 Trigonometric Functions

Q1. What is the focus of the Miscellaneous Exercise in Chapter 3?
The Miscellaneous Exercise in Chapter 3 focuses on applying the concepts of trigonometric functions, trigonometric identities, and solving trigonometric equations in various situations. It provides students with a variety of problems that test their understanding of the chapter.

Q2. What types of problems are included in this exercise?
The exercise includes problems that:

• Require simplifying complex trigonometric expressions using identities.

• Involve solving trigonometric equations for specific angles.

• Use sum and difference formulas to solve problems involving two angles.

• Test applications of identities like Pythagorean identity, reciprocal identities, and quotient identities.

Q3. Why is this exercise important for Class 11 exams?
This exercise is crucial as it consolidates all the trigonometric concepts covered in the chapter. It helps students practice the application of trigonometric ratios and identities in various forms, which is essential for solving problems in exams. Many board exam questions come from this type of problem-solving, making it vital for exam preparation.

Q4. How can students prepare effectively for this exercise?

• Review the basic trigonometric identities and understand how to apply them in solving problems.

• Practice simplifying trigonometric expressions and solving equations.

• Work through each type of problem carefully to build confidence in using trigonometric functions.

• Solve previous years’ exam questions related to this chapter to familiarize yourself with common problem types.

Q5. How do trigonometric functions help in solving real-life problems?
Trigonometric functions are widely used in various fields, including physics, engineering, architecture, and navigation. They help in solving problems involving angles of elevation and depression, wave motion, oscillations, and calculating distances in surveying and navigation.