NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

Komal MiglaniUpdated on 19 Aug 2025, 03:21 PM IST

Have you ever wondered how engineers design huge buildings, how sailors and pilots navigate their journey, or how shadows change their length throughout the day? You can find all these answers in trigonometry, a fascinating branch of mathematics. In class 11 Maths NCERT chapter 3, trigonometric functions, contains the advanced concepts of trigonometry like radian measure, relation between degree and radian, sign of trigonometric functions, graphs of trigonometric functions, domain and range of trigonometric functions, and trigonometric functions of the sum and difference of two angles. Understanding these concepts will make students more efficient in solving problems involving height, distance, and angles. NCERT solutions of class 11 of trigonometric functions will also build a strong foundation for more advanced trigonometric concepts, which have many practical, real-life applications like construction, navigation, architecture, etc.

This Story also Contains

  1. NCERT Solution for Class 11 Maths Chapter 3 Solutions: Download PDF
  2. Trigonometric Functions Class 11 NCERT Solutions (Exercises)
  3. Class 11 Maths NCERT Chapter 3: Extra Question
  4. Trigonometric Functions Class 11 Chapter 3: Topics
  5. Trigonometric Functions Class 11 Solutions: Important Formulae
  6. Approach to Solve Questions of Trigonometric Functions Class 11
  7. What Extra Should Students Study Beyond NCERT for JEE?
  8. NCERT Solutions for Class 11 Mathematics - Chapter Wise
NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions
Trigonometric Functions

This article on NCERT solutions for class 11 Maths Chapter 3 Trigonometric Functions offers clear and step-by-step solutions for the exercise problems in the NCERT Class 11 Maths Book. Students requiring trigonometric functions class 11 solutions will find this article useful. It covers all the important Class 11 Maths Chapter 3 question answers and concepts of trigonometric functions. According to the latest NCERT syllabus, the experienced Subject Matter Experts of Careers360 have curated these trigonometric functions class 11 NCERT solutions, ensuring that students can grasp the basic concepts effectively. For syllabus, notes, and PDF, refer to NCERT.

NCERT Solution for Class 11 Maths Chapter 3 Solutions: Download PDF

Careers360 provides NCERT Solutions for Class 11 Maths Chapter 3, curated by experts to make maths easy to understand and solve. A PDF version is available for students as well.

Download PDF

Trigonometric Functions Class 11 NCERT Solutions (Exercises)

NCERT Trigonometric Functions Class 11 Solutions: Exercise: 3.1

Page Number: 48-49

Total Questions: 7

Question 1: Find the radian measures corresponding to the following degree measures:

(i) $25 ^\circ$
(ii) $-47 ^\circ$ $30'$
(iii) $240^\circ$
(iv) $520^\circ$

Answer:

It is solved using the relation between degree and radian
(i) $25^\circ$
We know that $180^\circ$ = $\pi$ radian
So, $1^\circ = \frac{\pi }{180}$ radian
$25^\circ = \frac{\pi }{180}\times 25$ radian $=\frac{5\pi }{36}$ radian

(ii) $-47^\circ30'$
We know that
$-47^\circ30' = -47\frac{1}{2}^\circ = -\frac{95}{2}^\circ$
Now, we know that $180^\circ = \pi \Rightarrow 1^\circ = \frac{\pi}{180}$ radian
So, $-\frac{95}{2}^\circ = \frac{\pi}{180}\times \left (-\frac{95}{2} \right )$ radian $= \frac{-19\pi}{72}$ radian

(iii) $240^\circ$
We know that
$180^\circ = \pi \Rightarrow 1^\circ = \frac{\pi}{180}$ radian
So, $240^\circ = \frac{\pi}{180}\times 240 = \frac{4\pi}{3}$ radian

(iv) $520^\circ$
We know that
$180^\circ = \pi \Rightarrow 1^\circ = \frac{\pi}{180}$ radian
So, $520^\circ = \frac{\pi}{180}\times 520$ radian $= \frac{26\pi}{9}$ radian

Question 2: Find the degree measures corresponding to the following radian measures. (Use $\small \pi =\frac{22}{7}$)

$\small (i) \frac{11}{16}$
$\small (ii) -4$
$\small (iii) \frac{5\pi }{3}$
$\small (iv) \frac{7\pi }{6}$

Answer:
(i) $\frac{11}{16}$
We know that
$\pi$ radian $= 180^\circ \Rightarrow 1\ radian = \frac{180}{\pi}^\circ$
So, $\frac{11}{16}\ radian = \frac{180}{\pi}\times \frac{11}{16}^\circ$ (we need to take $\pi = \frac{22}{7}$ )
$\frac{11}{16}\ radian = \frac{180\times 7}{22}\times \frac{11}{16}^\circ = \frac{315}{8}^\circ$
(we use $1^\circ = 60'$ and 1' = 60'')
Here, 1' represents 1 minute and 60" represents 60 seconds
Now,
$\frac{315}{8}^\circ$
$=39\frac{3}{8}^\circ$
$ =39^\circ +\frac{3\times 60}{8}'$
$ = 39^\circ +22' + \frac{1}{2}'$
$ = 39^\circ +22' +30''= \frac{315}{8}^\circ = 39^\circ22'30''$

(ii) $-4$
We know that
$\pi$ radian $= 180^\circ \Rightarrow 1\ radian = \frac{180}{\pi}^\circ$ (we need to take $\pi = \frac{22}{7}$ )
So, $-4\ radian = \frac{-4\times 180}{\pi}= \frac{-4\times 180\times 7}{22} = -\frac{2520}{11}^\circ$
(we use $1^\circ = 60'$ and 1' = 60'')
$\Rightarrow \frac{-2520}{11}^\circ$
$= -229\frac{1}{11}^\circ $
$=-229^\circ + \frac{1\times 60}{11}' $
$= -229^\circ + 5' + \frac{5}{11}' $
$= -229^\circ +5' +27''= -229^\circ5'27''$

(iii) $\frac{5\pi}{3}$
We know that
$\pi$ radian $= 180^\circ \Rightarrow 1\ radian = \frac{180}{\pi} ^\circ$ (we need to take $\pi = \frac{22}{7}$ )
So, $\frac{5\pi}{3}\ radian = \frac{180}{\pi}\times \frac{5\pi}{3}^\circ = 300^\circ$

(iv) $\frac{7\pi}{6}$
We know that
$\pi$ radian $= 180^\circ \Rightarrow 1\ radian = \frac{180}{\pi} ^\circ$ (we need to take $\pi = \frac{22}{7}$ )
So, $\frac{7\pi}{6}\ radian = \frac{180}{\pi}\times \frac{7\pi}{6} = 210^\circ$

Question 3: A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

Answer:
Number of revolutions made by the wheel in 1 minute = 360
$\therefore$ Number of revolutions made by the wheel in 1 second = $\frac{360}{60} = 6$
($\because$ 1 minute = 60 seconds)
In one revolution, the wheel will cover $2\pi$ radians
So, in 6 revolutions it will cover = $6\times 2\pi = 12\pi$ radian
$\therefore$ In 1, the second wheel will turn $12\pi$ radians.

Question 4: Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm (use $\small \pi =\frac{22}{7}$ )

Answer:
We know that
$l = r\theta$ ( where $l$ is the length of the arc, $r$ is the radius of the circle and $\theta$ is the angle subtended)
here $r$ = 100 cm
and $l$ = 22 cm
Now,
$\theta = \frac{l}{r} = \frac{22}{100}$ radian
We know that
$\pi\ radian = 180^\circ$
So, 1 radian = $\frac{180}{\pi}^\circ$
$\therefore \frac{22}{100}\ radian = \frac{180}{\pi}\times\frac{22}{100}^\circ= \frac{180\times7}{22}\times\frac{22}{100} = \frac{63}{5}^\circ$
So, $\frac{63}{5}^\circ = 12\frac{3}{5}^\circ = 12^\circ + \frac{3\times60}{5}' = 12^\circ + 36' = 12 ^\circ36'$
Thus, the angle subtended at the centre of a circle $\theta = 12^\circ36'$.

Question 5: In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.

Answer:
Given: radius ($r$) of circle = $\frac{Diameter}{2} = \frac{40}{2} = 20$ cm
length of chord = 20 cm
We know that
$\theta = \frac{l}{r}$ ($r$ = 20 cm , $l$ = ? , $\theta$ = ?)
Now,
1654678632225
AB is the chord of length 20 cm, and OA and OB are radii of the circle, i.e. 20 cm each
The angle subtended by OA and OB at the centre = $\theta$
$\because$ OA = OB = AB
$\therefore$ $\Delta$ OAB is equilateral triangle
So, each angle is $60^\circ$
$\therefore$ $\theta = 60^\circ$ $= \frac{\pi}{3}$ radian
Now, we have $\theta$ and $r$
So, $l = r\theta = 20\times\frac{\pi}{3}=\frac{20\pi}{3}$
$\therefore$ the length of the minor arc of the chord ($l$) = $\frac{20\pi}{3}$ cm.

Question 6: If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.

Answer:
Given: $\theta_1 = 60^\circ, \theta_2 = 75^\circ\\$ and $l_1 = l_2$
We need to find the ratio of their radii $\frac{r_1}{r_2}$
We know that arc length $l = r \theta$
So, $l_1 = r_1 \theta_1$ and $l_2 = r_2\theta_2$
Now, $l_1 = l_2$
So, $\frac{ r_1 }{ r_2}= \frac{\theta_2}{\theta_1} = \frac {75}{60} = \frac{5}{4}$, which is the ratio of their radii.

Question 7: 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

Answer:
(i) We know that
$l = r \theta$
Now,
$r$ = 75 cm
$l$ = 10 cm
So,
$\theta = \frac{l}{r} = \frac{10}{75} = \frac{2}{15}$ radian

(ii) We know that
$l = r \theta$
Now,
$r$ = 75 cm
$l$ = 15 cm
So,
$\theta = \frac{l}{r} = \frac{15}{75} = \frac{1}{5}$ radian

(iii) We know that
$l = r \theta$
Now,
$r$ = 75 cm
$l$ = 21 cm
So,
$\theta = \frac{l}{r} = \frac{21}{75} = \frac{7}{25}$ radian

NCERT Trigonometric Functions Class 11 Solutions: Exercise: 3.2

Page Number: 57

Total Questions: 10

Question 1: Find the values of other five trigonometric functions as $\small \cos x = -\frac{1}{2}$ , $x$ lies in third quadrant.

Answer:$\cos x = -\frac {1}{2}$
$\because \sec x = \frac{1}{\cos x} = \frac{1}{-\frac {1}{2}} = -2$
$x$ lies in III quadrants. Therefore, sec x is negative
$\sin ^{2}x +\cos^{2}x = 1 ⇒ \sin^{2}x = 1 - \cos^{2}x$
$⇒ \sin^{2}x = 1 -\left ( -\frac{1}{2} \right )^{2}$
$⇒ \sin^{2}x = 1 - \frac{1}{4} = \frac{3}{4}$
$⇒ \sin x = \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}$
$x$ lies in III quadrants. Therefore, sin x is negative
$\therefore \sin x= - \frac{\sqrt{3}}{2}$
$\because cosec \ x = \frac {1}{\sin x}= \frac{1}{- \frac{\sqrt{3}}{2}} =- \frac{2}{\sqrt{3}}$
$x$ lies in III quadrants. Therefore, cosec x is negative
$\tan x = \frac{\sin x}{\cos x} = \frac {-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \sqrt{3}$
$x$ lies in III quadrants. Therefore, tan x is positive
$\cot x = \frac{1}{\tan x} = \frac{1}{\sqrt{3}}$
$x$ lies in III quadrants. Therefore, cot x is positive.

Question 2: Find the values of other five trigonometric functions as $\small \sin x = \frac{3}{5}$, $x$ lies in second quadrant.

Answer: $\sin x = \frac {3}{5}$
$cosec \ x = \frac{1}{\sin x}=\frac {1}{\frac {3}{5}} = \frac {5}{3}$
$x$ lies in the second quadrant. Therefore, cosec x is positive
$\sin^{2}x + \cos ^{2}x = 1$
$⇒ \cos ^{2}x = 1 - \sin ^{2}x$
$⇒ \cos ^{2}x = 1 - \left ( \frac{3}{5} \right )^{2}$
$⇒ \cos ^{2}x = 1 - \frac {9}{25} = \frac {16}{25}$
$⇒ \cos x = \sqrt{\frac {16}{25}} = \pm \frac {4}{5}$
$x$ lies in the second quadrant. Therefore, cos x is negative
$\therefore \cos x = - \frac {4}{5}$
$\sec x = \frac {1}{\cos x} = \frac{1}{- \frac {4}{5}} = -\frac {5}{4}$
$x$ lies in the second quadrant. Therefore, sec x is negative
$\tan x = \frac {\sin x}{\cos x} = \frac {\frac{3}{5}}{-\frac{4}{5}} = -\frac {3}{4}$
$x$ lies in the second quadrant. Therefore, tan x is negative
$\cot x = \frac {1}{\tan x} = \frac {1}{-\frac {3}{4}} = -\frac{4}{3}$
$x$ lies in the second quadrant. Therefore, cot x is negative.

Question 3: Find the values of other five trigonometric functions as $\small \cot x = \frac{3}{4}$, $x$ lies in third quadrant.

Answer:
$\cot x= \frac {3}{4}$
$\tan x = \frac{1}{\cot x}= \frac{1}{\frac {3}{4}} = \frac {4}{3}$
$1 + \tan ^ {2}x = \sec ^{2}x$
$⇒ 1+\frac{4^2}{3^2} = \sec ^{2}x$
$⇒ 1 + \frac {16}{9} = \sec ^{2}x$
$⇒ \frac {25}{9} = \sec ^{2}x$
$⇒ \sec x = \sqrt {\frac {25}{9}} = \pm \frac {5}{3}$
$x$ lies in x lies in third quadrant. therefore sec x is negative
$\sec x = -\frac{5}{3}$
$\cos x = \frac {1}{\sec x} = \frac {1}{-\frac{5}{3}} = -\frac{3}{5}$
$\sin ^{2 }x+ \cos ^{2}x = 1⇒ \sin ^{2 }x = 1 - \cos ^{2}x$
$⇒ \sin ^{2 }x = 1 -\left ( -\frac{3}{5} \right )^{2}$
$⇒ \sin ^{2 }x = 1 - \frac {9}{25}$
$⇒ \sin ^{2 }x = \frac{16}{25}$
$⇒ \sin x = \sqrt {\frac{16}{25}} = \pm \frac{4}{5}$
$x$ lies in the third quadrant. Therefore, sin x is negative
$\sin x = -\frac {4}{5}$
$cosec \ x = \frac {1}{\csc} = \frac {1}{-\frac{4}{5}} = - \frac{5}{4}$.

Question 4: Find the values of other five trigonometric functions as $\small \sec x = \frac{13}{5}$, $x$ lies in fourth quadrant.

Answer:
$\sec x = \frac {13}{5}$
$\cos x = \frac {1}{\sec x} = \frac{1}{\frac {13}{5}} = \frac {5}{13}$
$\sin^{2}x + \cos^{2}x = 1⇒ \sin^{2}x = 1 - \cos^{2}x$
$⇒ \sin^{2}x = 1 - (\frac {5}{13})^2$
$⇒ \sin^{2}x = 1 - \frac {25}{169} = \frac {144}{169}$
$⇒ \sin x = \sqrt { \frac {144}{169}} = \pm \frac {12}{13}$
$x$ lies in fourth quadrant. Therefore, $\sin x$ is negative.
$\sin x =- \frac {12}{13}$
$\csc x = \frac {1}{\sin x} = \frac {1}{-\frac {12}{13}} = -\frac {13}{12}$
$\tan x = \frac {\sin x}{\cos x} = \frac {-\frac{12}{13}}{\frac{5}{13}} = -\frac {12}{5}$
$\cot x = \frac {1}{\tan x} = \frac {1}{-\frac{12}{5}} = -\frac{5}{12}$.

Question 5: Find the values of the other five trigonometric functions as $\small \tan x = -\frac{5}{12}$, $x$ lies in second quadrant.

Answer:
$\tan x = -\frac {5}{12}$
$\cot x = \frac {1}{\tan x} = \frac {1}{-\frac{5}{12}} = -\frac {12}{5}$
$1 + \tan^{2}x = \sec^{2}x$
$⇒ 1 + \left ( -\frac{5}{12} \right )^{2} = \sec^{2}x$
$⇒ 1 + \frac {25}{144} = \sec^{2}x$
$⇒ \frac {169}{144} = \sec^{2}x$
$⇒ \sec x = \sqrt {\frac {169}{144}} = \pm \frac {13}{12}$
$x$ lies in the second quadrant. Therefore, the value of $\sec x$ is negative
$\sec x = - \frac {13}{12}$
$\cos x = \frac{1}{\sec x}= \frac{1}{-\frac{13}{12}} = -\frac {12}{13}$
$\sin^{2}x + \cos^{2}x = 1$
$⇒ \sin^{2}x = 1 - \cos^{2}x$
$⇒ \sin^{2}x = 1 - \left ( -\frac{12}{13} \right )^{2}$
$⇒ \sin^{2}x = 1 - \frac{144}{169}⇒\sin^{2}x = \frac {25}{169}$
$⇒ \sin x = \sqrt {\frac{25}{169}} = \pm \frac{5}{13}$
$x$ lies in the second quadrant.
Therefore, the value of $\sin x$ is positive.
$\sin x = \frac {5}{13}$
$\csc = \frac {1}{\sin x} = \frac {1}{\frac {5}{13}} = \frac {13}{5}$.

Question 6: Find the values of the trigonometric functions $\small \sin 765^\circ$

Answer:
We know that values of $\sin x$ repeat after an interval of $2\pi$ or $360^\circ$
$\sin765^\circ = \sin (2\times360^\circ + 45^\circ ) = \sin45^\circ = \frac {1}{\sqrt{2}}$.

Question 7: Find the values of the trigonometric functions $\small cosec \ (-1410^\circ)$

Answer:
We know that value of $\operatorname{cosec} x$ repeats after an interval of $2\pi$ or $360^\circ$.
$\operatorname{cosec} (-1410^\circ) = \operatorname{cosec} (360^\circ\times4+(-1410^\circ))= \operatorname{cosec}\ 30^\circ = 2$

Question 8: Find the values of the trigonometric functions $\small \tan \frac{19\pi }{3}$

Answer:
We know that $\tan x$ repeats after an interval of $\pi$ or 180$^\circ$.
$\tan (\frac{19\pi}{3}) = \tan (6\pi+\frac{\pi}{3})= \tan \frac{\pi}{3} =\tan 60^\circ = \sqrt{3}$

Question 9: Find the values of the trigonometric functions $\sin\left ( -\frac{11\pi}{3} \right )$

Answer:
We know that $\sin x$ repeats after an interval of $2\pi$ or $360^\circ$
$\sin \left ( -\frac{11\pi}{3} \right ) = \sin \left (4\pi +(-\frac{11\pi}{3}) \right ) = \sin \frac{\pi}{3} = \frac {\sqrt{3}}{2}$

Question 10: Find the values of the trigonometric functions $\small \cot \left ( -\frac{15\pi }{4} \right )$

Answer:
We know that $\cot x$ repeats after an interval of $\pi$ or 180$^\circ$.
$\cot \left ( -\frac{15\pi}{4} \right ) = \cot \left (4\pi +(-\frac {15\pi}{4}) \right ) = \cot \left ( \frac{\pi}{4} \right ) = 1$

NCERT Trigonometric Functions Class 11 Solutions: Exercise: 3.3

Page Number: 67-68

Total Questions: 25

Question 1: Prove that $\small \sin ^{2} \left ( \frac{\pi }{6} \right ) + \cos ^{2}\left ( \frac{\pi }{3} \right ) - \tan ^{2}\left ( \frac{\pi }{4} \right ) = -\frac{1}{2}$

Answer:
We know the values of sin 30°, cos 60°, and tan 45°.
$\sin \left ( \frac{\pi}{6} \right ) = \left ( \frac{1}{2} \right ), \\ \cos \left ( \frac{\pi}{3} \right ) = \left ( \frac{1}{2} \right ), \\ \tan \left ( \frac{\pi}{4} \right ) = 1$
L.H.S. $=\sin^{2}\frac{\pi}{6}+\cos^{2}\frac{\pi}{3}-\tan^{2}\frac{\pi}{4}=$ $\left ( \frac{1}{2} \right )^{2}+ \left ( \frac {1}{2} \right )^{2}-1^{2}$
$= \frac{1}{4}+\frac{1}{4}-1= -\frac{1}{2} =$ R.H.S.

Question 2: Prove that $\small 2\sin ^{2}\left ( \frac{\pi }{6} \right ) + \operatorname{cosec} ^{2}\left ( \frac{7\pi }{6} \right )\cos ^{2}\frac{\pi }{3} = \frac{3}{2}$

Answer:
$\sin\frac{\pi}{6} = \frac {1}{2}, \operatorname{cosec} \frac{7\pi}{6} = \operatorname{cosec}\left ( \pi + \frac{\pi}{6} \right ) = - \operatorname{cosec} \frac{\pi}{6}=-2, \\ \cos \frac{\pi}{3} = \frac{1}{2}$
L.H.S. = $2\sin^{2}\frac{\pi}{6} + \operatorname{cosec}^{2}\frac{7\pi}{6}\cos^{2}\frac{\pi}{3} = 2\left ( \frac{1}{2} \right )^{2}+\left ( -2 \right )^{2}\left ( \frac{1}{2} \right )^{2}= 2\times\frac{1}{4} + 4\times\frac{1}{4} = \frac {1}{2} + 1= \frac{3}{2}$
= R.H.S.

Question 3: Prove that $\small \cot ^{2}\left ( \frac{\pi }{6} \right ) + \csc \left ( \frac{5\pi }{6} \right ) + 3\tan ^{2}\left ( \frac{\pi }{6} \right ) = 6$

Answer:
We know the values of cot 30°, tan 30°, and cosec 30°.
$\cot \frac{\pi}{6} = \sqrt{3}, \operatorname{cosec}\frac{5\pi}{6} = \operatorname{cosec}\left ( \pi - \frac{\pi}{6} \right )=\operatorname{cosec}\frac{\pi}{6} = 2, \tan\frac{\pi}{6}= \frac{1}{\sqrt{3}}$
L.H.S. $=\cot^{2}\frac{\pi}{6} + \operatorname{cosec}\frac{5\pi}{6} +3\tan^{2}\frac{\pi}{6} = \left ( \sqrt{3} \right )^{2} + 2 + 3\times\left ( \frac{1}{\sqrt{3}} \right )^{2}= 3+2+1 = 6=$ R.H.S.

Question 4: Prove that $\small 2\sin ^{2}\left ( \frac{3\pi }{4} \right ) + 2\cos ^{2}\left ( \frac{\pi }{4} \right ) + 2\sec ^{2}\left ( \frac{\pi }{3} \right ) = 10$

Answer:
$\sin \frac{3\pi}{4} = \sin\left ( \pi-\frac{\pi}{4} \right ) = \sin \frac{\pi}{4}= \frac{1}{\sqrt{2}},\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}, \sec\frac{\pi}{3}= 2$
Using the above values
L.H.S. $=2\sin^{2}\frac{3\pi}{4} +2\cos^{2}\frac{\pi}{4}+2\sec^{2}\frac{\pi}{3} = 2\times\left ( \frac{1}{\sqrt{2}} \right )^{2}+2\times\left ( \frac{1}{\sqrt{2}} \right )^{2}+2\left ( 2 \right )^{2}\\ \\ \Rightarrow 1+1+8=10=$ R.H.S.

Question 5(i): Find the value of $\small (i) \sin 75^\circ$

Answer:
$\sin 75^\circ = \sin(45^\circ + 30^\circ)$
We know that
$\sin(x+y)=\sin x\cos y + \cos x\sin y$
Using this identity
$\sin 75^\circ = \sin(45^\circ + 30^\circ) = \sin45^\circ\cos30^\circ + \cos45^\circ\sin30^\circ$
$= \frac{1}{\sqrt{2}}\times\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\times\frac{1}{2}\\ \\ \Rightarrow \frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}} = \frac{\sqrt{3}+1}{2\sqrt{2}}$

Question 5 (ii): Find the value of
$\small (ii) \tan 15^\circ$

Answer:
$\tan 15^\circ = \tan (45^\circ - 30^\circ)$
We know that,
$\left [ \tan(x-y)= \frac{\tan x - \tan y}{1+\tan x\tan y} \right ]$
By using this, we can write,
$\tan (45^\circ - 30^\circ)= \frac{\tan 45^\circ - tan30^\circ}{1+\tan45^\circ\tan30^\circ}= \frac{1-\frac{1}{\sqrt{3}}}{1+1\left ( \frac{1}{\sqrt{3}} \right )} = \frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1}$
$=\frac{\left ( \sqrt{3}-1 \right )^{2}}{\left ( \sqrt{3}+1 \right )\left ( \sqrt{3} -1\right )}=\frac{3+1-2\sqrt{3}}{\left ( \sqrt{3} \right )^{2}-\left ( 1 \right )^{2}}\\ \\ \Rightarrow \frac {4-2\sqrt{3}}{3-1}=\frac{2\left ( 2-\sqrt{3} \right )}{2}= 2-\sqrt{3}$

Question 6: Prove the following: $\small \cos \left ( \frac{\pi }{4}-x \right )\cos \left ( \frac{\pi }{4}-y \right ) - \sin \left ( \frac{\pi }{4} -x\right )\sin \left ( \frac{\pi }{4}-y \right ) =\sin (x+y)$

Answer:
$\cos\left ( \frac{\pi}{4}-x \right )\cos\left ( \frac{\pi}{4}-y \right ) - \sin\left ( \frac{\pi}{4}-x \right )\sin\left ( \frac{\pi}{4}-y \right )$
Multiply and divide by 2 both the cos and the sin functions
We get,
$\frac{1}{2}\left [2 \cos\left ( \frac{\pi}{4}-x \right )\cos\left ( \frac{\pi}{4}-y \right ) \right ] + \frac{1}{2}\left [- 2\sin\left ( \frac{\pi}{4}-x \right )\sin\left ( \frac{\pi}{4}-y \right ) \right ]$
Now, we know that
2cosAcosB = cos(A+B) + cos(A-B) --------(i)
-2sinAsinB = cos(A+B) - cos(A-B) ----------(ii)
We use these two identities
In our question A = $\left (\frac{\pi}{4}-x \right )$
B = $\left (\frac{\pi}{4}-y \right )$
So,
$\frac{1}{2}\left [ \cos \left \{ \left ( \frac{\pi}{4}-x \right) +\left ( \frac{\pi}{4}-y \right ) \right \} + \cos \left \{ \left ( \frac{\pi}{4}-x \right) -\left ( \frac{\pi}{4}-y \right ) \right \} \right ] +\\ \\ \frac{1}{2}\left [ \cos \left \{ \left ( \frac{\pi}{4}-x \right) +\left ( \frac{\pi}{4}-y \right ) \right \} - \cos \left \{ \left ( \frac{\pi}{4}-x \right) +\left ( \frac{\pi}{4}-y \right ) \right \} \right ]$
$= 2 \times \frac{1}{2} \left [ \cos \left \{ \left ( \frac{\pi}{4}-x \right )+\left ( \frac{\pi}{4}-y \right ) \right \} \right ]$
$= \cos \left [ \frac{\pi}{2}-(x+y) \right ]$
As we know that
$(\cos \left ( \frac{\pi}{2} - A \right ) = \sin A)$
By using this
$= \cos \left [ \frac{\pi}{2}-(x+y) \right ]$ $=\sin(x+y)=$ R.H.S.

Question 7: Prove the following $\small \frac{\tan \left ( \frac{\pi }{4}+x \right )}{\tan \left ( \frac{\pi }{4} -x\right )} = \left ( \frac{1+\tan x}{1-\tan x} \right )^{2}$

Answer:
As we know that
$(\tan (A +B ) = \frac {\tan A + \tan B}{1- \tan A\tan B})$ and $\tan (A-B) = \frac {\tan A - \tan B }{1+ \tan A \tan B}$
So, by using these identities
L.H.S. $=\frac{\tan \left ( \frac{\pi}{4}+x \right )}{\tan \left ( \frac{\pi}{4}-x \right )} = \frac{\frac{\tan \frac {\pi}{4} + \tan x}{1- \tan \frac{\pi}{4}\tan x}} {\frac{\tan \frac {\pi}{4} - \tan x}{1+ \tan \frac{\pi}{4}\tan x}} =\frac{ \frac {1+\tan x }{1- \tan x}} { \frac {1-\tan x }{1+ \tan x}} = \left ( \frac{1 + \tan x}{1 - \tan x} \right )^{2}=$ R.H.S.

Question 8: Prove the following $\small \frac{\cos (\pi +x)\cos (-x)}{\sin (\pi -x)\cos \left ( \frac{\pi }{2}+x \right )} = \cot ^{2} x$

Answer:
As we know,
$\cos(\pi+x) = -\cos x$ , $\sin (\pi - x ) = \sin x$ , $\cos \left ( \frac{\pi}{2} + x\right ) = - \sin x$ and $\cos (-x) = \cos x$
By using these, our equation simplifies to
$\frac{\cos x \times -\cos x}{\sin x \times - \sin x} = \frac{- \cos^{2}x}{-\sin^{2}x} = \cot ^ {2}x$ $(\because \cot x = \frac {\cos x}{\sin x})=$ R.H.S.

Question 9: Prove the following $\small \cos \left ( \frac{3\pi }{2} +x\right )\cos (2\pi +x)\left [ \cot \left ( \frac{3\pi }{2}-x \right ) +\cot (2\pi +x)\right ] = 1$

Answer:
We know that
$\cos \left ( \frac{3\pi}{2}+x \right ) = \sin x, \cos (2\pi +x)= \cos x, \cot\left ( \frac{3\pi}{2} -x\right ) = \tan x, \cot (2\pi + x) = \cot x$
So, by using these, our equation simplifies to
$\cos \left ( \frac{3\pi }{2} +x\right )\cos (2\pi +x)\left [ \cot \left ( \frac{3\pi }{2}-x \right ) +\cot (2\pi +x)\right ]$
$=\sin x\cos x [\tan x + \cot x] = \sin x\cos x [\frac {\sin x}{\cos x} + \frac{\cos x}{\sin x}]$
$= \sin x\cos x\left [ \frac{\sin^{2}x+\cos^{2}x}{\sin x\cos x } \right ] =\sin^{2}x+\cos^{2}x = 1=$ R.H.S.

Question 10: Prove the following $\small \sin (n+1)x\sin(n+2)x + \cos(n+1)x\cos(n+2)x =\cos x$

Answer:
Multiply and divide by 2
$= \frac {2\sin(n+1)x \sin(n+2)x + 2\cos (n+1)x\cos(n+2)x}{2}$
Now by using identities
–2sinAsinB = cos(A+B) – cos(A–B)
2cosAcosB = cos(A+B) + cos(A–B)
Now, $\frac{\left \{ -\left (\cos(2n+3)x - \cos (-x) \right ) + \left ( \cos(2n+3) +\cos(-x) \right )\right \}}{2}\\ \\ \left ( \because \cos(-x) = \cos x \right )$
$= \frac{2\cos x}{2} = \cos x=$ R.H.S.

Question 11: Prove the following $\small \cos \left ( \frac{3\pi }{4}+x \right ) - \cos\left ( \frac{3\pi }{4} -x\right ) = -\sqrt{2} \sin x$

Answer:
We know that
[ cos(A+B) - cos (A-B) = -2sinAsinB ]
By using this identity
$\cos \left ( \frac {3\pi}{4}+x \right ) - \cos \left ( \frac {3\pi}{4}-x \right ) = -2\sin\frac{3\pi}{4}\sin x = -2\times \frac{1}{\sqrt{2}}\sin x\\ \\ = -\sqrt{2}\sin x=$ R.H.S.

Question 12: Prove the following $\small \sin^{2}6x - \sin^{2}4x = \sin2x\sin10x$

Answer:
We know that
$a^{2} - b^{2} = (a+b)(a-b)$
So, $\sin^{2}6x - \sin^{2}4x =(\sin6x + \sin4x)(\sin6x - \sin4x)$
Now, we know that
$\sin A + \sin B = 2\sin \left ( \frac{A+B}{2} \right )\cos\left ( \frac{A-B}{2} \right ), \sin A - \sin B = 2\cos \left ( \frac{A+B}{2} \right )\sin\left ( \frac{A-B}{2} \right )$
By using these identities
sin6x + sin4x = 2sin5x cosx
sin6x - sin4x = 2cos5x sinx
$\Rightarrow \sin^{2}6x - \sin^{2}4x = (2\cos5x\sin5x)(2\sin x\cos x)$
Now,
2sinAcosB = sin(A+B) + sin(A-B)
2cosAsinB = sin(A+B) - sin(A-B)
by using these identities
2cos5x sin5x = sin10x - 0
2sinx cosx = sin2x + 0
So, $\sin^{2}6x-\sin^{2}4x = \sin2x\sin10x$

Question 13: Prove the following $\small \cos^{2}2x - \cos^{2}6x = \sin4x\sin8x$

Answer:
As we know that
$a^{2}-b^{2} =(a-b)(a+b)$
$\therefore \cos^{2}2x -\cos^{2}6x = (\cos2x-\cos6x)(\cos2x+\cos6x)$
Now, $\cos A - \cos B = -2\sin\left ( \frac{A+B}{2} \right )\sin\left ( \frac{A-B}{2} \right )\\ \\ \cos A + \cos B = 2\cos\left ( \frac{A+B}{2} \right )\cos\left ( \frac{A-B}{2} \right )$
By using these identities
cos2x - cos6x = -2sin(4x)sin(-2x) = 2sin4xsin2x ( $\because$ sin(-x) = -sin x and cos(-x) = cosx)
cos2x + cos 6x = 2cos4xcos(-2x) = 2cos4xcos2x
So our equation becomes
$(\cos2x-\cos6x)(\cos2x+\cos6x)=(2\sin4x\sin2x)(2\cos4x\cos2x)=(2\sin2x\cos2x)(2\sin4x\cos4x)$
$=\sin4x\sin8x=$ R.H.S.

Question 14: Prove the following $\small \sin2x +2\sin4x + \sin6x = 4\cos^{2}x\sin4x$

Answer:
We know that
$\sin A+ \sin B = 2\sin \left ( \frac{A+B}{2} \right )\cos\left ( \frac{A-B}{2} \right )$
We are using this identity
sin2x + 2sin4x + sin6x = (sin2x + sin6x) + 2sin4x
sin2x + sin6x = 2sin4xcos(-2x) = 2sin4xcos(2x) ( $\because$ cos(-x) = cos x)
So, our equation becomes
sin2x + 2sin4x + sin6x = 2sin4xcos(2x) + 2sin4x
Now, take the 2sin4x common
sin2x + 2sin4x + sin6x = 2sin4x(cos2x +1) ( $\because \cos2x = 2\cos^{2}x - 1$ )
= $2\sin4x( 2\cos^{2}x - 1$ +1 )
= $2\sin4x( 2\cos^{2}x$ )
= $4\sin4x\cos^{2}x=$ R.H.S.

Question 15: Prove the following $\small \cot4x(\sin5x + \sin3x) = \cot x(\sin5x - \sin3x)$

Answer:
We know that
$\sin x + \sin y = 2\sin\left ( \frac{x+y}{2} \right )\cos\left (\frac{x-y}{2} \right )$
By using this, we get
sin5x + sin3x = 2sin4xcosx
$\frac{\cos4x}{\sin4x}\left ( 2\sin4x\cos x \right ) = 2\cos4x\cos x\\ \\$
Now multiply and divide by sin x
$\\\ \\ \frac{2\cos4x\cos x \sin x}{\sin x} =\cot x (2\cos4x\sin x) \left ( \because \frac{\cos x}{\ sin x} = \cot x \right )\\ \\$
Now we know that
$\\ 2\cos x\sin y = \sin(x+y) - \sin(x-y)\\ \\$
By using this, our equation becomes
$=\cot x (\sin5x - \sin3x)=$ R.H.S.

Question 16: Prove the following $\small \frac{\cos 9x - \cos 5x}{\sin17x - \sin3x} = -\frac{\sin2x}{\cos10x}$

Answer:
As we know that
$\\ \cos x - \cos y = -2\sin\frac{x+y}{2}\sin\frac{x-y}{2 }, \cos 9x - \cos 5x = -2\sin 7x \sin2x$
$\sin x - \sin y = 2\cos\frac{x+y}{2}\sin\frac{x-y}{2 }, \sin 17x - \sin 3x = 2\cos10x \sin7x$
Now, $\frac{\cos 9x - \cos 5x}{\sin 17x - \sin 3x} =\frac{-2\sin 7x \sin2x}{2\cos10x \sin7x} = -\frac{\sin 2x}{\cos10x}=$ R.H.S.

Question 17: Prove the following $\small \frac{\sin5x + \sin3x}{\cos5x + \cos3x} = \tan4x$

Answer:
We know that
$\\ \sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$ and $\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} \\$
We use these identities
$\\ \sin5x + \sin3x = 2\sin4x\cos x, \cos5 x + \cos 3x = 2\cos4x\cos x$
Now, $\frac{\sin5x + \sin3x}{\cos5 x + \cos 3x} = \frac{ 2\sin4x\cos x}{2\cos4x\cos x} = \frac{\sin4x}{\cos 4x} = \tan 4x=$ R.H.S.

Question 18: Prove the following $\small \frac{\sin x - \sin y}{\cos x+\cos y} = \tan \frac{(x-y)}{2}$

Answer:
We know that
$\sin x - \sin y = 2\cos\frac{x+y}{2 }\sin\frac{x-y}{2}$ and $\cos x +\cos y = 2\cos\frac{x+y}{2 }\cos\frac{x-y}{2}\\$
We use these identities
$\frac{\sin x - \sin y}{\cos x +\cos y} =\frac{2\cos\frac{x+y}{2 }\sin\frac{x-y}{2}}{ 2\cos\frac{x+y}{2 }\cos\frac{x-y}{2}} = \frac{\sin\frac{x-y}{2}}{\cos\frac{x-y}{2}} = \tan \frac{x-y}{2}=$ R.H.S.

Question 19: Prove the following $\small \frac{\sin x + \sin 3x}{\cos x + \cos3x} = \tan2x$

Answer:
We know that
$\sin x + \sin y = 2\sin\frac{x+y}{2}\cos\frac{x-y}{2}$
$\cos x + \cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2}$
We use these equations,
$\sin x + \sin3x = 2\sin2x\cos(-x) = 2\sin2x\cos x (\because \cos(-x) = \cos x)$
$\cos x + \cos3x = 2\cos2x\cos(-x) =2\cos2x\cos x (\because \cos(-x) = \cos x)$
Now, $\frac{\sin x + \sin3x}{\cos x + \cos3x} = \frac {2\sin2x\cos x}{2\cos2x\cos x}= \frac{\sin2x}{\cos2x} = \tan2x=$ R.H.S.

Question 20: Prove the following $\small \frac{\sin x - \sin 3x}{\sin^{2}x-\cos^{2}x} = 2\sin x$

Answer:
We know that
$\sin3x = 3\sin x - 4\sin^{3}x, \cos^{2}x-\sin^{2}x = \cos2x$ and $\cos2x = 1 - 2\sin^{2}x$
We use these identities
$\sin x - \sin3x = \sin x - (3\sin x - 4\sin^{3}x) = 4\sin^{3}x - 2\sin x = 2\sin x (2\sin^{2}x - 1)$
$\sin^{2}x-\cos^2x = \sin^{2}x-(1-\sin^2x) =2\sin^{2}x-1$
Now, $\frac{\sin x - \sin3x}{\sin^{2}-\cos^{2}x } = \frac{ 2\sin x (2\sin^{2}x - 1)}{ 2\sin^{2}x - 1} = 2\sin x=$ R.H.S.

Question 21: Prove the following $\small \frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x$

Answer:
We know that
$\cos x + \cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2}$ and $\sin x + \sin y = 2\sin\frac{x+y}{2}\cos\frac{x-y}{2}$
We use these identities
$\frac{(\cos4x + \cos2x) + \cos3x}{(\sin4x+\sin2x)+\sin3x} = \frac{2\cos3x\cos x + \cos3x}{2\sin3x\cos x+\sin3x} = \frac{2\cos3x(\cos x+1)}{2\sin3x(\cos x+1)}=\cot 3x=$ R.H.S.

Question 22: prove the following $\small \cot x \cot2x - \cot2x\cot3x - \cot3x\cot x =1$

Answer:
L.H.S.
= $\cot x \cot2x - \cot3x(\cot2x - \cot x)$
Now we can write $\cot3x = \cot(2x + x)$
⇒ $\cot(a+b) = \frac{\cot a \cot b - 1}{\cot a + \cot b}$
So, $\cot x \cot2x-\frac{\cot 2x \cot x - 1}{\cot 2x + \cot x}(\cot2x+\cot x)$
= $\cot x \cot2x - (\cot2x\cot x -1)$
= $\cot x \cot2x - \cot2x\cot x +1= 1 =$ R.H.S.

Question 23: Prove that $\small \tan4x = \frac{4\tan x(1-\tan^{2}x)}{1-6 \tan^{2}x+\tan^{4}x}$

Answer:
We know that
$\tan2A=\frac{2\tan A}{1 - \tan^{2}A}$
and we can write tan 4x = tan 2(2x)
So, $\tan4x=\frac{2\tan 2x}{1 - \tan^{2}2x}$ = $\frac{2( \frac{2\tan x}{1 - \tan^{2}x})}{1 - (\frac{2\tan x}{1 - \tan^{2}x})^{2}}$
= $\frac{2 (2\tan x)(1 - \tan^{2}x)}{(1-\tan x)^{2} - (4\tan^{2} x)}$
= $\frac{(4\tan x)(1 - \tan^{2}x)}{(1)^{2}+(\tan^{2} x)^{2} - 2 \tan^{2} x - (4\tan^{2} x)}$
= $\frac{(4\tan x)(1 - \tan^{2}x)}{1^{2}+\tan^{4} x - 6 \tan^{2} x }$ = R.H.S.

Question 24: Prove the following $\small \cos4x = 1 - 8\sin^{2}x\cos^{2}x$

Answer:
We know that
$\cos2x=1-2\sin^{2}x$
We use this in our problem
$\cos 4x = \cos 2(2x)$
= $1-2\sin^{2}2x$
= $1-2(2\sin x \cos x)^{2}$ $(\because \sin2x = 2\sin x \cos x)$
= $1-8\sin^{2}x\cos^{2}x=$ R.H.S.

Question 25: Prove the following $\small \cos6x = 32\cos^{6}x -48\cos^{4}x + 18\cos^{2}x-1$

Answer:
We know that
$\cos 3x = 4 \cos^{3}x - 3\cos x$
We use this in our problem
we can write $\cos 6x$ as $\cos 3(2x)$
$\cos 3(2x) = 4 \cos^{3}2x - 3\cos 2x$
= $4(2\cos^{2}x - 1)^{3} - 3(2\cos^{2}x - 1) (\because \cos 2x = 2\cos^{2}x - 1)$
= $4[(2\cos^{2}x)^{3} -(1)^{3}-3(2\cos^{2}x)^{2}(1) + 3(2\cos^{2}x)(1)^{2}]-6\cos^{2}x + 3[(a-b)^{3} = a^{3} - b^{3} - 3a^{2}b+ 3ab^{2}]$
= $32\cos^{6}x - 4 - 48 \cos^{4}x + 24 \cos^{2}x - 6\cos^{2}x + 3$
= $32\cos^{6}x - 48 \cos^{4}x + 18\cos^{2}x - 1 =$ R.H.S.

NCERT Trigonometric Functions Class 11 Solutions: Miscellaneous Exercise

Page Number: 71-72

Total Questions: 10

Question 1: Prove that $\small 2\cos\frac{\pi }{13}\cos\frac{9\pi }{13}+\cos\frac{3\pi }{13}+\cos\frac{5\pi }{13}=0$

Answer:
We know that
cos A+ cos B = $2\cos(\frac{A+B}{2})\cos(\frac{A-B}{2})$
We use this in our problem
$\small 2\cos\frac{\pi }{13}\cos\frac{9\pi }{13}+2\cos\frac{(\frac{3\pi }{13}+\frac{5\pi}{13})}{2}\cos\frac{(\frac{3\pi}{13}-\frac{5\pi }{13})}{2}$
= $\small 2\cos\frac{\pi }{13}\cos\frac{9\pi }{13}+2\cos\frac{4\pi }{13}\cos\frac{-\pi}{13}$ ( we know that cos(-x) = cos x )
= $\small 2\cos\frac{\pi }{13}\cos\frac{9\pi }{13}+2\cos\frac{4\pi }{13}\cos\frac{\pi}{13}$
= $\small 2\cos\frac{\pi }{13}(\cos\frac{9\pi }{13}+\cos\frac{4\pi }{13})$
Again, use the above identity
= $\small 2\cos\frac{\pi }{13}(2\cos(\frac{\frac{9\pi }{13}+\frac{4\pi }{13}}{2})\cos(\frac{\frac{9\pi }{13}-\frac{4\pi }{13}}{2})$
= $\small 2\cos\frac{\pi }{13}2\cos\frac{\pi }{2}\cos\frac{5\pi }{26}$
we know that $\small \cos\frac{\pi }{2}$ = 0
So, $\small 2\cos\frac{\pi }{13}2\cos\frac{\pi }{2}\cos\frac{5\pi }{26}$ = 0 = R.H.S.

Question 2: Prove that $\small (\sin 3x + \sin x)\sin x + (\cos 3x - \cos x )\cos x = 0$

Answer:
We know that
$\sin3x=3\sin x - 4\sin^{3}x$ and $\cos3x=4\cos^{3}x - 3\cos x$
We use this in our problem
$\small (\sin 3x + \sin x)\sin x + (\cos 3x - \cos x )\cos x$
= $(3\sin x - 4\sin^{3}x+ \sin x) \sin x$ + $(4\cos^{3}x - 3\cos x- \cos x)\cos x$
= (4sinx - 4 $\small \sin^{3}x$ )sin x + (4 $\small \cos^{3}x$ - 4cos x)cos x
Now take the 4sinx common from 1st term and -4cosx from 2nd term
= 4 $\small \sin^{2}x$ (1 - $\small \sin^{2}x$ ) - 4 $\small \cos^{2}x$ (1 - $\small \cos^{2}x$ )
= 4 $\small \sin^{2}x$ $\small \cos^{2}x$ - 4 $\small \cos^{2}x$ $\small \sin^{2}x$ ($\small \because \cos^{2}x = 1 - \sin^2x$ and $\sin^{2}x = 1 -\cos^{2}x$)
= 0 = R.H.S.

Question 3: Prove that $\small (\cos x + \cos y)^{2} + (\sin x - \sin y)^{2} = 4 \cos^{2}\left ( \frac{x+y}{2} \right )$

Answer:
We know that $(a+b)^{2} = a^{2} + 2ab + b^{2}$ and $(a-b)^{2} = a^{2} - 2ab + b^{2}$
We use these two in our problem
$(\sin x-\sin y)^{2} = \sin^{2}x - 2\sin x\sin y + \sin^{2}y$ and $(\cos x+\cos y)^{2} = \cos^{2}x + 2\cos x\cos y + \cos^{2}y$
$\small (\cos x + \cos y)^{2} + (\sin x - \sin y)^{2}$ = $\cos^{2}x + 2\cos x\cos y + \cos^{2}y$ + $\sin^{2}x - 2\sin x\sin y + \sin^{2}y$
= 1 + 2 cos x cos y + 1 - 2 sin x sin y $\left ( \because \sin^{2}x + \cos^{2}x = 1\ and \ \sin^{2}y + \cos^{2}y = 1 \right )$
= 2 + 2(cos x cos y - sin x sin y)
= 2 + 2cos(x + y)
= 2(1 + cos(x + y) )
Now we can write
$\cos(x + y) =2\cos^{2}\frac{(x + y)}{2} - 1$ $\left ( \because \cos2x = 2\cos^{2}x - 1 \ \Rightarrow \cos x = 2\cos^{2}\frac{x}{2} - 1\right )$
= $2(1 + 2\cos^{2}\frac{(x + y)}{2} - 1)$
= $4\cos^{2}\frac{(x + y)}{2}$
= R.H.S.

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

Answer:
We know that $(a+b)^{2} = a^{2} + 2ab + b^{2}$ and $(a-b)^{2} = a^{2} - 2ab + b^{2}$
We use these two in our problem
$(\sin x-\sin y)^{2} = \sin^{2}x - 2\sin x\sin y + \sin^{2}y$ and $(\cos x-\cos y)^{2} = \cos^{2}x - 2\cos x\cos y + \cos^{2}y$
$\small (\cos x - \cos y)^{2} + (\sin x - \sin y)^{2}$ = $\cos^{2}x - 2\cos x\cos y + \cos^{2}y$ + $\sin^{2}x - 2\sin x\sin y + \sin^{2}y$
= 1 - 2cos x cos y + 1 - 2sin x sin y $\left ( \because \sin^{2}x + \cos^{2}x = 1\ and \ \sin^{2}y + \cos^{2}y = 1 \right )$
= 2 - 2(cos x cos y + sin x sin y)
= 2 - 2cos(x - y) $\small (\because \cos(x-y) =\cos x \cos y + \sin x \sin y)$
= 2(1 - cos(x - y) )
Now we can write
$\cos(x - y) = 1 -2\sin^{2}\frac{(x - y)}{2}$ $\left ( \because \cos2x = 1 - 2\sin^{2}x \ \Rightarrow \cos x = 1 - 2\sin^{2}\frac{x}{2} \right )$
So, $2(1 - \cos(x - y) ) = 2(1 - ( 1 -2\sin^{2}\frac{(x - y)}{2}))$
= $4\sin^{2}\frac{(x - y)}{2}$ = R.H.S.

Question 5: Prove that $\small \sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x\cos2x \sin4x$

Answer:
we know that $\sin A + \sin B =2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$
We use this identity in our problem
If we notice we need sin4x in our final result so it is better if we need a combination of sin7x and sin x, sin3x and sin5x to get sin4x.
$(\sin7x + \sin x) + (\sin5x + \sin3x) = 2\sin\frac{7x+x}{2}\cos\frac{7x-x}{2}$ $+2\sin\frac{5x+3x}{2}\cos\frac{5x-3x}{2}$
$=$ $2\sin4x\cos3x + 2\sin4x\cos x$
take 2sin4x common
= 2sin4x(cos3x + cosx)
We know that
$\cos A + \cos B =2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$
We use this
$\cos3x + \cos x =2\cos\frac{3x+x}{2}\cos\frac{3x-x}{2}$
= $2\cos2x\cos x$
Now 2sin4x(cos3x + cosx) = 2sin4x( $2\cos2x\cos x$)
= $4\cos x \cos2x\sin4x$ = R.H.S.

Question 6: Prove that $\small \frac{(\sin 7x + \sin 5x) + (\sin9x + \sin 3x)}{(\cos7x + \cos5x) + (\cos9x + \cos3x)} = \tan6x$

Answer:
We know that
$\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$
$\cos A + \cos B =2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$
We use these two identities in our problem
sin7x + sin5x = $2\sin\frac{7x+5x}{2}\cos\frac{7x-5x}{2}$ = $2\sin6x\cos x$
sin 9x + sin 3x = $2\sin\frac{9x+3x}{2}\cos\frac{9x-3x}{2}$ = $2\sin6x\cos 3x$
cos 7x + cos5x = $2\cos\frac{7x+5x}{2}\cos\frac{7x-5x}{2}$ = $2\cos6x\cos x$
cos 9x + cos3x = $2\cos\frac{9x+3x}{2}\cos\frac{9x-3x}{2}$ = $2\cos6x\cos 3x$
$\small \frac{(\sin 7x + \sin 5x) + (\sin9x + \sin 3x)}{(\cos7x + \cos5x) + (\cos9x + \cos3x)}$ = $\small \frac{(2\sin 6x\cos x) + (2\sin6x \cos3x)}{(2\cos6x cos x) + (2\cos6x \cos3x)}$
= $\small \frac{2\sin6x(\cos x + \cos3x)}{2\cos6x (\cos x + \cos3x)} = \tan6x$ = R.H.S. $\small \left ( \because \frac{\sin x}{\cos x} = \tan x\right )$

Question:7: Prove that $\small \sin3x + \sin2x - \sin x = 4\sin x \cos\frac{x}{2}\cos\frac{3x}{2}$

Answer:
We know that
$\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$
$\sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$
We use these identities
$\sin3x - \sin x = 2\cos\frac{3x+x}{2}\sin\frac{3x-x}{2}$
$= 2\cos2x\sin x$
sin2x + $2\cos2x\sin x$ = 2sinx cosx + $2\cos2x\sin x$
Take 2 sinx common
$2\sin x ( \cos x + \cos2x) = 2\sin x(2\cos\frac{2x+x}{2}\cos\frac{2x-x}{2})$
$= 2\sin x(2\cos\frac{3x}{2}\cos\frac{x}{2})$
$= 4\sin x\cos\frac{3x}{2}\cos\frac{x}{2} =$ R.H.S.

Question 8: Find $\small \sin\frac{x}{2} , \cos\frac{x}{2} , and \tan\frac{x}{2}$ in $\small \tan x = - \frac{4}{3}$ , x in quadrant II.

Answer:
tan x = $-\frac{4}{3}$
We know that,
$\sec^{2}x = 1 + \tan^{2}x$
$= 1 +\left ( -\frac{4}{3} \right )^{2}$
$= 1 + \frac{16}{9}$ = $\frac{25}{9}$
$\sec x = \sqrt{\frac{25}{9}}$ = $\pm\frac{5}{3}$
x lies in the II quadrant that's why sec x is -ve
So, $\sec x =-\frac{5}{3}$
Now, $\cos x = \frac{1}{\sec x}$ = $-\frac{3}{5}$
We know that,
$\cos x = 2\cos^{2}\frac{x}{2}- 1$ ( $\because \cos2x = 2\cos^{2}x - 1 \Rightarrow \cos x = 2\cos^{2}\frac{x}{2} - 1$ )
⇒ $-\frac{3}{5}+ 1 = 2$ $\cos^{2}\frac{x}{2}$
⇒ $\frac{-3+5}{5}$ = $2\cos^{2}\frac{x}{2}$
⇒ $\frac{2}{5}$ = $2\cos^{2}\frac{x}{2}$
⇒ $\cos^{2}\frac{x}{2}$ = $\frac{1}{5}$
⇒ $\cos\frac{x}{2}$ = $\sqrt{\frac{1}{5}}$ = $\pm\frac{1}{\sqrt5}$
x lies in II quadrant so value of $\cos\frac{x}{2}$ is +ve
$\cos\frac{x}{2}$ = $\frac{1}{\sqrt5} = \frac{\sqrt5}{5}$
We know that
$\cos x =1 - 2\sin^{2}\frac{x}{2}$
⇒ $2\sin^{2}\frac{x}{2}$ = 1 - $(-\frac{3}{5})$ = $\frac{8}{5}$
⇒ $\sin^{2}\frac{x}{2} = \frac{4}{5}\\ \\=\sin\frac{x}{2} = \sqrt{ \frac{4}{5}} = \pm \frac{2}{\sqrt{5}}$
x lies in the II quadrant so the value of sin x is +ve
$\sin\frac{x}{2} = \frac{2}{\sqrt{}5} = \frac{2\sqrt5}{5}$
$\tan \frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{2\sqrt5}{5}}{\left ( \frac{\sqrt5}{5} \right )} = 2$

Question 9: Find $\small \sin\frac{x}{2} , \cos\frac{x}{2} , and \tan\frac{x}{2}$ in $\small \cos x = -\frac{1}{3}$ , x in quadrant III

Answer:
$\pi < x < \frac{3\pi}{2}⇒ \frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$
We know that
cos x = $2\cos^{2}\frac{x}{2} - 1$
$2\cos^{2}\frac{x}{2} =$ cos x + 1 = $\left ( -\frac{1}{3} \right )$ + 1 = $\left ( \frac{-1+3}{3} \right )$ = $\frac{2}{3}$
⇒ $\cos\frac{x}{2} = \sqrt{ \frac{1}{3}} = \pm \frac{1}{\sqrt3}$
⇒ $\cos\frac{x}{2} = - \frac{1}{\sqrt3} = - \frac{\sqrt3}{3}$ (As x lies in 3rd quadrant)
We know that
cos x = $1 - 2\sin^{2}\frac{x}{2}$
⇒ $2\sin^{2}\frac{x}{2} = 1 - \cos x$ = 1 - $\left ( -\frac{1}{3} \right )$ = $\frac{3+1}{3}$ = $\frac{4}{3}$
⇒ $2\sin^{2}\frac{x}{2} = \frac{4}{3}⇒ \sin^{2}\frac{x}{2} = \frac{2}{3}⇒ \sin\frac{x}{2} = \pm \sqrt{ \frac{2}{3}} = \frac{\sqrt6}{3}$
Because $\sin\frac{x}{2}$ is +ve in given quadrant
$\tan\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{\sqrt6}{3}}{\frac{-\sqrt3}{3}} = - \sqrt2$

Question 10: Find $\small \sin\frac{x}{2} , \cos\frac{x}{2} , and \tan\frac{x}{2}$ in $\small \sin x = \frac{1}{4}$ ,x in quadrant II

Answer:
$\frac{\pi}{2} < x < \pi⇒ \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$ all functions are positive in this range
We know that
$\cos^{2}x = 1 - \sin^{2}x$ = 1 - $\left ( \frac{1}{4} \right )^{2}$ = $1 - \frac{1}{16}$ = $\frac{15}{16}$
cos x = $\sqrt\frac{15}{16} = \pm \frac{\sqrt15}{4} = - \frac{\sqrt15}{4}$ (cos x is -ve in II quadrant)
We know that
cosx = $2\cos^{2}\frac{x}{2} - 1$
$2\cos^{2}\frac{x}{2} = \cos x + 1 = -\frac{\sqrt15}{4} + 1 = \frac{-\sqrt15+4}{4}$
$\cos^{2}\frac{x}{2} = \frac{-\sqrt15+4}{8}$
$\cos\frac{x}{2} = \pm \sqrt\frac{-\sqrt15+4}{8} = \frac{\sqrt{-\sqrt15+4}}{2\sqrt2} = \frac{\sqrt{8-2\sqrt15}}{4}$ (because all functions are posititve in given range)
Similarly,
cos x = $1-2\sin^{2}\frac{x}{2}$
$2\sin^{2}\frac{x}{2} = 1 - \cos x⇒ 2\sin^{2}\frac{x}{2} = 1 -\left (\frac{-\sqrt15}{4} \right ) = \frac{4+\sqrt15}{4}$
$\sin\frac{x}{2} = \pm \sqrt\frac{\sqrt15+4}{8} = \frac{\sqrt{\sqrt15+4}}{2\sqrt2} = \frac{\sqrt{8+2\sqrt15}}{4}$ (because all functions are posititve in given range)
$\tan\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{\sqrt{8+2\sqrt15}}{4}}{\frac{\sqrt{8-2\sqrt15}}{4}} = \frac{{8+2\sqrt15}}{\sqrt{64 - 15\times4}} = \frac{{8+2\sqrt15}}{\sqrt{4}} = 4 + \sqrt15$

Also, read,

Class 11 Maths NCERT Chapter 3: Extra Question

Question:
The solution of $\frac{\sqrt{3}+1}{\sin x}+\frac{\sqrt{3}-1}{\cos x}=4 \sqrt{2}$ in the interval $\left(0, \frac{\pi}{2}\right)$ is:

Solution:

$\begin{aligned} & \frac{\sqrt{3}+1}{\sin x}+\frac{\sqrt{3}-1}{\cos x}=4 \sqrt{2} \\ \Rightarrow & \left(\frac{\sqrt{3}+1}{2 \sqrt{2}}\right) \frac{1}{\sin x}+\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right) \frac{1}{\cos x}=2 \\ \Rightarrow & \frac{\cos \frac{\pi}{12}}{\sin x}+\frac{\sin \frac{\pi}{12}}{\cos x}=2 \\ \Rightarrow & \cos \left(x-\frac{\pi}{12}\right)=\sin 2 x \\ & \Rightarrow \sin \left(\frac{\pi}{2}-\left(x-\frac{\pi}{12}\right)\right)=\sin 2 x \\ \Rightarrow & \frac{\pi}{2}-\left(x-\frac{\pi}{12}\right)=n \pi+(-1)^{\mathrm{n}} 2 x \\ &⇒ x=\frac{7 \pi}{36}, \frac{5 \pi}{12}\end{aligned}$

Hence, the correct answer is $\frac{7 \pi}{36}, \frac{5 \pi}{12}$.

Trigonometric Functions Class 11 Chapter 3: Topics

The topics discussed in the NCERT Solutions for class 11 chapter 3 Trigonometric Functions are:

Trigonometric Functions Class 11 Solutions: Important Formulae

Angle Conversion:

Radian Measure = $\frac{π}{180}$ × Degree Measure

Degree Measure = $\frac{180}{π}$ × Radian Measure

Trigonometric Ratios:

  • sin θ = P / H

  • cos θ = B / H

  • tan θ = P / B

  • cosec θ = H / P

  • sec θ = H / B

  • cot θ = B / P

Reciprocal Trigonometric Ratios:

  • sin θ = 1 / cosec θ

  • cosec θ = 1 / sin θ

  • cos θ = 1 / sec θ

  • sec θ = 1 / cos θ

  • tan θ = 1 / cot θ

  • cot θ = 1 / tan θ

Trigonometric Ratios of Complementary Angles:

  • sin (90° – θ) = cos θ

  • cos (90° – θ) = sin θ

  • tan (90° – θ) = cot θ

  • cot (90° – θ) = tan θ

  • sec (90° – θ) = cosec θ

  • cosec (90° – θ) = sec θ

Periodic Trigonometric Ratios:

  • sin(π/2-θ) = cos θ

  • cos(π/2-θ) = sin θ

  • sin(π-θ) = sin θ

  • cos(π-θ) = -cos θ

  • sin(π+θ) = -sin θ

  • cos(π+θ) = -cos θ

  • sin(2π-θ) = -sin θ

  • cos(2π-θ) = cos θ

Trigonometric Identities:

  • sin² θ + cos² θ = 1

  • cosec² θ – cot² θ = 1

  • sec² θ – tan² θ = 1

Product to Sum Formulas:

  • sin x sin y = 1/2 [cos(x–y) − cos(x+y)]

  • cos x cos y = 1/2[cos(x–y) + cos(x+y)]

  • sin x cos y = 1/2[sin(x+y) + sin(x−y)]

  • cos x sin y = 1/2[sin(x+y) – sin(x−y)]

Sum to Product Formulas:

  • sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]

  • sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]

  • cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]

  • cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]

General Trigonometric Formulas:

  • sin (x+y) = sin x × cos y + cos x × sin y

  • cos(x+y) = cosx × cosy − sinx × siny

  • cos(x–y) = cosx × cosy + sinx × siny

  • sin(x–y) = sinx × cosy − cosx × siny

Sum and Difference Formulas for tan:

  • tan (x+y) = (tan x + tan y) / (1 − tan x tan y)

  • tan (x−y) = (tan x − tan y) / (1 + tan x tan y)

Double Angle Formulas for tan:

  • tan 2θ = (2 tan θ) / (1 - tan²θ)

Triple Angle Formulas for sin, cos, and tan:

  • sin 3θ = 3sin θ – 4sin³θ

  • cos 3θ = 4cos³θ – 3cos θ

  • tan 3θ = [3tan θ – tan³θ] / [1 − 3tan²θ]

Approach to Solve Questions of Trigonometric Functions Class 11

Here are some approaches that students can use to approach the questions related to trigonometric functions.

  • Identify the perpendicular, base, and hypotenuse in a right-angled triangle efficiently to find the trigonometric ratios of sine, cosine, tangent, secant, cosecant and cotangent.
  • Learn the relationship between degrees and radians because this chapter includes many questions requiring conversions between them. Remember that π radians = 180°
  • Draw a table to understand the values of all the trigonometric ratios of standard angles (0°, 30°, 45°, 60°, 90°) in both degrees and radians.
  • Use graphs wherever necessary to understand the maximum, minimum values of trigonometric functions over intervals.
  • Be comfortable with the various trigonometric identities to simplify expressions and master the fundamental identities.
    sin²x + cos²x = 1
    1 + tan²x = sec²x
    1 + cot²x = cosec²x
  • Practice all the formulas, such as the sine rule, cosine rule, projection rule and areas related to triangles, including their applications.
  • Memorise the ASTC rule (All, Sine, Tangent, Cosine) to determine the sign of a trigonometric ratio in each quadrant.

What Extra Should Students Study Beyond NCERT for JEE?

Here is a comparison list of the concepts in Trigonometric Functions that are covered in JEE and NCERT, to help students understand what extra they need to study beyond the NCERT for JEE:

Concept Name

JEE

NCERT

Measuring Angles

Trigonometric Ratios

Trigonometric Identities

Sign of Trigonometric Functions

Graphs of General Trigonometric Functions

Trigonometric Ratios of Allied Angles

Trigonometric Ratios of Compound Angles

Sum-to-Product and Product-to-Sum Formulas

Product To Sum Formulas

Double Angle Formulas

Triple Angle Identities

Half Angle Formula

Trigonometric Ratio of a Submultiple of an Angle

Trigonometric Ratios of Some Specific Angles

Summation Of Series In Trigonometry

Conditional Trigonometric Identities

Maximum and Minimum Value of Trigonometric Function

General Solution of Trigonometric Equations

Simultaneous Trigonometric Equations

Trigonometric Equation using Minimum and Maximum Value of a Function

Trigonometric Inequality

Law of Sines

Law of Cosines

Law Of Tangents

Projection Formula

Semiperimeter and Half Angle Formulae

Area of a Triangle

Circumcircle of a Triangle

In-Circle and In-Centre

Escribed Circle of Triangle

Important Solutions of Triangle Formulas

Height and Distance

Inverse Trigonometric Functions

Domain and Range of Trigonometric Functions

Inverse Functions

Graph of Inverse Trigonometric Function

Trigonometric Ratios of Complementary Angles

Sum and difference of angles in terms of arctan

Multiple Angles

NCERT Solutions for Class 11 Mathematics - Chapter Wise

Given below is the chapter-wise list of the NCERT Class 11 Maths solutions with their respective links:

Also, read,

NCERT Solutions for Class 11 - Subject Wise

Here are the subject-wise links for the NCERT solutions of class 11:

NCERT Books and NCERT Syllabus

Here are some useful links for NCERT books and the NCERT syllabus for class 11:

Frequently Asked Questions (FAQs)

Q: How to derive the sine and cosine function graphs?
A:

To derive the sine and cosine function graphs, plot $y=\sin\theta$ and $y=\cos\theta$ for the values of $\theta$ like 0°, 30°, 60°, 90°, 180°, etc..

The sine graph starts at (0, 0) and goes up to 1 at 90°, drops to –1 at 270°, and comes to 0 again at 360° then repeats.

Q: What are the domain and range of trigonometric functions in Class 11?
A:

In class 11 chapter 3 trigonometric functions, a domain basically denotes the input values and range denotes the output values.

For example: The domain of $\sin\theta$ and $\cos\theta$ is $(-\infty,\infty)$ and range is (–1, 1).

Q: How to find the general solution of a trigonometric equation?
A:

To find the general solution of a trigonometric equation, use the following standard equations:

$\sin \theta=\sin \alpha \Rightarrow \theta=n \pi+(-1)^n \alpha, n \in \mathbb{Z}$

$\cos \theta=\cos \alpha \Rightarrow \theta=2 n \pi \pm \alpha, n \in \mathbb{Z}$

$\tan \theta=\tan \alpha \Rightarrow \theta=n \pi+\alpha, n \in \mathbb{Z}$

Q: What are the key formulas of Trigonometric Functions in NCERT Class 11?
A:

The key formulas of Trigonometric Functions in NCERT Class 11 are:

$1) \cos^2x+\sin^2x=1$

$2) 1+\tan^2x =\sec^2x$

$3)1+\cot^2x =\operatorname{cosec}^2x$

$4)\cos (2n\pi + x)= \cos x$

$5)\sin (2n\pi + x) = \sin x$

$6) \sin (-x) = -\sin x$

$7) \cos (-x)= \cos x$ 

$8)\cos (x + y) = \cos x \cos y - \sin x \sin y$

$9)\cos (x - y) = \cos x \cos y + \sin x \sin y$

$10)\sin (x + y) = \sin x \cos y + \cos x \sin y$

$11)\sin (x - y) = \sin x \cos y - \cos x \sin y$  

$12) \tan(x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$

$13) \tan(x-y)=\frac{\tan x-\tan y}{1+\tan x\tan y}$

$14) \cot(x+y)=\frac{\cot x\cot y-1}{\cot x+\cot y}$

$15) \cot(x-y)=\frac{1+\cot x\cot y}{\cot y-\cot x}$  

Q: How to solve trigonometric equations in Class 11 Maths Chapter 3?
A:

To solve trigonometric equations in Class 11 Maths Chapter 3, express the given terms in basic trigonometric functions like sin, cos, and tan, then try to apply proper trigonometric identities and use simplification.

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