NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions

 

NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions: Trigonometry has various real-time applications. It is used to solve height and distance problems. This cahpter gives an introduction to basic properties and identities of trigonometric functions and questions based on the same are answered in NCERT solutions for class 11 maths chapter 3 trigonometric functions. You will use the trigonometric identities in other chapters of mathematics and throughout the NCERT physics for both class 11 and 12 also. So it is very important to memorize and understand the basic identities and properties of trigonometric functions. CBSE NCERT solutions for class 11 maths chapter 3 trigonometric functions will help you for the same. Command on this chapter is required to understand the concepts of inverse trigonometric functions which will be taught in class 12. You should practice more in order to get clarity of the concepts. First, try to solve NCERT problems. If you getting difficulties in doing so, you can take help from solutions of NCERT for class 11 maths chapter 3 trigonometric functions. Check all NCERT solutions at a single place which will be helpful when you are not able to solve the NCERT questions.

Let's understand this chapter with one example.

If we want to measure the height of a building such that we are standing at 50m away from building at point P and the angle of elevation made with the ground at P is 45 degree (as shown in the below figure). Then what will be the height of the building?

Let the height of the building be QR and distance from the base of the building to point P is 50 meter

Using the trigonometric function 

tan(45)=\frac{QR}{QP}\\\Rightarrow 1=\frac{QR}{50}\\\Rightarrow QR=50\ m

There are many other examples such as trigonometry is used in electric circuit analysis, predicting the heights of tides in the ocean, analyzing a musical tone and in seismology, etc.

 

The main topics of this chapter are listed below:

3.1 Introduction

3.2 Angles

3.3 Trigonometric Functions

3.4 Trigonometric Functions of Sum and Difference of Two Angles

3.5 Trigonometric Equations

 

NCERT solutions for class 11 maths chapter 3 trigonometric functions-Exercise: 3.1

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

(i) 25 \degree
(ii)-47 \degree30'
(iii) 240\degree
(iv)520\degree

Answer:

It is solved using relation between degree and radian

(i) 25\degree
We know that  180\degree = \pi radian 

So,      1\degree = \frac{\pi }{180}      radian


 25\degree = \frac{\pi }{180}\times 25   radian    =\frac{5\pi }{36}   radian      
(ii)   -47\degree30'
 We know that
          -47\degree30' = -47\frac{1}{2}degree = -\frac{95}{2}\degree

Now, we know that       180\degree = \pi \Rightarrow 1\degree = \frac{\pi}{180}   radian
   
So,    -\frac{95}{2}\degree = \frac{\pi}{180}\times \left (-\frac{95}{2} \right ) radian  \Rightarrow \frac{-19\pi}{72}  radian        
(iii)   240\degree
We know that 
        
      180\degree = \pi \Rightarrow 1\degree = \frac{\pi}{180}   radian

So, 240\degree = \frac{\pi}{180}\times 240 \Rightarrow \frac{4\pi}{3}   radian      
(iv)   520\degree
 We know that 

               180\degree = \pi \Rightarrow 1\degree = \frac{\pi}{180}   radian

So, 520\degree \Rightarrow \frac{\pi}{180}\times 520  radian \Rightarrow \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:

(1)   \frac{11}{16}

We know that 
    \pi radian   = 180\degree \Rightarrow 1 radian = \frac{180}{\pi} degree

So,   \frac{11}{16} radian = \frac{180}{\pi}\times \frac{11}{16}degree                            (we need to take \pi = \frac{22}{7} )

   \frac{11}{16}radian = \frac{180\times 7}{22}\times \frac{11}{16}degree \Rightarrow \frac{315}{8}degree 
 
                                                                                             (we use 1\degree = 60' and 1' = 60'')

Here 1' represents 1 minute and 60" represents 60 seconds
Now, 

\frac{315}{8}degree=39\frac{3}{8}degree\\ \\ =39\degree +\frac{3\times 60}{8}minutes \Rightarrow 39\degree +22' + \frac{1}{2}minutes \Rightarrow 39\degree +22' +30''\\ \\ \Rightarrow \frac{315}{8}degree = 39\degree22'30''                                                                                                             

(ii)  -4
We know that

\pi radian   = 180\degree \Rightarrow 1 radian = \frac{180}{\pi} degree             (we need to take \pi = \frac{22}{7} )


So,  -4 radian =  \frac{-4\times 180}{\pi} \Rightarrow \frac{-4\times 180\times 7}{22} \Rightarrow \frac{-2520}{11}degree                                                                                                                   


                                                                                            (we use 1\degree = 60' and 1' = 60'')

\Rightarrow \frac{-2520}{11}degree = -229\frac{1}{11}degree =-229\degree + \frac{1\times 60}{11}minutes \\ \\ \Rightarrow -229\degree + 5' + \frac{5}{11}minutes = -229\degree +5' +27''\\ \\ -\frac{2520}{11} = -229\degree5'27''

(iii)   \frac{5\pi}{3}

We know that 
    \pi radian   = 180\degree \Rightarrow 1 radian = \frac{180}{\pi} degree    (we need to take \pi = \frac{22}{7} )


So,  \frac{5\pi}{3}radian = \frac{180}{\pi}\times \frac{5\pi}{3}degree = 300\degree       
(iv)  \frac{7\pi}{6}

We know that 
    \pi radian   = 180\degree \Rightarrow 1 radian = \frac{180}{\pi} degree        (we need to take \pi = \frac{22}{7} )


So, \frac{7\pi}{6}radian = \frac{180}{\pi}\times \frac{7\pi}{6} = 210\degree           

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 revolutions wheel will cover 2\pi radian
So, in 6 revolutions it will cover =6\times 2\pi = 12\piradian 

\therefore   In 1 the second  wheel will turn 12\pi radian  

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\degree\\ \\So, 1radian = \frac{180}{\pi}degree\\ \\ \therefore \frac{22}{100}radian = \frac{180}{\pi}\times\frac{22}{100}degree\Rightarrow \frac{180\times7}{22}\times\frac{22}{100} = \frac{63}{5}degree \\ \\ So, \\ \\\frac{63}{5}degree = 12\frac{3}{5}degree = 12\degree + \frac{3\times60}{5}minute = 12\degree + 36'\\ \\ \therefore \frac{63}{5}degree = 12 \degree36' 
So, 
            Angle subtended at the centre of a circle       \Theta = 12\degree36'

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{40cm}{2} = 20 cm
length of chord = 20 cm 

We know that 
                          \theta = \frac{l}{r}                                      (r = 20cm , l = ? , \theta = ?)

Now, 
                                                             
AB is the chord of length 20cm and OA and OB are radii of circle 20 cm each
The angle subtended by OA and OB at centre = \theta
\because  OA = OB = AB 
\therefore   \DeltaOAB is equilateral triangle
So, each angle equilateral is 60\degree
\therefore \theta = 60\degree  = \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\degree\\ \theta_2 = 75\degree\\        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\\l_2 = r_2\theta_2         
Now, 
               \frac{l_1}{l_2}=\frac{ r_1 \theta_1}{ r_2\theta_2}                           ( l_1 = l_2)
So,
         \frac{ r_1 }{ r_2}= \frac{\theta_2}{\theta_1} = \frac {75}{60} = \frac{5}{4}         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 = 75cm
                    l = 10cm

So, 
         \theta = \frac{l}{r} = \frac{10}{75} = \frac{2}{15}radian

(ii)   We know that

   l = r \theta
Now, 
                    r = 75cm
                    l = 15cm

So,
            \theta = \frac{l}{r} = \frac{15}{75} = \frac{1}{5}radian 

(iii)  We know that

   l = r \theta
Now, 
                    r = 75cm
                    l = 21cm

So,    
           \theta = \frac{l}{r} = \frac{21}{75} = \frac{7}{25}radian

Solutions for class 11 maths chapter 3 trigonometric functions-Exercise: 3.2

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

Answer:

Solution 
\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 \small \sin x = \frac{3}{5}  x lies in second quadrant.

Answer:

Solution 

\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 \small \cot x = \frac{3}{4}  , x lies in third quadrant.

Answer:

Solution 

\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  x lies in  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 \small \sec x = \frac{13}{5}  , x lies in fourth quadrant.

Answer:

Solution
\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}\\ \sin^{2}x = 1 - \frac {25}{169} = \frac {144}{169}\\ \sin x = \sqrt { \frac {144}{169}} = \pm \frac {12}{13}
 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 \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 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\degree

Answer:
We know that values of sin x repeat after an interval of  2\pi\ or\ 360\ degree

\sin765\degree = \sin (2\times360\degree + 45\degree ) = \sin45\degree\\ sin45\degree = \frac {1}{\sqrt{2}}

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

Answer:

We know that value of cosec x repeats after an interval of 2\pi \ or \ 360\degree
cosec (-1410\degree) = cosec (-1410\degree + 360\degree\times4)\\ cosec\ 30\degree = 2

or

cosec(-1410\degree)= - cosec(1410\degree)\\= -cosec(4 \times 360\degree - 30\degree)= - cosec(-30\degree) = 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 degree
\tan (\frac{19\pi}{3}) = \tan (6\pi+\frac{\pi}{3})= \tan \frac{\pi}{3} =\tan 60\degree = \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\degree
\sin \left ( -\frac{11\pi}{3} \right ) = \sin \left ( -4\pi +\frac{\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\degree
\cot \left ( -\frac{15\pi}{4} \right ) = \cot \left ( -4\pi +\frac {\pi}{4} \right ) = \cot \left ( \frac{\pi}{4} \right ) = 1

CBSE NCERT solutions for class 11 maths chapter 3 trigonometric functions-Exercise: 3.3

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 degree), cos (60 degree) and tan (45 degree). That is: 

   
\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
\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 ) + cosec ^{2}\left ( \frac{7\pi }{6} \right )\cos ^{2}\frac{\pi }{3} = \frac{3}{2}

Answer:

The solutions for the given problem is done as follows.

\sin\frac{\pi}{6} = \frac {1}{2}\\ \\ cosec\frac{7\pi}{6} = cosec\left ( \pi + \frac{\pi}{6} \right ) = -cosec \frac{\pi}{6}=-2\\ \\ \cos \frac{\pi}{3} = \frac{1}{2}
2\sin^{2}\frac{\pi}{6} +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}\\ \\ \Rightarrow 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 degree), tan (30 degree) and cosec (30 degree)

\cot \frac{\pi}{6} = \sqrt{3}\\ \\ cosec\frac{5\pi}{6} = cosec\left ( \pi - \frac{\pi}{6} \right )=cosec\frac{\pi}{6} = 2\\ \\ \tan\frac{\pi}{6}= \frac{1}{\sqrt{3}}

\cot^{2}\frac{\pi}{6} + 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}\\ \\ \Rightarrow 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

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\degree

Answer:

\sin 75\degree = \sin(45\degree + 30\degree)
We know that 
          (sin(x+y)=sinxcosy + cosxsiny)
Using this idendity

\sin 75\degree = \sin(45\degree + 30\degree) = \sin45\degree\cos30\degree + \cos45\degree\sin30\degree\\ \\ \Rightarrow \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\degree

Answer:

\tan 15\degree = \tan (45\degree - 30\degree)
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\degree - 30\degree)= \frac{\tan 45\degree - tan30\degree}{1+\tan45\degree\tan30\degree}\\ \\ \Rightarrow \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 cos and 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 ]

\Rightarrow 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 

\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 that,
\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 simplify 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}]\\ \\ \Rightarrow \sin x\cos x\left [ \frac{\sin^{2}x+\cos^{2}x}{\sin x\cos x } \right ] =\sin^{2}x+\cos^{2}x = 1R.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)

\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

hence 
 \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
                                                                                                    cos(-x) = cosx)
cos2x + cos 6x = 2cos4xcos(-2x) = 2cos4xcos2x

So our equation becomes 


                                                                       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 )
                                     =2sin4x(2\cos^{2}x - 1 +1 )
                                     =2sin4x(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 nultiply 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\\ \\ \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 \\ \\ \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 

\\ 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}\\and\\ \\ \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)\\ \\ \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}-\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)\\ \\ \cos^{2}x- \sin^{2} = \cos2x\\ \cos2x = 1 - 2\sin^{2}x\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}-\cos^{2}x = - \cos2x \ \ \ \ \ \ \ \ \ \ (\cos2x = 1 - 2\sin^{2}x)\\ \sin^{2}-\cos^{2}x = -( 1 - 2\sin^{2}x) = 2\sin^(2)x - 1\\ \\ \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\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}-\cos^{2}x = - \cos2x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because \cos2x = 1 - 2\sin^{2}x)\\ \sin^{2}-\cos^{2}x = -( 1 - 2\sin^{2}x) = 2\sin^(2)x - 1\\ \\ \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(1+\cos x)}{2\sin3x(1+\cos x)}\\ \ \ \\ \ \ \ \ \ \ \ = cot 3x                   

=RHS

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

Answer:

cot x cot2x - cot3x(cot2x - cotx)
Now we can write cot3x = cot(2x + x)

and we know that

cot(a+b) = \frac{\cot a \cot b - 1}{\cot a + \cot b}
 So,    
             cotx\ cot2x-\frac{\cot 2x \cot x - 1}{\cot 2x + \cot x}(cot2x+cotx)   
           =   cotx cot2x - (cot2xcotx -1)
           =  cotx cot2x - cot2xcotx +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 - 3cos 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[(2cos^{2}x)^{3} -(1)^{3} -3(2cos^{2}x)^{2}(1) + 3(2cos^{2}x)(1)^{2}]    -6\cos^{2}x + 3             (\because (a-b)^{3} = a^{3} - b^{3} - 3a^{2}b+ 3ab^{2})
                         = 32cos^{6}x - 4 - 48cos^{4}x + 24cos^{2}x  - 6\cos^{2}x + 3
                         =   32cos^{6}x  - 48cos^{4}x + 18cos^{2}x - 1  = R.H.S.

NCERT solutions for class 11 maths chapter 3 trigonometric function-Exercise: 3.4

Question:1 Find the principal and general solutions of the following equations: \tan x= \sqrt{3}

Answer:

It is given that given 
\tan x= \sqrt{3}
Now, we know  that \tan\frac{\pi}{3}= \sqrt3 and \tan\frac{4\pi}{3}= \tan \left ( \pi+\frac{\pi}{3} \right )=\sqrt3

Therefore,
the principal solutions of the equation are x = \frac{\pi}{3},\frac{4\pi}{3}
Now,
The general solution is \tan x =\tan \frac{\pi}{3}

x =n{\pi} + \frac{\pi}{3}  where n \ \epsilon \ Z and Z denotes sets of integer

Therefore,  the general solution of the equation is x =n{\pi} + \frac{\pi}{3}  where n \ \epsilon \ Z and Z denotes sets of integer

Question:2 Find the principal and general solutions of the following equations: \small \sec x = 2

Answer:

We know that value of  \sec\frac{\pi}{3} = 2 and \sec\frac{5\pi}{3} = \sec\left ( 2\pi -\frac{\pi}{3} \right ) = \sec\frac{\pi}{3} = 2

 Therefore the principal solutions are x = \frac{\pi}{3} and \frac{5\pi}{3}
 \sec x = \sec\frac{\pi}{3}
We know that value of sec x repeats after an interval of 2\pi
So, by this we can say that 

the general solution is x = 2n\pi \pm \frac{\pi}{3}  where n \epsilon Z 

Question:3 Find the principal and general solutions of the following equations: \small \cot x = - \sqrt{3}

Answer:

we know that    \ cot\frac{\pi}{6} = \sqrt{3}   and we know that \ \cot\frac{5\pi}{6} = \cot\left ( \pi -\frac{\pi}{6} \right ) = -cot\frac{\pi}{6} = -\sqrt{3}

Similarly , the value for \ \cot\frac{11\pi}{6} = \cot\left ( 2\pi -\frac{\pi}{6} \right ) = -cot\frac{\pi}{6} = -\sqrt{3}
 Therefore, principal solution is x = \frac{5\pi}{6} \ and \ \frac{11\pi}{6}


We also  know that the value of cot x repeats after an interval of \pi 
There the general solution is x  = n\pi \pm \frac{5\pi}{6} \ where \ n \ \epsilon \ Z  

Question:4 Find the principal and general solutions of the following equations: \small cosec x = -2

Answer:

We know that
                    cosec \frac{\pi}{6} = 2
                  
                   cosec (\pi + \frac{\pi}{6}) = -cosec\frac{\pi}{6} = -2               and also                cosec (2\pi - \frac{\pi}{6}) = cosec\frac{11\pi}{6} = -2
So,
                   cosec x= cosec\frac{7\pi}{6}                                  and                                   cosec x= cosec\frac{11\pi}{6}

So, the principal solutions are x = \frac{7\pi}{6} \ and \ \frac{11\pi}{6} 

 

 

Now,
             cosec x= cosec\frac{7\pi}{6}
             
              \sin x = \sin\frac{7\pi}{6}                                                                                     \left ( \because \sin x = \frac{1}{cosec x} \right )

          x = n\pi + (-1)^{n}\frac{7\pi}{6}
Therefore, the general solution is

  x = n\pi + (-1)^{n}\frac{7\pi}{6}     

where n \ \epsilon \ Z

Question:5 Find the general solution for each of the following equation \small \cos 4x = \cos 2x

Answer:

cos4x = cos2x
cos4x - cos2x = 0
We know that
\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}
We use this identity
\therefore  cos 4x - cos 2x  = -2sin3xsinx
\Rightarrow -2sin3xsinx = 0    \Rightarrow   sin3xsinx=0
So, by this we can that either 
sin3x = 0     or    sinx = 0
3x = n\pi                 x = n\pi
  x = \frac{n\pi}{3}                x = n\pi 

Therefore, the general solution is

 x=\frac{n\pi}{3}\ or\ n\pi \ where \ n\in Z

Question:6 Find the general solution of the following equation \small \cos 3x + \cos x -\cos 2x = 0

Answer:

We know that 
\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} \\ and \\ \cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}
We use these identities
(cos3x + cosx) - cos2x = 2cos2xcosx -cos2x = 0
                                     = cos2x(2cosx-1) = 0
So, either 
 cos2x = 0                        or                                 cosx=\frac{1}{2}
        2x=(2n+1)\frac{\pi}{2}                                         cosx =\cos\frac{\pi}{3}
           x=(2n+1)\frac{\pi}{4}                                              x =2n\pi \pm \frac{\pi}{3}

\therefore the general solution is 

 x=(2n+1)\frac{\pi}{4}   \ or \ 2n\pi \pm \frac{\pi}{3}

Question:7 Find the general solution of the following equation \small \sin 2x + \cos x = 0

Answer:

sin2x + cosx = 0
We know that 
sin2x = 2sinxcosx
So, 
2sinxcosx + cosx = 0
cosx(2sinx + 1) = 0
So, we can say that either 

cosx = 0                                       or                           2sinx + 1 = 0
    x=(2n+1)\frac{\pi}{2}                                                          sinx =\sin\frac{7\pi}{6}
                                                                                               x=n\pi +(-1)^{n}\frac{7\pi}{6}
Therefore, the general solution is

 x=(2n+1)\frac{\pi}{2}    or   n\pi +(-1)^{n}\frac{7\pi}{6} \ where \ n\in Z 

Question:8 Find the general solution of the following equation  \small \sec^{2}2x = 1 - \tan2x

Answer:

We know that 
\sec^{2}x = 1 + \tan^{2}x
So,
       1 + \tan^{2}2x = 1 -\tan2x
       \tan^{2}2x + \tan2x = 0\\ \\ \tan2x(\tan2x+1) = 0
 either
      tan2x = 0             or                     tan2x = -1                                                  (     \tan x = \tan \left ( \pi - \frac{\pi}{4} \right ) = \tan\frac{3\pi}{4} )
            2x = n\pi                       2x=n\pi + \frac{3\pi}{4}
               x=\frac{n\pi}{2}                        x=\frac{n\pi}{2} + \frac{3\pi}{8}
  Where n \epsilon Z

Question:9 Find the general solution of the following equation \small \sin x + \sin 3x + \sin 5x = 0

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
                      \sin 5x + \sin x = 2\sin\frac{5x+x}{2}\cos\frac{5x-x}{2} =2\sin3x\cos2x
Now our problem simplifeis to
                           2\sin3x\cos2x+ \sin3x = 0
take sin3x common
                           \sin3x(2\cos2x+ 1) = 0
So, either 
            sin3x = 0                                or                           \cos2x = -\frac{1}{2}                \left ( \cos2x = -\cos\frac{\pi}{3} = \cos\left ( \pi - \frac{\pi}{3} \right ) = \cos\frac{2\pi}{3} \right )
                 3x = n\pi                                                             2x = 2n\pi \pm \frac{2\pi}{3}                                                        
                    x = \frac{n\pi}{3}                                                               x = n\pi \pm \frac{\pi}{3}
Where n \ \epsilon \ Z                         

Solutions of NCERT for class 11 maths chapter 3 trigonometric functions-Miscellaneous Exercise

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) sinx  + (4\cos^{3}x - 3\cos x- cos x)cos x 
  =   (4sinx - 4\small \sin^{3}x)sinx + (4\small \cos^{3}x - 4cos x)cosx
  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 + 2cosxcosy + 1 - 2sinxsiny                           \left ( \because \sin^{2}x + \cos^{2}x = 1\ and \ \sin^{2}y + \cos^{2}y = 1 \right )
                                                                  = 2 + 2(cosxcosy - sinxsiny)
                                                                  = 2 + 2cos(x + y)          
                                                                  =  2(1 + cos(x + y) )
    Now we can write
                             cos(x + y) =2cos^{2}\frac{(x + y)}{2} - 1                                             \left ( \because \cos2x = 2cos^{2}x - 1 \ \Rightarrow \cos x = 2\cos^{2}\frac{x}{2} - 1\right )

                                              =   2(1 + 2cos^{2}\frac{(x + y)}{2} - 1)  
                                              =4cos^{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 - 2cosxcosy + 1 - 2sinxsiny                           \left ( \because \sin^{2}x + \cos^{2}x = 1\ and \ \sin^{2}y + \cos^{2}y = 1 \right )
                                                                  = 2 - 2(cosxcosy + sinxsiny)
                                                                  = 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 -2sin^{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 -2sin^{2}\frac{(x + y)}{2})) 
                  
                                             
                                           = 4sin^{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 
                     sinA + sinB =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 made a combination of sin7x and sin x , sin3x and  sin5x tp get sin4x

 (sin7x + sinx) + (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)
 Now, 
We know that 
                     cosA + cosB =2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
We use this
                  cos3x + cosx =2\cos\frac{3x+x}{2}\cos\frac{3x-x}{2}
                                          = 2\cos2x\cos x
                                          = 2sin4x(2\cos2x\cos x)
                                          = 4cosxcos2xsin4x = 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 

        sinA + sinB = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}
        and 
         cosA + cosB =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 
 cosA + cosB = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
 sinA - sinB = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}
           
we use these identities
                                      sin3x - sinx = 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 
                       2sinx ( cosx + cos2x) = 2sinx(2\cos\frac{2x+x}{2}\cos\frac{2x-x}{2})
             
                                                            = 2sinx(2\cos\frac{3x}{2}\cos\frac{x}{2})
                                                           = 4sinx\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 II quadrant  thats 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 II quadrant So 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}
Now,
      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} = \sqrt{ \frac{2}{3}} = \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    

NCERT solutions for class 11 mathematics

chapter-1

NCERT solutions for class 11 maths chapter 1 Sets

chapter-2

Solutions of NCERT for class 11 chapter 2 Relations and Functions

chapter-3

CBSE NCERT solutions for class 11 maths chapter 3 Trigonometric Functions

chapter-4

NCERT solutions for class 11 chapter 4 Principle of Mathematical Induction

chapter-5

Solutions of NCERT for class 11 chapter 5 Complex Numbers and Quadratic equations

chapter-6

CBSE NCERT solutions for class 11 maths chapter 6 Linear Inequalities

chapter-7

NCERT solutions for class 11 maths chapter 7 Permutation and Combinations

chapter-8

Solutions of NCERT for class 11 maths chapter 8 Binomial Theorem

chapter-9

CBSE NCERT solutions for class 11 maths chapter 9 Sequences and Series

chapter-10

NCERT solutions for class 11 maths chapter 10 Straight Lines

chapter-11

Solutions of NCERT for class 11 maths chapter 11 Conic Section

chapter-12

CBSE NCERT solutions for class 11 maths chapter 12 Introduction to Three Dimensional Geometry

chapter-13

NCERT solutions for class 11 maths chapter 13 Limits and Derivatives

chapter-14

Solutions of NCERT for class 11 maths chapter 14 Mathematical Reasoning

chapter-15

CBSE NCERT solutions for class 11 maths chapter 15 Statistics

chapter-16

NCERT solutions for class 11 maths chapter 16 Probability

NCERT solutions for class 11- Subject wise

Solutions of NCERT for class 11 biology

CBSE NCERT solutions for class 11 maths

NCERT solutions for class 11 chemistry

Solutions of NCERT for Class 11 physics

The basic identities used in NCERT solutions for class 11 maths chapter 3 Trigonometric Functions are listed below

 \\1) cos^2x+sin^2x=1\\2)\ 1+tan^2x\ \ \ \ =sec^2x\\3)1+cot^2x\ \ \ \ \ \ =cosec^2x\\4)cos (2n\pi + x) \ = cos x \\5)sin (2n\pi + x) \ = sin x \\6) sin (-x) \ \ \ \ \ \ \ = -sinx \\7) cos (-x) \ \ \ \ \ \ \ = cos x

The above identities you may have studied in your high school classes also. Here are a few more identities that you have to remember and understand from the NCERT solutions for class 11 maths chapter 3 trigonometric functions

\\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

Some conditional identities from the  NCERT solutions for class 11 maths chapter 3 Trigonometric Functions

If  angles x, y and (x ± y) is not an odd multiple of π 2, then

\\a) tan(x+y)=\frac{tanx+tany}{1-tanxtany}\\b)tan(x-y)=\frac{tanx-tany}{1+tanxtany}

If  angles x, y and (x ± y) is not a multiple of π, then

\\a) cot(x+y)=\frac{cotxcoty-1}{cotx+coty}\\b)cot(x-y)=\frac{1+cotxcoty}{coty-cotx}

There are a few more identities used in the NCERT solutions for class 11 maths chapter 3 trigonometric functions which can be derived using the above identities. Try to derive it by your self.

Happy Reading !!!

 

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