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    Home » NCERT Solutions for Math Class 11 Chapter 3 Trigonometric Functions
    Class 11 Math

    NCERT Solutions for Math Class 11 Chapter 3 Trigonometric Functions

    AdminBy AdminUpdated:August 11, 202311 Mins Read
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    Exercise 3.1 : Solutions of Questions on Page Number : 54

    NCERT Solutions for Math Class 11 Chapter 3 Trigonometric Functions

    Q1 :Find the radian measures corresponding to the following degree measures:
    (i) 25° (ii) – 47° 30′ (iii) 240° (iv) 520°
    Answer :
    (i)
    25°
    We know that 180° = π radian

    (ii) -47° 30′
    -47° 30′ = degree [1° = 60′]
    degree
    Since 180° = π radian

    (iii) 240°
    We know that 180° = π radian

    (iv) 520°
    We know that 180° = π radian


    Q2 :Find the degree measures corresponding to the following radian measures
    (i) (ii) – 4
    (iii) (iv)

    Answer :
    (i)We know that π radian = 180°

    (ii)– 4
    We know that π radian = 180°

    (iii)
    We know that π radian = 180°

    (iv)
    We know that π radian = 180°


    Q3 :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
    ∴Number of revolutions made by the wheel in 1 second = 
    In one complete revolution, the wheel turns an angle of 2π radian.
    Hence, in 6 complete revolutions, it will turn an angle of 6 × 2π radian, i.e.,
    12 π radian
    Thus, in one second, the wheel turns an angle of 12π radian.


    Q4 :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.
    Answer :
    We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then 
    Therefore, forr = 100 cm, l = 22 cm, we have

    Thus, the required angle is 12°36“².


    Q5 :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 :
    Diameter of the circle = 40 cm
    ∴Radius (r) of the circle = 
    Let AB be a chord (length = 20 cm) of the circle.

    In ΔOAB, OA = OB = Radius of circle = 20 cm
    Also, AB = 20 cm
    Thus, ΔOAB is an equilateral triangle.
    ∴θ = 60° = 
    We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then 
    Thus, the length of the minor arc of the chord is


    Q6 :If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.
    Answer :
    Let the radii of the two circles be and. Let an arc of length l subtend an angle of 60° at the centre of the circle of radius r1, while let an arc of length l subtend an angle of 75° at the centre of the circle of radius r2.
    Now, 60° =  and 75° = 
    We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then 

    Thus, the ratio of the radii is 5:4.


    Q7 :Find the angle in radian though 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 :
    We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then  
    It is given that r = 75 cm
    (i) Here, l = 10 cm

    (ii) Here, l = 15 cm

    (iii) Here, l = 21 cm


    Exercise 3.2 : Solutions of Questions on Page Number : 63
    Q1 :Find the values of other five trigonometric functions if , x lies in third quadrant.
    Answer :

    Since x lies in the 3rd quadrant, the value of sin x will be negative.


    Q2 :Find the values of other five trigonometric functions if  , x lies in second quadrant.
    Answer :

    Since x lies in the 2nd quadrant, the value of cos x will be negative


    Q3 :Find the values of other five trigonometric functions if  , x lies in third quadrant.
    Answer :

    Since x lies in the 3rd quadrant, the value of sec x will be negative.


    Q4 :Find the values of other five trigonometric functions if  , x lies in fourth quadrant.
    Answer :

    Since x lies in the 4th quadrant, the value of sin x will be negative.


    Q5 :Find the values of other five trigonometric functions if  , x lies in second quadrant.
    Answer :

    Since x lies in the 2nd quadrant, the value of sec x will be negative.
    ∴sec x = 


    Q6 :Find the value of the trigonometric function sin 765°
    Answer :
    It is known that the values of sin x repeat after an interval of 2π or 360°.


    Q7 :Find the value of the trigonometric function cosec (-1410°)
    Answer :
    It is known that the values of cosec x repeat after an interval of 2π or 360°.


    Q8 :Find the value of the trigonometric function 
    Answer :
    It is known that the values of tan x repeat after an interval of π or 180°.


    Q9 :Find the value of the trigonometric function 
    Answer :
    It is known that the values of sin x repeat after an interval of 2π or 360°.


    Q10 :Find the value of the trigonometric function  
    Answer :
    It is known that the values of cot x repeat after an interval of π or 180°.


    Exercise 3.3 : Solutions of Questions on Page Number : 73
    Q1 : 
    Answer :
    L.H.S. = 


    Q2 :Prove that 
    Answer :
    L.H.S. = 


    Q3 :Prove that 
    Answer :
    L.H.S. = 


    Q4 :Prove that  
    Answer :
    L.H.S = 


    Q5 :Find the value of:
    (i) sin 75°
    (ii) tan 15°
    Answer :
    (i) sin 75° = sin (45° + 30°)
    = sin 45° cos 30° + cos 45° sin 30°
    [sin (x + y) = sin x cos y + cos x sin y]

    (ii) tan 15° = tan (45° – 30°)


    Q6 :Prove that:  
    Answer :


    Q7 :Prove that: 
    Answer :
    It is known that 
    ∴L.H.S. =


    Q8 :Prove that 
    Answer : 


    Q9 : 
    Answer :
    L.H.S. = 


    Q10 :Prove that sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x
    Answer :
    L.H.S. = sin (n + 1)x sin(n + 2)x + cos (n + 1)x cos(n + 2)x
    4


    Q11 :Prove that 
    Answer :
    It is known that 
    ∴L.H.S. = 


    Q12 :Prove that sin2 6x – sin2 4x = sin 2x sin 10x
    Answer :
    It is known that 
    ∴L.H.S. = sin26x – sin24x
    = (sin 6x + sin 4x) (sin 6x – sin 4x) 
    = (2 sin 5x cos x) (2 cos 5x sin x)
    = (2 sin 5x cos 5x) (2 sin x cos x)
    = sin 10x sin 2x
    = R.H.S.


    Q13 :Prove that cos2 2x – cos2 6x = sin 4x sin 8x
    Answer :
    It is known that 
    ∴L.H.S. = cos2 2x – cos2 6x
    = (cos 2x + cos 6x) (cos 2x – 6x)

    = [2 cos 4x cos 2x] [-2 sin 4x (-sin 2x)]
    = (2 sin 4x cos 4x) (2 sin 2x cos 2x)
    = sin 8x sin 4x
    = R.H.S.


    Q14 :Prove that sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x
    Answer :
    L.H.S. = sin 2x + 2 sin 4x + sin 6x
    = [sin 2x + sin 6x] + 2 sin 4x

    = 2 sin 4x cos (- 2x) + 2 sin 4x
    = 2 sin 4x cos 2x + 2 sin 4x
    = 2 sin 4x (cos 2x + 1)
    = 2 sin 4x (2 cos2 x – 1 + 1)
    = 2 sin 4x (2 cos2 x)
    = 4cos2 x sin 4x
    = R.H.S.


    Q15 :Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
    Answer :
    L.H.S = cot 4x (sin 5x + sin 3x)

    = 2 cos 4x cos x
    R.H.S. = cot x (sin 5x – sin 3x)

    = 2 cos 4x. cos x
    L.H.S. = R.H.S.


    Q16 :Prove that 
    Answer :
    It is known that 
    ∴L.H.S = 


    Q17 :Prove that 
    Answer :
    It is known that 
    ∴L.H.S. = 


    Q18 :Prove that 
    Answer :
    It is known that 
    ∴L.H.S. = 


    Q19 :Prove that 
    Answer :
    It is known that 
    ∴L.H.S. = 


    Q20 :Prove that 
    Answer :
    It is known that 
    ∴L.H.S. = 


    Q21 :Prove that 
    Answer :
    L.H.S. = 


    Q22 :Prove that cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1
    Answer :
    L.H.S. = cot x cot 2x – cot 2x cot 3x – cot 3x cot x
    = cot x cot 2x – cot 3x (cot 2x + cot x)
    = cot x cot 2x – cot (2x + x) (cot 2x + cot x)

    = cot x cot 2x – (cot 2x cot x – 1)
    = 1 = R.H.S.


    Q23 :Prove that 
    Answer :
    It is known that 
    ∴L.H.S. = tan 4x = tan 2(2x)


    Q24 :Prove that cos 4x = 1 – 8sin2 x cos2 x
    Answer :
    L.H.S. = cos 4x
    = cos 2(2x)
    = 1 – 2 sin2 2x [cos 2A = 1 – 2 sin2 A]
    = 1 – 2(2 sin x cos x)2 [sin2A = 2sin A cosA]
    = 1 – 8 sin2x cos2x
    = R.H.S.


    Q25 :Prove that: cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1
    Answer :
    L.H.S. = cos 6x
    = cos 3(2x)
    = 4 cos3 2x – 3 cos 2x [cos 3A = 4 cos3 A – 3 cos A]
    = 4 [(2 cos2 x – 1)3 – 3 (2 cos2 x – 1) [cos 2x = 2 cos2 x – 1]
    = 4 [(2 cos2 x)3 – (1)3 – 3 (2 cos2 x)2 + 3 (2 cos2 x)] – 6cos2 x + 3
    = 4 [8cos6x – 1 – 12 cos4x + 6 cos2x] – 6 cos2x + 3
    = 32 cos6x – 4 – 48 cos4x + 24 cos2 x – 6 cos2x + 3
    = 32 cos6x – 48 cos4x + 18 cos2x – 1
    = R.H.S


    Exercise 3.4 : Solutions of Questions on Page Number : 78
    Q1 :Find the principal and general solutions of the equation 
    Answer :

    Therefore, the principal solutions are x =  and  

    Therefore, the general solution is 


    Q2 :Find the principal and general solutions of the equation 
    Answer :

    Therefore, the principal solutions are x =  and 
    Therefore, the general solution is,  where n ∈ Z


    Q3 :Find the principal and general solutions of the equation 
    Answer :

    Therefore, the principal solutions are x =  and 
    Therefore, the general solution is 


    Q4 :Find the general solution of cosec x = -2
    Answer :
    cosec x= -2

    Therefore, the principal solutions are x = 
    Therefore, the general solution is 


    Q5 :Find the general solution of the equation cos 4x = cos 2x
    Answer :


    Q6 :Find the general solution of the equation cos 3x + cos x – cos 2x = 0
    Answer :
    cos 3x + cos x – cos 2x = 0


    Q7 :Find the general solution of the equation sin 2x + cos x = 0
    Answer :
    sin 2x + cos x = 0

    Therefore, the general solution is 


    Q8 :Find the general solution of the equation sec2 2x = 1 – tan 2x
    Answer :
    sec2 2x = 1 – tan 2x

    Therefore, the general solution is 


    Q9 :Find the general solution of the equation  sin x + sin 3x + sin 5x =0
    Answer :
    sin x + sin 3x + sin 5x =0


    Therefore, the general solution is 


    Exercise Miscellaneous : Solutions of Questions on Page Number : 81
    Q1 :Prove that: 
    Answer :

    L.H.S.

    = 0 = R.H.S


    Q2 :Prove that: (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0
    Answer :
    L.H.S.= (sin 3x + sin x) sin x + (cos 3x – cos x) cos x
    = RH.S.


    Q3 :Prove that: 
    Answer :
    L.H.S. =


    Q4 :Prove that: 
    Answer :
    L.H.S. = 


    Q5 :Prove that: 
    Answer :
    It is known that 
    ∴L.H.S. = 


    Q6 :Prove that: 
    Answer :
    It is known that 
    L.H.S. = 
    = tan 6x
    = R.H.S.


    Q7 :Prove that: 
    Answer :
    L.H.S. = 


    Q8 :  , x in quadrant II
    Answer :
    Here, x is in quadrant II.
    i.e., 

    Therefore,  are all positive.

    As x is in quadrant II, cosx is negative.
    ∴ 




    Thus, the respective values of   are  


    Q9 :Find  for  , x in quadrant III
    Answer :
    Here, x is in quadrant III.

    Therefore,  and  are negative, where as  is positive.


    Now, 

    Thus, the respective values of  are 


    Q10 :Find  for  , x in quadrant II
    Answer :
    Here, x is in quadrant II.

    Therefore,,  and  are all positive.

    [cosx is negative in quadrant II]



    Thus, the respective values of  are   

     

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    • Chapter 1 – Sets
    • Chapter 2 – Relations and Functions
    • Chapter 3 – Trigonometric Functions
    • Chapter 4 – Principle of Mathematical Induction
    • Chapter 5 – Complex Numbers and Quadratic Equations
    • Chapter 6 – Linear Inequalities
    • Chapter 7 – Permutation and Combinations
    • Chapter 8 – Binomial Theorem
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    • Chapter 10 – Straight Lines
    • Chapter 11 – Conic Sections
    • Chapter 12 – Introduction to three Dimensional Geometry
    • Chapter 13 – Limits and Derivatives
    • Chapter 14 – Mathematical Reasoning
    • Chapter 15 – Statistics
    • Chapter 16 – Probability
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