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NCERT Solutions for Math Class 11 Chapter-3 Trigonometric Functions


Exercise 3.1 : Solutions of Questions on Page Number : 54

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 sec2x = 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