NCERT Solutions Class 11 Mathematics Chapter 3 Trigonometric Functions

Exercise 3.1

Question. 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 
∴ 25° = π/180° ×25 radian = 5π/36 radian  

(ii) -47° 30' 
- 47° 30' = -47.5 degree [1° = 60']
= -95/2 degree 
Since, 180° = π radian 

(iii) 240° 
We know that 180° = π radian
∴ 240° = (π/180) ×240 radian = (4/3)π radian

(iv) 520° 
We know that 180° = π radian
∴ 520° = π/180 ×520 radian = 26π/9 radian

Question. Find the degree measures corresponding to the following radian measures (use π = 22/7). 
(i) 11/16    
(ii) -4
(iii) 5π/3 
(iv) 7π/6 
Answer : 
(i) 11/16

(ii) -4
We know that π radian = 180° 

(iii) 5π/3
We know that π radian = 180° 
∴ 5π/3 radian = 180/π × 5π/3 degree = 300° 

(iv) 7π/6 
We know that π radian = 180° 
∴ 7π/6 radian = 180/π × 7π/6 = 210°

Question. 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 = 360/60 = 6
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.

Question. 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 π = 22/7) . 
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
θ = 1/r
Therefore, for = 100 cm, l = 22 cm, we have

Question. 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 = 40/2 cm = 20 cm
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° = π/3 radian
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 θ = l/r. 

Question. 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° = π/3 radian and 75° = 5π/12 radian 
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 θ = 1/r or l = rθ . 

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

Question. 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 θ = l/r.
It is given that r = 75 cm
(i) Here, l = 10 cm
θ = 10/75 radian = 2/15 radian
(ii) Here, l = 15 cm  
θ = 15/75 radian = 1/5 radian
(iii) Here, l = 21 cm 
θ = 21/75  radian = 7/25 radian

Exercise 3.2

Question. Find the values of other five trigonometric functions in cosx = −1/2 , x lies in third quadrant.
Answer : 
Given:
cosx = −1/2 ,
sec x = 1/cosx
Substituting the values
 =  1/ (-1/2) = -2

Question. Find the values of other five trigonometric functions if sin x = 3/5 x lies in quadrant 
Answer : 
sin x = 3/5 
cosec x = 1/sin x = 1/(3/5) = 5/3 
sin²x + cos² x = 1 
⇒ cos² x = 1 - sin² x 
⇒ cos² x = 1 - (3/5)² 
⇒ cos² x = 1 - 9/25 
⇒ cos² x = 16/25 
⇒ cos x = ± 4/5 
Since x lies in the 2nd quadrant, the value of cos x will be negative 
∴ cos x = -4/5 

Question. Find the values of other five trigonometric functions if cot x = 3/4, x lies in third quadrant.  
Answer : 
cot x = 3/4 
tan = 1/cos x = 1/ (3/4) = 4/3
1 + tan² x = sec² x 
⇒ 1 + (4/3)² = sec² x 
⇒ 1 + 16/9 = sec² x 
⇒ 25/9 = sec² x 
⇒ sec x = ± 5/3 
Since x lies in the 3rd quadrant, the value of sec x will be negative. 
∴ sec x = -5/3 

Question. Find the values of other five trigonometric functions if sec x = 13/5 , x lies in fourth quadrant. 
Answer : 
sec x = 13/5 

Question. Find the values of other five trigonometric functions if tan x = -5/12, x lies in second quadrant. 
Answer : 
tan x = -5/12 
cot x = 1/tan x = 1/(-5/12) = -(12/5)

Question. 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°.
∴ sin 765° = sin(2×360° + 45°) = sin 45° = 1/√2.

Question. 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° . 
∴ cosec (-1410°) = cosec (-1410° + 4×360°) 
= cosec (-1410° + 1440°)
= cosec 30° = 2

Question. Find the value of the trigonometric function tan(19π/3) .
Answer : 
It is known that the values of tan x repeat after an interval of π or  180° . 

Question. Find the value of the trigonometric function sin(-11π/3)
Answer : 
It is known that the values of sin x repeat after an interval of 2π or 360° . 

Question. Find the value of the trigonometric function cot(-15π/4) 
Answer : 
It is known that the values of cot x repeat after an interval of π or  180° . 

Exercise 3.3

Question. sin² (π/6) + cos² (π/3) – tan² (π/4) = -1/2 
Answer : 

Question. Prove that 2sin² (π/6) + cosec² (7π/6) cos² (π/3) = 3/2 
Answer : 

Question. Prove that cot² (π/6) + cosec (5π/6) + 3 tan² (π/6) = 6 
Answer : 

Question. Prove that 2sin² (3π/4) + 2cos² (π/4) + 2sec² (π/3) = 10 
Answer : 

Question. 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]

Question. Prove that : cos(π/4 - x) cos (π/4 - y) - sin(π/4 - x) sin(π/4 - y) = sin(x + y) 
Answer : 

Question. Prove that : tan(π/4 + x)/tan(π/4 - x) = [(1+ tan x)/(1 - tan x)² 
Answer : 

Question. Prove that  [cos(π + x) cos (-x)]/[sin(π - x) cos(π/2 + x)] = cot² x 
Answer : 

Question. cos(3π/2 + x) cos(2π + x) [cot(3π/2 - x) + cot(2π + x)] = 1
Answer : 

L.H.S = cos(3π/2 + x) cos(2π + x) [cot(3π/2 - x) + cot(2π + x)] 
= sin x cos x[tan x + cot x] 

Question. 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 
= (1/2)[2sin(n + 1)x sin(n + 2)x + 2cos(n + 1)x cos(n + 2)x]

Question. Prove that cos(3π/4 + x) - cos(3π/4 - x) = -√2 sin x 
Answer : 

= -2sin(3π/4) sin x
= -2sin(π - π/4) sin x 
= -2 sin (π/4) sin x 
= -2 × 1/√2 × sin x 
= -√2 sin x 
= R.H.S. 

Question. Prove that sin² 6x – sin² 4x = sin 2x sin 10x 
Answer : 
It is known that sin A + sin B = 2 sin (A + B)/2 cos (A - B)/2 , sin A - sin B = 2 cos (A + B)/2 sin (a - B)/2  . 
∴ L.H.S. = sin² 6x - sin² 4x 
= (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.

Question. Prove that cos² 2x – cos² 6x = sin 4x sin 8x 
Answer : 
It is known that cos A + cos B = [2 cos(A+B)/2][cos(A-B)/2], cos A - cos B = -[2 sin(A+B)/2][sin (A-B)/2].
∴ 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 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. 

Question. Prove that sin 2x + 2sin 4x + sin 6x = 4cos² x sin 4x
Answer : 
L.H.S = sin 2x + 2sin 4x + sin 6x 
= [sin 2x + sin 6x]  + 2sin4x 

[∵ sin A + sin B = 2sin (A + B)/2. cos(A - B)/2]
= 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.

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

= (cos x/sin x) [2 cos 4x sin x] 
= 2 cos 4x. cos x
= L.H.S = R.H.S 

Question. Prove that (cos 9x - cos 5x)/(sin 17x - sin 3x) = -sin 2x/cos 10x  
Answer : 
It is known that  
cos A - cos B = -2 sin(A+B)/2 .sin(A-B)/2, sin A - sin B = 2 cos(A+B)/2 .sin(A-B)/2

Question. Prove that (sin 5x + sin 3x)/(cos 5x + cos 3x) = tan 4x 
Answer : 
It is known that 
cos A + cos B = 2cos(A+B)/2 .cos(A-B)/2, sin A + sin B = 2sin(A+B)/2 .cos(A-B)/2 

Question. Prove that (sin x - sin 3x)/(sin² x - cos² x) = 2 sin x 
Answer : 
It is known that  
sin A - sin B = 2cos(A+B)/2 .sin (A-B)/2, cos²A - sin²A = cos2A 

Question. Prove that (cos 4x + cos 3x + cos 2x)/(sin 4x + sin 3x + sin 2x) = cot 3x
Answer : 

Question. 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)

Question. Prove that tan 4x = [4 tan x(1 - tan² x)]/(1 - 6 tan² x + tan⁴ x) 
Answer : 
It is known that tan 2A = 2 tan A/(1 - tan² A
∴ L.H.S = tan 4x = tan 2(2x) 

Question. 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.

Question. 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 cos2x [cos 3A = 4 cos³ A – 3 cosA]
= 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

Question. Find the principal and general solutions of the equation tan x = √x 
Answer : 
tan x = √x 
It is known that tan(π/3) = √3 and tan (4π/3) = tan(π + π/3) = tan (π/3) = √3 
Therefore, the principal solutions are x = π/3 and 4π/3. 
Now, tan x = tan (π/3) 
⇒ x = nπ + π/3, where n ∈ Z 
Therefore, the general solution is x = nπ + π/3, where n ∈ Z

Question. Find the principal and general solutions of the equation sec x = 2 
Answer : 
sec x = 2 
It is known that sec(π/3) = 2 and sec(5π/3) = sec(2π - π/3) = sec(π/3) = 2 
Therefore, the principal solutions are x = π/3 and 5π/3. 
Now, sec π = sec(π/3) 
⇒ cos x = cos (π/3)  [sec x = 1/cos x] 
⇒ x = 2nπ ± π/3, where n ∈ Z 
Therefore, the general solution is x = 2nπ ± π/3, where n ∈ Z

Question. Find the principal and general solutions of the equation cot x = -√3 
Answer :  
cos x = -√3 
It is known that cot (π/6) = √3

Question. Find the general solution of cosec x = –2
Answer : 
cosec x = -2 
It is known that 
cosec (π/6) = 2 
∴ cosec(π + π/2) = -cosec(π/2) = -2 and cosec(2π - π/6) = -cosec (π/6) = -2 
i.e., cosec (7π/6) = -2 and cosec(11x/6) = -2 
Therefore, the principal solution are x = 7π/6 and 11π/6. 
Now, cosec x = cosec(7π/6) 
⇒ sin x = sin(7π/6)  [cosec x = 1/sin x] 
⇒ x = nπ + (-1)ⁿ (7π/6) , where n ∈ Z 
Therefore, the general solution is x = nπ + (-1)ⁿ (7π/6) where n ∈ Z.

Question. Find the general solution of the equation cos 4x = cos 2x 
Answer : 
cos 4x = cos 2x 
⇒ cos 4x - cos 2x = 0 

⇒ sin 3x sin x = 0 
⇒ sin 3x = 0 or sin x = 0 
∴ 3x = nπ or x = nπ, where n ∈ Z
⇒ x = nπ/3  or x = nπ, where n ∈ Z

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

⇒ 2 cos 2x cos x - cos 2x = 0 
⇒ cos 2x (2 cos x - 1) = 0 
⇒ cos 2x = 0 or 2 cos x - 1 = 0 
⇒ cos 2x = 0 or cos x = 1/2 
∴ 2x = (2n + 1)π/2 or cos x = cos π/3, where n ∈ Z 
⇒ x = (2n + 1)π/4 or x = 2nπ ± π/3, where n ∈ Z

Question. Find the general solution of the equation sin 2x + cos x = 0
Answer : 
sin 2x + cos x = 0 
⇒ 2 sin x cos x + cos x = 0 
⇒ cos x = 0 or  2 sin x + 1 = 0 
Now, cos x = 0
⇒ cos x = (2n+1)π/2, where n ∈ Z 
2 sin x + 1 = 0

Therefore, the general solution is (2n + 1)π/2 or nπ + (-1)ⁿ 7π/6, n ∈ Z.

Question. Find the general solution for each of the following equations sec² 2x = 1– tan 2x
Answer : 
sec² 2x = 1– tan 2x 
⇒ 1 + tan² 2x = 1 - tan 2x 
⇒ tan² 2x + tan 2x = 0 
⇒ tan 2x(tan 2x + 1) = 0 
⇒ tan 2x = 0   or tan 2x + 1 = 0 
Now, tan 2x = 0 
⇒ tan 2x = tan 0 
⇒ 2x = nπ + 0 , where n ∈ Z 
⇒ x = nπ/2 , where n ∈ Z 
tan 2x + 1 = 0 
⇒ tan 2x = -1 = -tan(π/4) = tan(π - π/4) = tan(3π/4) 
⇒ 2x = nπ + 3π/4, where n ∈ Z 
⇒ x = nπ/2 + 3π/8, where n ∈ Z 
Therefore, the general solution is nπ/2 or nπ/2 + 3π/8 , n ∈ Z.

Question. Find the general solution of the equation sin x + sin 3x + sin 5x = 0
Answer : 
sin x + sin 3x + sin 5x = 0 
(sin x + sin 5x) + sin 3x = 0 

⇒ 2 sin 3x cos(-2x) + sin 3x  = 0
⇒ 2 sin 3x cos 2x + sin 3x = 0 
⇒ sin 3x(2 cos 2x + 1) = 0 
⇒ sin 3x = 0 or 2 cos 2x + 1 = 0 
Now, sin 3x = 0 ⇒ 3x = nπ , where n ∈ Z 
i.e., x = nπ/3, where n ∈ Z 
i.e., x = nπ/3, where n ∈ Z 
2 cos 2x + 1 = 0 

Question. Find, sin(x/2) cos(x/2) and tan(x/2) for sin x = 1/4 , x in quadrant II 
Answer : 
Here, x is in quadrant II. 
i.e., π/2 < x < π 
⇒ π/4 < x/2 < π/2 
Therefore, sin (x/2), cos (x/2), and tan(x/2) are all positive. 
It is given that sin x = 1/4. 
cos²x = 1 - sin²x  = 1 - (1/4)² = 1 - 1/16 = 15/16 
⇒ cos x = -√15/4  [cos x is negative in quadrant II]