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NCERT Class 11 Mathematics Chapter 3 Trigonometric Functions
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Trigonometric Functions
Chapter – 3
| Exercise 3.1 |
1. Find the radian measures corresponding to the following degree measures.
(i) 25°
Ans: 25°
We know that 180⁰ = π radian
(ii) -47°30
Ans: -47°30′
-47°30′ =
(iii) 240°
Ans: 240°
We know that 180° = π radian
(iv) 520°
Ans:
2. Find the degree measures corresponding to the following radian measures
Ans: 11/16
We know that ㅠ radian = 180°
(ii) -4
Ans:
Ans: 5π/3
We know that ㅠ radian = 180°
Ans: 7π/6
We know that ㅠ radian = 180°
3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
Ans: 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 2 ㅠ radian.
Hence, in 6 complete revolutions, it will turn an angle of 6 × 2 ㅠ radian, i.e.,
12 ㅠ radian
Thus, in one second, the wheel turns an angle of 12ㅠ radian.
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
Ans: We know that in a circle of radius r unit, if an arc of length / unit subtends an angle θ radian at the centre, then
Thus, the required angle is 12°36″².
5. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor are of the chord.
Ans: Diameter of the circle = 40 cm
∴ Radius (r) of the circle =
Let AB be a chord (length = 20 cm) of the circle.
In ∆OAB , ΟΑ = OB = Radius of circle = 20 cm
Also, AB = 20cm
Thus, ∆OAB is an equilateral triangle.
We know that in a circle of radius r unit, if an arc of length / unit subtends an angle θ radian at the centre, then θ = l/r.
Thus, the length of the minor arc of the chord is 20ㅠ /3 cm.
6. If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.
Ans: Let the radii of the two circles be r₁and r₂. Let an arc of length / subtend an angle of 60° at the centre of the circle of radius r,, while let an arc of length / subtend an angle of 75° at the centre of the circle of radius r₂.
We know that in a circle of radius r unit, if an arc of length / unit subtends an angle θ radian at the centre, then
Thus, the ratio of the radii is 5:4.
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.
Ans:
(ii) 15 cm.
Ans:
(iii) 21 cm.
Ans:
| Exercise 3.2 |
1. Find the values of the other five trigonometric functions in Exercises 1 to 5.
Ans:
Since x lies in the 3ʳᵈ quadrant, the value of sin x will be negative.
Ans:
Since x lies in the 2ⁿᵈ quadrant, the value of cos x will be negative
Ans
Since x lies in the 3ʳᵈ quadrant, the value of sec x will be negative.
Ans:
Since x lies in the 4ᵗʰ quadrant, the value of sin x will be negative.
Ans:
Since x lies in the 2ⁿᵈ quadrant, the value of sec x will be negative.
6. Find the values of the trigonometric functions in Exercises 6 to 10.
Ans: It is known that the values of sin x repeat after an interval of 2ㅠ or 360°.
7. cosec (- 1410⁰).
Ans: It is known that the values of cosec x repeat after an interval of 2ㅠ or 360°.
∴ cosec (-1410°) = cosec(-1410°+4×360°)
= cosec (-1410°+1440°)
= cosec30° = 2.
Ans: It is known that the values of tan x repeat after an interval of ㅠ or 180°.
Ans: It is known that the values of sin x repeat after an interval of 2 ㅠ or 360°.
Ans: It is known that the values of cot x repeat after an interval of ㅠ or 180°.
| Exercise 3.3 |
Prove that:
Ans:
Ans:
Ans:
Ans:
5. Find the value of:
(i) sin 75°.
Ans: 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°.
Ans: tan 15° = tan (45° – 30°)
Prove the following:
Ans:
Ans:
Ans:
Ans:
10. sin(n + 1)x sin(n + 2)x + cos(n + 1)x cos(n + 2) x = cos x.
Ans:
Ans:
12. sin² 6x – sin² 4x = sin 2x sin 10x.
Ans:
It is known
∴ L.H.S.= sin² 6x – sin² 4x
= (sin 6x + sin 4x) (sin 6x – sin4x)
= (2 sin 5x cos x)(2 cos 5x sin x)
= (2sin 5x cos 5x)(2sin x cos x)
= sin 10x sin 2x
= R.H.S.
13. cos² 2x – cos² 6x = sin 4x sin 8x.
Ans: It is known
that
∴ L.H.S.= cos² 2x – cos² 6x
= (cos 2x + cos 6x) (cos 2x – 6x)
= [2cos 4x cos 2x] [ – 2 sin 4x(- sin 2x)]
= (2sin 4x cos 4x)(2sin 2x cos 2x)
= sin 8x sin 4x
= R .H.S.
14. sin2 x + 2 sin 4x + sin 6x = 4cos² x sin 4x.
Ans: L.H.S.= sin 2x + 2sin 4x + sin 6x
= [sin 2x + sin 6x] + 2sin 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( 2cos² x – 1+1)
= 2 sin 4x (2cos² x)
= 4cos² x sin 4x.
15. cot 4x(sin 5x + sin 3x) = cot x (sin 5x – sin 3x).
Ans: L.H. S = cot 4x(sin 5x + sin 3x)
= 2cos 4x cos x
R.H.S. = cot x (sin 5x – sin 3x)
= 2 cos 4x. cos x
L.H.S. = R.H.S.
Ans: It is known that
Ans: It is known that
Ans: It is known that
Ans: It is known that
= tan 2x
= R.H.S.
Ans: It is known that
= – 2×(- sin x)
= 2 sin x = R.H.S.
Ans:
22. cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1.
Ans: L.H.S.= cot x cot 2x – cot 2x cot 3x – cot 3x cot x
=cot × cot 2x – cot 3x(cot 2x + cot x)
=cot × cot 2x – cot(2x + x) (cot 2x + cot x)
= cot x cot 2x – (cot 2x cot × – 1)
= 1 =R.H.S
Ans: It is known that
∴ L.H.S.= tan 4x = tan 2 (2x)
24. cos 4x = 1 – 8sin² x cos² x.
Ans: L.H.S.= cos 4x
= cos 2(2x)
= 1 – 2sin² 2x [cos 2A = 1 – 2sin² A]
= 1 – 2 (2sin x cos x)² [sin 2A = 2sin A cos A]
= 1 – 8 sin² x cos² x
= R.H.S.
25. cos 6x = 32cos⁶ x – 48cos⁴ x + 18cos² x – 1.
Ans: L.H.S.= cos 6x
= cos 3 (2x)
= 4 cos³ 2x – 3 cos 2x[cos 3A = 4cos³ A – 3 cos A]
=4[(2cos² x – 1)³ – 3(2 cos² x – 1)[cos 2x = 2cos² x – 1]
= 4[(2 cos² x)³ – (1)³ – 3(2cos² x)² + 3(2cos² x)] – 6cos² x + 3
= 4[8 cos⁶ x – 1 – 12 cos⁴ x + 6 cos² x] – 6 cos²x + 3
= 32 cos⁶ x – 4 – 48 cos⁴ x + 24cos² x – 6 cos² x + 3
= 32 cos⁶ x – 48 cos⁴ x + 18 cos² x – 1
= R .H.S.
Miscellaneous Exercise on Chapter 3
Ans:
2. (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0.
Ans: L.H.S.
= (sin 3x + sin x) sin x +(cos 3x – cos x)cos x
= sin 3x sin x + sin² x + cos 3x cos x – cos²x
= cos 3x cos x + sin 3x sin x – (cos² x – sin² x)
= cos(3x – x) – cos 2x [cos(A – B) = cos A cos B + sin A sin B]
= cos 2x – cos 2x
= 0
= RH.S.
Ans: L.H.S. = (cos x + cos y)² + (sin x – sin y)²
= cos² x + cos² y + 2cos x cos y + sin² x + sin² y – 2sin x sin y
= (cos² x + sin² x) + (cos² y + sin² y) + 2(cos x cos y – sin x sin y)
= 1 + 1 + 2cos(x + y) [cos(A + B) = (cos A cos B – sin A sin B)]
= 2 + 2cos(x + y)
= 2[1 + cos(x + y)]
Ans: L.H.S.= (cos x – cos y)² + (sin x – sin y)²
= cos² x + cos² y – 2cos x cos y + sin² x + sin² y – 2sin x sin y
= (cos² x + sin² x) + (cos² y + sin² y) – 2[cos x cos y + sin x sin y]
= 1 + 1 – 2[cos(x – y)] [cos(A – B) = cos A cos B + sin A sin B]
= 2[1 – cos(x – y)]
5. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x.
Ans:
Ans:
Ans: L.H.S.= sin 3x + sin 2x – sin x
Ans:
Ans:
Ans:
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