Plus One Maths Chapter Wise Questions and Answers Chapter 3 Trigonometric Functions

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Kerala Plus One Maths Chapter Wise Questions and Answers Chapter 3 Trigonometric Functions

Plus One Maths Trigonometric Functions Three Mark Questions and Answers

Question 1.
Prove the following
Plus One Maths Trigonometric Functions Three Mark Questions and Answers 1
Answer:
i) LHS
Plus One Maths Trigonometric Functions Three Mark Questions and Answers 2

ii) LHS
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Plus One Maths Chapter Wise Questions and Answers Chapter 3 Trigonometric Functions

iii) LHS = sin 2x + 2 sin 4x + sin 6x
= 2 sin 4xcos2x + 2sin 4x
= 2 sin 4x(cos2x + 1) = 4 cos2 x sin 4x

iv) LHS
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v) LHS
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vi) LHS
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vii) LHS = sin2 6x – sin2 4x
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= 2 sin 10x sin(-2x)
= 2 sin 10x sin2x

viii) LHS
Plus One Maths Trigonometric Functions Three Mark Questions and Answers 8

Plus One Maths Chapter Wise Questions and Answers Chapter 3 Trigonometric Functions

Question 2.
Find the general solution of the following equations.

  1. cos4x = cos2x
  2. sin 2x +cosx = 0
  3. cos3x + cosx – cos2x = 0

Answer:
1. Given; cos 4x = cos 2x
⇒ cos4x – cos 2x = 0
⇒ -2 sin 3x sin x = 0
General solution is
⇒ sin3x = 0; ⇒ 3x = nπ ⇒ x = \(\frac{n \pi}{3}\), ∈ Z
Again we have;
⇒ sinx = 0; ⇒ x = nπ; n ∈ Z

2. Given; sin 2x + cosx = 0
⇒ 2sin xcosx + cosx = 0
⇒ cosx(2sin x + 1) = 0
General solution is
⇒ cosx = 0 ⇒ x = (2n + 1) \(\frac{\pi}{2}\), n ∈ Z
Again we have; 2sin x + 1 = 0
Plus One Maths Trigonometric Functions Three Mark Questions and Answers 9

3. Given; cos3x +cosx – cos2x = 0
⇒ 2 cos2x cosx – cos2x = 0
⇒ cos2x(2cosx – 1) = 0
General solution is
Plus One Maths Trigonometric Functions Three Mark Questions and Answers 10
Again we have; 2cosx -1 = 0
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Plus One Maths Chapter Wise Questions and Answers Chapter 3 Trigonometric Functions

Question 3.
In Triangle ABC, if a = 25, b = 52 and c = 63, find cos A and sin A.
Answer:
Plus One Maths Trigonometric Functions Three Mark Questions and Answers 12

Question 4.
For any ΔABC, prove that a(b cosC – c cosB) = b2 – c2
Answer:
LHS = ab cos C – ac cos B
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Question 5.
For any ΔABC, prove that, \(\frac{\sin (B-C)}{\sin (B+C)}=\frac{b^{2}-c^{2}}{a^{2}}\).
Answer:
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Plus One Maths Chapter Wise Questions and Answers Chapter 3 Trigonometric Functions

Question 6.

  1. Convert \(\frac{2 \pi}{3}\) radian measure into degree measure. (1)
  2. Prove that \(\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x\) (2)

Answer:
1. \(\frac{2 \pi}{3}=\frac{2 \pi}{3} \times \frac{180}{\pi}=120^{\circ}\)

2. LHS
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Plus One Maths Trigonometric Functions Four Mark Questions and Answers

Question 1.
For any ΔABC, prove that
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Answer:
Plus One Maths Trigonometric Functions Three Mark Questions and Answers 17
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Plus One Maths Chapter Wise Questions and Answers Chapter 3 Trigonometric Functions

Question 2.
For any ΔABC, prove that \(\sin \frac{B-C}{2}=\frac{b-c}{a} \cos \frac{A}{2}\).
Answer:
Plus One Maths Trigonometric Functions Three Mark Questions and Answers 19
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Plus One Maths Chapter Wise Questions and Answers Chapter 3 Trigonometric Functions

Question 3.
(i) Which of the following is not possible. (1)
(a) sin x = \(\frac{1}{2}\)
(b) cos x = \(\frac{2}{3}\)
(c) cosec x = \(\frac{1}{3}\)
(d) tan x = 8
(ii) Find the value of sin 15°. (2)
(iii) Hence write the value of cos 75° (1)
Answer:
(i) (c) cosec x = \(\frac{1}{3}\)

(ii) sin 15° = sin(45° – 30°)
= sin45°cos30°- cos45°sin30°
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(iii) sin 15° = sin(90° – 75°) = cos 75°

Plus One Maths Trigonometric Functions Six Mark Questions and Answers

Question 1.
The angle of elevation of the top point P of the vertical tower PQ of height h from a point A is 45° and from a point B, the angle of elevation is 60°, where B is a point at a distance d from the point A measured along the line AB which makes angle 30° with AQ. Prove that d = h(\(\sqrt{3}\) – 1).
Answer:
Plus One Maths Trigonometric Functions Three Mark Questions and Answers 22
From the figure we have ∠PAQ = 45°, ∠BAQ = 30°and ∠PBH = 60°
in right ∆AQP
Clearly ∠APQ = 45°, ∠BPH = 30° , giving ∠APB = 15° ⇒ ∠PAB = 15°
In ∆APQ ,PQ = AQ = h
AP2 = h2 + h2 = 2h2 ⇒ AP = \(\sqrt{2}\)h
From ∆ABP,
Plus One Maths Trigonometric Functions Three Mark Questions and Answers 23

Plus One Maths Chapter Wise Questions and Answers Chapter 3 Trigonometric Functions

Question 2.
A tree stands vertically on a hill side which makes an angle of 15° with the horizontal. From a point on the ground 35m down the hill from the base of the tree, the angle of elevation of the top of the tree is 60°. Find the height of the tree.
Answer:
Plus One Maths Trigonometric Functions Three Mark Questions and Answers 24
Let BC represent the tree, A be the point 35m down the hill from the base of the tree and h be the height of the tree.
Clearly in ∆ABC
∠BAC = 60°- 15° =45°;
∠ACB = 30°; ∠ABC = 105°
Plus One Maths Trigonometric Functions Three Mark Questions and Answers 25

Question 3.
(i) If sin x = cos x, x ∈ [0, π] then is
(a) 0
(b) \(\frac{\pi}{4}\)
(c) \(\frac{\pi}{3}\)
(d) π
(ii) Write the following in ascending order of tits values, sin 100°, sin 0°, sin 50°, sin 200°
(iii) Solve: sin2x – sin4x + sin6x = 0
Answer:
(i) (b) \(\frac{\pi}{4}\)

(ii) sin 100° = sin(l 80 – 80) = sin 80°
sin 200° = sin(l 80° + 20°) = -sin 20°
The ascending order is
sin 200°, sin 0°, sin 50°, sin 100°

(iii) sin2x + sin6x – sin4x = 0
⇒ 2sin 4x cos2x – sin 4x = 0
⇒ sin 4x(2 cos 2x – 1) = 0
⇒ sin4x = 0 or (2cos2x – 1) = 0
⇒ 4x = nπ or cos2x = \(\frac{1}{2}\)
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Plus One Maths Chapter Wise Questions and Answers Chapter 3 Trigonometric Functions

Plus One Maths Trigonometric Functions Practice Problems Questions and Answers

Question 1.
Convert the following degree measure into radian measure.
i)  45°
ii) 25°
iii) 240°
iv) 40°20′
v) -47°30′
Answer:
Plus One Maths Trigonometric Functions Three Mark Questions and Answers 27

Plus One Maths Chapter Wise Questions and Answers Chapter 3 Trigonometric Functions

Question 2.
Convert the following radian measure into degree measure,
i)   6
ii) -4
iii) \(\frac{5 \pi}{3}\)
iv) \(\frac{7 \pi}{6}\)
v) \(\frac{11}{16}\)
Answer:
Plus One Maths Trigonometric Functions Three Mark Questions and Answers 28

Question 3.
The minute hand of a watch is 1.5 cm long. How far does its tip move in 40 minutes? (Use π = 3.14)
Answer:
60 minutes = 360 degrees.
1 minutes = 6 degrees.
40 minutes = 240 degrees.
240° = 240 × \(\frac{\pi}{180}=\frac{4 \pi}{3}\)
The required distance travelled = l = rθ
= 1.5 × \(\frac{4 \pi}{3}\) = 2 × 3.14 = 6.28 cm

Question 4.
In a circle of diameter 40 cm, the length of a cord is 20 cm. Find the length of minor arc of the chord.
Answer:
The radius and chord join to form a equilateral triangle. Therefore
l = rθ = 20 × \(\frac{\pi}{3}\)
= 20 × \(\frac{3.14}{3}\) = 20.933.
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Plus One Maths Chapter Wise Questions and Answers Chapter 3 Trigonometric Functions

Question 5.
If the arcs of the same lengths in the two circles subtend angles 65° and 110° at the centre, find the ratio of their radii.
Answer:
We have l = rθ, the radius and angle are inversely proportional. Therefore;
Plus One Maths Trigonometric Functions Three Mark Questions and Answers 30

Question 6.
Find the values of the other five trigonometric functions in the following; (2 score each)

  1. cos x = \(-\frac{3}{5}\), x lies in the third quadrant.
  2. cot x = \(-\frac{5}{12}\), x lies in the second quadrant.
  3. sin x = \(\frac{1}{4}\), x lies in the second quadrant.

Answer:
1. Given;
cos x = \(-\frac{3}{5}\)
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2. Given;
cot x = \(-\frac{5}{12}\)
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Plus One Maths Trigonometric Functions Three Mark Questions and Answers 33

3. Given;
sin x = \(\frac{1}{4}\); cosecx = 4
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Plus One Maths Chapter Wise Questions and Answers Chapter 3 Trigonometric Functions

Question 7.
Find the value of the trigonometric functions. (2 score each)
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Answer:
Plus One Maths Trigonometric Functions Three Mark Questions and Answers 37
Plus One Maths Trigonometric Functions Three Mark Questions and Answers 38

Question 8.
Find the value of the following.
Plus One Maths Trigonometric Functions Three Mark Questions and Answers 39
iv) sin 75°
v) tan 15°
Answer:
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Plus One Maths Chapter Wise Questions and Answers Chapter 3 Trigonometric Functions
iv) sin 75° = sin(45° + 35°)
= sin 45° cos30° + cos45° sin 30°
Plus One Maths Trigonometric Functions Three Mark Questions and Answers 43

v) tan 15° = tan(45° – 30°) = \(\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}}\)
Plus One Maths Trigonometric Functions Three Mark Questions and Answers 44

Plus One Maths Chapter Wise Questions and Answers Chapter 3 Trigonometric Functions

Question 9.
Find the principal and general solution of the following.

  1. sin x = \(\frac{\sqrt{3}}{2}\)
  2. cosx = \(\frac{1}{2}\)
  3. tan x = \(\sqrt{3}\)
  4. cos ecx = -2

Answer:
1. Given; sin x = \(\frac{\sqrt{3}}{2}\) = sin \(\frac{\pi}{3}\)
General solution is; x = nπ + (-1)n\(\frac{\pi}{3}\),
n ∈ Z
Put n = 0, 1 we get principal solution; x = \(\frac{\pi}{3} ; \frac{2 \pi}{3}\).

2. Given; cosx = \(\frac{1}{2}\) = cos \(\frac{\pi}{3}\)
General solution is; x = 2nπ ± \(\frac{\pi}{3}\), n ∈ Z
Put n = 0, 1 we get principal solution;
n = 0 ⇒ x = \(\frac{\pi}{3}\); n = 1 ⇒ x = 2π – \(\frac{\pi}{3}\) = \(\frac{5\pi}{3}\).

3. Given; tan x = \(\sqrt{3}\) = tan\(\frac{\pi}{3}\)
General solution is; ⇒ x = nπ + \(\frac{\pi}{3}\), n ∈ Z
Put n = 0, 1 we get principal solution;
n = 0 ⇒ x = \(\frac{\pi}{3}\); n = 1 ⇒ x = π + \(\frac{\pi}{3}\) = \(4\frac{\pi}{3}\).

4. Given; cosecx = -2
⇒ sin x = \(-\frac{1}{2}\) = – sin \(\frac{\pi}{6}\) = sin(-\(\frac{\pi}{6}\) )
General solution is; x = nπ – (-1)n \(\frac{\pi}{6}\), n ∈ Z
Put n = 1, 2 we get principal solution;
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