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

Students can Download Chapter 3 Trigonometric Functions Questions and Answers, Plus One Maths Chapter Wise Questions and Answers helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.

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

Plus One Maths Trigonometric Functions Three Mark Questions and Answers

Plus One Maths Chapter Wise Questions And Answers Pdf Question 1.
Prove the following

Answer:
i) LHS

ii) LHS

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

v) LHS

vi) LHS

vii) LHS = sin² 6x – sin² 4x

= 2 sin 10x sin(-2x)
= 2 sin 10x sin2x

viii) LHS

Plus One Maths 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

3. Given; cos3x +cosx – cos2x = 0
⇒ 2 cos2x cosx – cos2x = 0
⇒ cos2x(2cosx – 1) = 0
General solution is

Again we have; 2cosx -1 = 0

Plus One Maths Trigonometry Equations Question 3.
In Triangle ABC, if a = 25, b = 52 and c = 63, find cos A and sin A.
Answer:

Plus One Maths Text Book Questions And Answers Question 4.
For any ΔABC, prove that a(b cosC – c cosB) = b² – c²
Answer:
LHS = ab cos C – ac cos B

Hsslive Maths Textbook Answers Plus One Question 5.
For any ΔABC, prove that, \(\frac{\sin (B-C)}{\sin (B+C)}=\frac{b^{2}-c^{2}}{a^{2}}\).
Answer:

Plus One Chapter Wise Questions And Answers 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

Plus One Maths Trigonometric Functions Four Mark Questions and Answers

Plus One Maths Chapter Wise Questions And Answers Question 1.
For any ΔABC, prove that

Answer:

Plus One Maths Questions And Answers Question 2.
For any ΔABC, prove that \(\sin \frac{B-C}{2}=\frac{b-c}{a} \cos \frac{A}{2}\).
Answer:

Plus One Trigonometry Questions And Answers 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°

(iii) sin 15° = sin(90° – 75°) = cos 75°

Plus One Maths Trigonometric Functions Six Mark Questions and Answers

Kerala Sslc Maths Chapter Wise 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:

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,

Important Questions For Class 11 Maths Trigonometry 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:

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°

Trigonometric Functions Class 11 Pdf State Board 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 = \(\frac{1}{2}\)

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:

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:

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.

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;

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 = \(-\frac{3}{5}\)

2. Given;
cot x = \(-\frac{5}{12}\)

3. Given;
sin x = \(\frac{1}{4}\); cosecx = 4

Question 7.
Find the value of the trigonometric functions. (2 score each)

Answer:

Question 8.
Find the value of the following.

iv) sin 75°
v) tan 15°
Answer:

iv) sin 75° = sin(45° + 35°)
= sin 45° cos30° + cos45° sin 30°

v) tan 15° = tan(45° – 30°) = \(\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}}\)

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;