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

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 cos^{2} x sin 4x

iv) LHS

v) LHS

vi) LHS

vii) LHS = sin^{2} 6x – sin^{2} 4x

= 2 sin 10x sin(-2x)

= 2 sin 10x sin2x

viii) LHS

Question 2.

Find the general solution of the following equations.

- cos4x = cos2x
- sin 2x +cosx = 0
- 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

Question 3.

In Triangle ABC, if a = 25, b = 52 and c = 63, find cos A and sin A.

Answer:

Question 4.

For any ΔABC, prove that a(b cosC – c cosB) = b^{2} – c^{2}

Answer:

LHS = ab cos C – ac cos B

Question 5.

For any ΔABC, prove that, \(\frac{\sin (B-C)}{\sin (B+C)}=\frac{b^{2}-c^{2}}{a^{2}}\).

Answer:

Question 6.

- Convert \(\frac{2 \pi}{3}\) radian measure into degree measure. (1)
- 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

Question 1.

For any ΔABC, prove that

Answer:

Question 2.

For any ΔABC, prove that \(\sin \frac{B-C}{2}=\frac{b-c}{a} \cos \frac{A}{2}\).

Answer:

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

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

AP^{2} = h^{2} + h^{2} = 2h^{2} ⇒ AP = \(\sqrt{2}\)h

From ∆ABP,

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°

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)

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

- sin x = \(\frac{\sqrt{3}}{2}\)
- cosx = \(\frac{1}{2}\)
- tan x = \(\sqrt{3}\)
- 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;