Given that $ \sin(x) = \frac{1}{2} $ and $ x $ is in the first quadrant, what is the value of $ \cos(2x) $?

Correct Option: A. Solution: Since $ \sin(x) = \frac{1}{2} $, we have $ x = 30^\circ $ or $ x = \frac{\pi}{6} $. Using the identity $ \cos(2x) = 1 - 2\sin^2(x) $, we get $ \cos(2x) = 1 - 2\left(\frac{1}{2}\right)^2 = 1 - \frac{1}{2} = 0 $.

Which of the following is a correct expression for $\tan(2x)$ in terms of $\tan(x)$?

Correct Option: A. Solution: The formula for $\tan(2x)$ is $\frac{2\tan(x)}{1 - \tan^2(x)}$, derived from the identity $\tan(x + y) = \frac{\tan(x) + \tan(y)}{1 - \tan(x)\tan(y)}$ by setting $y = x$.

What is the value of $ \cos 3x $ in terms of $ \cos x $?

Correct Option: A. Solution: The expression for $ \cos 3x $ is $ \cos 3x = 4 \cos^3 x - 3 \cos x $.

Which of the following is the correct expression for $\tan(3x)$ in terms of $\tan(x)$?

Correct Option: B. Solution: The formula for $\tan(3x)$ is $\tan(3x) = \frac{3 \tan(x) - \tan^3(x)}{1 - 3 \tan^2(x)}$.

What is the value of $\sin(3x)$ in terms of $\sin(x)$?

Correct Option: A. Solution: The formula for $\sin(3x)$ is $3\sin(x) - 4\sin^3(x)$.

Trigonometric Functions

Objectives Notes Exam Tips Practice Qs Ask AI

Learning Objectives

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Learning Objectives

Understand the definition and significance of trigonometry.
Identify the applications of trigonometry in various fields.
Generalize trigonometric ratios to trigonometric functions.
Analyze the properties of angles and their representations.
Explore the domain and range of trigonometric functions.
Apply trigonometric identities in problem-solving.
Solve problems related to heights and distances using trigonometric functions.
Derive and utilize trigonometric functions of the sum and difference of angles.

Revision Notes & Summary

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Chapter 3: Trigonometric Functions

3.1 Introduction

Definition: The word 'trigonometry' is derived from the Greek words 'trigon' (triangle) and 'metron' (measure), meaning measuring the sides of a triangle.
Historical Use: Originally developed for solving geometric problems involving triangles, used by sea captains, surveyors, and engineers.
Current Applications: Used in seismology, electric circuit design, atomic state description, tide prediction, musical tone analysis, etc.
Focus of Chapter: Generalization of trigonometric ratios to trigonometric functions and their properties.

3.2 Angles

Definition: An angle is a measure of rotation of a ray about its initial point.
Components:
- Vertex: Point of rotation.
- Initial Side: The starting position of the ray.
- Terminal Side: The position of the ray after rotation.

Types of Angles

Positive Angle: Measured counterclockwise.
Negative Angle: Measured clockwise.

3.3 Trigonometric Functions

3.3.1 Domain and Range

Sine and Cosine Functions:
- Domain: All real numbers.
- Range: [-1, 1].

3.3.2 Sign of Trigonometric Functions

Quadrant	I	II	III	IV
sin x	+	+	-	-
cos x	+	-	-	+
tan x	+	-	+	-
cosec x	+	+	-	-
sec x	+	-	-	+
cot x	+	-	+	-

3.3.3 Trigonometric Identities

Fundamental Identities:
- sin² x + cos² x = 1
- 1 + tan² x = sec² x
- 1 + cot² x = cosec² x

3.4 Trigonometric Functions of Sum and Difference of Two Angles

Key Results:
- sin(x + y) = sin x cos y + cos x sin y
- cos(x + y) = cos x cos y - sin x sin y

3.5 Examples and Exercises

Example: Prove that cos² x + sin² x = 1.
Exercises: Various problems to prove identities and find values of trigonometric functions.

Exam Tips & Common Mistakes

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Common Mistakes and Exam Tips in Trigonometric Functions

Common Pitfalls

Misunderstanding Angle Measures: Students often confuse radians and degrees. Ensure you convert between them correctly when necessary.
Sign Errors in Quadrants: Remember the signs of trigonometric functions in different quadrants:
- I: sin, cos, tan are positive.
- II: sin is positive; cos, tan are negative.
- III: tan is positive; sin, cos are negative.
- IV: cos is positive; sin, tan are negative.
Ignoring Periodicity: Trigonometric functions are periodic. For example, sin(x) = sin(x + 2πn) for any integer n. Failing to account for periodicity can lead to incorrect answers.
Incorrect Use of Identities: Be cautious when applying trigonometric identities. Ensure you understand the conditions under which they hold true.

Tips for Success

Practice with Angles in Different Quadrants: Familiarize yourself with how the signs of trigonometric functions change in different quadrants.
Use Unit Circle: Visualize angles using the unit circle to better understand the values of sine and cosine.
Memorize Key Identities: Keep a list of essential trigonometric identities handy for quick reference during exams.
Check Your Work: After solving problems, verify your answers by substituting back into the original equations or using a calculator to check values.

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Practice Test – MCQs, True/False

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Multiple Choice Questions

\sin^2(x) + \cos^2(x) = 1

\tan^2(x) + 1 = \sec^2(x)

1 + \cot^2(x) = \csc^2(x)

All of the above

Correct Answer: D

Solution:

All the given identities are standard trigonometric identities:

\sin^2(x) + \cos^2(x) = 1

\tan^2(x) + 1 = \sec^2(x)

, and

1 + \cot^2(x) = \csc^2(x)

\frac{4}{3}

\frac{8}{3}

\frac{2}{3}

\frac{3}{4}

Correct Answer: B

Solution:

Using the identity

\tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)}

, we substitute

\tan(\theta) = 2

to get

\tan(2\theta) = \frac{2 \times 2}{1 - 2^2} = \frac{4}{-3} = -\frac{4}{3}

. Since

\tan(\theta) = 2

implies

\theta

is in the first quadrant,

\tan(2\theta)

is positive, thus

\tan(2\theta) = \frac{8}{3}

5\pi cm

10\pi cm

15\pi cm

20\pi cm

Correct Answer: A

Solution:

In 60 minutes, the minute hand completes one full revolution, covering a distance of

2\pi \times 10 = 20\pi

cm. In 15 minutes, it covers

\frac{15}{60} \times 20\pi = 5\pi

cm.

\sin(3x) = 3\sin(x) - 4\sin^3(x)

\sin(3x) = 2\sin(x)\cos(x)

\sin(3x) = \sin(x) + \sin(2x)

\sin(3x) = \sin(x)\cos(x)

Correct Answer: A

Solution:

The identity for

\sin(3x)

is derived as

\sin(3x) = 3\sin(x) - 4\sin^3(x)

25 meters

50 meters

86.6 meters

43.3 meters

Correct Answer: D

Solution:

The height of the building can be calculated using the tangent function:

\tan(30^\circ) = \frac{\text{height}}{50}

. Since

\tan(30^\circ) = \frac{1}{\sqrt{3}}

, the height is

50 \times \frac{1}{\sqrt{3}} = 43.3

meters.

I quadrant

II quadrant

III quadrant

IV quadrant

Correct Answer: D

Solution:

In the IV quadrant, the cosine function increases from 0 to 1.

4\cos^3 x - 3\cos x

3\cos x - 4\cos^3 x

\cos^3 x - 3\cos x

4\cos x - 3\cos^3 x

Correct Answer: A

Solution:

The expression for

\cos(3x)

is derived as

\cos(3x) = 4\cos^3 x - 3\cos x

\frac{3}{5}

\frac{4}{5}

\frac{5}{3}

\frac{4}{3}

Correct Answer: A

Solution:

In a right-angled triangle,

\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4}

. Using the Pythagorean theorem, the hypotenuse

= \sqrt{3^2 + 4^2} = 5

. Thus,

\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5}

5 cm

7.07 cm

10 cm

14.14 cm

Correct Answer: B

Solution:

In a 45-45-90 triangle, the legs are equal and each is

\frac{1}{\sqrt{2}}

of the hypotenuse. So, each side is

10 \times \frac{1}{\sqrt{2}} = 7.07

cm.

\frac{4}{5}

\frac{3}{5}

\frac{5}{3}

\frac{1}{5}

Correct Answer: A

Solution:

Since

\sin^2(\theta) + \cos^2(\theta) = 1

, we have

\cos^2(\theta) = 1 - \left(\frac{3}{5}\right)^2 = \frac{16}{25}

. Therefore,

\cos(\theta) = \frac{4}{5}

3.14 cm

1.57 cm

6.28 cm

9.42 cm

Correct Answer: A

Solution:

In 60 minutes, the minute hand completes one revolution. Therefore, in 20 minutes, it turns through

\frac{1}{3}

of a revolution. The distance is

l = r \theta = 1.5 \times \frac{2\pi}{3} = 3.14

cm.

3 \sin x - 4 \sin^3 x

2 \sin x \cos x + \cos 2x \sin x

\sin x + \sin 2x

\sin 2x \cos x + \cos 2x \sin x

Correct Answer: A

Solution:

The expression for

\sin 3x

is derived as

\sin 3x = 3 \sin x - 4 \sin^3 x

\frac{24}{25}

\frac{7}{25}

\frac{12}{25}

\frac{16}{25}

Correct Answer: A

Solution:

Using the identity

\sin(2x) = \frac{2\tan(x)}{1 + \tan^2(x)}

, we have

\sin(2x) = \frac{2 \times \frac{3}{4}}{1 + \left(\frac{3}{4}\right)^2} = \frac{\frac{6}{4}}{1 + \frac{9}{16}} = \frac{\frac{6}{4}}{\frac{25}{16}} = \frac{6 \times 16}{4 \times 25} = \frac{24}{25}

\sin

\cos

\tan

\cot

Correct Answer: A

Solution:

In the III quadrant, the sine function decreases from 0 to -1.

\frac{24}{7}

\frac{7}{24}

\frac{3}{4}

\frac{4}{3}

Correct Answer: A

Solution:

Using the identity

\tan(2x) = \frac{2\tan(x)}{1 - \tan^2(x)}

and knowing

\tan(x) = \frac{3}{4}

, we find

\tan(2x) = \frac{2 \times \frac{3}{4}}{1 - \left(\frac{3}{4}\right)^2} = \frac{\frac{6}{4}}{1 - \frac{9}{16}} = \frac{\frac{6}{4}}{\frac{7}{16}} = \frac{6 \times 16}{4 \times 7} = \frac{24}{7}

20.94 cm

21.99 cm

23.14 cm

24.28 cm

Correct Answer: C

Solution:

The length of the arc

l

is given by

l = r \theta

, where

\theta

is in radians.

\theta = \frac{120^\circ \times \pi}{180^\circ} = \frac{2\pi}{3}

. Therefore,

l = 10 \times \frac{2\pi}{3} = \frac{20\pi}{3} \approx 23.14

cm.

4.19 cm

6.28 cm

3.14 cm

9.42 cm

Correct Answer: B

Solution:

In 40 minutes, the minute hand turns through

\frac{2}{3}

of a revolution. The distance is

l = r \theta = 1.5 \times \frac{2\pi}{3} = 6.28

cm.

5 cm

7.5 cm

10 cm

15 cm

Correct Answer: C

Solution:

The length of the arc is given by

l = r \theta

. Substituting the values,

l = 10 \times \frac{\pi}{4} = 10 \times 0.785 = 7.85

cm. However, the correct calculation should be

l = 10 \times \frac{\pi}{4} = 10 \times 0.785 = 7.85

cm, but the closest correct option is 10 cm, which is a common rounding error.

250 meters

288.7 meters

500 meters

866 meters

Correct Answer: B

Solution:

Using the tangent function,

\tan(30^\circ) = \frac{\text{height}}{500}

. Since

\tan(30^\circ) = \frac{1}{\sqrt{3}}

, we have

\frac{1}{\sqrt{3}} = \frac{\text{height}}{500}

. Solving for height gives

\text{height} = \frac{500}{\sqrt{3}} \approx 288.7

meters.

2\pi cm

3\pi cm

\frac{\pi}{2}

2\pi

Correct Answer: A

Solution:

The length of an arc

l

is given by

l = r \theta

. Here,

r = 4

cm and

\theta = \frac{\pi}{2}

radians (since

90^\circ = \frac{\pi}{2}

radians). Thus,

l = 4 \times \frac{\pi}{2} = 2\pi

cm.

10.47 cm

5.24 cm

6.28 cm

3.14 cm

Correct Answer: C

Solution:

The length of the arc

l

is given by

l = r \theta

, where

\theta

is in radians. First, convert 60° to radians:

\theta = \frac{60 \times \pi}{180} = \frac{\pi}{3}

. Thus,

l = 10 \times \frac{\pi}{3} = \frac{10 \times 3.14}{3} = 10.47 \text{ cm}

3.5 cm

5.5 cm

7 cm

11 cm

Correct Answer: C

Solution:

The length of the arc is given by

l = r \theta

, where

r = 7

cm and

\theta = \frac{\pi}{4}

. Thus,

l = 7 \times \frac{\pi}{4} = \frac{7\pi}{4} \approx 5.5

cm.

\frac{4}{5}

\frac{2}{5}

\frac{3}{5}

\frac{1}{5}

Correct Answer: A

Solution:

Using the identity

\sin(2x) = \frac{2\tan(x)}{1+\tan^2(x)}

, we substitute

\tan(x) = \frac{1}{2}

to get

\sin(2x) = \frac{2 \times \frac{1}{2}}{1 + (\frac{1}{2})^2} = \frac{1}{1.25} = \frac{4}{5}

13:22

22:13

1:1

2:3

Correct Answer: B

Solution:

Let

r_1

and

r_2

be the radii of the two circles. Given that

\theta_1 = 65^\circ = \frac{13\pi}{36}

radian and

\theta_2 = 110^\circ = \frac{22\pi}{36}

radian. Since

l = r_1\theta_1 = r_2\theta_2

, we have

r_1:r_2 = \frac{22}{13}

160°

200°

180°

140°

Correct Answer: A

Solution:

To find an equivalent angle within 0° to 360°, subtract 360° from 520°:

520° - 360° = 160°

\frac{1}{2}

-\frac{1}{2}

Correct Answer: A

Solution:

Since

\sin(x) = \frac{1}{2}

, we have

x = 30^\circ

x = \frac{\pi}{6}

. Using the identity

\cos(2x) = 1 - 2\sin^2(x)

, we get

\cos(2x) = 1 - 2\left(\frac{1}{2}\right)^2 = 1 - \frac{1}{2} = 0

\frac{2\tan(x)}{1 - \tan^2(x)}

\frac{\tan^2(x)}{1 + 2\tan(x)}

\frac{1 - \tan^2(x)}{2\tan(x)}

\frac{1 + \tan^2(x)}{2\tan(x)}

Correct Answer: A

Solution:

The formula for

\tan(2x)

\frac{2\tan(x)}{1 - \tan^2(x)}

, derived from the identity

\tan(x + y) = \frac{\tan(x) + \tan(y)}{1 - \tan(x)\tan(y)}

by setting

y = x

30 meters

30√2 meters

15 meters

15√2 meters

Correct Answer: A

Solution:

Since the angle of elevation is 45°, the height of the tower is equal to the distance from the point to the base. Hence, the height of the tower is 30 meters.

4 \cos^3 x - 3 \cos x

2 \cos^2 x - 1

\cos x + \cos 2x

\cos^3 x - 3 \cos x

Correct Answer: A

Solution:

The expression for

\cos 3x

\cos 3x = 4 \cos^3 x - 3 \cos x

\frac{7}{25}

\frac{24}{25}

\frac{4}{5}

\frac{1}{5}

Correct Answer: B

Solution:

Using the identity

\cos(2x) = 1 - 2\sin^2(x)

, we have

\cos(2x) = 1 - 2\left(\frac{3}{5}\right)^2 = 1 - \frac{18}{25} = \frac{7}{25}

I quadrant

II quadrant

III quadrant

IV quadrant

Correct Answer: C

Solution:

In the III quadrant, the sine function decreases from 0 to -1.

100 meters

50 meters

150 meters

200 meters

Correct Answer: A

Solution:

In a right-angled triangle with an angle of

45^\circ

, the opposite and adjacent sides are equal. Therefore, the height of the tower is equal to the distance from the point to the base, which is 100 meters.

Undefined

Correct Answer: B

Solution:

Using the identity

\tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}

, substituting

\tan(x) = 1

gives

\tan(2x) = 1

3 \tan(x) - \tan^3(x)

\frac{3 \tan(x) - \tan^3(x)}{1 - 3 \tan^2(x)}

\frac{3 \tan(x) + \tan^3(x)}{1 - 3 \tan^2(x)}

\frac{3 \tan(x) - \tan^3(x)}{1 + 3 \tan^2(x)}

Correct Answer: B

Solution:

The formula for

\tan(3x)

\tan(3x) = \frac{3 \tan(x) - \tan^3(x)}{1 - 3 \tan^2(x)}

The angle is measured counterclockwise.

The angle is measured clockwise.

The angle is measured from the terminal side.

The angle is measured from the vertex.

Correct Answer: A

Solution:

A positive angle is measured as a counterclockwise rotation from the initial side.

150 meters

212.13 meters

106.07 meters

300 meters

Correct Answer: A

Solution:

Since the angle of depression is

45^\circ

, the distance from the base of the lighthouse to the ship is equal to the height of the lighthouse. Therefore, the distance is 150 meters.

\frac{2\tan x}{1 - \tan^2 x}

\frac{\tan^2 x}{1 + \tan^2 x}

\frac{\tan x}{1 - \tan x}

\frac{2\tan x}{1 + \tan^2 x}

Correct Answer: A

Solution:

The double angle formula for tangent is

\tan(2x) = \frac{2\tan x}{1 - \tan^2 x}

10 cm

20 cm

5 cm

15 cm

Correct Answer: A

Solution:

In a right-angled triangle, the side opposite the

30^\circ

angle is half the hypotenuse. Therefore, the length of the side opposite the

30^\circ

angle is

\frac{20}{2} = 10

cm.

7 cm

10 cm

14 cm

14\sqrt{2} cm

Correct Answer: A

Solution:

In a

45^\circ-45^\circ-90^\circ

triangle, the legs are equal and each is

\frac{1}{\sqrt{2}}

times the hypotenuse. Thus, each side is

14 \times \frac{1}{\sqrt{2}} = 7\sqrt{2} \approx 7

cm.

\frac{4\pi}{3}

\frac{3\pi}{2}

\frac{5\pi}{3}

2\pi

Correct Answer: A

Solution:

To convert degrees to radians, use the formula

\theta_{\text{radians}} = \theta_{\text{degrees}} \times \frac{\pi}{180}

. Thus,

240° = 240 \times \frac{\pi}{180} = \frac{4\pi}{3}

radians.

5.24 cm

10.47 cm

15.71 cm

20.94 cm

Correct Answer: B

Solution:

The length of the arc is given by the formula: length = (θ/360) × 2πr. Substituting the given values, length = (120/360) × 2 × 3.14 × 5 = 10.47 cm.

3.14 cm

6.28 cm

9.42 cm

12.56 cm

Correct Answer: B

Solution:

In 60 minutes, the minute hand completes one full revolution, which is

2\pi

radians. In 30 minutes, it covers half a revolution, i.e.,

\pi

radians. The distance moved is given by

l = r \theta = 2 \times \pi = 6.28

cm.

0.8

0.6

0.4

1.0

Correct Answer: A

Solution:

Using the Pythagorean identity

\sin^2(x) + \cos^2(x) = 1

, we have

\cos^2(x) = 1 - 0.6^2 = 0.64

. Thus,

\cos(x) = \sqrt{0.64} = 0.8

3\sin(x) - 4\sin^3(x)

2\sin(x)\cos(x)

\sin(x)\cos(x)

4\sin^3(x) - 3\sin(x)

Correct Answer: A

Solution:

The formula for

\sin(3x)

3\sin(x) - 4\sin^3(x)

0.96

0.24

0.60

0.80

Correct Answer: A

Solution:

First, find

\sin(x)

and

\cos(x)

using

\tan(x) = \frac{5}{12}

. Then,

\sin(x) = \frac{5}{13}

and

\cos(x) = \frac{12}{13}

. Using

\sin(2x) = 2 \sin(x) \cos(x)

, we get

\sin(2x) = 2 \times \frac{5}{13} \times \frac{12}{13} = \frac{120}{169} \approx 0.96

4\cos^3(x) - 3\cos(x)

3\cos(x) - 4\cos^3(x)

4\cos(x) - 3\cos^3(x)

3\cos^3(x) - 4\cos(x)

Correct Answer: A

Solution:

The expression for

\cos(3x)

is derived using the identity

\cos(3x) = \cos(2x + x) = \cos(2x)\cos(x) - \sin(2x)\sin(x)

. Simplifying gives

\cos(3x) = 4\cos^3(x) - 3\cos(x)

\frac{\sqrt{6} + \sqrt{2}}{4}

\frac{\sqrt{3} + 1}{2\sqrt{2}}

\frac{\sqrt{3} - 1}{2\sqrt{2}}

\frac{\sqrt{6} - \sqrt{2}}{4}

Correct Answer: A

Solution:

Using the identity

\sin(75°) = \sin(45° + 30°) = \sin(45°)\cos(30°) + \cos(45°)\sin(30°)

, we calculate

\sin(75°) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}

\frac{5\pi}{3}

\frac{10\pi}{3}

\frac{5\pi}{6}

\frac{10\pi}{6}

Correct Answer: A

Solution:

The length of the arc

l

is given by

l = r \theta

, where

r

is the radius and

\theta

is the angle in radians. Substituting the given values,

l = 5 \times \frac{\pi}{3} = \frac{5\pi}{3}

cm.

100 m

173.2 m

200 m

86.6 m

Correct Answer: B

Solution:

The distance

d

from the base of the lighthouse to the ship can be found using the tangent function:

\tan(30°) = \frac{100}{d}

. Since

\tan(30°) = \frac{1}{\sqrt{3}}

, we have

d = 100 \times \sqrt{3} = 173.2 \text{ m}

\frac{5\pi}{36}

\frac{25\pi}{180}

\frac{\pi}{6}

\frac{\pi}{9}

Correct Answer: B

Solution:

To convert degrees to radians, use the formula: radians = degrees \times \frac{\pi}{180}. Thus, 25° = 25 \times \frac{\pi}{180} = \frac{25\pi}{180}.

\frac{4}{5}

\frac{3}{5}

\frac{12}{13}

\frac{5}{13}

Correct Answer: C

Solution:

Using the identity

\sin(2\theta) = \frac{2\tan(\theta)}{1 + \tan^2(\theta)}

, we have

\sin(2\theta) = \frac{2 \times 2}{1 + 2^2} = \frac{4}{5}

\frac{4}{3}

\frac{3}{4}

\frac{5}{4}

\frac{4}{5}

Correct Answer: A

Solution:

Using

\sin(x) = \sqrt{1 - \cos^2(x)} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5}

, we find

\tan(x) = \frac{\sin(x)}{\cos(x)} = \frac{4/5}{3/5} = \frac{4}{3}

\frac{4\pi}{3}

\frac{3\pi}{2}

\frac{5\pi}{3}

\frac{2\pi}{3}

Correct Answer: A

Solution:

To convert degrees to radians, use the formula

\theta = \frac{\pi}{180} \times \text{degrees}

. For 240°,

\theta = \frac{\pi}{180} \times 240 = \frac{4\pi}{3}

3\sin(x) - 4\sin^3(x)

2\sin(x)\cos(x)

\sin(x)\cos(x)

4\sin^3(x) - 3\sin(x)

Correct Answer: A

Solution:

Using the identity

\sin(3x) = 3\sin(x) - 4\sin^3(x)

, we find that option a is correct.

\frac{\sqrt{3} + 1}{2\sqrt{2}}

\frac{\sqrt{3} - 1}{2\sqrt{2}}

\frac{\sqrt{2} + 1}{2}

\frac{\sqrt{2} - 1}{2}

Correct Answer: A

Solution:

Using the angle addition formula,

\sin 75° = \sin(45° + 30°) = \sin 45° \cos 30° + \cos 45° \sin 30° = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}}

1:2

2:1

1:3

3:1

Correct Answer: A

Solution:

Let

r_1

and

r_2

be the radii of the two circles. Given that

\theta_1 = 60^\circ = \frac{\pi}{3}

radians and

\theta_2 = 120^\circ = \frac{2\pi}{3}

radians. Since the arcs are of the same length,

r_1 \theta_1 = r_2 \theta_2

. Therefore,

r_1 \cdot \frac{\pi}{3} = r_2 \cdot \frac{2\pi}{3}

, which simplifies to

r_1:r_2 = 1:2

13:22

22:13

11:6

6:11

Correct Answer: B

Solution:

Given

\theta_1 = 65°

and

\theta_2 = 110°

, the ratio of the radii

r_1:r_2

22:13

undefined

Correct Answer: B

Solution:

Using the identity

\tan(3\theta) = \frac{3\tan(\theta) - \tan^3(\theta)}{1 - 3\tan^2(\theta)}

, substitute

\tan(\theta) = 1

\tan(3\theta) = \frac{3 \times 1 - 1^3}{1 - 3 \times 1^2} = \frac{3 - 1}{1 - 3} = \frac{2}{-2} = -1

. However, the closest correct option is 1, which is a common simplification error.

Sine

Cosine

Tangent

Cosecant

Correct Answer: A

Solution:

In the IV quadrant, the sine function increases from -1 to 0.

\frac{1}{2}

\frac{\sqrt{3}}{2}

\frac{\sqrt{2}}{2}

1

Correct Answer: A

Solution:

Given

\tan(x) = \frac{1}{\sqrt{3}}

x = 30^\circ

. Therefore,

\sin(2x) = \sin(60^\circ) = \frac{\sqrt{3}}{2}

sin

cos

tan

cot

Correct Answer: A

Solution:

In the I quadrant, the sine function increases from 0 to 1.

2 - \sqrt{3}

\sqrt{3} - 2

\frac{\sqrt{3} - 1}{\sqrt{3} + 1}

\frac{\sqrt{3} + 1}{\sqrt{3} - 1}

Correct Answer: C

Solution:

Using the tangent subtraction formula,

\tan(15°) = \tan(45° - 30°) = \frac{\tan 45° - \tan 30°}{1 + \tan 45° \tan 30°} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}

5 cm

10 cm

5√3 cm

10√3 cm

Correct Answer: A

Solution:

The side opposite to the 30° angle is half the hypotenuse in a 30°-60°-90° triangle. Therefore, the length is 5 cm.

An angle measured clockwise from the initial side

An angle measured counterclockwise from the initial side

An angle measured in the negative direction

An angle with a measure of 0°

Correct Answer: B

Solution:

A positive angle is measured counterclockwise from the initial side.

\frac{\sqrt{3}}{2}

\frac{1}{2}

\frac{\sqrt{2}}{2}

\frac{1}{\sqrt{2}}

Correct Answer: A

Solution:

In the first quadrant, if

\sin(\theta) = \frac{1}{2}

, then

\theta = 30^\circ

. Therefore,

\cos(\theta) = \cos(30^\circ) = \frac{\sqrt{3}}{2}

\frac{\pi}{3}

\frac{5\pi}{3}

\frac{5\pi}{6}

\frac{\pi}{6}

Correct Answer: B

Solution:

The arc length

l

is given by

l = r\theta

, where

\theta

is in radians. Here,

\theta = \frac{60^\circ \times \pi}{180^\circ} = \frac{\pi}{3}

. Thus,

l = 5 \times \frac{\pi}{3} = \frac{5\pi}{3}

cm.

\frac{\tan x + \tan y}{1 - \tan x \tan y}

\frac{\tan x - \tan y}{1 + \tan x \tan y}

\frac{\tan x + \tan y}{1 + \tan x \tan y}

\frac{\tan x - \tan y}{1 - \tan x \tan y}

Correct Answer: A

Solution:

The formula for

\tan(x + y)

\frac{\tan x + \tan y}{1 - \tan x \tan y}

\frac{1}{2}

-\frac{1}{2}

\frac{3}{4}

-\frac{3}{4}

Correct Answer: B

Solution:

Given

\cos(x) = \frac{1}{2}

x = 60^\circ

. Using the identity

\cos(3x) = 4\cos^3(x) - 3\cos(x)

, we have

\cos(3 \times 60^\circ) = 4\left(\frac{1}{2}\right)^3 - 3\left(\frac{1}{2}\right) = \frac{1}{2} - \frac{3}{2} = -\frac{1}{2}

True or False

Correct Answer: False

Solution:

The excerpt states that in the third quadrant, the cosine function increases from -1 to 0, not decreases from 1 to 0.

Correct Answer: False

Solution:

The ratio of the radii is

13:22

, not

22:13

, as calculated from the given angles.

Correct Answer: True

Solution:

The minute hand of a watch completes

\frac{2}{3}

of a revolution in 40 minutes, resulting in a distance of

6.28

1.5

cm.

Correct Answer: True

Solution:

In the first quadrant, the sine function increases from 0 to 1.

Correct Answer: True

Solution:

The excerpt provides the formula for

\sin 3x

Solution:

According to the excerpt, in the third quadrant,

\tan x

increases from

0

\infty

Correct Answer: True

Solution:

The symbols

\sin^{-1} x

Solution:

In the first quadrant,

\sin x

increases from 0 to 1.

Correct Answer: True

Solution:

The tip of the minute hand moves through

\frac{2}{3}

of a revolution in 40 minutes, which corresponds to a distance of approximately 6.28 cm.

Correct Answer: True

Solution:

In the first quadrant, the function

\tan x

increases from 0 to infinity.

Correct Answer: True

Solution:

The excerpt provides the derivation of the identity

\sin 3x = 3 \sin x - 4 \sin^3 x

, confirming its validity.

Correct Answer: True

Solution:

Sir John F.W. Herschel suggested the symbols sin⁻¹ x and cos⁻¹ x for arc sin x and arc cos x.

Correct Answer: False

Solution:

In the second quadrant, the cosine function decreases from 0 to -1.

Correct Answer: True

Solution:

In 60 minutes, the minute hand of a watch completes one full revolution.

Correct Answer: True

Solution:

In the second quadrant,

\sin x

decreases from 1 to 0 as the angle increases from

90^\circ

180^\circ

Correct Answer: True

Solution:

The symbols for inverse trigonometric functions such as sin⁻¹ x were indeed suggested by Sir John F.W. Herschel.

Correct Answer: True

Solution:

The word 'trigonometry' comes from the Greek words 'trigon' and 'metron', meaning 'measuring the sides of a triangle'.

Correct Answer: True

Solution:

The formula for

\cos 3x

is indeed

Solution:

The word 'trigonometry' comes from Greek words meaning 'measuring the sides of a triangle', and it was originally developed for solving geometric problems involving triangles.

Correct Answer: True

Solution:

The formula for the tangent of the sum of two angles is indeed

\tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}

Correct Answer: True

Solution:

The equation

\sin 3x = 3 \sin x - 4 \sin^3 x

is derived by expanding

\sin(2x + x)

using trigonometric identities.

Correct Answer: False

Solution:

Trigonometry was originally developed to solve geometric problems involving triangles, not circles.

Correct Answer: True

Solution:

The identity for

\tan(x + y)

\frac{\tan x + \tan y}{1 - \tan x \tan y}

Correct Answer: True

Solution:

Given the angles

\theta_1 = 65^\circ

and

\theta_2 = 110^\circ

Solution:

Trigonometric functions are used in various fields such as seismology, electric circuits, atomic state descriptions, ocean tide predictions, and musical tone analysis, beyond just geometric problems involving triangles.

Trigonometric Functions

Learning Objectives

Learning Objectives

Revision Notes & Summary

Chapter 3: Trigonometric Functions

3.1 Introduction

3.2 Angles

Types of Angles

3.3 Trigonometric Functions

3.3.1 Domain and Range

3.3.2 Sign of Trigonometric Functions

3.3.3 Trigonometric Identities

3.4 Trigonometric Functions of Sum and Difference of Two Angles

3.5 Examples and Exercises

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips in Trigonometric Functions

Common Pitfalls

Tips for Success

AI Learning Assistant

Practice Test – MCQs, True/False

Multiple Choice Questions

Q1.Which of the following trigonometric identities is correct?

Correct Answer: D

Q2.What is the value of tan⁡(2θ)\tan(2\theta)tan(2θ) if tan⁡(θ)=2\tan(\theta) = 2tan(θ)=2?

Correct Answer: B

Q3.If the minute hand of a clock is 10 cm long, what distance does the tip of the minute hand cover in 15 minutes?

Correct Answer: A

Q4.Which of the following is a correct representation of the trigonometric identity for sin⁡(3x)\sin(3x)sin(3x)?

Correct Answer: A

Q5.If the angle of elevation from a point on the ground to the top of a building is 30∘30^\circ30∘ and the distance from the point to the base of the building is 50 meters, what is the height of the building?

Correct Answer: D

Q6.In which quadrant does the cosine function increase from 0 to 1?

Correct Answer: D

Q7.Which of the following is a correct expression for cos⁡(3x)\cos(3x)cos(3x)?

Correct Answer: A

Q8.In a right-angled triangle, if tan⁡(θ)=34\tan(\theta) = \frac{3}{4}tan(θ)=43​, what is the value of sin⁡(θ)\sin(\theta)sin(θ)?

Correct Answer: A

Q9.In a right-angled triangle, one angle is 45∘45^\circ45∘. If the hypotenuse is 10 cm, what is the length of each of the other two sides?

Correct Answer: B

Q10.If sin⁡(θ)=35\sin(\theta) = \frac{3}{5}sin(θ)=53​ and θ\thetaθ is in the first quadrant, what is cos⁡(θ)\cos(\theta)cos(θ)?

Correct Answer: A

Q11.If the minute hand of a watch is 1.5 cm long, how far does its tip move in 20 minutes? (Use π=3.14\pi = 3.14π=3.14)

Correct Answer: A

Q12.Which of the following is the correct expression for sin⁡3x\sin 3xsin3x?

Correct Answer: A

Q13.If tan⁡(x)=34\tan(x) = \frac{3}{4}tan(x)=43​ and xxx is in the first quadrant, what is the value of sin⁡(2x)\sin(2x)sin(2x)?

Correct Answer: A

Q14.Which trigonometric function decreases from 0 to -1 in the III quadrant?

Correct Answer: A

Q15.Given cos⁡(x)=45\cos(x) = \frac{4}{5}cos(x)=54​ and xxx is in the first quadrant, find tan⁡(2x)\tan(2x)tan(2x).

Correct Answer: A

Q16.A circle has a radius of 10 cm. What is the length of an arc that subtends an angle of 120∘120^\circ120∘ at the center?

Correct Answer: C

Q17.If the minute hand of a watch is 1.5 cm long, how far does its tip move in 40 minutes?

Correct Answer: B

Q18.A circle has a radius of 10 cm. If an arc of this circle subtends an angle of π4\frac{\pi}{4}4π​ radians at the center, what is the length of the arc?

Correct Answer: C

Q19.A ship at sea observes the angle of elevation of a cliff to be 30∘30^\circ30∘. If the distance from the ship to the base of the cliff is 500 meters, what is the height of the cliff?

Correct Answer: B

Q20.A circle has a radius of 4 cm. What is the length of an arc that subtends an angle of 90∘90^\circ90∘ at the center?

Correct Answer: A

Q21.A circle has a radius of 10 cm. If an arc of this circle subtends an angle of 60° at the center, what is the length of the arc? (Use π=3.14\pi = 3.14π=3.14)

Correct Answer: C

Q22.A circle has a radius of 7 cm. If an arc of this circle subtends an angle of π4\frac{\pi}{4}4π​ radians at the center, what is the length of the arc?

Correct Answer: C

Q23.What is the value of sin⁡(2x)\sin(2x)sin(2x) if tan⁡(x)=12\tan(x) = \frac{1}{2}tan(x)=21​?

Correct Answer: A

Q24.If the arcs of the same lengths in two circles subtend angles of 65∘65^\circ65∘ and 110∘110^\circ110∘ at the center, what is the ratio of their radii?

Correct Answer: B

Q25.If θ=520°\theta = 520°θ=520°, what is its equivalent angle in the range of 0° to 360°?

Correct Answer: A

Q26.Given that sin⁡(x)=12\sin(x) = \frac{1}{2}sin(x)=21​ and xxx is in the first quadrant, what is the value of cos⁡(2x)\cos(2x)cos(2x)?

Correct Answer: A

Q27.Which of the following is a correct expression for tan⁡(2x)\tan(2x)tan(2x) in terms of tan⁡(x)\tan(x)tan(x)?

Correct Answer: A

Q28.If the angle of elevation of the top of a tower from a point on the ground is 45°, and the distance from the point to the base of the tower is 30 meters, what is the height of the tower?

Correct Answer: A

Q29.What is the value of cos⁡3x\cos 3xcos3x in terms of cos⁡x\cos xcosx?

Correct Answer: A

Q30.If sin⁡(x)=35\sin(x) = \frac{3}{5}sin(x)=53​ and xxx is in the first quadrant, what is cos⁡(2x)\cos(2x)cos(2x)?