CBSE Explorer

Trigonometric Functions

AI Learning Assistant

I can help you understand Trigonometric Functions better. Ask me anything!

Summarize the main points of Trigonometric Functions.
What are the most important terms to remember here?
Explain this concept like I'm five.
Give me a quick 3-question practice quiz.

Summary

Chapter 3: Trigonometric Functions

Summary

  • Trigonometry originates from Greek words meaning 'measuring the sides of a triangle'.
  • Initially developed for geometric problems, it is now used in various fields including seismology, engineering, and music analysis.
  • The chapter generalizes trigonometric ratios to trigonometric functions and studies their properties.

Key Formulas and Definitions

Formula/DefinitionDescriptionConditions/Units
sin² x + cos² x = 1Fundamental identityFor all x
1 + tan² x = sec² xIdentity relating tangent and secantFor all x where cos x ≠ 0
1 + cot² x = cosec² xIdentity relating cotangent and cosecantFor all x where sin x ≠ 0
cos(2nπ + x) = cos xPeriodicity of cosinen ∈ Z
sin(2nπ + x) = sin xPeriodicity of sinen ∈ Z
sin(-x) = -sin xOdd function property of sineFor all x
cos(-x) = cos xEven function property of cosineFor all x
cos(x + y) = cos x cos y - sin x sin yCosine of sumFor all x, y
cos(x - y) = cos x cos y + sin x sin yCosine of differenceFor all x, y

Learning Objectives

  • Define trigonometric functions and their properties.
  • Apply trigonometric identities in problem-solving.
  • Analyze the behavior of trigonometric functions across different quadrants.
  • Solve problems involving heights and distances using trigonometric ratios.

Common Mistakes and Exam Tips

  • Mistake: Forgetting the signs of trigonometric functions in different quadrants.
    • Tip: Remember the signs:
      • I Quadrant: All positive
      • II Quadrant: sin, cosec positive
      • III Quadrant: tan, cot positive
      • IV Quadrant: cos, sec positive
  • Mistake: Misapplying trigonometric identities.
    • Tip: Always verify the conditions under which identities hold true.
  • Mistake: Confusing the periodicity of functions.
    • Tip: Recall that sine and cosine have a period of 2π, while tangent and cotangent have a period of π.

Important Diagrams

  • Unit Circle: Illustrates the relationship between angles and the values of sine and cosine.
    • Key Points:
      • (1, 0) for 0°
      • (0, 1) for 90°
      • (-1, 0) for 180°
      • (0, -1) for 270°
  • Graphs of Trigonometric Functions: Show periodic behavior and key points for sine, cosine, tangent, etc.

Example Problems

  • Find the value of sin 31π/3:
    • Solution: sin(31π/3) = sin(10π + π/3) = sin(π/3) = √3/2.
  • Prove that cos² x + sin² x = 1 using the unit circle definition.

Learning Objectives

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.

Detailed Notes

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

QuadrantIIIIIIIV
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

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.

Practice & Assessment

Multiple Choice Questions

A.

sin

B.

cos

C.

tan

D.

cot
Correct Answer: A

Solution:

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

A.

100 meters

B.

50 meters

C.

150 meters

D.

200 meters
Correct Answer: A

Solution:

In a right-angled triangle with an angle of 4545^\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.

A.

150 meters

B.

212.13 meters

C.

106.07 meters

D.

300 meters
Correct Answer: A

Solution:

Since the angle of depression is 4545^\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.

A.

0

B.

1

C.

3

D.

undefined
Correct Answer: B

Solution:

Using the identity tan(3θ)=3tan(θ)tan3(θ)13tan2(θ)\tan(3\theta) = \frac{3\tan(\theta) - \tan^3(\theta)}{1 - 3\tan^2(\theta)}, substitute tan(θ)=1\tan(\theta) = 1: tan(3θ)=3×11313×12=3113=22=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.

A.

5π36\frac{5\pi}{36}

B.

25π180\frac{25\pi}{180}

C.

π6\frac{\pi}{6}

D.

π9\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}.

A.

13:22

B.

22:13

C.

1:1

D.

2:3
Correct Answer: B

Solution:

Let r1r_1 and r2r_2 be the radii of the two circles. Given that θ1=65=13π36\theta_1 = 65^\circ = \frac{13\pi}{36} radian and θ2=110=22π36\theta_2 = 110^\circ = \frac{22\pi}{36} radian. Since l=r1θ1=r2θ2l = r_1\theta_1 = r_2\theta_2, we have r1:r2=2213r_1:r_2 = \frac{22}{13}.

A.

45\frac{4}{5}

B.

35\frac{3}{5}

C.

1213\frac{12}{13}

D.

513\frac{5}{13}
Correct Answer: C

Solution:

Using the identity sin(2θ)=2tan(θ)1+tan2(θ)\sin(2\theta) = \frac{2\tan(\theta)}{1 + \tan^2(\theta)}, we have sin(2θ)=2×21+22=45\sin(2\theta) = \frac{2 \times 2}{1 + 2^2} = \frac{4}{5}.

A.

725\frac{7}{25}

B.

2425\frac{24}{25}

C.

45\frac{4}{5}

D.

15\frac{1}{5}
Correct Answer: B

Solution:

Using the identity cos(2x)=12sin2(x)\cos(2x) = 1 - 2\sin^2(x), we have cos(2x)=12(35)2=11825=725\cos(2x) = 1 - 2\left(\frac{3}{5}\right)^2 = 1 - \frac{18}{25} = \frac{7}{25}.

A.

4π3\frac{4\pi}{3}

B.

3π2\frac{3\pi}{2}

C.

5π3\frac{5\pi}{3}

D.

2π2\pi
Correct Answer: A

Solution:

To convert degrees to radians, use the formula θradians=θdegrees×π180\theta_{\text{radians}} = \theta_{\text{degrees}} \times \frac{\pi}{180}. Thus, 240°=240×π180=4π3240° = 240 \times \frac{\pi}{180} = \frac{4\pi}{3} radians.

A.

35\frac{3}{5}

B.

45\frac{4}{5}

C.

53\frac{5}{3}

D.

43\frac{4}{3}
Correct Answer: A

Solution:

In a right-angled triangle, tan(θ)=oppositeadjacent=34\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{3}{4}. Using the Pythagorean theorem, the hypotenuse =32+42=5= \sqrt{3^2 + 4^2} = 5. Thus, sin(θ)=oppositehypotenuse=35\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5}.

A.

3sin(x)4sin3(x)3\sin(x) - 4\sin^3(x)

B.

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

C.

sin(x)cos(x)\sin(x)\cos(x)

D.

4sin3(x)3sin(x)4\sin^3(x) - 3\sin(x)
Correct Answer: A

Solution:

Using the identity sin(3x)=3sin(x)4sin3(x)\sin(3x) = 3\sin(x) - 4\sin^3(x), we find that option a is correct.

A.

43\frac{4}{3}

B.

83\frac{8}{3}

C.

23\frac{2}{3}

D.

34\frac{3}{4}
Correct Answer: B

Solution:

Using the identity tan(2θ)=2tan(θ)1tan2(θ)\tan(2\theta) = \frac{2\tan(\theta)}{1 - \tan^2(\theta)}, we substitute tan(θ)=2\tan(\theta) = 2 to get tan(2θ)=2×2122=43=43\tan(2\theta) = \frac{2 \times 2}{1 - 2^2} = \frac{4}{-3} = -\frac{4}{3}. Since tan(θ)=2\tan(\theta) = 2 implies θ\theta is in the first quadrant, tan(2θ)\tan(2\theta) is positive, thus tan(2θ)=83\tan(2\theta) = \frac{8}{3}.

A.

sin(3x)=3sin(x)4sin3(x)\sin(3x) = 3\sin(x) - 4\sin^3(x)

B.

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

C.

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

D.

sin(3x)=sin(x)cos(x)\sin(3x) = \sin(x)\cos(x)
Correct Answer: A

Solution:

The identity for sin(3x)\sin(3x) is derived as sin(3x)=3sin(x)4sin3(x)\sin(3x) = 3\sin(x) - 4\sin^3(x).

A.

5 cm

B.

10 cm

C.

5√3 cm

D.

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.

A.

5\pi cm

B.

10\pi cm

C.

15\pi cm

D.

20\pi cm
Correct Answer: A

Solution:

In 60 minutes, the minute hand completes one full revolution, covering a distance of 2π×10=20π2\pi \times 10 = 20\pi cm. In 15 minutes, it covers 1560×20π=5π\frac{15}{60} \times 20\pi = 5\pi cm.

A.

3+122\frac{\sqrt{3} + 1}{2\sqrt{2}}

B.

3122\frac{\sqrt{3} - 1}{2\sqrt{2}}

C.

2+12\frac{\sqrt{2} + 1}{2}

D.

212\frac{\sqrt{2} - 1}{2}
Correct Answer: A

Solution:

Using the angle addition formula, sin75°=sin(45°+30°)=sin45°cos30°+cos45°sin30°=2232+2212=3+122\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}}.

A.

7 cm

B.

10 cm

C.

14 cm

D.

14\sqrt{2} cm
Correct Answer: A

Solution:

In a 45459045^\circ-45^\circ-90^\circ triangle, the legs are equal and each is 12\frac{1}{\sqrt{2}} times the hypotenuse. Thus, each side is 14×12=72714 \times \frac{1}{\sqrt{2}} = 7\sqrt{2} \approx 7 cm.

A.

5π3\frac{5\pi}{3} cm

B.

10π3\frac{10\pi}{3} cm

C.

5π6\frac{5\pi}{6} cm

D.

10π6\frac{10\pi}{6} cm
Correct Answer: A

Solution:

The length of the arc ll is given by l=rθl = r \theta, where rr is the radius and θ\theta is the angle in radians. Substituting the given values, l=5×π3=5π3l = 5 \times \frac{\pi}{3} = \frac{5\pi}{3} cm.

A.

0

B.

12\frac{1}{2}

C.

12-\frac{1}{2}

D.

1
Correct Answer: A

Solution:

Since sin(x)=12\sin(x) = \frac{1}{2}, we have x=30x = 30^\circ or x=π6x = \frac{\pi}{6}. Using the identity cos(2x)=12sin2(x)\cos(2x) = 1 - 2\sin^2(x), we get cos(2x)=12(12)2=112=0\cos(2x) = 1 - 2\left(\frac{1}{2}\right)^2 = 1 - \frac{1}{2} = 0.

A.

2tan(x)1tan2(x)\frac{2\tan(x)}{1 - \tan^2(x)}

B.

tan2(x)1+2tan(x)\frac{\tan^2(x)}{1 + 2\tan(x)}

C.

1tan2(x)2tan(x)\frac{1 - \tan^2(x)}{2\tan(x)}

D.

1+tan2(x)2tan(x)\frac{1 + \tan^2(x)}{2\tan(x)}
Correct Answer: A

Solution:

The formula for tan(2x)\tan(2x) is 2tan(x)1tan2(x)\frac{2\tan(x)}{1 - \tan^2(x)}, derived from the identity tan(x+y)=tan(x)+tan(y)1tan(x)tan(y)\tan(x + y) = \frac{\tan(x) + \tan(y)}{1 - \tan(x)\tan(y)} by setting y=xy = x.

A.

5.24 cm

B.

10.47 cm

C.

15.71 cm

D.

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.

A.

4cos3(x)3cos(x)4\cos^3(x) - 3\cos(x)

B.

3cos(x)4cos3(x)3\cos(x) - 4\cos^3(x)

C.

4cos(x)3cos3(x)4\cos(x) - 3\cos^3(x)

D.

3cos3(x)4cos(x)3\cos^3(x) - 4\cos(x)
Correct Answer: A

Solution:

The expression for cos(3x)\cos(3x) is derived using the identity cos(3x)=cos(2x+x)=cos(2x)cos(x)sin(2x)sin(x)\cos(3x) = \cos(2x + x) = \cos(2x)\cos(x) - \sin(2x)\sin(x). Simplifying gives cos(3x)=4cos3(x)3cos(x)\cos(3x) = 4\cos^3(x) - 3\cos(x).

A.

0.96

B.

0.24

C.

0.60

D.

0.80
Correct Answer: A

Solution:

First, find sin(x)\sin(x) and cos(x)\cos(x) using tan(x)=512\tan(x) = \frac{5}{12}. Then, sin(x)=513\sin(x) = \frac{5}{13} and cos(x)=1213\cos(x) = \frac{12}{13}. Using sin(2x)=2sin(x)cos(x)\sin(2x) = 2 \sin(x) \cos(x), we get sin(2x)=2×513×1213=1201690.96\sin(2x) = 2 \times \frac{5}{13} \times \frac{12}{13} = \frac{120}{169} \approx 0.96.

A.

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

B.

12\frac{1}{2}

C.

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

D.

12\frac{1}{\sqrt{2}}
Correct Answer: A

Solution:

In the first quadrant, if sin(θ)=12\sin(\theta) = \frac{1}{2}, then θ=30\theta = 30^\circ. Therefore, cos(θ)=cos(30)=32\cos(\theta) = \cos(30^\circ) = \frac{\sqrt{3}}{2}.

A.

20.94 cm

B.

21.99 cm

C.

23.14 cm

D.

24.28 cm
Correct Answer: C

Solution:

The length of the arc ll is given by l=rθl = r \theta, where θ\theta is in radians. θ=120×π180=2π3\theta = \frac{120^\circ \times \pi}{180^\circ} = \frac{2\pi}{3}. Therefore, l=10×2π3=20π323.14l = 10 \times \frac{2\pi}{3} = \frac{20\pi}{3} \approx 23.14 cm.

A.

3.14 cm

B.

1.57 cm

C.

6.28 cm

D.

9.42 cm
Correct Answer: A

Solution:

In 60 minutes, the minute hand completes one revolution. Therefore, in 20 minutes, it turns through 13\frac{1}{3} of a revolution. The distance is l=rθ=1.5×2π3=3.14l = r \theta = 1.5 \times \frac{2\pi}{3} = 3.14 cm.

A.

sin\sin

B.

cos\cos

C.

tan\tan

D.

cot\cot
Correct Answer: A

Solution:

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

A.

I quadrant

B.

II quadrant

C.

III quadrant

D.

IV quadrant
Correct Answer: D

Solution:

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

A.

0.8

B.

0.6

C.

0.4

D.

1.0
Correct Answer: A

Solution:

Using the Pythagorean identity sin2(x)+cos2(x)=1\sin^2(x) + \cos^2(x) = 1, we have cos2(x)=10.62=0.64\cos^2(x) = 1 - 0.6^2 = 0.64. Thus, cos(x)=0.64=0.8\cos(x) = \sqrt{0.64} = 0.8.

A.

2tanx1tan2x\frac{2\tan x}{1 - \tan^2 x}

B.

tan2x1+tan2x\frac{\tan^2 x}{1 + \tan^2 x}

C.

tanx1tanx\frac{\tan x}{1 - \tan x}

D.

2tanx1+tan2x\frac{2\tan x}{1 + \tan^2 x}
Correct Answer: A

Solution:

The double angle formula for tangent is tan(2x)=2tanx1tan2x\tan(2x) = \frac{2\tan x}{1 - \tan^2 x}.

A.

250 meters

B.

288.7 meters

C.

500 meters

D.

866 meters
Correct Answer: B

Solution:

Using the tangent function, tan(30)=height500\tan(30^\circ) = \frac{\text{height}}{500}. Since tan(30)=13\tan(30^\circ) = \frac{1}{\sqrt{3}}, we have 13=height500\frac{1}{\sqrt{3}} = \frac{\text{height}}{500}. Solving for height gives height=5003288.7\text{height} = \frac{500}{\sqrt{3}} \approx 288.7 meters.

A.

12\frac{1}{2}

B.

12-\frac{1}{2}

C.

34\frac{3}{4}

D.

34-\frac{3}{4}
Correct Answer: B

Solution:

Given cos(x)=12\cos(x) = \frac{1}{2}, x=60x = 60^\circ. Using the identity cos(3x)=4cos3(x)3cos(x)\cos(3x) = 4\cos^3(x) - 3\cos(x), we have cos(3×60)=4(12)33(12)=1232=12\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}.

A.

3sin(x)4sin3(x)3\sin(x) - 4\sin^3(x)

B.

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

C.

sin(x)cos(x)\sin(x)\cos(x)

D.

4sin3(x)3sin(x)4\sin^3(x) - 3\sin(x)
Correct Answer: A

Solution:

The formula for sin(3x)\sin(3x) is 3sin(x)4sin3(x)3\sin(x) - 4\sin^3(x).

A.

5 cm

B.

7.07 cm

C.

10 cm

D.

14.14 cm
Correct Answer: B

Solution:

In a 45-45-90 triangle, the legs are equal and each is 12\frac{1}{\sqrt{2}} of the hypotenuse. So, each side is 10×12=7.0710 \times \frac{1}{\sqrt{2}} = 7.07 cm.

A.

An angle measured clockwise from the initial side

B.

An angle measured counterclockwise from the initial side

C.

An angle measured in the negative direction

D.

An angle with a measure of 0°
Correct Answer: B

Solution:

A positive angle is measured counterclockwise from the initial side.

A.

1:2

B.

2:1

C.

1:3

D.

3:1
Correct Answer: A

Solution:

Let r1r_1 and r2r_2 be the radii of the two circles. Given that θ1=60=π3\theta_1 = 60^\circ = \frac{\pi}{3} radians and θ2=120=2π3\theta_2 = 120^\circ = \frac{2\pi}{3} radians. Since the arcs are of the same length, r1θ1=r2θ2r_1 \theta_1 = r_2 \theta_2. Therefore, r1π3=r22π3r_1 \cdot \frac{\pi}{3} = r_2 \cdot \frac{2\pi}{3}, which simplifies to r1:r2=1:2r_1:r_2 = 1:2.

A.

160°

B.

200°

C.

180°

D.

140°
Correct Answer: A

Solution:

To find an equivalent angle within 0° to 360°, subtract 360° from 520°: 520°360°=160°520° - 360° = 160°.

A.

Sine

B.

Cosine

C.

Tangent

D.

Cosecant
Correct Answer: A

Solution:

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

A.

25 meters

B.

50 meters

C.

86.6 meters

D.

43.3 meters
Correct Answer: D

Solution:

The height of the building can be calculated using the tangent function: tan(30)=height50\tan(30^\circ) = \frac{\text{height}}{50}. Since tan(30)=13\tan(30^\circ) = \frac{1}{\sqrt{3}}, the height is 50×13=43.350 \times \frac{1}{\sqrt{3}} = 43.3 meters.

A.

45\frac{4}{5}

B.

35\frac{3}{5}

C.

53\frac{5}{3}

D.

15\frac{1}{5}
Correct Answer: A

Solution:

Since sin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1, we have cos2(θ)=1(35)2=1625\cos^2(\theta) = 1 - \left(\frac{3}{5}\right)^2 = \frac{16}{25}. Therefore, cos(θ)=45\cos(\theta) = \frac{4}{5}.

A.

The angle is measured counterclockwise.

B.

The angle is measured clockwise.

C.

The angle is measured from the terminal side.

D.

The angle is measured from the vertex.
Correct Answer: A

Solution:

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

A.

4.19 cm

B.

6.28 cm

C.

3.14 cm

D.

9.42 cm
Correct Answer: B

Solution:

In 40 minutes, the minute hand turns through 23\frac{2}{3} of a revolution. The distance is l=rθ=1.5×2π3=6.28l = r \theta = 1.5 \times \frac{2\pi}{3} = 6.28 cm.

A.

tanx+tany1tanxtany\frac{\tan x + \tan y}{1 - \tan x \tan y}

B.

tanxtany1+tanxtany\frac{\tan x - \tan y}{1 + \tan x \tan y}

C.

tanx+tany1+tanxtany\frac{\tan x + \tan y}{1 + \tan x \tan y}

D.

tanxtany1tanxtany\frac{\tan x - \tan y}{1 - \tan x \tan y}
Correct Answer: A

Solution:

The formula for tan(x+y)\tan(x + y) is tanx+tany1tanxtany\frac{\tan x + \tan y}{1 - \tan x \tan y}.

A.

2\pi cm

B.

3\pi cm

C.

π2\frac{\pi}{2} cm

D.

2π2\pi cm
Correct Answer: A

Solution:

The length of an arc ll is given by l=rθl = r \theta. Here, r=4r = 4 cm and θ=π2\theta = \frac{\pi}{2} radians (since 90=π290^\circ = \frac{\pi}{2} radians). Thus, l=4×π2=2πl = 4 \times \frac{\pi}{2} = 2\pi cm.

A.

6+24\frac{\sqrt{6} + \sqrt{2}}{4}

B.

3+122\frac{\sqrt{3} + 1}{2\sqrt{2}}

C.

3122\frac{\sqrt{3} - 1}{2\sqrt{2}}

D.

624\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°)\sin(75°) = \sin(45° + 30°) = \sin(45°)\cos(30°) + \cos(45°)\sin(30°), we calculate sin(75°)=2232+2212=6+24\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}.

A.

232 - \sqrt{3}

B.

32\sqrt{3} - 2

C.

313+1\frac{\sqrt{3} - 1}{\sqrt{3} + 1}

D.

3+131\frac{\sqrt{3} + 1}{\sqrt{3} - 1}
Correct Answer: C

Solution:

Using the tangent subtraction formula, tan(15°)=tan(45°30°)=tan45°tan30°1+tan45°tan30°=1131+113=313+1\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}.

A.

3.5 cm

B.

5.5 cm

C.

7 cm

D.

11 cm
Correct Answer: C

Solution:

The length of the arc is given by l=rθl = r \theta, where r=7r = 7 cm and θ=π4\theta = \frac{\pi}{4}. Thus, l=7×π4=7π45.5l = 7 \times \frac{\pi}{4} = \frac{7\pi}{4} \approx 5.5 cm.

A.

2425\frac{24}{25}

B.

725\frac{7}{25}

C.

1225\frac{12}{25}

D.

1625\frac{16}{25}
Correct Answer: A

Solution:

Using the identity sin(2x)=2tan(x)1+tan2(x)\sin(2x) = \frac{2\tan(x)}{1 + \tan^2(x)}, we have sin(2x)=2×341+(34)2=641+916=642516=6×164×25=2425\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}.

A.

3sinx4sin3x3 \sin x - 4 \sin^3 x

B.

2sinxcosx+cos2xsinx2 \sin x \cos x + \cos 2x \sin x

C.

sinx+sin2x\sin x + \sin 2x

D.

sin2xcosx+cos2xsinx\sin 2x \cos x + \cos 2x \sin x
Correct Answer: A

Solution:

The expression for sin3x\sin 3x is derived as sin3x=3sinx4sin3x\sin 3x = 3 \sin x - 4 \sin^3 x.

A.

0

B.

1

C.

2

D.

Undefined
Correct Answer: B

Solution:

Using the identity tan(2x)=2tan(x)1tan2(x)\tan(2x) = \frac{2\tan(x)}{1-\tan^2(x)}, substituting tan(x)=1\tan(x) = 1 gives tan(2x)=1\tan(2x) = 1.

A.

43\frac{4}{3}

B.

34\frac{3}{4}

C.

54\frac{5}{4}

D.

45\frac{4}{5}
Correct Answer: A

Solution:

Using sin(x)=1cos2(x)=1(35)2=45\sin(x) = \sqrt{1 - \cos^2(x)} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5}, we find tan(x)=sin(x)cos(x)=4/53/5=43\tan(x) = \frac{\sin(x)}{\cos(x)} = \frac{4/5}{3/5} = \frac{4}{3}.

A.

I quadrant

B.

II quadrant

C.

III quadrant

D.

IV quadrant
Correct Answer: C

Solution:

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

A.

4π3\frac{4\pi}{3}

B.

3π2\frac{3\pi}{2}

C.

5π3\frac{5\pi}{3}

D.

2π3\frac{2\pi}{3}
Correct Answer: A

Solution:

To convert degrees to radians, use the formula θ=π180×degrees\theta = \frac{\pi}{180} \times \text{degrees}. For 240°, θ=π180×240=4π3\theta = \frac{\pi}{180} \times 240 = \frac{4\pi}{3}.

A.

sin2(x)+cos2(x)=1\sin^2(x) + \cos^2(x) = 1

B.

tan2(x)+1=sec2(x)\tan^2(x) + 1 = \sec^2(x)

C.

1+cot2(x)=csc2(x)1 + \cot^2(x) = \csc^2(x)

D.

All of the above
Correct Answer: D

Solution:

All the given identities are standard trigonometric identities: sin2(x)+cos2(x)=1\sin^2(x) + \cos^2(x) = 1, tan2(x)+1=sec2(x)\tan^2(x) + 1 = \sec^2(x), and 1+cot2(x)=csc2(x)1 + \cot^2(x) = \csc^2(x).

A.

100 m

B.

173.2 m

C.

200 m

D.

86.6 m
Correct Answer: B

Solution:

The distance dd from the base of the lighthouse to the ship can be found using the tangent function: tan(30°)=100d\tan(30°) = \frac{100}{d}. Since tan(30°)=13\tan(30°) = \frac{1}{\sqrt{3}}, we have d=100×3=173.2 md = 100 \times \sqrt{3} = 173.2 \text{ m}.

A.

4cos3x3cosx4 \cos^3 x - 3 \cos x

B.

2cos2x12 \cos^2 x - 1

C.

cosx+cos2x\cos x + \cos 2x

D.

cos3x3cosx\cos^3 x - 3 \cos x
Correct Answer: A

Solution:

The expression for cos3x\cos 3x is cos3x=4cos3x3cosx\cos 3x = 4 \cos^3 x - 3 \cos x.

A.

12\frac{1}{2}

B.

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

C.

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

D.

11
Correct Answer: A

Solution:

Given tan(x)=13\tan(x) = \frac{1}{\sqrt{3}}, x=30x = 30^\circ. Therefore, sin(2x)=sin(60)=32\sin(2x) = \sin(60^\circ) = \frac{\sqrt{3}}{2}.

A.

π3\frac{\pi}{3} cm

B.

5π3\frac{5\pi}{3} cm

C.

5π6\frac{5\pi}{6} cm

D.

π6\frac{\pi}{6} cm
Correct Answer: B

Solution:

The arc length ll is given by l=rθl = r\theta, where θ\theta is in radians. Here, θ=60×π180=π3\theta = \frac{60^\circ \times \pi}{180^\circ} = \frac{\pi}{3}. Thus, l=5×π3=5π3l = 5 \times \frac{\pi}{3} = \frac{5\pi}{3} cm.

A.

45\frac{4}{5}

B.

25\frac{2}{5}

C.

35\frac{3}{5}

D.

15\frac{1}{5}
Correct Answer: A

Solution:

Using the identity sin(2x)=2tan(x)1+tan2(x)\sin(2x) = \frac{2\tan(x)}{1+\tan^2(x)}, we substitute tan(x)=12\tan(x) = \frac{1}{2} to get sin(2x)=2×121+(12)2=11.25=45\sin(2x) = \frac{2 \times \frac{1}{2}}{1 + (\frac{1}{2})^2} = \frac{1}{1.25} = \frac{4}{5}.

A.

247\frac{24}{7}

B.

724\frac{7}{24}

C.

34\frac{3}{4}

D.

43\frac{4}{3}
Correct Answer: A

Solution:

Using the identity tan(2x)=2tan(x)1tan2(x)\tan(2x) = \frac{2\tan(x)}{1 - \tan^2(x)} and knowing tan(x)=34\tan(x) = \frac{3}{4}, we find tan(2x)=2×341(34)2=641916=64716=6×164×7=247\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}.

A.

3tan(x)tan3(x)3 \tan(x) - \tan^3(x)

B.

3tan(x)tan3(x)13tan2(x)\frac{3 \tan(x) - \tan^3(x)}{1 - 3 \tan^2(x)}

C.

3tan(x)+tan3(x)13tan2(x)\frac{3 \tan(x) + \tan^3(x)}{1 - 3 \tan^2(x)}

D.

3tan(x)tan3(x)1+3tan2(x)\frac{3 \tan(x) - \tan^3(x)}{1 + 3 \tan^2(x)}
Correct Answer: B

Solution:

The formula for tan(3x)\tan(3x) is tan(3x)=3tan(x)tan3(x)13tan2(x)\tan(3x) = \frac{3 \tan(x) - \tan^3(x)}{1 - 3 \tan^2(x)}.

A.

5 cm

B.

7.5 cm

C.

10 cm

D.

15 cm
Correct Answer: C

Solution:

The length of the arc is given by l=rθl = r \theta. Substituting the values, l=10×π4=10×0.785=7.85l = 10 \times \frac{\pi}{4} = 10 \times 0.785 = 7.85 cm. However, the correct calculation should be l=10×π4=10×0.785=7.85l = 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.

A.

10 cm

B.

20 cm

C.

5 cm

D.

15 cm
Correct Answer: A

Solution:

In a right-angled triangle, the side opposite the 3030^\circ angle is half the hypotenuse. Therefore, the length of the side opposite the 3030^\circ angle is 202=10\frac{20}{2} = 10 cm.

A.

4cos3x3cosx4\cos^3 x - 3\cos x

B.

3cosx4cos3x3\cos x - 4\cos^3 x

C.

cos3x3cosx\cos^3 x - 3\cos x

D.

4cosx3cos3x4\cos x - 3\cos^3 x
Correct Answer: A

Solution:

The expression for cos(3x)\cos(3x) is derived as cos(3x)=4cos3x3cosx\cos(3x) = 4\cos^3 x - 3\cos x.

A.

13:22

B.

22:13

C.

11:6

D.

6:11
Correct Answer: B

Solution:

Given θ1=65°\theta_1 = 65° and θ2=110°\theta_2 = 110°, the ratio of the radii r1:r2r_1:r_2 is 22:1322:13.

A.

30 meters

B.

30√2 meters

C.

15 meters

D.

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.

A.

3.14 cm

B.

6.28 cm

C.

9.42 cm

D.

12.56 cm
Correct Answer: B

Solution:

In 60 minutes, the minute hand completes one full revolution, which is 2π2\pi radians. In 30 minutes, it covers half a revolution, i.e., π\pi radians. The distance moved is given by l=rθ=2×π=6.28l = r \theta = 2 \times \pi = 6.28 cm.

A.

10.47 cm

B.

5.24 cm

C.

6.28 cm

D.

3.14 cm
Correct Answer: C

Solution:

The length of the arc ll is given by l=rθl = r \theta, where θ\theta is in radians. First, convert 60° to radians: θ=60×π180=π3\theta = \frac{60 \times \pi}{180} = \frac{\pi}{3}. Thus, l=10×π3=10×3.143=10.47 cml = 10 \times \frac{\pi}{3} = \frac{10 \times 3.14}{3} = 10.47 \text{ cm}.

True or False

Correct Answer: True

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 term 'trigonometry' comes from the Greek words 'trigon', meaning triangle, and 'metron', meaning measure.

Correct Answer: False

Solution:

The symbols sin⁻¹ x and cos⁻¹ x were suggested by the astronomer Sir John F.W. Hersehel, not Arya Bhatt.

Correct Answer: True

Solution:

The symbols such as sin⁻¹ x for arc sin x were suggested by Sir John F.W. Herschel.

Correct Answer: True

Solution:

In the first quadrant, sinx\sin x increases from 0 to 1.

Correct Answer: False

Solution:

In the first quadrant, cosx\cos x decreases from 1 to 0.

Correct Answer: True

Solution:

Trigonometric functions are indeed used in the design of electric circuits, among other applications.

Correct Answer: False

Solution:

The minute hand of a watch completes one revolution in 60 minutes.

Correct Answer: False

Solution:

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

Correct Answer: False

Solution:

Trigonometry was originally developed to solve geometric problems involving triangles.

Correct Answer: False

Solution:

Trigonometric functions are used in various fields such as seismology, electric circuits, atomic state descriptions, ocean tide predictions, and musical tone analysis, not just navigation.

Correct Answer: False

Solution:

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

Correct Answer: True

Solution:

Sir John F.W. Herschel is credited with suggesting the symbols sin⁻¹ x and cos⁻¹ x for arc sin x and arc cos x.

Correct Answer: True

Solution:

The excerpt states that trigonometry was originally developed for solving geometric problems involving triangles and is currently used in various fields such as seismology and electric circuit design.

Correct Answer: True

Solution:

In 40 minutes, the minute hand covers 23\frac{2}{3} of a full revolution, resulting in a distance of 6.286.28 cm when the radius is 1.51.5 cm.

Correct Answer: True

Solution:

The minute hand of a watch completes 23\frac{2}{3} of a revolution in 40 minutes, resulting in a distance of 6.286.28 cm.

Correct Answer: False

Solution:

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

Correct Answer: True

Solution:

Trigonometry was initially developed to solve geometric problems involving triangles, which were crucial for navigation and surveying.

Correct Answer: True

Solution:

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

Correct Answer: True

Solution:

Thales is credited with determining the height of a pyramid using the proportionality of shadows.

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: True

Solution:

Given the angles θ1=65\theta_1 = 65^\circ and θ2=110\theta_2 = 110^\circ, the ratio of the radii is 22:1322:13.

Correct Answer: True

Solution:

Thales is credited with calculating the distance of a ship at sea through the proportionality of sides of similar triangles.

Correct Answer: True

Solution:

The formula for cos3x\cos 3x is indeed 4cos3x3cosx4 \cos^3 x - 3 \cos x, as derived from the angle addition formulas.

Correct Answer: True

Solution:

In 40 minutes, the minute hand turns through 23\frac{2}{3} of a revolution, and the distance traveled is calculated as 6.28 cm.

Correct Answer: True

Solution:

An angle is defined as the measure of rotation of a ray about its initial point, as described in the excerpt.

Correct Answer: True

Solution:

In 60 minutes, the minute hand completes one full revolution. Therefore, in 40 minutes, it moves through 23\frac{2}{3} of a revolution.

Correct Answer: False

Solution:

The symbols sin⁻¹ x, cos⁻¹ x, etc., for arc sin x, arc cos x, etc., were suggested by the astronomer Sir John F.W. Herschel.

Correct Answer: True

Solution:

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

Correct Answer: True

Solution:

The excerpt mentions that Thales is credited with determining the height of a pyramid in Egypt by measuring shadows and using the proportionality of sides of similar triangles.

Correct Answer: True

Solution:

Thales determined the height of a great pyramid in Egypt by measuring shadows and comparing ratios.

Correct Answer: False

Solution:

The ratio of the radii is 13:2213:22, not 22:1322:13, as calculated from the given angles.

Correct Answer: False

Solution:

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

Correct Answer: True

Solution:

The tip of the minute hand moves through 23\frac{2}{3} of a revolution in 40 minutes, which corresponds to a distance of approximately 6.28 cm.

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 equation sin3x=3sinx4sin3x\sin 3x = 3 \sin x - 4 \sin^3 x is derived by expanding sin(2x+x)\sin(2x + x) using trigonometric identities.

Correct Answer: True

Solution:

According to the excerpt, in the third quadrant, tanx\tan x increases from 00 to \infty.

Correct Answer: True

Solution:

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

Correct Answer: True

Solution:

In the first quadrant, the function cosx\cos x decreases from 1 to 0.

Correct Answer: True

Solution:

The tangent function is undefined at angles where the cosine function is zero because it involves division by zero.

Correct Answer: True

Solution:

Sir John F.W. Herschel suggested the symbols sin⁻¹ x, cos⁻¹ x, etc., for inverse trigonometric functions.

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:

An angle is defined as the measure of rotation of a given ray about its initial point.

Correct Answer: True

Solution:

The word 'trigonometry' is derived from the Greek words 'trigon' and 'metron', meaning measuring the sides of a triangle. It was originally developed to solve geometric problems involving triangles.

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: True

Solution:

Trigonometry is used in many areas such as the science of seismology, designing electric circuits, describing the state of an atom, predicting the heights of tides in the ocean, analysing a musical tone and in many other areas.

Correct Answer: True

Solution:

The identity for tan(x+y)\tan(x + y) is tanx+tany1tanxtany\frac{\tan x + \tan y}{1 - \tan x \tan y}.

Correct Answer: True

Solution:

Thales is associated with height and distance problems and is credited with determining the height of a pyramid by comparing the ratios of shadows.

Correct Answer: True

Solution:

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

Correct Answer: True

Solution:

The excerpt provides the formula for sin3x\sin 3x as 3sinx4sin3x3 \sin x - 4 \sin^3 x, which is correct.

Correct Answer: True

Solution:

The formula for the tangent of the sum of two angles is given by tan(x+y)=tanx+tany1tanxtany\tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}, which is a valid trigonometric identity.

Correct Answer: False

Solution:

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

Correct Answer: True

Solution:

In the second quadrant, sinx\sin x decreases from 1 to 0 as the angle increases from 9090^\circ to 180180^\circ.

Correct Answer: False

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.

Correct Answer: False

Solution:

Trigonometry is used in various fields including seismology, electric circuit design, and music analysis, not just navigation and surveying.

Correct Answer: True

Solution:

The formula for the tangent of the sum of two angles is indeed tan(x+y)=tanx+tany1tanxtany\tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}.

Correct Answer: True

Solution:

In the first quadrant, the function tanx\tan x increases from 0 to infinity.

Correct Answer: True

Solution:

The minute hand of a watch completes one full revolution, or 360 degrees, in 60 minutes.

Correct Answer: True

Solution:

The term 'trigonometry' is derived from the Greek words 'trigon' and 'metron', which mean 'measuring the sides of a triangle'.

Correct Answer: True

Solution:

Thales determined the height of a pyramid in Egypt by comparing the shadows of the pyramid and an auxiliary staff of known height.

Correct Answer: False

Solution:

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

Correct Answer: True

Solution:

The symbols sin1x\sin^{-1} x, cos1x\cos^{-1} x, etc., were suggested by Sir John F.W. Hersehel.

Correct Answer: True

Solution:

The excerpt provides the derivation of the identity sin3x=3sinx4sin3x\sin 3x = 3 \sin x - 4 \sin^3 x, confirming its validity.