Trigonometric Functions — Class 11 Maths Notes

Trigonometric Functions · Class 11 Maths · 11 topics.

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Topics covered in Trigonometric Functions

  1. 1.Introduction of Trigonometric Functions

    Brief Introduction

    Trigonometric functions are mathematical functions that relate the angles of a triangle to the lengths of its sides. These functions are crucial in various fields like engineering, physics, architecture, and even in daily life activities like navigation and measuring heights.

    Real-Life Example

    Imagine you are standing near a tall tree and want to know its height. By using a trigonometric function, you can measure the angle of elevation from your position to the top of the tree and the distance from you to the tree, and then calculate its height.

    Basic Trigonometric Functions

    There are six primary trigonometric functions:

    1. Sine (sin): It is the ratio of the opposite side to the hypotenuse.
    2. Cosine (cos): It is the ratio of the adjacent side to the hypotenuse.
    3. Tangent (tan): It is the ratio of the opposite side to the adjacent side.
    4. Cosecant (csc): It is the reciprocal of sine.
    5. Secant (sec): It is the reciprocal of cosine.
    6. Cotangent (cot): It is the reciprocal of tangent.

    Understanding through an Example

    Example: Finding the Height of a Tree

    Step-by-Step Explanation:

    1. Measure the distance from the base of the tree to your standing point (let’s say 10 meters).
    2. Measure the angle of elevation from your standing point to the top of the tree using a protractor (let’s say the angle is 30 degrees).
    3. Use the tangent function because it relates the opposite side (height of the tree) to the adjacent side (distance from the tree).

    tan⁡(𝜃)=oppositeadjacenttan(θ)=adjacentopposite​

    tan⁡(30∘)=height10tan(30∘)=10height​

    1. Calculate the height:

    height=10×tan⁡(30∘)height=10×tan(30∘)

    Since tan⁡(30∘)=13tan(30∘)=3​1​,

    height=10×13≈5.77 metersheight=10×3​1​≈5.77 meters

    Application in Real Life and Careers

    1. Architecture and Engineering: Designing buildings, bridges, and other structures.
    2. Astronomy: Calculating distances between celestial bodies.
    3. Navigation: Finding the position of ships and aircraft.
    4. Physics: Analyzing wave patterns and oscillations.

    Activity for Practice

    1. Measure the height of a pole or building near you using the method described above.
    2. Use different angles and distances to see how the calculations change.

      ------------------------------------------------------------------------------------------------------------- Angles: Degree Measure and Radian Measure

      Brief Introduction

      Angles are fundamental in understanding and applying trigonometric functions. They help us measure the rotation between two intersecting lines. There are two common units for measuring angles: degrees and radians.

      Real-Life Example

      Consider a clock. The hands of a clock move in a circular path, and the angle between the minute and hour hands can be measured in degrees or radians to determine the exact time.

      Degree Measure

      A degree is a unit for measuring angles, denoted by the symbol "°". A full circle is 360 degrees.

      Radian Measure

      A radian is another unit for measuring angles, based on the radius of a circle. One radian is the angle created when the arc length is equal to the radius of the circle. A full circle is 2𝜋2π radians.

      Converting Between Degrees and Radians

      Conversion Formulas:

      1. From Degrees to Radians:

        radians=degrees×𝜋180radians=degrees×180π​

      2. From Radians to Degrees:

        degrees=radians×180𝜋degrees=radians×π180​

      Understanding through an Example

      Example 1: Converting Degrees to Radians

      Convert 90 degrees to radians.

      Step-by-Step Explanation:

      1. Use the conversion formula:radians=90∘×𝜋180radians=90∘×180π​
      2. Simplify the expression:radians=90𝜋180=𝜋2radians=18090π​=2π​

      So, 90 degrees is equal to 𝜋22π​ radians.

      Example 2: Converting Radians to Degrees

      Convert 𝜋44π​ radians to degrees.

      Step-by-Step Explanation:

      1. Use the conversion formula:degrees=𝜋4×180𝜋degrees=4π​×π180​
      2. Simplify the expression:degrees=1804=45∘degrees=4180​=45∘

      So, 𝜋44π​ radians is equal to 45 degrees.

      Real-Life Applications and Careers

      1. Engineering: Designing mechanical parts and structures often involves measuring angles.
      2. Physics: Understanding rotational motion and wave mechanics.
      3. Computer Graphics: Creating animations and simulations requires precise angle measurements.
      4. Navigation and Astronomy: Calculating positions and paths of celestial bodies.

      Activity for Practice

      1. Convert the following angles from degrees to radians: 30°, 45°, 120°.
      2. Convert the following angles from radians to degrees: 𝜋66π​, 𝜋π, 2𝜋2π.

  2. 2.Relation between Radian and Real Numbers

    Brief Introduction

    Radians and real numbers are intrinsically related in trigonometry. The radian measure of an angle is defined in terms of the radius of a circle and can be expressed as a real number. This relationship helps in connecting angles with the properties of circles and understanding periodic functions like sine and cosine.

    Real-Life Example

    Consider a circular track. If you run around the track, the angle you sweep out can be measured in radians. The distance you run along the track (the arc length) can be directly related to this angle, and both can be represented as real numbers.

    Understanding Radian Measure

    Definition

    A radian is the angle subtended at the center of a circle by an arc whose length is equal to the circle's radius. Mathematically, one radian is approximately 57.3 degrees.

    Relationship with Circle

    • Full Circle: A full circle is 2𝜋2π radians.
    • Half Circle: A half circle is 𝜋π radians.
    • Quarter Circle: A quarter circle is 𝜋22π​ radians.

    Conversion

    Radians are real numbers that can be related to the circumference of a circle. The circumference 𝐶C of a circle is given by: 𝐶=2𝜋𝑟C=2πr where 𝑟r is the radius.

    Real Numbers and Radians

    Real numbers encompass all the rational and irrational numbers. Radians, as real numbers, can be positive, negative, or zero:

    • Positive Radians: Measured counterclockwise from the positive x-axis.
    • Negative Radians: Measured clockwise from the positive x-axis.
    • Zero Radians: The angle of zero radians corresponds to the starting point on the x-axis.

    Periodicity and Real Numbers

    Trigonometric functions like sine and cosine are periodic functions that repeat their values over regular intervals. The period for these functions is 2𝜋2π radians.

    Example of Periodicity

    The sine function, sin⁡(𝜃)sin(θ), repeats every 2𝜋2π radians: sin⁡(𝜃+2𝜋)=sin⁡(𝜃)sin(θ+2π)=sin(θ)

    Applying the Concept

    Example: Calculating Arc Length

    Suppose we have a circle with radius 5 units. If we sweep an angle of 𝜋33π​ radians, we can find the arc length 𝑠s.

    Step-by-Step Explanation:

    1. Formula for Arc Length: 𝑠=𝑟𝜃s=rθ where 𝑟r is the radius and 𝜃θ is the angle in radians.

    2. Substitute the values: 𝑠=5×𝜋3s=5×3π​

    3. Calculate the arc length: 𝑠=5𝜋3≈5.24 unitss=35π​≈5.24 units

    Application in Real Life and Careers

    1. Engineering: Designing mechanical parts involving rotational motion.
    2. Physics: Analyzing oscillatory and wave phenomena.
    3. Computer Science: Programming simulations involving circular motion.
    4. Astronomy: Calculating orbital paths of celestial bodies.

    Activity for Practice

    1. Find the arc length for a circle with radius 7 units and an angle of 𝜋44π​ radians.
    2. Calculate the sine and cosine of angles 2𝜋2π, 𝜋π, and 𝜋22π​ and observe their periodic nature.

  3. 3.Relation between Degree and Radian

    Brief Introduction

    Degrees and radians are two units used to measure angles. Understanding the relationship between these units is essential for converting angles from one unit to another, especially in fields such as trigonometry, physics, and engineering.

    Real-Life Example

    Think of a pizza divided into slices. If a pizza is cut into 6 equal slices, each slice forms an angle at the center. The angle can be measured in degrees (60° per slice) or radians (𝜋33π​ radians per slice). Both units describe the same angle but in different ways.

    Understanding Degrees

    A degree is a unit of angular measure. One complete revolution around a circle is 360 degrees (360°). Therefore, an angle of 90° represents one-quarter of a complete revolution.

    Understanding Radians

    A radian is another unit of angular measure. One radian is defined as the angle subtended at the center of a circle by an arc whose length is equal to the circle's radius. One complete revolution around a circle is 2𝜋2π radians. Therefore, an angle of 𝜋π radians represents half of a complete revolution.

    Conversion Between Degrees and Radians

    Conversion Formulas:

    1. From Degrees to Radians:

      radians=degrees×𝜋180radians=degrees×180π​

    2. From Radians to Degrees:

      degrees=radians×180𝜋degrees=radians×π180​

    Understanding through Examples

    Example 1: Converting Degrees to Radians

    Convert 180 degrees to radians.

    Step-by-Step Explanation:

    1. Use the conversion formula:radians=180∘×𝜋180radians=180∘×180π​
    2. Simplify the expression:radians=𝜋radians=π

    So, 180 degrees is equal to 𝜋π radians.

    Example 2: Converting Radians to Degrees

    Convert 𝜋66π​ radians to degrees.

    Step-by-Step Explanation:

    1. Use the conversion formula:degrees=𝜋6×180𝜋degrees=6π​×π180​
    2. Simplify the expression:degrees=1806=30∘degrees=6180​=30∘

    So, 𝜋66π​ radians is equal to 30 degrees.

    Real-Life Applications and Careers

    1. Engineering: Converting angles in mechanical and civil engineering designs.
    2. Physics: Understanding rotational motion and wave mechanics.
    3. Computer Graphics: Animating objects in simulations and games.
    4. Astronomy: Measuring angles between celestial bodies.

    Activity for Practice

    1. Convert the following angles from degrees to radians: 45°, 90°, 270°.
    2. Convert the following angles from radians to degrees: 𝜋22π​, 2𝜋2π, 3𝜋443π​.

  4. 4.Notational Convention in Trigonometry

    Brief Introduction

    Notational conventions are standardized methods of writing mathematical expressions. In trigonometry, these conventions help ensure clarity and consistency when dealing with angles, functions, and their measurements. Understanding these conventions is crucial for accurately interpreting and solving trigonometric problems.

    Common Notational Conventions in Trigonometry

    1. Angle Measurement

    • Degrees (°): Angles are often measured in degrees. For example, 90 degrees is written as 90°.
    • Radians (rad): Angles can also be measured in radians. For example, π/2 radians is written as π/2 or sometimes as 1.57 rad.

    2. Trigonometric Functions

    • Sine: Represented as sin⁡𝜃sinθ
    • Cosine: Represented as cos⁡𝜃cosθ
    • Tangent: Represented as tan⁡𝜃tanθ
    • Cosecant: Represented as csc⁡𝜃cscθ
    • Secant: Represented as sec⁡𝜃secθ
    • Cotangent: Represented as cot⁡𝜃cotθ

    3. Angle Symbols

    • Greek letters like 𝜃θ, 𝛼α, 𝛽β are commonly used to represent angles.
    • For example, sin⁡𝜃sinθ means the sine of the angle theta.

    4. Function Arguments

    • The argument of a trigonometric function (the angle) is usually placed within parentheses, such as sin⁡(𝜃)sin(θ) or cos⁡(𝛼)cos(α).
    • Sometimes, the argument may be written without parentheses for simplicity, such as sin⁡𝜃sinθ.

    5. Inverse Trigonometric Functions

    • Arcsine: Represented as sin⁡−1𝜃sin−1θ or arcsin⁡𝜃arcsinθ
    • Arccosine: Represented as cos⁡−1𝜃cos−1θ or arccos⁡𝜃arccosθ
    • Arctangent: Represented as tan⁡−1𝜃tan−1θ or arctan⁡𝜃arctanθ

    Understanding through Examples

    Example 1: Writing Trigonometric Functions

    Consider an angle of 45 degrees. The sine and cosine of this angle can be written as:

    • sin⁡45∘sin45∘
    • cos⁡45∘cos45∘

    Example 2: Converting Between Units

    An angle of 90 degrees can be converted to radians and written as:

    • 90∘=𝜋2 rad90∘=2π​ rad

    Example 3: Using Inverse Functions

    To find the angle whose sine is 0.5, you can write:

    • 𝜃=sin⁡−1(0.5)θ=sin−1(0.5) or 𝜃=arcsin⁡(0.5)θ=arcsin(0.5)

    Real-Life Applications and Careers

    1. Engineering: Standardized notation helps engineers communicate designs and calculations accurately.
    2. Physics: Consistent notation is crucial for understanding wave functions and oscillations.
    3. Mathematics Education: Helps in teaching and learning trigonometric concepts clearly.
    4. Computer Science: Used in algorithms for graphics and simulations.

    Activity for Practice

    1. Write the cosine of 30 degrees using both degrees and radians.
    2. Convert an angle of 120 degrees to radians and write the sine of this angle.
    3. Find the angle whose tangent is 1 using inverse trigonometric notation.

  5. 5.Exercise Questions

    Solving the Given Problems

    Problem 1: Convert Degree Measures to Radian Measures

    1. Convert 25° to radians:

      radians=25∘×𝜋180radians=25∘×180π​radians=25𝜋180=5𝜋36radians=18025π​=365π​

    2. Convert −47∘30′−47∘30′ to radians: First, convert the angle to decimal degrees.

      −47∘30′=−47.5∘−47∘30′=−47.5∘

      Then convert to radians:

      radians=−47.5∘×𝜋180radians=−47.5∘×180π​radians=−47.5𝜋180=−95𝜋360=−19𝜋72radians=180−47.5π​=360−95π​=−7219π​

    3. Convert 240° to radians:

      radians=240∘×𝜋180radians=240∘×180π​radians=240𝜋180=4𝜋3radians=180240π​=34π​

    4. Convert 520° to radians:

      radians=520∘×𝜋180radians=520∘×180π​radians=520𝜋180=26𝜋9radians=180520π​=926π​

    Problem 2: Convert Radian Measures to Degree Measures

    1. Convert 11161611​ rad to degrees:

      degrees=1116×180𝜋degrees=1611​×π180​

      Using 𝜋≈227π≈722​:

      degrees=1116×180×722degrees=1611​×22180×7​degrees=11×1260352=13860352≈39.38∘degrees=35211×1260​=35213860​≈39.38∘

    2. Convert -4 rad to degrees:

      degrees=−4×180𝜋degrees=−4×π180​

      Using 𝜋≈227π≈722​:

      degrees=−4×180×722degrees=−4×22180×7​degrees=−4×126022=−4×57.27≈−229.09∘degrees=−4×221260​=−4×57.27≈−229.09∘

    3. Convert 5𝜋335π​ rad to degrees:

      degrees=5𝜋3×180𝜋degrees=35π​×π180​degrees=5×1803=300∘degrees=35×180​=300∘

    4. Convert 7𝜋667π​ rad to degrees:

      degrees=7𝜋6×180𝜋degrees=67π​×π180​degrees=7×1806=210∘degrees=67×180​=210∘

    Problem 3: Revolutions to Radians

    A wheel makes 360 revolutions in one minute. Find the number of radians it turns in one second.

    1. One complete revolution is 2𝜋2π radians.
    2. In one minute (60 seconds), the wheel makes 360 revolutions:360×2𝜋 radians360×2π radians
    3. In one second:360×2𝜋60=12𝜋 radians60360×2π​=12π radians

    Problem 4: Degree Measure of Angle from Arc Length

    Find the degree measure of the angle subtended at the center of a circle of radius 100 cm by an arc of length 22 cm.

    1. Use the formula:𝜃=𝑠𝑟θ=rs​where 𝑠s is the arc length and 𝑟r is the radius.𝜃=22 cm100 cm=0.22 radiansθ=100 cm22 cm​=0.22 radians
    2. Convert radians to degrees:degrees=0.22×180𝜋degrees=0.22×π180​Using 𝜋≈227π≈722​:degrees=0.22×180×722degrees=0.22×22180×7​degrees=0.22×57.27≈12.6∘degrees=0.22×57.27≈12.6∘

    Summary of Answers

    1. Degree to Radians:

      • (i) 25° = 5𝜋36365π​
      • (ii) −47∘30′=−19𝜋72−47∘30′=−7219π​
      • (iii) 240° = 4𝜋334π​
      • (iv) 520° = 26𝜋9926π​

    2. Radian to Degrees:

      • (i) 11161611​ rad ≈ 39.38°
      • (ii) -4 rad ≈ -229.09°
      • (iii) 5𝜋335π​ rad = 300°
      • (iv) 7𝜋667π​ rad = 210°

    3. Revolutions to Radians:

      • 12𝜋12π radians in one second.

    4. Angle from Arc Length:

      • 12.6°

  6. 6.Exercise Questions

    Question 1

    In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of the minor arc of the chord.

    Solution:

    1. Find the radius:

      Radius(𝑟)=Diameter2=40 cm2=20 cmRadius(r)=2Diameter​=240 cm​=20 cm

    2. Determine the angle subtended by the chord at the center: Let's denote the angle subtended by the chord as 𝜃θ. We use the cosine rule for the central angle in the isosceles triangle formed by the radii and the chord.

      Using the chord length formula:

      cos⁡(𝜃2)=Chord length2×Radiuscos(2θ​)=2×RadiusChord length​cos⁡(𝜃2)=20 cm2×20 cm=2040=12cos(2θ​)=2×20 cm20 cm​=4020​=21​

    3. Find 𝜃/2θ/2:

      𝜃2=cos⁡−1(12)=𝜋32θ​=cos−1(21​)=3π​

    4. Calculate 𝜃θ:

      𝜃=2×𝜋3=2𝜋3 radiansθ=2×3π​=32π​ radians

    5. Calculate the length of the minor arc:

      Arc length=𝑟×𝜃=20 cm×2𝜋3Arc length=r×θ=20 cm×32π​Arc length=40𝜋3≈41.89 cmArc length=340π​≈41.89 cm

    Question 2

    If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.

    Solution:

    1. Convert angles to radians:

      60∘=60×𝜋180=𝜋3 radians60∘=18060×π​=3π​ radians75∘=75×𝜋180=5𝜋12 radians75∘=18075×π​=125π​ radians

    2. Let 𝑟1r1​ and 𝑟2r2​ be the radii of the circles.

    3. Use the arc length formula 𝑠=𝑟𝜃s=rθ:

      Arc length in first circle=𝑟1×𝜋3Arc length in first circle=r1​×3π​Arc length in second circle=𝑟2×5𝜋12Arc length in second circle=r2​×125π​

    4. Since the arc lengths are the same:

      𝑟1×𝜋3=𝑟2×5𝜋12r1​×3π​=r2​×125π​

    5. Solve for the ratio of the radii:

      𝑟1×12𝜋×13=𝑟2×5𝜋r1​×π12​×31​=r2​×π5​𝑟1×4=𝑟2×5r1​×4=r2​×5𝑟1𝑟2=54r2​r1​​=45​

    Question 3

    Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length:

    • (i) 10 cm
    • (ii) 15 cm
    • (iii) 21 cm

    Solution:

    1. Use the arc length formula 𝜃=𝑠𝑟θ=rs​, where 𝑠s is the arc length and 𝑟r is the radius (pendulum length).

    (i) Arc length = 10 cm

    𝜃=10 cm75 cm=215 radiansθ=75 cm10 cm​=152​ radians

    (ii) Arc length = 15 cm

    𝜃=15 cm75 cm=15 radiansθ=75 cm15 cm​=51​ radians

    (iii) Arc length = 21 cm

    𝜃=21 cm75 cm=725 radiansθ=75 cm21 cm​=257​ radians

  7. 7.Trigonometric Functions, their signs, and their domains and ranges.

    Trigonometric Functions

    Trigonometric functions relate the angles of a triangle to the lengths of its sides. The main trigonometric functions are:

    1. Sine (sin θ)
    2. Cosine (cos θ)
    3. Tangent (tan θ)
    4. Cosecant (csc θ)
    5. Secant (sec θ)
    6. Cotangent (cot θ)

    These functions are defined using a right-angled triangle or the unit circle.

    Unit Circle Definition

    In the unit circle, the coordinates of a point (x, y) on the circle can be used to define the trigonometric functions as follows:

    • sin⁡𝜃=𝑦sinθ=y
    • cos⁡𝜃=𝑥cosθ=x
    • tan⁡𝜃=𝑦𝑥tanθ=xy​ (where 𝑥≠0x=0)
    • csc⁡𝜃=1𝑦cscθ=y1​ (where 𝑦≠0y=0)
    • sec⁡𝜃=1𝑥secθ=x1​ (where 𝑥≠0x=0)
    • cot⁡𝜃=𝑥𝑦cotθ=yx​ (where 𝑦≠0y=0)

    Sign of Trigonometric Functions

    The sign of trigonometric functions depends on the quadrant in which the angle lies. The unit circle is divided into four quadrants:

    1. Quadrant I (0° to 90° or 0 to π/2 radians)

      • sin⁡𝜃>0sinθ>0
      • cos⁡𝜃>0cosθ>0
      • tan⁡𝜃>0tanθ>0

    2. Quadrant II (90° to 180° or π/2 to π radians)

      • sin⁡𝜃>0sinθ>0
      • cos⁡𝜃<0cosθ<0
      • tan⁡𝜃<0tanθ<0

    3. Quadrant III (180° to 270° or π to 3π/2 radians)

      • sin⁡𝜃<0sinθ<0
      • cos⁡𝜃<0cosθ<0
      • tan⁡𝜃>0tanθ>0

    4. Quadrant IV (270° to 360° or 3π/2 to 2π radians)

      • sin⁡𝜃<0sinθ<0
      • cos⁡𝜃>0cosθ>0
      • tan⁡𝜃<0tanθ<0

    Domain and Range of Trigonometric Functions

    1. Sine and Cosine

      • Domain: All real numbers (−∞,∞)(−∞,∞)
      • Range: [−1,1][−1,1]

    2. Tangent and Cotangent

      • Domain: All real numbers except (𝜋2+𝑛𝜋)(2π​+nπ), where 𝑛n is an integer.
      • Range: (−∞,∞)(−∞,∞)

    3. Secant and Cosecant

      • Domain: All real numbers except 𝑛𝜋nπ (for secant) and (𝜋2+𝑛𝜋)(2π​+nπ) (for cosecant), where 𝑛n is an integer.
      • Range: (−∞,−1]∪[1,∞)(−∞,−1]∪[1,∞)

    Examples

    1. Finding the value of trigonometric functions at specific angles:

      • sin⁡30°=12sin30°=21​
      • cos⁡60°=12cos60°=21​
      • tan⁡45°=1tan45°=1

    2. Solving a trigonometric equation:

      Find 𝜃θ if sin⁡𝜃=12sinθ=21​.

      Solution:

      • sin⁡𝜃=12sinθ=21​ at 𝜃=30°θ=30° or 𝜃=150°θ=150° in the first and second quadrants.

    Problem Solving Example

    Example: Solve for 𝑥x in the equation 2sin⁡𝑥−1=02sinx−1=0.

    Solution:

    1. 2sin⁡𝑥−1=02sinx−1=0
    2. 2sin⁡𝑥=12sinx=1
    3. sin⁡𝑥=12sinx=21​

    sin⁡𝑥=12sinx=21​ at 𝑥=30°x=30° or 𝑥=150°x=150°.

  8. 8.Exercise Questions

    Exercise 1

    Question: Find the values of other five trigonometric functions if cos⁡𝑥=−12cosx=−21​, 𝑥x lies in the third quadrant.

    Solution: In the third quadrant:

    • cos⁡𝑥=−12cosx=−21​
    • sin⁡𝑥sinx is negative
    • tan⁡𝑥tanx is positive
    • cot⁡𝑥cotx is positive
    • sec⁡𝑥secx is negative
    • csc⁡𝑥cscx is negative

    Using the Pythagorean identity, sin⁡2𝑥+cos⁡2𝑥=1sin2x+cos2x=1: sin⁡2𝑥+(−12)2=1sin2x+(−21​)2=1 sin⁡2𝑥+14=1sin2x+41​=1 sin⁡2𝑥=1−14sin2x=1−41​ sin⁡2𝑥=34sin2x=43​ sin⁡𝑥=−34=−32sinx=−43​​=−23​​

    Therefore: tan⁡𝑥=sin⁡𝑥cos⁡𝑥=−32−12=3tanx=cosxsinx​=−21​−23​​​=3​ cot⁡𝑥=1tan⁡𝑥=13=33cotx=tanx1​=3​1​=33​​ sec⁡𝑥=1cos⁡𝑥=−2secx=cosx1​=−2 csc⁡𝑥=1sin⁡𝑥=−23=−233cscx=sinx1​=−3​2​=−323​​

    Exercise 2

    Question: Find the values of other five trigonometric functions if sin⁡𝑥=35sinx=53​, 𝑥x lies in the second quadrant.

    Solution: In the second quadrant:

    • sin⁡𝑥=35sinx=53​
    • cos⁡𝑥cosx is negative
    • tan⁡𝑥tanx is negative
    • cot⁡𝑥cotx is negative
    • sec⁡𝑥secx is negative
    • csc⁡𝑥cscx is positive

    Using the Pythagorean identity, sin⁡2𝑥+cos⁡2𝑥=1sin2x+cos2x=1: (35)2+cos⁡2𝑥=1(53​)2+cos2x=1 925+cos⁡2𝑥=1259​+cos2x=1 cos⁡2𝑥=1−925cos2x=1−259​ cos⁡2𝑥=1625cos2x=2516​ cos⁡𝑥=−45cosx=−54​

    Therefore: tan⁡𝑥=sin⁡𝑥cos⁡𝑥=35−45=−34tanx=cosxsinx​=−54​53​​=−43​ cot⁡𝑥=1tan⁡𝑥=−43cotx=tanx1​=−34​ sec⁡𝑥=1cos⁡𝑥=−54secx=cosx1​=−45​ csc⁡𝑥=1sin⁡𝑥=53cscx=sinx1​=35​

    Exercise 3

    Question: Find the values of other five trigonometric functions if cot⁡𝑥=−34cotx=−43​, 𝑥x lies in the third quadrant.

    Solution: In the third quadrant:

    • cot⁡𝑥=−34cotx=−43​
    • tan⁡𝑥tanx is positive
    • sin⁡𝑥sinx is negative
    • cos⁡𝑥cosx is negative
    • sec⁡𝑥secx is negative
    • csc⁡𝑥cscx is negative

    Given cot⁡𝑥=−34cotx=−43​: tan⁡𝑥=−1cot⁡𝑥=43tanx=−cotx1​=34​

    Using the identity 1+cot⁡2𝑥=csc⁡2𝑥1+cot2x=csc2x: 1+(−34)2=csc⁡2𝑥1+(−43​)2=csc2x 1+916=csc⁡2𝑥1+169​=csc2x csc⁡2𝑥=2516csc2x=1625​ csc⁡𝑥=−54cscx=−45​ (since sin⁡𝑥sinx is negative)

    sin⁡𝑥=−1csc⁡𝑥=−45sinx=−cscx1​=−54​

    Using the identity 1+tan⁡2𝑥=sec⁡2𝑥1+tan2x=sec2x: 1+(43)2=sec⁡2𝑥1+(34​)2=sec2x 1+169=sec⁡2𝑥1+916​=sec2x sec⁡2𝑥=259sec2x=925​ sec⁡𝑥=−53secx=−35​ (since cos⁡𝑥cosx is negative)

    cos⁡𝑥=−1sec⁡𝑥=−35cosx=−secx1​=−53​

    Therefore: tan⁡𝑥=sin⁡𝑥cos⁡𝑥=−45−35=43tanx=cosxsinx​=−53​−54​​=34​

    Exercise 4

    Question: Find the values of other five trigonometric functions if sec⁡𝑥=135secx=513​, 𝑥x lies in the fourth quadrant.

    Solution: In the fourth quadrant:

    • sec⁡𝑥=135secx=513​
    • cos⁡𝑥cosx is positive
    • sin⁡𝑥sinx is negative
    • tan⁡𝑥tanx is negative
    • cot⁡𝑥cotx is negative
    • csc⁡𝑥cscx is negative

    Given sec⁡𝑥=135secx=513​: cos⁡𝑥=1sec⁡𝑥=513cosx=secx1​=135​

    Using the identity sin⁡2𝑥+cos⁡2𝑥=1sin2x+cos2x=1: sin⁡2𝑥+(513)2=1sin2x+(135​)2=1 sin⁡2𝑥+25169=1sin2x+16925​=1 sin⁡2𝑥=1−25169sin2x=1−16925​ sin⁡2𝑥=144169sin2x=169144​ sin⁡𝑥=−1213sinx=−1312​

    Therefore: tan⁡𝑥=sin⁡𝑥cos⁡𝑥=−1213513=−125tanx=cosxsinx​=135​−1312​​=−512​ cot⁡𝑥=1tan⁡𝑥=−512cotx=tanx1​=−125​ csc⁡𝑥=1sin⁡𝑥=−1312cscx=sinx1​=−1213​

    Exercise 5

    Question: Find the values of other five trigonometric functions if tan⁡𝑥=−512tanx=−125​, 𝑥x lies in the second quadrant.

    Solution: In the second quadrant:

    • tan⁡𝑥=−512tanx=−125​
    • \sin x is positive
    • \cos x is negative
    • \cot x is negative
    • \sec x is negative
    • \csc x is positive

    Using the identity 1+tan⁡2𝑥=sec⁡2𝑥1+tan2x=sec2x: 1+(−512)2=sec⁡2𝑥1+(−125​)2=sec2x 1+25144=sec⁡2𝑥1+14425​=sec2x sec⁡2𝑥=169144sec2x=144169​ sec⁡𝑥=−1312secx=−1213​ (since cos⁡𝑥cosx is negative)

    cos⁡𝑥=−1sec⁡𝑥=−1213cosx=−secx1​=−1312​

    Using the identity sin⁡2𝑥+cos⁡2𝑥=1sin2x+cos2x=1: sin⁡2𝑥+(−1213)2=1sin2x+(−1312​)2=1 sin⁡2𝑥+144169=1sin2x+169144​=1 sin⁡2𝑥=1−144169sin2x=1−169144​ sin⁡2𝑥=25169sin2x=16925​ sin⁡𝑥=513sinx=135​

    Therefore: cot⁡𝑥=1tan⁡𝑥=−125cotx=tanx1​=−512​ csc⁡𝑥=1sin⁡𝑥=135cscx=sinx1​=513​

    Exercise 6

    Question: Find the value of sin⁡765°sin765°.

    Solution: First, reduce the angle: 765°−2⋅360°=765°−720°=45°765°−2⋅360°=765°−720°=45° sin⁡765°=sin⁡45°=22sin765°=sin45°=22​​

    Exercise 7

    Question: Find the value of csc⁡(−1410°)csc(−1410°).

    Solution: First, reduce the angle: −1410°+4⋅360°=−1410°+1440°=30°−1410°+4⋅360°=−1410°+1440°=30° csc⁡(−1410°)=csc⁡30°=2csc(−1410°)=csc30°=2

    Exercise 8

    Question: Find the value of tan⁡(19𝜋3)tan(319π​).

    Solution: First, reduce the angle: 19𝜋3=18𝜋3+𝜋3=6𝜋+𝜋3319π​=318π​+3π​=6π+3π​ Since tan⁡tan is periodic with period 𝜋π: tan⁡(19𝜋3)=tan⁡(𝜋3)=3tan(319π​)=tan(3π​)=3​

    Exercise 9

    Question: Find the value of sin⁡(−11𝜋3)sin(3−11π​).

    Solution: First, reduce the angle: −11𝜋3=−12𝜋3+𝜋3=−4𝜋+𝜋33−11π​=−312π​+3π​=−4π+3π​ Since sin⁡sin is periodic with period 2𝜋2π: sin⁡(−11𝜋3)=sin⁡(𝜋3)=32sin(3−11π​)=sin(3π​)=23​​

    Exercise 10

    Question: Find the value of cot⁡(−15𝜋4)cot(4−15π​).

    Solution: First, reduce the angle: −15𝜋4=−4𝜋+𝜋44−15π​=−4π+4π​ Since cot⁡cot is periodic with period 𝜋π: cot⁡(−15𝜋4)=cot⁡(𝜋4)=1cot(4−15π​)=cot(4π​)=1

  9. 9.Trigonometric Functions of Sum and Difference of Two Angles

    The trigonometric functions of the sum and difference of two angles are important identities in trigonometry. These identities help us to simplify expressions and solve trigonometric equations involving the sum or difference of angles.

    Sum and Difference Formulas

    1. Sine of Sum and Difference

      • sin⁡(𝐴+𝐵)=sin⁡𝐴cos⁡𝐵+cos⁡𝐴sin⁡𝐵sin(A+B)=sinAcosB+cosAsinB
      • sin⁡(𝐴−𝐵)=sin⁡𝐴cos⁡𝐵−cos⁡𝐴sin⁡𝐵sin(A−B)=sinAcosB−cosAsinB

    2. Cosine of Sum and Difference

      • cos⁡(𝐴+𝐵)=cos⁡𝐴cos⁡𝐵−sin⁡𝐴sin⁡𝐵cos(A+B)=cosAcosB−sinAsinB
      • cos⁡(𝐴−𝐵)=cos⁡𝐴cos⁡𝐵+sin⁡𝐴sin⁡𝐵cos(A−B)=cosAcosB+sinAsinB

    3. Tangent of Sum and Difference

      • tan⁡(𝐴+𝐵)=tan⁡𝐴+tan⁡𝐵1−tan⁡𝐴tan⁡𝐵tan(A+B)=1−tanAtanBtanA+tanB​
      • tan⁡(𝐴−𝐵)=tan⁡𝐴−tan⁡𝐵1+tan⁡𝐴tan⁡𝐵tan(A−B)=1+tanAtanBtanA−tanB​

    These formulas allow us to find the sine, cosine, and tangent of the sum or difference of two angles using the trigonometric functions of the individual angles.

    Examples

    Let's solve some problems using these identities.

    Example 1: Find sin⁡(75°)sin(75°)

    Solution: We can express 75°75° as the sum of 45°45° and 30°30°.

    sin⁡(75°)=sin⁡(45°+30°)sin(75°)=sin(45°+30°)

    Using the sine sum formula:

    sin⁡(45°+30°)=sin⁡45°cos⁡30°+cos⁡45°sin⁡30°sin(45°+30°)=sin45°cos30°+cos45°sin30°

    Now, substitute the known values:

    sin⁡45°=22,cos⁡30°=32,cos⁡45°=22,sin⁡30°=12sin45°=22​​,cos30°=23​​,cos45°=22​​,sin30°=21​

    sin⁡(75°)=22⋅32+22⋅12sin(75°)=22​​⋅23​​+22​​⋅21​

    sin⁡(75°)=64+24sin(75°)=46​​+42​​

    sin⁡(75°)=6+24sin(75°)=46​+2​​

    Example 2: Find cos⁡(15°)cos(15°)

    Solution: We can express 15°15° as the difference of 45°45° and 30°30°.

    cos⁡(15°)=cos⁡(45°−30°)cos(15°)=cos(45°−30°)

    Using the cosine difference formula:

    cos⁡(45°−30°)=cos⁡45°cos⁡30°+sin⁡45°sin⁡30°cos(45°−30°)=cos45°cos30°+sin45°sin30°

    Now, substitute the known values:

    cos⁡45°=22,cos⁡30°=32,sin⁡45°=22,sin⁡30°=12cos45°=22​​,cos30°=23​​,sin45°=22​​,sin30°=21​

    cos⁡(15°)=22⋅32+22⋅12cos(15°)=22​​⋅23​​+22​​⋅21​

    cos⁡(15°)=64+24cos(15°)=46​​+42​​

    cos⁡(15°)=6+24cos(15°)=46​+2​​

    Example 3: Find tan⁡(75°)tan(75°)

    Solution: We can express 75°75° as the sum of 45°45° and 30°30°.

    tan⁡(75°)=tan⁡(45°+30°)tan(75°)=tan(45°+30°)

    Using the tangent sum formula:

    tan⁡(45°+30°)=tan⁡45°+tan⁡30°1−tan⁡45°tan⁡30°tan(45°+30°)=1−tan45°tan30°tan45°+tan30°​

    Now, substitute the known values:

    tan⁡45°=1,tan⁡30°=13tan45°=1,tan30°=3​1​

    tan⁡(75°)=1+131−1⋅13tan(75°)=1−1⋅3​1​1+3​1​​

    tan⁡(75°)=1+131−13tan(75°)=1−3​1​1+3​1​​

    tan⁡(75°)=3+133−13tan(75°)=3​3​−1​3​3​+1​​

    tan⁡(75°)=3+13−1tan(75°)=3​−13​+1​

    Multiplying numerator and denominator by 3+13​+1:

    tan⁡(75°)=(3+1)2(3−1)(3+1)tan(75°)=(3​−1)(3​+1)(3​+1)2​

    tan⁡(75°)=3+23+13−1tan(75°)=3−13+23​+1​

    tan⁡(75°)=4+232tan(75°)=24+23​​

    tan⁡(75°)=2+3tan(75°)=2+3​

    Example 4: Find cos⁡(105°)cos(105°)

    Solution: We can express 105°105° as the sum of 60°60° and 45°45°.

    cos⁡(105°)=cos⁡(60°+45°)cos(105°)=cos(60°+45°)

    Using the cosine sum formula:

    cos⁡(60°+45°)=cos⁡60°cos⁡45°−sin⁡60°sin⁡45°cos(60°+45°)=cos60°cos45°−sin60°sin45°

    Now, substitute the known values:

    cos⁡60°=12,cos⁡45°=22,sin⁡60°=32,sin⁡45°=22cos60°=21​,cos45°=22​​,sin60°=23​​,sin45°=22​​

    cos⁡(105°)=12⋅22−32⋅22cos(105°)=21​⋅22​​−23​​⋅22​​

    cos⁡(105°)=24−64cos(105°)=42​​−46​​

    cos⁡(105°)=2−64cos(105°)=42​−6​​

  10. 10.Exercise Questions

    Problem 1

    Prove that:

    sin⁡2𝜋6+cos⁡2𝜋3−tan⁡2𝜋4=12sin26π​+cos23π​−tan24π​=21​

    Solution:

    1. Calculate each trigonometric value:

      • sin⁡𝜋6=12sin6π​=21​
      • cos⁡𝜋3=12cos3π​=21​
      • tan⁡𝜋4=1tan4π​=1

    2. Substitute these values into the expression:

    sin⁡2𝜋6=(12)2=14sin26π​=(21​)2=41​

    cos⁡2𝜋3=(12)2=14cos23π​=(21​)2=41​

    tan⁡2𝜋4=12=1tan24π​=12=1

    1. Combine these results:

    sin⁡2𝜋6+cos⁡2𝜋3−tan⁡2𝜋4=14+14−1=12−1=−12sin26π​+cos23π​−tan24π​=41​+41​−1=21​−1=−21​

    Thus, the left-hand side is indeed equal to the right-hand side:

    −12=−12−21​=−21​

    Therefore, the identity is proved.

    Problem 2

    Prove that:

    2sin⁡2𝜋6+csc⁡27𝜋6cos⁡2𝜋3=322sin26π​+csc267π​cos23π​=23​

    Solution:

    1. Calculate each trigonometric value:

      • sin⁡𝜋6=12sin6π​=21​
      • csc⁡7𝜋6=−2csc67π​=−2 (since sin⁡7𝜋6=−12sin67π​=−21​)
      • cos⁡𝜋3=12cos3π​=21​

    2. Substitute these values into the expression:

    2sin⁡2𝜋6=2(12)2=2⋅14=122sin26π​=2(21​)2=2⋅41​=21​

    csc⁡27𝜋6cos⁡2𝜋3=(−2)2(12)2=4⋅14=1csc267π​cos23π​=(−2)2(21​)2=4⋅41​=1

    1. Combine these results:

    2sin⁡2𝜋6+csc⁡27𝜋6cos⁡2𝜋3=12+1=12+22=322sin26π​+csc267π​cos23π​=21​+1=21​+22​=23​

    Thus, the left-hand side is indeed equal to the right-hand side:

    32=3223​=23​

    Therefore, the identity is proved.

    Problem 3

    Prove that:

    sin⁡𝑥−sin⁡𝑦cos⁡𝑥+cos⁡𝑦=tan⁡𝑥−𝑦2cosx+cosysinx−siny​=tan2x−y​

    Solution:

    1. Use the sum-to-product identities for the numerator and the denominator:

      • sin⁡𝑥−sin⁡𝑦=2cos⁡(𝑥+𝑦2)sin⁡(𝑥−𝑦2)sinx−siny=2cos(2x+y​)sin(2x−y​)
      • cos⁡𝑥+cos⁡𝑦=2cos⁡(𝑥+𝑦2)cos⁡(𝑥−𝑦2)cosx+cosy=2cos(2x+y​)cos(2x−y​)

    2. Substitute these identities into the expression:

    sin⁡𝑥−sin⁡𝑦cos⁡𝑥+cos⁡𝑦=2cos⁡(𝑥+𝑦2)sin⁡(𝑥−𝑦2)2cos⁡(𝑥+𝑦2)cos⁡(𝑥−𝑦2)cosx+cosysinx−siny​=2cos(2x+y​)cos(2x−y​)2cos(2x+y​)sin(2x−y​)​

    1. Simplify the expression:

    sin⁡𝑥−sin⁡𝑦cos⁡𝑥+cos⁡𝑦=sin⁡(𝑥−𝑦2)cos⁡(𝑥−𝑦2)=tan⁡(𝑥−𝑦2)cosx+cosysinx−siny​=cos(2x−y​)sin(2x−y​)​=tan(2x−y​)

    Thus, the left-hand side is indeed equal to the right-hand side:

    sin⁡𝑥−sin⁡𝑦cos⁡𝑥+cos⁡𝑦=tan⁡(𝑥−𝑦2)cosx+cosysinx−siny​=tan(2x−y​)

    Therefore, the identity is proved.

    Problem 4

    Prove that:

    sin⁡𝑥−sin⁡3𝑥=2sin⁡𝑥sinx−sin3x=2sinx

    Solution:

    1. Use the trigonometric identity for sin⁡3𝑥sin3x:

    sin⁡3𝑥=3sin⁡𝑥−4sin⁡3𝑥sin3x=3sinx−4sin3x

    1. Substitute this identity into the expression:

    sin⁡𝑥−sin⁡3𝑥=sin⁡𝑥−(3sin⁡𝑥−4sin⁡3𝑥)sinx−sin3x=sinx−(3sinx−4sin3x)

    1. Simplify the expression:

    sin⁡𝑥−sin⁡3𝑥=sin⁡𝑥−3sin⁡𝑥+4sin⁡3𝑥sinx−sin3x=sinx−3sinx+4sin3x

    sin⁡𝑥−sin⁡3𝑥=−2sin⁡𝑥+4sin⁡3𝑥sinx−sin3x=−2sinx+4sin3x

  11. 11.Exercise Questions

    Problem 1

    Prove that:

    tan⁡4𝑥=4tan⁡𝑥(1−tan⁡2𝑥)1−6tan⁡2𝑥+tan⁡4𝑥tan4x=1−6tan2x+tan4x4tanx(1−tan2x)​

    Solution:

    To prove this, we'll use the tangent double-angle and sum formulas.

    1. Double Angle Formula for Tangent:

    tan⁡2𝑥=2tan⁡𝑥1−tan⁡2𝑥tan2x=1−tan2x2tanx​

    1. Applying Double Angle Formula Twice:

    tan⁡4𝑥=tan⁡(2⋅2𝑥)tan4x=tan(2⋅2x) tan⁡4𝑥=2tan⁡2𝑥1−tan⁡22𝑥tan4x=1−tan22x2tan2x​

    1. Substitute tan⁡2𝑥=2tan⁡𝑥1−tan⁡2𝑥tan2x=1−tan2x2tanx​:

    tan⁡2𝑥=2tan⁡𝑥1−tan⁡2𝑥tan2x=1−tan2x2tanx​

    So,

    tan⁡4𝑥=2(2tan⁡𝑥1−tan⁡2𝑥)1−(2tan⁡𝑥1−tan⁡2𝑥)2tan4x=1−(1−tan2x2tanx​)22(1−tan2x2tanx​)​

    1. Simplify the expression:

    tan⁡4𝑥=4tan⁡𝑥1−4tan⁡2𝑥(1−tan⁡2𝑥)2tan4x=1−(1−tan2x)24tan2x​4tanx​

    tan⁡4𝑥=4tan⁡𝑥(1−tan⁡2𝑥)2(1−tan⁡2𝑥)2−4tan⁡2𝑥tan4x=(1−tan2x)2−4tan2x4tanx(1−tan2x)2​

    1. Expand the denominator:

    (1−tan⁡2𝑥)2=1−2tan⁡2𝑥+tan⁡4𝑥(1−tan2x)2=1−2tan2x+tan4x

    (1−tan⁡2𝑥)2−4tan⁡2𝑥=1−2tan⁡2𝑥+tan⁡4𝑥−4tan⁡2𝑥(1−tan2x)2−4tan2x=1−2tan2x+tan4x−4tan2x

    =1−6tan⁡2𝑥+tan⁡4𝑥=1−6tan2x+tan4x

    So,

    tan⁡4𝑥=4tan⁡𝑥(1−tan⁡2𝑥)1−6tan⁡2𝑥+tan⁡4𝑥tan4x=1−6tan2x+tan4x4tanx(1−tan2x)​

    Thus, the identity is proved.

    Problem 2

    Prove that:

    cos⁡4𝑥=1−8sin⁡2𝑥cos⁡2𝑥cos4x=1−8sin2xcos2x

    Solution:

    We'll use the double-angle identities for cosine and sine.

    1. Double Angle Formulas:

    cos⁡2𝑥=2cos⁡2𝑥−1cos2x=2cos2x−1 sin⁡2𝑥=2sin⁡𝑥cos⁡𝑥sin2x=2sinxcosx

    1. Express cos⁡4𝑥cos4x in terms of cos⁡2𝑥cos2x:

    cos⁡4𝑥=2cos⁡22𝑥−1cos4x=2cos22x−1

    1. Substitute cos⁡2𝑥=1−2sin⁡2𝑥cos2x=1−2sin2x:

    cos⁡2𝑥=1−2sin⁡2𝑥cos2x=1−2sin2x

    1. Express cos⁡4𝑥cos4x using cos⁡2𝑥cos2x:

    cos⁡4𝑥=2(2cos⁡2𝑥−1)2−1cos4x=2(2cos2x−1)2−1

    1. Expand the expression:

    cos⁡4𝑥=2(4cos⁡4𝑥−4cos⁡2𝑥+1)−1cos4x=2(4cos4x−4cos2x+1)−1

    cos⁡4𝑥=8cos⁡4𝑥−8cos⁡2𝑥+2−1cos4x=8cos4x−8cos2x+2−1

    cos⁡4𝑥=8cos⁡4𝑥−8cos⁡2𝑥+1cos4x=8cos4x−8cos2x+1

    1. Using the identity sin⁡2𝑥=2sin⁡𝑥cos⁡𝑥sin2x=2sinxcosx:

    sin⁡22𝑥=4sin⁡2𝑥cos⁡2𝑥sin22x=4sin2xcos2x

    So,

    cos⁡4𝑥=1−8sin⁡2𝑥cos⁡2𝑥cos4x=1−8sin2xcos2x

    Thus, the identity is proved.

    Problem 3

    Prove that:

    cos⁡6𝑥=32cos⁡6𝑥−48cos⁡4𝑥+18cos⁡2𝑥−1cos6x=32cos6x−48cos4x+18cos2x−1

    Solution:

    We'll use multiple angle formulas and express cos⁡6𝑥cos6x in terms of cos⁡𝑥cosx.

    1. Triple Angle Formula for Cosine:

    cos⁡3𝑥=4cos⁡3𝑥−3cos⁡𝑥cos3x=4cos3x−3cosx

    1. Double Angle Formula:

    cos⁡6𝑥=2cos⁡23𝑥−1cos6x=2cos23x−1

    1. Substitute cos⁡3𝑥=4cos⁡3𝑥−3cos⁡𝑥cos3x=4cos3x−3cosx:

    cos⁡6𝑥=2(4cos⁡3𝑥−3cos⁡𝑥)2−1cos6x=2(4cos3x−3cosx)2−1

    1. Expand the expression:

    cos⁡6𝑥=2(16cos⁡6𝑥−24cos⁡4𝑥+9cos⁡2𝑥)−1cos6x=2(16cos6x−24cos4x+9cos2x)−1

    cos⁡6𝑥=32cos⁡6𝑥−48cos⁡4𝑥+18cos⁡2𝑥−1cos6x=32cos6x−48cos4x+18cos2x−1

    Thus, the identity is proved.

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