Math 4 part 1

3. ∠B = 75° a = 6.5 b = 24.14. ∠B = 45° a= 7 b=75. ∠A = 34° 5’ a = 8.9 b = 13.3Lesson 41. ∠B = 14° b = 3.2 c = 13.42. ∠B = 67° 38’ a = 9.1 c = 23.83. ∠A = 60° a = 19.1 c = 224. ∠A = 72° b = 1.4 c = 4.55. ∠B = 13° a = 181.9 c = 186.7Lesson 51. ∠A = 36° 57’ ∠B = 53° 3’ 22

c = 26.32. ∠A = 30° 15’ ∠B = 59° 45’ c = 13.93. ∠A = 15° 57’ ∠B = 74° 3’ c = 7.34. ∠A = 60° ∠B = 30° c = 3.55. ∠A = 45° ∠B = 45° c = 353.6Lesson 61. Figure:tan B = b 150 ft a 210 fttan B = 150 23 210tan B = 0.7143 B = 350 32’

2. Figure: 250 m 170 m tan A = a b tan A = 250 170 tan A = 1.4706 A = 550 48’1. Figure: a 530 b = 12 ft tan A = a b tan 53o = a 12 a = 12 tan 53o a = 12 (1.3270) a = 15.92 ft height of the flagpole 24

4. Figure: c = 65 m a 70° Sin A = a c Sin 70° = a 65 a = 65 sin 70° a = 65 ( 0.9397) a = 61.08 height of the kite5. Figure: h 45 19 yFirst solve for y and x: Solve for y: sin 19° = y 25 y = 25 sin 19° 25

y = 25 ( 0.3256) y = 8.14 Solve for x: cos 19° = x 25 x = 25 cos 19° x = 25 (0.9455) x = 23.6 Solve for h: tan 45° = h + y x h + y = x tan 45° h = x tan 45° – y h = 23.6(1) – 8.1 h = 23.6 – 8.1 h = 15.5 height of the mango treeWhat have you learned 1. YZ or x 2. XY or z 3. XZ or y 4. v = 17 tan 48° 5. w = 17 cos 48° 6. v = 13 cos 23° 7. m = 13 sin 23° 8. p = 25.6 9. m = 23.9 10. ∠M = 43° 26

Module 1 Circular Functions and Trigonometry What this module is about This module is about the unit circle. From this module you will learn thetrigonometric definition of an angle, angle measurement, converting degreemeasure to radian and vice versa. The lessons were presented in a very simpleway so it will be easy for you to understand and be able to solve problems alonewithout difficulty. Treat the lesson with fun and take time to go back if you thinkyou are at a loss. What you are expected to learnThis module is designed for you to: 1. define a unit circle, arc length, coterminal and reference angles. 2. convert degree measure to radian and radian to degree. 3. visualize rotations along the unit circle and relate these to angle measures. 4. illustrate angles in standard position, coterminal angles and reference angles.How much do you knowA. Write the letter of the correct answer.1. What is the circumference of a circle in terms of π ?a. π b. 2π c. 3π d. 4π2. An acute angle between the terminal side and the x-axis is called ______a. coterminal b. reference c. quadrantal d. right

3. 60 ° in radian measure is equal toa. π b. π c. π d. π 2 3 4 64. 2.5 rad express to the nearest seconds is equal toa.143° 14′ 24″ b. 143° 14′ 26″ c. 43° 14′ 26″ d. 43° 14′ 27″5. What is the measure of an angle subtended by an arc that is 7 cm if the radius of the circle is 5 cm?a. 1. 4 rad b. 1.5 rad c. 1.6 rad d. 1.7 rad6. Point M ( 24 , y) lie on the unit circle and M is in Q II. What is the value of 25y?a. 6 b. − 6 c. 7 d. − 7 25 25 25 257. What is the measure of the reference angle of a 315o angle?a. 45 o b. 15 o c. -45 o d. -15 o8. In which quadrant does the terminal side of 5π lie? 6a. I b. II c. III d. IV9. A unit circle is divided into 10 congruent arcs. What is the length of each arc?a. π b. π c. 2π d. 10π 10 5 5B. Solve:10. The minute hand of the clock is 12 cm long. Find the length of the arc traced by the minute hand as it moved from its position at 3:00 to 3:40. 2

What you will do Lesson 1 The Unit Circle A unit circle is defined as a circle whose radius is equal to one unit andwhose center is at the origin. Every point on the unit circle satisfies the equationx2 + y2 + 1.The figure below shows a circle with radius equal to 1 unit. If thecircumference of a circle is defined bythe formula c = 2πr and r = 1, thenc = 2π or 360° or 1 revolution. r =1If 2π = 360°, then π = 180° or ºone-half revolution.Example:1. Imagine the Quezon Memorial Circle as a venue for morning joggers. The maintainers have placed stopping points where they could relax. If each jogger starts at C BPoint A, the distance he would Atravel at each terminal pointis shown in table below. D Stopping B C D A Point Distance or π 3π Arclength 2 π 2π 2 This illustrates the circumference of the unit circle 2π when divided by 4:will give 2π = π , the measure of each arc. 42 3

Similarly, the measure of each arc of a unit circle divided into:a. 6 congruent arcs = 2π = π 63b. 8 congruent arcs = 2π = π 84c. 12 congruent arcs = π 6These measurements are called arclengths. Let’s go back to the unit circle which we divided into 4 congruent arcs.From A, the length of each arc in each terminal points is given as: π BB: + 2C: 2π = π CA 2 -D: 3π D 2A: 4π = 2π 2 This is true in a counterclockwise rotation. If the rotation goes clockwise,the arclengths would be negative. Thus, the arclengths of the terminal points in a clockwise direction wouldyield: π B = - 3πD=- 2 2C = -π A = -2πWe call these measurements as directed arclengths. 4

2. Suppose a point is allowed to move around the circle starting from point A,find the arclength of each terminal point.The unit circle is divided into C 8 congruent arcs. Therefore, DBeach arc measures π . EA 4 F H GA counterclockwise move that A clockwise move that terminates at: terminates at:Terminal pt. Arclength Terminal pt. Arclength B π H -π C G 4 D 4 F E 2π or π E − 2π or − π F 42 D 42 G C - 3π H 3π B 4 A A 4 − 4π or − π 4π orπ 4 4 - 5π 4 5π − 6π or − 3π 4 42 6π or 3π - 7π 42 4 7π − 8π or − 2π 4 4 8π or2π 4 5

A rotation can be repeated. For example a two complete rotation is equalto 4π. A one and a half revolution is equal to 3π. An arclength of 9π will also be at terminal point B. This is also equal to 42π + π . 4Try this outA. Find the length of each arc of a unit circle divided into:1. 10 4. 182. 14 5. 203. 16 6. 24B. Given the unit circle: Identify the terminal points of each arclength:1. π 6. 5π 11. -2π E D 3 6 F C 12. − π B2. 7π 7. − 5π 3 G A 6 6 L 13. − 2π H3. 11π 8. 3π 3 I JK 6 2 14. π4. − 7π 9. π 2 65. −11π 10. -π 15. - π 6 2 6

Lesson 2Conversion of Degree to Radian and Vice Versa Before discussing conversion of angle measures, you have to understandthat there are two unit of angle measure that are commonly used:1. Degree measure2. Radian measure. A complete revolution is divided into 360 equal parts. A degree issubdivided to minutes and seconds.1 rev = 360° ° is the symbol for degrees 1° = 60′ ′ is the symbol for minutes 1′ = 60″ ″ is the symbol for seconds For all circles, the radian measure of the circumference is 2πradians. But the angle has a measure of 360°. hence, 2π rad = 360° π rad = 180° 1 rad = 180 or 57.296° π 1° = π rad or 0.017453 rad 180 Now, you are to convert degrees to radians. To convert from degrees toradians, multiply the number of degrees by π . Then simplify. 180Examples: Convert the measure of the following angles from degrees to radians. 7

1. 70° = 70° x π = 7π rad 180 182. -225° = -225° x π = − 5π rad 180 43. 90° = 90° x π = π rad 180 24. 135° = 135° x π = 3π rad 180 45. 270° = 270° x π 3π = rad 180 2 To convert from radians to degree, multiply the number of radians by 180 . πThen simplify.Examples:Express each radian measure in degrees1. 2π = 2π x 180 = 90°44 π2. 5π = 5π x 180 = 300°33 π3. 2π = 2π x 180 = 120°33 π4. −11π = −11π x 180 = -330°66 π5. − 23π = − 23π x 180 = 1380° 33 πTry this out 8

A. Convert the following to radian measure:1. 60° 6. -366°2. 150° 7. 22.5°3. 240° 8. 720°4. 780° 9. 225°5. -300° 10. 612°B. Express each radian measure in degrees:1. 7π 6. − 7π 2 52. 13π 7. − 5π 6 93. 20π 8. − 23π 3 34. 12π 9. − 4π 5 55. 7π 10. − 7π 2 4 Lesson 3 Angles Intercepting an Arc A radian is defined as the measure of an angle intercepting an arc whoselength is equal to the radius of the circle. An arc length is the distance betweentwo points along a circle expressed in linear units. 9

angle in radian = arclength radius of the circle rs θ)or θ = s rYou can now use this knowledge to solve problems.Examples: 1. A wheel of radius 80 cm rolls along the ground without slipping and rotates through an angle of 45°. How far does the wheel move? Solution: Use the formula θ = s to solve for the distance s. r Let: radius = 80 cm θ = 45° Convert 45° to π radians: 45° x π rad = π rad 180 4 θ= s r π rad = s 4 80 s = π rad x 80 4 s = 20π 2. The minute hand of a clock is 5 cm long. How far does the tip of the hand travel in 35 min? 10

Solution: Arc length formula = deg ree ( 2πr ) 180 360° in 60 min time or 360 = 6° 60 35 min ⇒ 35 x 6° = 120° L = 120 ( 2 ) ( 3.1416 ) ( 5 cm ) 360 = 18.33 cm Try this out Solve the following: 1. The pendulum of a clock swings through an angle of 0.15 rad. If it swings a distance of 30 cm, what is the length of the pendulum? 2. The minute hand of the clock is 10 cm long. How far does the tip of the hand move after 12 minutes? 3. An arc 15 cm long is measured on the circumference of a circle of radius 10 cm. Find an angle subtended at the center. Lesson 4 Rotations Along the Unit Circle An angle can be thought of as the amount of rotation generated when a rayis rotated about its endpoints. The initial position of the ray is called the initialside of the angle and the position of the ray at the endpoint is called terminalside. A clockwise rotation generates a negative angle while a counterclockwiserotation generates a positive angle. Imagine the terminal side of an angle whose terminal side is on thepositive x-axis being rotated along the unit circle. 11

Positive angle Negative angle Y YX XExample 1:Illustrate 1. 5π radians 5. 30° 2 2. 9π radians 6. -90° 4 3. 3π radians 7. -500° 4. −13π radians 8. 270° 4The positive side of the x-axis is the initial side1. 5π radians 2. 9π radians 2Y 4YX X 12

3. 3π radians 4. −13π radians Y 4 Y X X 5. Y 6. Y 30° X X -90° 7. Y 8.X Y -500° X Example 2: 270° 13

How many degrees is the angle formed when the rotating ray makesa. 3 complete counterclockwise turns?b. 2 5 complete clockwise turns? 6Solutions:a. 3 ( 360 )° = 1080° b. 2 5 ( -360°) = -1020° 6Try this outA. Draw an arc whose length is: 1. 4π units 2. 5π units 4 3. -3π units 4. − 3π units 2 5. 7π units 12B. Draw the following angle measures.1. 115° 3. -620°2. -250° 4. 300° Lesson 5 Angles 14

An angle whose vertex lies at the origin of the rectangular coordinatesystem and whose initial side is positive along the positive x-axis is said to be instandard position. Y Terminal side X Vertex initial sideAngles in standard position. YExamples: b. Y X a. O XS d. Yc. Y X X N AAngles not in standard position: Y Y b. a. 15

JA XXc. Y d. YXS X NQuadrantal Angles: A quadrantal angle is an angle in standard position and whose terminalside lies on the x-axis or y-axis.Example: YY F XG X ∠F is not a quadrantal angle, since the terminal side does not lie on the x – axis or y - axis. ∠G is a quadrantal angle since its terminal side lie on the x - axis.Coterminal Angles: Coterminal angles are angles having the same initial side and the sameterminal side. YY 16

C B A DX XExamples:Determine the measure of the smallest positive angle coterminal with: a. 65° b. 128° c. -213° d. 654°Solution: Angles coterminal with a given angle θ may be derived using theformula θ + 360n for all integers n. 17

a. 65° + 360° = 425° b. 128° + 360° = 488° Y Y 128o 65o X X 488o 425o d. 654° - 360° = 294°c. -213° + 360° = 147° Y Y 294o 147o X X 654o -213oFinding coterminal angle less than 360°Examples:a. 750° = 755° - 360°(20) c. 660° = 660° - 360° = 35° = 300°b. 380° = 380° - 360° d. 820° = 820° - 360°(2) = 20° = 100°Reference Angles A reference angle (A) is a positive acute angle formed between the x-axis and the terminal side of a given angle. 18

Examples: b. Y Let: A = reference angle A a. Y X A Xc. Y d. Y X A X AExamples: In each of the following determine the quadrant in which the angle liesand determine the reference angle. a. 73° b. 135° c. 300° d. 920° 19

Solution: The reference angle can be derived using the formula 180°n ± θ. a. 73° Y 180°n ± θ 73°. 73° terminates in Ql, hence X 180°( 0 ) - θ = 73° θ = 73°, the reference angle itself b. 135° 135° terminates in Qll, hence Y 135° 180°( 1 ) - θ = 135° 45° θ = 180° - 135° X θ = 45° is the reference angle c. 300° Y 300° 300° terminates in Q1V , hence 180°( 2 ) - θ = 300° X θ = 360° - 300° 60° θ = 60° is the reference angle 20

d. 920°First find the number of multiples of 180° in 920° 900° has 4 multiples of 180° and a remainder of 200° The terminal side of 200° is in Qlll. 180°( 1 ) - θ = 200° θ = 200° - 180° θ = 20° is the reference angleIllustration is left for you.Try this outA. Determine the smallest positive coterminal angle with the given angle.1. 57° 6. -349°2. -250° 7. 100°3. 94° 8. 207°4. -175° 9. 185°5. 116° 10. 409°B. Determine the quadrant in which the angle lies and find the reference angle.1. 84° 6. 480°2. -140° 7. -650°3. 355° 8. 740°4. -365° 9. 330°5. 290° 10. 204 21

Let’s summarize The circle of radius one with center at origin is called the unit circle To convert from degrees to radians, multiply the number of degreesby π . Then simplify. 180 To convert from radians to degree, multiply the number of radiansby 180 . Then simplify. π An angle is the amount of rotation where one side is called the initialside and the other is the terminal side. An angle is in standard position if it is constructed in a rectangularcoordinate system with vertex at the origin and the initial side on thepositive side of the x-axis. Coterminal angle are angles having the same initial side and thesame terminal side. Reference angle is an acute angle between the terminal side andthe x-axis. To find the reference angle, write the angle in the form 180n ± θwhere θ is the reference angle. 22

What have you learnedAnswer the following correctly: 1. A circle is divided into 6 congruent arcs. What is the measure of each arc? 2. Express 120° in radian measure. 3. What is the reference angle of -380°? 4. The coterminal angle less than 360° of 820° is __________. 5. Convert − 7π rad to degree measure. 6 6. On a circle of radius 20cm, the arc intercepts a central angle of 1 rad. 5 What is the arclength? 7. At what quadrant is the terminal side of - 1080° located? 8. How many degrees is the angle formed by a ray that makes 3 1 complete 5 rotations counterclockwise? 9. How many degrees is the angle formed by a ray that makes 2 2 complete 3 rotations clockwise? 10. A minute hand of a clock is 5 cm long. How far does the tip of the hand travel in 50 min? 23

Answer Key 6. c 7. a How much do you know 8. Qll 1. b 9. 5π 2. b 3. d 20 4. a 10. 20.57 cm 5. a 4. πTry this out 9Lesson 1A. 1. π 5. π 10 5 2. π 6. π 12 7 3. π 8B. 1. C 6. F 11. A 2. H 7. H 12. K 3. L 8. J 13. I 4. F 9. G 14. D 5. B 10. G 15. J 24

Lesson 2 6. 61π rad 30A. 1. π rad 3 7. π rad 8 2. 5π rad 3 8. 4π 3. 4π rad 9. 5π 3 4 4. 13π rad 10. 17π 3 5 5. − 5π rad 6. -252° 3 7. -100° 8. -1300°B. 1. 63° 9. -144° 2. 390° 10. -315° 3. 1200° 4. 432° Y Y 5. 105° 2. 5πLesson 3 4 1. 200 cm X 2. 12.57 cm 3. 1.5 radLesson 4A. 1. 4πX 25

Y 4. − 3π Y 2 3. 3π XX 5. 7π Y 12 XB. 1. Y 2. Y X 115° X 3. -250o Y 4.X Y -620° 300° X 26

Lesson 5 6. 11° 7. 460°A. 1. 417° 8. 567° 2. 110° 9. 545° 3. 454° 10. 769° 4. 185° 6. Q1, 60° 5. 476° 7. Qll, 10° 8. Q ll, 20° B. 1. Ql , 84° 9. Q1, 30° 2. Qlll, 35° 10. Q lll, 24° 3. Q lV, 5° 4. QlV, 5° 6. s = 4 cm 5. Q lll, 70° 7. Q1What have you learned 8. -960° 1. π 9. 1152° 3 10. 26.18 cm 2. 2π rad 3 3. 20° 4. 100° 5. -21π rad 27

Module 2Circular Functions and Trigonometry What this module is about This module is about determining the coordinates of angles in standard position in aunit circle; the six circular functions and finding the six circular functions of special angles;As you go over the discussion, examples and exercises, you will understand what circularfunctions are all about. Anytime you feel you are at a loss, do not hesitate to go back to thediscussion and examples.What you are expected to learnThis module is designed for you to:1. determine the coordinates of the terminal side of an angle in standard position in a unit circle1.1 when one coordinate is given (apply the Pythagorean Theorem and the properties of special right triangles)1.2 when the angle is of the form:• 180ºn ± 30º • 180ºn ± 45º• 180ºn ± 60º • 90ºn2. define the six circular functions• sine • cotangent• cosine • secant• tangent • cosecant3. find the six circular functions of angles with special values 1

How much do you know1. The x-coordinate of an angle in the along the unit circle is 3 . If the terminal side of 4 the angle is located in the fourth quadrant, what is its y-coordinate?a. 1 b. 7 c. - 7 d. - 1 4 4 4 42. An angle measuring 30o is in standard position along the unit circle. What are its coordinates?a.  1 , 3  b.  1 ,− 3  c.  3 , 1  d.  − 3 , 1  2 2 2 2 2 2 2 23. What are the coordinates of the point of intersection of the terminal side of a 420o angle and the unit circle?a.  3 , 1  b.  1 , 3  c.  2, 2  d.  1 ,− 3  2 2 2 2 2 2 2 24. What is the y-coordinate of a 225o angle along the unit circle?a. − 1 b. - 3 c. - 2 d. -1 2 2 25. It is the relationship between the arc length and the x-coordinate. a. Circular function c. Cosine function b. Sine function d. Tangent function6. What is cos 120oa. 1 b. − 1 c. 3 d. − 3 2 2 2 2 2

What you will do Lesson 1 Coordinates of Points on the Unit Circle In the previous module you have learned about the measures of arcs on aunit circle. Now, let us find the coordinates of the point where the terminal side ofan angle in standard position lies.Consider a circle whose center is at (0,0). The circle of radius one withcenter at origin is called the unit circle. Every point on the unit circle satisfies theequation x2 + y2 = 1. YIt intersects at the points: B A: ( 1,0 ) XC A B: ( 0, 1 ) C: ( -1, 0 ) D: ( 0, -1 ) D You can determine whether a point is on the unit circle if the equation x2 +y2 = 1 is satisfied.Examples:Determine whether each point lie on the unit circle. a. ( 12 , − 5 ) c. ( 1 , 1 ) 13 13 32 b.  − 2, 2  d. ( 1 , 3 ) 2 2 22Solution:a. ( 12 , − 5 ) implies that x = 12 and y = − 5 13 13 13 13 Substitute the values of x and y in x2 + y2 = 1 3

( 12 )2 + ( − 5 )2 = 1 13 13 144 + 25 = 1 169 169 169 = 1 True 169The point ( 12 , − 5 ) is on the unit circle. 13 13b.  − 2, 2  implies that x = - 2 and y = 2 2 2 2 2 Substitute the values of x and y in x2 + y2 = 1(- 2 )2 + ( 2 )2 = 1 22 2 + 2 =1 44 4 = 1 True 4 The point  − 2, 2  lie on the unit circle. 2 2c. ( 1 , 1 ) 32 ( 1 , 1 ) implies that x = 1 , and y = 1 32 32 Substitute the values of x and y in x2 + y2 = 1 ( 1 )2 + ( 1 )2 = 1 32 1 + 1 =1 94 4

4+9 = 1 36 13 = 1 False 36The point ( 1 , 1 ) is not on the unit circle. 32d. ( 1 , 3 ) 22( 1 , 3 ) implies that x = 1 and y = 322 22Substitute the values of x and y inx2 + y2 = 1( 1 )2 + ( 3 )2 = 1 22 1+3 =1 44 4 = 1 True 4 The point ( 1 , 3 ) is on the unit circle. 22 Knowing this equation, the other coordinate of a point of intersection of theunit circle and the terminal side of an angle in standard position can be obtainedwhen one of its coordinates is given.Example 1: If the x-coordinate of an angle in standard position is 1 , what is the y- 2coordinate?Solution: Use the equation of the unit circle, x2 + y2 = 1. Substitute the given valueof x in the equation of the unit circle to obtain the value of y. 5

 1 2 + y 2 = 1 Substitute the given x-coordinate 2 1 + y2 =1  1 2 = 1 4 2 4 y2 =1- 1 Addition property of equality 1− 1 = 4 −1 = 3 4 y2 = 3 444 Take the square root of both numerator and 4 denominator y=± 3 4 =2 4 y=± 3 2 ∴y = ± 3 . 2Example 2: The y-coordinate of an angle in standard position is 1 . If the terminal side 7of the angle lies between 90o and 180o, what is its x-coordinate?Solution: Use the equation of the unit circle x2 + y2 = 1. Substitute the given y-coordinate to find the x-coordinate. x2 +  1 2 = 1 Substitute the given y-coordinate 7 x2 + 1 =1  1 2 = 1 49  7  49 x2 = 1− 1 Addition Property of Equality 49 x2 = 48 1− 1 = 49 − 1 = 48 49 49 49 49 6

x = ± 48 Take the square root of both sides 49 x = ± 16 ⋅ 3 49 x=±4 3 7 ∴ x = − 4 3 , since 90o < θ <180o which means that the terminal side of 7the angle lies in the second quadrant where the x-coordinate is negative.Try this outA. Let B be a point on the first quadrant of the unit circle. The x-coordinate of the point of intersection of the unit circle and the terminal side at B is 1 . 2 Y C• B ( 1 , y) 2 X OF A •• DE1. Name the angle in standard position.2. What special kind of triangle is formed out of points O, B, and F?3. How will you find the y-coordinate of B?4. Find the y-coordinate of ∠AOB.5. What is the sign of the y-coordinate of B?6. What then are the coordinates of B?7. What are the coordinates of point C which is a reflection of point B in the second quadrant?8. If D is a reflection of C in the third quadrant, what are the coordinates of the terminal point of ∠AOD? 7

9. If E is a reflection of B in the fourth quadrant, what are the coordinates of the terminal point of ∠AOE? 10. In what quadrant is the abscissa or x-coordinate of the terminal point of an angle positive? Negative? 11. What about the y-coordinate or ordinate? In what quadrant is it positive? Negative?B. Determine whether each of the following points lie on the unit circle. 1. ( 3 , 4 ) 55 2. ( 0.8, -0.6 ) 3. ( 2 2 , −1 ) 334. ( 2 , -1 )5. ( − 8 , −15 ) 17 17C. One of the coordinates of the point of intersection of the unit circle and the terminal side of an angle in standard position is given. Find the other coordinate.1. x = - 1 6. y = 1 , 0 < θ < 90° 2 22. y = − 2 7. y = 1 , 90° < θ < 180° 5 73. y = 1 8. x = − 2 ,180° < θ < 270° 10 34. x = 1 9. y = − 1 , 270° < θ < 360° 3 55. x = 2 10. x = 5 , 0 < θ < 90° 2 7 8

D. Find the missing coordinate of the point of intersection of the unit circle andthe terminal ray of an angle in standard position.1.  3 , __  θ is in QI 7.  __,− 3  θ is in QIII 2  42. ( __, -0.6) θ is in QIII 8.  __,− 3  180 < θ < 270° 53.  __, 5  θ is in QII  13  9.  2 , __  0° < θ < 90° 7 4. ( __, 1) θ = 90°5. (0, __) θ = 270° 10.  __, 5  90° < θ < 180°  12 6.  − 12 , __  θ is in QII  13  Lesson 2The Coordinates of Points on the Unit Circle in the form 180on ± AIn geometry, you have learned that angles whose measures are 30o, 45o, and60o are called special angles. Now, let these special angles be amount of rotations onthe unit circle as you can see in the figures below. Angle measures 30, 45 and 60 arealso the arclengths π , π and π respectively 64 3 YY Y π π X π 16 14 3X 30o X 45o 1 60oFigure 1 Figure 2 Figure 3What do you think are the coordinates of each of the terminal points? 9

Consider figure 1. Let P(x, y) be the terminal point of the angle. Drop aperpendicular from P to the x-axis and call the point of intersection M. Notice thattriangle OPM is a 30o-60o-90o triangle. Y Recall that in a 30o-60o-90o triangle, the 1 P(x,y) length of the leg opposite the 30o angle is 1 the 2X o y 30 length of the hypotenuse. Hence, y = 1 . 2 O xM The x-coordinate of P can be obtained using the equation of the unit circle x2 + y2 = 1. Figure 1 By substitution, x2 = 3 x2 +  1 2 = 1 4 2 x= ± 3 4 x2 + 1 = 1 4 x2 = 1 - 1 x= ± 3 4 2Since P is in QI, then x = 3 . Thus, P ( π )has coordinates  3 , 1  . 26 2 2 Consider this time figure 2. Let P(x, y) be the terminal point of the angle. Drop aperpendicular from P to the x-axis and call the point of intersection M. Notice thattriangle OPM is a 45o-45o-90o triangle and that a 45o angle in standard position has itsterminal side in the first quadrant. Y Recall that in a 45o-45o-90o, the lengths of P(x, y) the two legs are equal. Thus, in Figure 2, x = y. 1y The coordinates of P can be obtained using the equation of the unit circle, x2 + y2 = 1X 45o where x = y. Ox M Figure 2 x2 + x2 = 1 x=± 1 2x2 = 1 2 x2 = 1 x = ± 1 or ± 2 2 2 2 22 10

But P is in quadrant I. Hence, x = 2 and y = 2 . 22 ∴P ( π ) has coordinates  2, 2  . 4 2 2 Now, consider figure 3. Let P(x, y) be the terminal point of the angle. Drop aperpendicular from P to the x-axis and call the point of intersection M. Notice that ∆POMis a right triangle where ∠POM = 60o. Y P(x,y) To determine the coordinates of P, use the same relation as in Figure 1 since ∠OPM = 30o. 1 Knowing that the side opposite 30o is 1 the y 2X 60o length of the hypotenuse, then in figure 3, x = 1 . 2 O xM To find y, use the equation of the unit Figure 3 circle, x2 + y2 = 1. Thus,  1 2 + y2 = 1 2 y2 = 1− 1 4 y=3 4 y=± 3 4 y=± 3 2 Since P is in the first quadrant y = 3 . 2 Hence, P( π )has coordinates  1 , 3  . 3 2 2 11

The coordinates of the terminal points of special angles on a unit circle can nowbe used to determine the coordinates of points of intersection of angles in the form180on ± A where A is a special angle. This can be done by determining the location ofthe terminal point of the given angle and the reference angle, A.Example: 1. Determine the coordinates of the point of intersection of the terminal side of a135o angle on the unit circle.Solution: The terminal point of 135o or 3π is located in the second quadrant. Its reference 4angle can be determined using the form 180on – A. Y 135 = 180o(1) – A 135o = 180o – AP(x,y) A = 180o – 135o A = 45o 135OX 3π or 135o is a reflection of π or 45o in QII. 44 Thus, the terminal point P of 135o has coordinates  − 2, 2  . 2 22. Determine the coordinates of the point of intersection of the terminal side of a 210o angle on the unit circle.Solution: The terminal point of 210o or 7π is located in QIII. Its reference angle can be 6determined using the form 180on + A. Y 180o(1) + A = 210o 180o + A = 210oX A = 210o – 180o P(x,y) A = 30o Notice that 210o is a reflection of 30o in the third quadrant. Hence, its coordinates can be obtained using the coordinates of 30o but following the signs of the ordered pairs in the third quadrant. 12

Hence, the terminal point P of 210o or 7π has coordinates  − 3 ,− 1  . 6 2 23. Determine the coordinates of the point of intersection of the terminal side of a 660o angle on the unit circle.Solution:The terminal side of 660o or 11π is in the fourth quadrant. It can be written in the 3form 180on - A. Y 180o(4) – A = 660o – A = 660o – 720o -A = -60o A = 60oX Observe that 660o is a reflection of 60o on the fourth quadrant. Hence, the coordinates of P(x, y) the terminal point can be obtained from the coordinates of 60o but following the signs of coordinates in the fourth quadrant. Then, P( 11π ) has coordinates  1 ,− 3  . 3 2 24. Determine the coordinates of the point of intersection of the terminal side of a -750o angle on the unit circle.Solution: The direction of rotation of -750o or − 25π is clockwise starting from (1, 0). Its 6terminal side is located in the fourth quadrant and be expressed in the form 180on ± Awhere A is a special angle. Y 180o(-4) – A = -750o -720o – A = -750o -A = -750 + 720oX -A = -30o P(x, y) A = 30o Thus, the reference angle of -750o is 30o. The coordinates of the terminal pointcan be obtained from the coordinates of the points of intersection of the terminal side of30o and the unit circle. 13

Therefore, the coordinates of the point of intersection of the terminal side of -750oor − 25π and the unit circle are  3 ,− 1  . 6 2 2 Observe from the examples that the coordinates of the point of intersection of theunit circle and the terminal side of an angle in standard position are as follows: 1. If an angle is of the form 180°n ± 30°, the coordinates are  ± 3 ,± 1  . 2 2 2. If an angle is of the form 180°n ± 60°, the coordinates are  ± 1 ,± 3  . 2 2 3. If an angle is of the form 180°n ± 45°, the coordinates are  ± 2 ,± 2  . 2 2 The signs of the coordinates depend upon the position of the terminal side of theangle.Try this out Find the reference angle for each of the following angles on a unit circle anddetermine the coordinates of its terminal point. 1. 120o 6. 315o 11. 765o 16. -225o 2. 150o 7. 330o 12. -120o 17. -300o 3. 225o 8. 480o 13. -135o 18. -480o 4. 240o 9. 510o 14. -150o 19. -600o 5. 300o 10. 585o 15. -210o 20. -1020o Lesson 3 The Sine and Cosine Functions In the previous lesson, you have learned that each terminal point P of special angleson the unit circle corresponds to coordinates of a point (x, y) that satisfies the equation ofthe unit circle x2 + y2 = 1. These coordinates of points have special names. The relationbetween the angle and the y-coordinate is called the sine function while the relationbetween the angle and the x-coordinate is called the cosine function. 14

Y P(x, y) = (cos θ, sin θ) In the figure at the left, the terminal side of angle θ intersected the unit circle at point P(x, y).Xθ Thus, the x-coordinate of P is called the cosine function of θ and can be expressed as cos θ = x, and the y-coordinate of P is called the sine function of θ and can be expressed as sin θ = y. Hence, for each value of θ in the unit circle, the x-coordinate of the terminal point isthe cosine of θ and the y-coordinate is the sine of θ.In symbols, x = cos θ and y = sin θExamples:1. Evaluate sin 0o and cos 0o.Solution: When θ = 0o is set in standard position, it intersects Y the unit circle at the point (1, 0). By definition, sin θ = y and cos θ = x.X P(1, 0) Hence, in P(1,0) where x = 1 and y = 0 sin 0o = 0 and cos 0o = 1.2. Evaluate sin 90o and cos 90o. ( Note: 90o = π ) 2Solution: θ = 90o is set in standard position on the unit circle Y as shown in the figure. Its terminal side intersects the unit (0, 1) circle at the point (0, 1). By definition, sin θ = y and cos θ = x.X 90o Hence, sin 90o = 1 and cos 90o = 0 or sin π = 1and cos π = 0. 22 Note that the sine and cosine functions of angles which are integral multiples of 90ocan be easily evaluated. 15

The table below shows the sine and cosine functions of the integral multiples of 90ofor 0o ≤ θ ≤ 360o. θ 0 90o or π 180o or π 270o or 3π 360o or 2π 22 P(θ) (1, 0) (0, 1) (-1, 0) (0, -1) (1, 0) sin θ 0 1 0 -1 0cos θ 1 0 -1 0 1 To evaluate angles that are not multiples of 90o, use the concept of reference anglesas in the previous lesson.3. Evaluate sin 30o and cos 30o. (Note: 30o = π ) 6Solution: Y θ = 30o is set in standard position, as in the ( )1 figure at the left, the terminal side of the angle 3 , 1 ( )intersects the unit circle at the point P P 2 2 ,3 1Y 30o 22 . By definition, sin 30o = 1 and cos 30o = 3 or 22 sin π = 1 and cos π = 3 62 62Similarly, angles of multiples of 30o or π can be found in the other quadrants: 150o 6or 5π in QII, 210o or 7π in QIII and 330o or 11π in QIV.66 6 16

4. Evaluate sin -60o and cos -60o. ( Note: 60o = π 3Solution: Y If θ = -60o or - π is in standard position as in the 3 figure, its terminal side intersects the unit circle at the ( )pointX 1 ,− 3 . 2 2 -60o ( )1P 1 ,− 3 Hence, by definition, 2 2 sin -60o = 1 and cos -60o = - 3 or 22 sin - π = 1 and cos -60o = - 3 . 32 25. Evaluate sin 570o and cos 570o. (Note: 570o = 19π ) 6Solution: Y The figure at the left shows 570o in standard position. 570o ( )Its reference angle is 30o and the terminal side intersects the−3,−1 2 2X unit circle at the point .−( )P3,− 1 Hence, by the definition of sine and cosine functions, 2 2 sin 570o = - 1 and cos 570o = - 3 or 22 sin 19π = - 1 and cos 19π = - 3 . 62 62 17

Try this outA. The unit circle below is divided into 8 congruent arcs. Complete the table. Y Terminal Degree Coordinates Sin θ Cos θ Point Measure C BD A BXE A C H D F E G F G HB. Evaluate the following: 6. sin 420o 7. sin -45o 1. sin 60o 8. cos -90o 2. cos 120o 9. sin -180o 3. cos 135o 10. cos -330o 4. sin 150o 5. cos 270oC. Identify the quadrant/quadrants where the angle is/are located:1. sin θ > 0 5. sin θ > 0 and cos θ < 02. cos θ < 0 6. sin θ > 0 and cos θ > 03. cos θ > 0 7. sin θ < 0 and cos θ > 04. sin θ < 0 8. sin θ < 0 and cos θ < 0 18