6.3 Trigonometric Functions of Any Angle

  • Find angles that are coterminal with a given angle, and find the complement and the supplement of a given angle.

  • Determine the six trigonometric function values for any angle in standard position when the coordinates of a point on the terminal side are given.

  • Find the function values for any angle whose terminal side lies on an axis.

  • Find the function values for an angle whose terminal side makes an angle of 30°30°, 45°45°, or 60°60° with the x-axis.

  • Use a calculator to find function values and angles.

Angles, Rotations, and Degree Measure

An angle is a familiar figure in the world around us.

An angle is the union of two rays with a common endpoint called the vertex. In trigonometry, we often think of an angle as a rotation. To do so, think of locating a ray along the positive x-axis with its endpoint at the origin. This ray is called the initial side of the angle. Though we leave that ray fixed, think of making a copy of it and rotating it. A rotation counterclockwise is a positive rotation, and a rotation clockwise is a negative rotation. The ray at the end of the rotation is called the terminal side of the angle. The angle formed is said to be in standard position.

The measure of an angle or a rotation may be given in degrees. The Babylonians developed the idea of dividing the circumference of a circle into 360 equal parts, or degrees. If we let the measure of one of these parts be 1°1°, then one complete positive revolution or rotation has a measure of 360°360°. One half of a revolution has a measure of 180°180°, one fourth of a revolution has a measure of 90°90°, and so on. We can also speak of an angle of measure 60°60°,135°330°135°330°, or 420°420°. The terminal sides of these angles lie in quadrants I, II, IV, and I, respectively. The negative rotations -30°30°, -110°110°, and -225°225° represent angles with terminal sides in quadrants IV, III, and II, respectively.

If two or more angles have the same terminal side, the angles are said to be coterminal. To find angles coterminal with a given angle, we add or subtract multiples of 360°360°. For example, 420°420°, shown above, has the same terminal side as 60°60°, since 420°=60°+360°420°=60°+360°. Thus we say that angles of measure 60°60° and 420°420° are coterminal. The negative rotation that measures -300°300° is also coterminal with 60°60° because 60°-360°=-300°60°360°=300°. The set of all angles coterminal with 60°60° can be expressed as 60°+n360°60°+n360°, where nn is an integer. Other examples of coterminal angles shown above are 90°90° and -270°270°, -90°90° and 270°270°,135°135° and -225°225°, -30°30° and 330°330°, and -110°110° and 610°610°.

Example 1

Find two positive angles and two negative angles that are coterminal with (a) 51°51° and (b) -7°7°.

Solution

  1. a) We add and subtract multiples of 360°360°. Many answers are possible.

    Thus angles of measure 411°411°, 1131°1131°, -309°309°, and -669°669° are coterminal with 51°51°.

  2. b) We have the following:

    -7°+360°=353°,-7°+2(360°)=713°7°+360°=353°,7°+2(360°)=713°,

    -7°-360°=-367°,-7°-10(360°)=-3607°7°360°=367°,7°10(360°)=3607°.

    Thus angles of measure 353°353°, 713°713°, -367°367°, and -3607°3607° are coterminal with -7°7°.

Now Try Exercise 13.

Angles can be classified by their measures, as seen in the following figures.

Recall that two acute angles are complementary if the sum of their measures is 90°90°. For example, angles that measure 10°10° and 80°80° are complementary because 10°+80°=90°10°+80°=90°. Two positive angles are supplementary if the sum of their measures is 180°180°. For example, angles that measure 45°45° and 135°135° are supplementary because 45°+135°=180°45°+135°=180°.

Example 2

Find the complement and the supplement of 71.46°71.46°.

Solution

We have 90°-71.46°=18.54°and180°-71.46°=108.54°.

90°71.46°=18.54°and180°71.46°=108.54°.

Thus the complement of 71.46°71.46° is 18.54°18.54°, and the supplement is 108.54°108.54°.

Now Try Exercise 19.

  Trigonometric Functions of Angles or Rotations

Many applied problems in trigonometry involve the use of angles that are not acute. Thus we need to extend the domains of the trigonometric functions defined in Section 6.1 to angles, or rotations, of any size. To do this, we first consider a right triangle with one vertex at the origin of a coordinate system and one vertex on the positive x-axis. (See the figure at left.) The other vertex is at PP, a point on the circle whose center is at the origin and whose radius r is the length of the hypotenuse of the triangle. This triangle is a reference triangle for angle θθ, which is in standard position. Note that yy is the length of the side opposite θθ and x is the length of the side adjacent to θθ.

Recalling the definitions in Section 6.1, we note that three of the trigonometric functions of angle θθ are defined as follows:

sinθ=opphyp=yr,cosθ=adjhyp=xr,tanθ=oppadj=yx.
sinθ=opphyp=yr,cosθ=adjhyp=xr,tanθ=oppadj=yx.

Since x and yy are the coordinates of the point PP and the length of the radius is the length of the hypotenuse, we can also define these functions as follows:

sin θ=y-coordinateradius,  cos θ=x-coordinateradius, tan θ=y-coordinatex-coordinate.
sin θ=ycoordinateradius,  cos θ=xcoordinateradius, tan θ=ycoordinatexcoordinate.

We will use these definitions for functions of angles of any measure. The following figures show angles whose terminal sides lie in quadrants II, III, and IV.

A reference triangle can be drawn for angles in any quadrant, as shown on the preceding page. Note that the angle is in standard position; that is, it is always measured from the positive half of the x-axis. The point P(x,y)P(x,y) is a point, other than the vertex, on the terminal side of the angle. Each of its two coordinates may be positive, negative, or zero, depending on the location of the terminal side. The length of the radius, which is also the length of the hypotenuse of the reference triangle, is always considered positive. (Note that x2+y2=r2x2+y2=r2, or r=x2+y2r=x2+y2.) Regardless of the location of PP, we have the following definitions. We now extend the domain of the six trigonometric functions to include all angles, or rotations.

Values of the trigonometric functions can be positive, negative, or zero, depending on where the terminal side of the angle lies. Since the length of the radius is always positive, the signs of the function values depend only on the coordinates of the point P on the terminal side of the angle. In the first quadrant, all function values are positive because both coordinates are positive. In the second quadrant, first coordinates are negative and second coordinates are positive; thus only the sine values and the cosecant values are positive. Similarly, we can determine the signs of the function values in the third and the fourth quadrants. Because of the reciprocal relationships, we need learn only the signs for the sine, cosine, and tangent functions.

Example 3

Find the six trigonometric function values for each angle shown.

  1. a)

  2. b)

  3. c)

Solution

  1. a) We first determine r, the distance from the origin (0, 0) to the point (-4,-3). The distance between (0, 0) and any point (x, y) on the terminal side of the angle is

    r=(x-0)2+(y-0)2=x2+y2.

    Substituting -4 for x and -3 for y, we find

    r=(-4)2+(-3)2=16+9=25=5.

    Using the definitions of the trigonometric functions, we can now find the function values of θ. We substitute -4 for x, -3 for y, and 5 for r :

    sin θ=yr=-35=-35,csc θ=ry=5-3=-53,cos θ=xr=-45=-45,sec θ=rx=5-4=-54,tan θ=yx=-3-4=34,cot θ=xy=-4-3=43.

    As expected, the tangent value and the cotangent value are positive and the other four values are negative. This is true for all angles in quadrant III.

  2. b) We first determine r, the distance from the origin to the point (1,-1):

    r=12+(-1)2=1+1=2.

    Substituting 1 for x, -1 for y, and 2 for r, we find

    sin θ=yr=-12=-22,csc θ=ry=2-1=-2,cos θ=xr=12=22,sec θ=rx=21=2,tan θ=yx=-11=-1,cot θ=xy=1-1=-1.
  3. c) We determine r, the distance from the origin to the point (-1,3):

    r=(-1)2+(3)2=1+3=4=2.

    Substituting -1 for x, 3 for y, and 2 for r, we find that the trigonometric function values of θ are

    sin θ=32,csc θ=23=233,cos θ=-12=-12,sec θ=2-1=-2,tan θ=3-1=-3,cot θ=-13=-33.

Now Try Exercise 29.

Any point other than the origin on the terminal side of an angle in standard position can be used to determine the trigonometric function values of that angle. The function values are the same regardless of which point is used. To illustrate this, let’s consider an angle θ in standard position whose terminal side lies on the line y=-12x. We can determine two second-quadrant solutions of the equation, find the length r for each point, and then compare the sine, cosine, and tangent function values using each point.

  • If x=-4, then y=-12(-4)=2.

  • If x=-8, then y=-12(-8)=4.

  • For (-4,2),r=(-4)2+22=20=25.

  • For (-8,4),r=(-8)2+42=80=45.

Using (-4,2) and r=25, we find that

sin θ=yr=225=15=55,cos θ=xr=-425=-25=-255,andtan θ=yx=2-4=-12.

Using (-8,4) and r=45, we find that

sin θ=yr=445=15=55,cos θ=xr=-845=-25=-255 ,andtan θ=yx=4-8=-12.

We see that the function values are the same using either point. Any point other than the origin on the terminal side of an angle can be used to determine the trigonometric function values.

The Six Functions Related

When we know one of the function values of an angle, we can find the other five if we know the quadrant in which the terminal side lies. The procedure is to sketch a reference triangle in the appropriate quadrant, use the Pythagorean equation as needed to find the lengths of its sides, and then find the ratios of the sides.

Example 4

Given that tan θ=-23 and θ is in the second quadrant, find the other function values.

Solution

We first sketch a second-quadrant angle. Since

tan θ=yx=-23=2-3 , Expressing-23as2-3 since θ is in quadrant II

we make the legs’ lengths 3 and 2. We measure off the 3 units in the negative direction since θ is in quadrant II. The hypotenuse must then have length 32+22, or 13. Now we read off the appropriate ratios:

sin θ=213,or21313,csc θ=132,cos θ=-313,or-31313,sec θ=-133,tan θ=-23,cot θ=-32.

Now Try Exercise 35.

Terminal Side on an Axis

An angle whose terminal side falls on one of the axes is a quadrantal angle. One of the coordinates of any point on that side is 0. The definitions of the trigonometric functions still apply, but in some cases, function values will not be defined because a denominator will be 0.

Example 5

Find the sine, cosine, and tangent values for 90°, 180°, 270°, and 360°.

Solution

We first make a drawing of each angle in standard position and label a point on the terminal side. Since the function values are the same for all points on the terminal side, we choose (0,1), (-1,0), (0,-1), and (1,0) for convenience. Note that r=1 for each choice.

Then by the definitions we get

sin 90°=11=1,sin 180°=01=0,sin 270°=-11=-1,sin 360°=01=0,cos 90°=01=0,cos 180°=-11=-1,cos 270°=01=0,cos 360°=11=1,tan 90°=10,Not definedtan 180°=0-1=0,tan 270°=-10,Not definedtan 360°=01=0.

In Example 5, all the values can be found using a calculator, but you will find that it is convenient to be able to compute them mentally. It is also helpful to note that coterminal angles have the same function values. For example, 0° and 360° are coterminal; thus, sin 0°=0, cos 0°=1, and tan 0°=0.

Example 6

Find each of the following.

  1. a) sin (-90°)

  2. b) csc 540°

Solution

  1. a) We note that -90° is coterminal with 270°. Thus,

    sin (-90°)=sin 270°=-1.
  2. b) Since 540°=180°+360°, 540° and 180° are coterminal. Thus,

    csc 540°=csc 180°=1sin 180°=10,which is not defined.

Now Try Exercises 45 and 55.

Reference Angles: 30°, 45° and 60°

We can also mentally determine trigonometric function values whenever the terminal side makes a 30°, 45°, or 60° angle with the x-axis. Consider, for example, an angle of 150°. The terminal side makes a 30° angle with the x-axis, since 180°-150°=30°.

As the figure shows, ΔONP is congruent to ΔONP; therefore, the ratios of the sides of the two triangles are the same. Thus the trigonometric function values are the same except perhaps for the sign. We could determine the function values directly from ΔONP, but this is not necessary. If we remember that in quadrant II, the sine is positive and the cosine and the tangent are negative, we can simply use the function values of 30° that we already know and prefix the appropriate sign. Thus,

sin 150°=sin 30°=12,cos 150°=-cos 30°=-32,andtan 150°=-tan 30°=-13,or-33.

Triangle ONP is the reference triangle, and the acute angle NOP is called a reference angle.

Example 7

Find the sine, cosine, and tangent function values for each of the following.

  1. a) 225°

  2. b) -780°

Solution

  1. a) We draw a figure showing the terminal side of a 225° angle. The reference angle is 225°-180°, or 45°.

    Recall from Section 6.1 that sin 45°=2/2, cos 45°=2/2, and tan 45°=1. Also note that in the third quadrant, the sine and the cosine are negative and the tangent is positive. Thus we have

    sin 225°=-22,cos 225°=-22, and tan 225°=1.
  2. b) We draw a figure showing the terminal side of a -780° angle. Since -780°+2(360°)=-60°, we know that -780° and -60° are coterminal.

    The reference angle for -60° is the acute angle formed by the terminal side of the angle and the x-axis. Thus the reference angle for -60° is 60°. We know that since -780° is a fourth-quadrant angle, the cosine is positive and the sine and the tangent are negative. Recalling that sin 60°=3/2, cos 60°=1/2, and tan 60°=3, we have

    sin  (-780°)=-32,cos  (-780°)=12,andtan  (-780°)=-3.

Now Try Exercises 47 and 51.

Function Values for Any Angle

When the terminal side of an angle falls on one of the axes or makes a 30°, 45°, or 60° angle with the x-axis, we can find exact function values without the use of a calculator. But this group is only a small subset of all angles. Using a calculator, we can approximate the trigonometric function values of any angle. In fact, we can approximate or find exact function values of all angles without using a reference angle.

Example 8

Find each of the following function values using a calculator and round the answer to four decimal places, where appropriate.

  1. a) cos 112°

  2. b) sec 500°

  3. c) tan (-83.4°)

  4. d) csc 351.75°

  5. e) cos 2400°

  6. f) sin 175°4090

  7. g) cot (-135°)

Solution

Using a calculator set in DEGREE mode, we find the values.

  1. a) cos 112°-0.3746

  2. b) sec 500°=1cos 500°-1.3054

  3. c) tan (-83.4°)-8.6427

  4. d) csc 351.75°=1sin 351.75°-6.9690

  5. e) cos 2400°=-0.5

  6. f) sin 175°4090.0755

  7. g) cot (-135°)=1tan (-135°)=1

Now Try Exercises 85 and 91.

In many applications, we have a trigonometric function value and want to find the measure of a corresponding angle. When only acute angles are considered, there is only one angle for each trigonometric function value. This is not the case when we extend the domain of the trigonometric functions to the set of all angles. For a given function value, there is an infinite number of angles that have that function value. There can be two such angles for each value in the range from 0° to 360°. To determine a unique answer in the interval (0°, 360°), the quadrant in which the terminal side lies must be specified.

The calculator gives the reference angle as an output for each function value that is entered as an input. Knowing the reference angle and the quadrant in which the terminal side lies, we can find the specified angle.

Example 9

Given the function value and the quadrant restriction, find θ.

  1. a) sin θ=0.2812, 90°<θ<180°

  2. b) cot θ=-0.1611, 270°<θ<360°

Solution

  1. a) We first sketch the angle in the second quadrant. We use the calculator to find the acute angle (reference angle) whose sine is 0.2812. The reference angle is approximately 16.33°. We find the angle θ by subtracting 16.33° from 180°:

    180°-16.33°=163.67°.

    Thus, θ163.67°.

  2. b) We begin by sketching the angle in the fourth quadrant. Because the tangent and cotangent values are reciprocals, we know that

    tan θ1-0.1611-6.2073.

    We use the calculator to find the acute angle (reference angle) whose tangent is 6.2073, ignoring the fact that θ is negative. The reference angle is approximately 80.85°. We find angle θ by subtracting 80.85° from 360°:

    360°-80.85°=279.15°.

    Thus, θ279.15°.

Now Try Exercise 97.

Bearing: Second-Type

In aerial navigation, directions are given in degrees clockwise from north. Thus east is 90°, south is 180°, and west is 270°. Several aerial directions, or bearings, are given below.

Example 10 Aerial Navigation.

An airplane flies 218 mi from an airport in a direction of 245°. How far south of the airport is the plane then? How far west?

Solution

We first find the measure of ABC:

ABC=270°-245°=25°.

From the figure shown at left, we see that the distance south of the airport b and the distance west of the airport a are legs of a right triangle. We have

b218=sin 25°b=218 sin 25°92 mi

and

a218=cos 25°a=218 cos 25°198 mi.

The airplane is about 92 mi south and about 198 mi west of the airport.

Now Try Exercise 105.

6.3 Exercise Set

For angles of the following measures, state in which quadrant the terminal side lies. It helps to sketch the angle in standard position.

  1. 1. 187°

  2. 2. -14.3°

  3. 3. 245°15

  4. 4. -120°

  5. 5. 800°

  6. 6. 1075°

  7. 7. -460.5°

  8. 8. 315°

  9. 9. -912°

  10. 10. 13°1558

  11. 11. 537°

  12. 12. -345.14°

Find two positive angles and two negative angles that are coterminal with the given angle. Answers may vary.

  1. 13. 74°

  2. 14. -81°

  3. 15. 115.3°

  4. 16. 275°10

  5. 17. -180°

  6. 18. -310°

Find the complement and the supplement.

  1. 19. 17.11°

  2. 20. 47°38

  3. 21. 12°314

  4. 22. 9.038°

  5. 23. 45.2°

  6. 24. 67.31°

Find the six trigonometric function values for the angle shown.

  1. 25.

  2. 26.

  3. 27.

  4. 28.

  5. 29.

  6. 30.

The terminal side of angle θ in standard position lies on the given line in the given quadrant. Find θ, cosθ and tanθ.

  1. 31. 2x+3y=0; quadrant IV

  2. 32. 4x+y=0; quadrant II

  3. 33. 5x-4y=0; quadrant I

  4. 34. y=0.8x; quadrant III

A function value and a quadrant are given. Find the other five trigonometric function values. Give exact answers.

  1. 35. sin θ=-13; quadrant III

  2. 36. tan β=5; quadrant I

  3. 37. cot θ=-2; quadrant IV

  4. 38. cos α=-45; quadrant II

  5. 39. cos ϕ=35; quadrant IV

  6. 40. sin θ=-513; quadrant III

Find the reference angle and the exact function value if it exists.

  1. 41. cos 150°

  2. 42. sec (-225°)

  3. 43. tan(-135°)

  4. 44. sin(-45°)

  5. 45. sin7560°

  6. 46. tan 270°

  7. 47. cos 495°

  8. 48. tan 675°

  9. 49. csc (-210°)

  10. 50. sin 300°

  11. 51. cot 570°

  12. 52. cos (-120°)

  13. 53. tan 330°

  14. 54. cot 855°

  15. 55. sec (-90°)

  16. 56. sin 90°

  17. 57. cos (-180°)

  18. 58. csc 90°

  19. 59. tan 240°

  20. 60. cot (-180°)

  21. 61. sin 495°

  22. 62. sin 1050°

  23. 63. csc 225°

  24. 64. sin (-450°)

  25. 65. cos 0°

  26. 66. tan 480°

  27. 67. cot (-90°)

  28. 68. sec 315°

  29. 69. cos 90°

  30. 70. sin (-135°)

  31. 71. cos 270°

  32. 72. tan 0°

In Exercises 7380, find the signs of the six trigonometric function values for the given angles.

  1. 73. 319°

  2. 74. -57°

  3. 75. 194°

  4. 76. -620°

  5. 77. -215°

  6. 78. 290°

  7. 79. -272°

  8. 80. 91°

Use a calculator in Exercises 8184, but do not use the trigonometric function keys.

  1. 81. Given that

    sin 41°=0.6561,cos 41°=0.7547,tan 41°=0.8693,

    find the trigonometric function values for 319°.

  2. 82. Given that

    sin 27°=0.4540,cos 27°=0.8910,tan 27°=0.5095,

    find the trigonometric function values for 333°.

  3. 83. Given that

    sin 65°=0.9063,cos 65°=0.4226,tan 65°=2.1445,

    find the trigonometric function values for 115°.

  4. 84. Given that

    sin 35°=0.5736,cos 35°=0.8192,tan 35°=0.7002,

    find the trigonometric function values for 215°.

Find the function value. Round to four decimal places.

  1. 85. tan 310.8°

  2. 86. cos 205.5°

  3. 87. cot 146.15°

  4. 88. sin (-16.4°)

  5. 89. sin 118°42

  6. 90. cos 273°45

  7. 91. cos (-295.8°)

  8. 92. tan 1086.2°

  9. 93. cos 5417°

  10. 94. sec 240°55

  11. 95. csc 520°

  12. 96. sin 3824°

Given the function value and the quadrant restriction, find θ.

  1. Function Value Interval θ
    97.sinθ=-0.9956 (270°, 360°)
    98.tanθ=0.2460 (180°, 270°)
    99.cos θ=-0.9388 (180°, 270°)
    100.secθ=-1.0485 (90°, 180°)
    101.tanθ=-3.0545 (270°, 360°)
    102.sinθ=-0.4313 (180°, 270°)
    103.csc θ=1.0480 (0°, 90°)
    104.cosθ=-0.0990 (90°, 180°)
  2. 105. Aerial Navigation. An airplane flies 150 km from an airport in a direction of 120°. How far east of the airport is the plane then? How far south?

  3. 106. Aerial Navigation. An airplane leaves an airport and travels for 100 mi in a direction of 300°. How far north of the airport is the plane then? How far west?

  4. 107. Aerial Navigation. An airplane travels at 150 km/h for 2 hr in a direction of 138° from Omaha. At the end of this time, how far south of Omaha is the plane?

  5. 108. Aerial Navigation. An airplane travels at 120 km/h for 2 hr in a direction of 319° from Chicago. At the end of this time, how far north of Chicago is the plane?

Skill Maintenance

Graph the function. Sketch and label any vertical asymptotes.

  1. 109. f(x)=1x2-25[4.5]

  2. 110. g(x)=x3-2x+1[4.2]

Determine the domain and the range of the function.

  1. 111. f(x)=x-4x+2 [1.2], [4.5]

  2. 112. g(x)=x2-92x2-7x-15 [1.2], [4.1], [4.5]

Find the zeros of the function.

  1. 113. f(x)=12-x[1.5]

  2. 114. g(x)=x2-x-6[3.2]

Find the x-intercept(s) of the graph of the function.

  1. 115. f(x)=12-x[1.5]

  2. 116. g(x)=x2-x-6[3.2]

Synthesis

  1. 117. Tallest Ferris Wheel. On March 31, 2014, the world’s tallest ferris wheel, the High Roller, opened in Las Vegas at the LINQ. Each cabin, which can fit up to 40 passengers, is 220 ft from the center of the wheel. The wheel makes one revolution in approximately 30 min. (Source: Nancy Trejos, USA Today, April 7, 2014). When you board, you are 6 ft above the ground. After you have rotated through an angle of 315°, how far above the ground are you?

  2. 118. Valve Cap on a Bicycle. The valve cap on a bicycle wheel is 12.5 in. from the center of the wheel. From the position shown, the wheel starts to roll. After the wheel has turned 390°, how far above the ground is the valve cap? Assume that the outer radius of the tire is 13.375 in.

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