1. Trigonometric Functions
Angles
Trigonometric Functions
Evaluating Trigonometric Functions
Solving Right Triangles
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2. Angles
Basic Terminology ▪ Degree Measure ▪ Standard Position ▪ Coterminal Angles
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3. 5.1 Example 1 Finding the Complement and the Supplement of an Angle (page 495)
For an angle measuring 55°, find the measure of its complement and its supplement.
Complement: 90° − 55° = 35°
Supplement: 180° − 55° = 125°
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4. Find the measure of each angle.
5.1 Example 2(a) Finding Measures of Complementary and Supplementary Angles (page 495)
Find the measure of each angle.
The two angles form a right angle, so they are complements.
The measures of the two angles are
and
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5. Find the measure of each angle.
5.1 Example 2(b) Finding Measures of Complementary and Supplementary Angles (page 495)
Find the measure of each angle.
The two angles form a straight angle, so they are supplements.
The measures of the two angles are
and
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6. Trigonometric Functions
Trigonometric Functions ▪ Quadrantal Angles ▪ Reciprocal Identities ▪ Signs and Ranges of Function Values ▪ Pythagorean Identities ▪ Quotient Identities
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7. 5.2 Example 1 Finding Function Values of an Angle (page 503)
The terminal side of an angle θ in standard position passes through the point (12, 5). Find the values of the six trigonometric functions of angle θ.
x = 12 and y = 5. 13
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8. 5.2 Example 2 Finding Function Values of an Angle (page 503)
The terminal side of an angle θ in standard position passes through the point (8, –6). Find the values of the six trigonometric functions of angle θ.
x = 8 and y = –6. 6 10
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9. 5.2 Example 2 Finding Function Values of an Angle (cont.)
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10. 5.2 Example 3 Finding Function Values of an Angle (page 505)
Find the values of the six trigonometric functions of angle θ in standard position, if the terminal side of θ is defined by 3x – 2y = 0, x ≤ 0.
Since x ≤ 0, the graph of the line 3x – 2y = 0 is shown to the left of the y-axis.
Find a point on the line: Let x = –2. Then
A point on the line is (–2, –3).
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11. 5.2 Example 3 Finding Function Values of an Angle (cont.)
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12. Find the values of the six trigonometric functions of a 360° angle.
5.2 Example 4(a) Finding Function Values of Quadrantal Angles (page 506)
Find the values of the six trigonometric functions of a 360° angle.
The terminal side passes through (2, 0). So x = 2 and y = 0 and r = 2.
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13. 5.2 Example 4(b) Finding Function Values of Quadrantal Angles (page 506)
Find the values of the six trigonometric functions of an angle θ in standard position with terminal side through (0, –5).
x = 0 and y = –5 and r = 5.
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14. 5.2 Example 5 Using the Reciprocal Identities (page 508)
Find each function value.
(a) tan θ, given that cot θ = 4.
tan θ is the reciprocal of cot θ.
(b) sec θ, given that
sec θ is the reciprocal of cos θ.
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15. 5.2 Example 6 Identifying the Quadrant of an Angle (page 509)
Identify the quadrant (or possible quadrants) of an angle θ that satisfies the given conditions.
(a) tan θ > 0, csc θ < 0
tan θ > 0 in quadrants I and III, while csc θ < 0 in quadrants III and IV. Both conditions are met only in quadrant III.
(b) sin θ > 0, csc θ > 0
sin θ > 0 in quadrants I and II, as is csc θ. Both conditions are met in quadrants I and II.
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16. Decide whether each statement is possible or impossible.
5.2 Example 7 Deciding Whether a Value is in the Range of a Trigonometric Function (page 510)
Decide whether each statement is possible or impossible.
(a) cot θ = – (b) cos θ = –1.7 (c) csc θ = 0
(a) cot θ = –.999 is possible because the range of cot θ is
(b) cos θ = –1.7 is impossible because the range of cos θ is [–1, 1].
(c) csc θ = 0 is impossible because the range of csc θ is
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17. Find cot θ and csc θ given that cos θ and θ is in quadrant II.
5.2 Example 8 Finding Other Function Values Given One Value and the Quadrant (page 511)
Find cot θ and csc θ given that cos θ and θ is in quadrant II.
Since θ is in quadrant II, cot θ < 0 and csc θ > 0.
Reject the negative root:
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18. 5.2 Example 7 Finding Other Function Values Given One Value and the Quadrant (cont.)
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19. Evaluating Trigonometric Functions
Right-Triangle-Based Definitions of the Trigonometric Functions ▪ Cofunctions ▪ Trigonometric Function Values of Special Angles ▪ Reference Angles ▪ Special Angles as Reference Angles ▪ Finding Angle Measures Using a Calculator ▪ Finding Angle Measures
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20. 5.3 Example 1 Finding Trigonometric Function Values of An Acute Angle (page 516)
Find the sine, cosine, and tangent values for angles D and E in the figure.
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21. 5.3 Example 1 Finding Trigonometric Function Values of An Acute Angle (cont.)
Find the sine, cosine, and tangent values for angles D and E in the figure.
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22. 5.3 Example 2 Writing Functions in Terms of Cofunctions (page 517)
Write each function in terms of its cofunction.
(a) sin 9°
= cos (90° – 9°) = cos 81°
(b) cot 76°
= tan (90° – 76°) = tan 14°
(c) csc 45°
= sec (90° – 45°) = sec 45°
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23. 5.3 Example 3 Finding Trigonometric Values for 30° (page 518)
Find the six trigonometric function values for a 30° angle.
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24. 5.3 Example 3 Finding Trigonometric Values for 30° (cont.)
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25. 5.3 Example 4(a) Finding Reference Angles (page 519)
Find the reference angle for 294°.
294 ° lies in quadrant IV.
The reference angle is 360° – 294° = 66°.
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26. 5.3 Example 4(b) Finding Reference Angles (page 519)
Find the reference angle for 883°.
Find a coterminal angle between 0° and 360° by dividing 883° by 360°. The quotient is about 2.5.
883° is coterminal with 163°.
The reference angle is 180° – 163° = 17°.
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27. Find the values of the six trigonometric functions for 135°.
5.3 Example 5 Finding Trigonometric Functions of a Quadrant II Angle (page 520)
Find the values of the six trigonometric functions for 135°.
The reference angle for 135° is 45°.
Choose point P on the terminal side of the angle. The coordinates of P are (1, –1).
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28. 5.3 Example 5 Finding Trigonometric Functions of a Quadrant II Angle (cont.)
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29. Find the exact value of sin(–150°).
5.3 Example 6(a) Finding Trigonometric Function Values Using Reference Angles (page 521)
Find the exact value of sin(–150°).
An angle of –150° is coterminal with an angle of –150° + 360° = 210°.
The reference angle is 210° – 180° = 30°.
Since an angle of –150° lies in quadrant III, its sine is negative.
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30. Find the exact value of cot(780°).
5.3 Example 6(b) Finding Trigonometric Function Values Using Reference Angles (page 521)
Find the exact value of cot(780°).
An angle of 780° is coterminal with an angle of 780° – 2 ∙ 360° = 60°.
The reference angle is 60°.
Since an angle of 780° lies in quadrant I, its cotangent is positive.
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31. 5.3 Example 7 Finding Function Values with a Calculator (page 522)
Approximate the value of each expression.
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32. 5.3 Example 7 Finding Function Values with a Calculator (cont.)
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33. 5.3 Example 8 Using Inverse Trigonometric Functions to Find Angles (page 523)
Use a calculator to find an angle θ in the interval [0°, 90°] that satisfies each condition.
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34. Find all values of θ, if θ is in the interval [0°, 360°) and sin θ =
5.3 Example 9 Finding Angle Measures Given an Interval and a Function Value (page 523)
Find all values of θ, if θ is in the interval [0°, 360°) and sin θ =
Since sin θ is negative, θ must lie in quadrants III or IV.
The absolute value of sin θ is so the reference angle is 60°.
The angle in quadrant III is 60° + 180° = 240°.
The angle in quadrant IV is 360° – 60° = 300°.
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35. 5.3 Example 10 Finding Grade Resistance (page 523)
The force F in pounds when an automobile travels uphill or downhill on a highway is called grade resistance and is modeled by the equation , where θ is the grade and W is the weight of the automobile.
If the automobile is moving uphill, then θ > 0°; if it is moving downhill, then θ < 0°.
(a) Calculate F to the nearest 10 pounds for a 5500-lb car traveling an uphill grade with θ = 3.9°.
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36. 5.3 Example 10 Finding Grade Resistance (cont.)
(b) Calculate F to the nearest 10 pounds for a 2800-lb car traveling a downhill grade with θ = –4.8°.
(c) A 2400-lb car traveling uphill has a grade resistance of 288 pounds. What is the angle of the grade?
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37. Solving Right Triangles
Significant Digits ▪ Solving Triangles ▪ Angles of Elevation or Depression ▪ Bearing ▪ Further Applications
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38. Solve right triangle ABC, if B = 28°40′ and a = 25.3 cm.
5.4 Example 1 Solving a Right Triangle Given an Angle and a Side (page 532)
Solve right triangle ABC, if B = 28°40′ and a = 25.3 cm.
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39. Three significant digits
5.4 Example 1 Solving a Right Triangle Given an Angle and a Side (cont.)
Three significant digits
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40. Three significant digits
5.4 Example 1 Solving a Right Triangle Given an Angle and a Side (cont.)
Three significant digits
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41. 5.4 Example 1 Solving a Right Triangle Given an Angle and a Side (cont.)
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42. 5.4 Example 2 Solving a Right Triangle Given Two Sides (page 533)
Solve right triangle ABC, if a = cm and b = cm.
Use the Pythagorean theorem to find c:
Four significant digits
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43. 5.4 Example 2 Solving a Right Triangle Given Two Sides (cont.)
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44. 5.4 Example 2 Solving a Right Triangle Given Two Sides (cont.)
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45. The angle of elevation of the sun is 63.39°.
5.4 Example 3 Finding a Length When the Angle of Elevation is Known (page 534)
The length of a shadow of a flagpole ft tall is ft. Find the angle of elevation of the sun.
Four significant digits
The angle of elevation of the sun is 63.39°.
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46. Angle C is a right angle because angles CAB and CBA are complementary.
5.4 Example 4 Solving Problem Involving Bearing (First Method) (page 535)
Radar stations A and B are on an east-west line, 8.6 km apart. Station A detects a plane at C, on a bearing of 53°. Stations B simultaneously detects the same plane, on a bearing of 323°. Find the distance from B to C.
A line drawn due north is perpendicular to an east-west line, so right angles are formed at A and B, and angles CAB and CBA can be found as shown in the figure.
Angle C is a right angle because angles CAB and CBA are complementary.
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47. 5.4 Example 4 Solving Problem Involving Bearing (First Method) (cont.)
Find distance a by using the cosine function for angle A.
The distance from B to C is about 5.2 km.
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48. The information in the example gives and
5.4 Example 5 Solving Problem Involving Bearing (Second Method) (page 536)
The bearing from A to C is N 64° W. The bearing from A to B is S 82° W. The bearing from B to C is N 26° E. A plane flying at 350 mph take 1.8 hours to go from A to B. Find the distance from B to C.
The information in the example gives and
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49. 5. 4 Example 5 Solving Problem Involving Bearing (Second Method) (cont
The sum of the measures of angles EAB and FBA is 180º because they are interior angles on the same side of a transversal.
90°
34°
56°
98°
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50. The distance from B to C is about 350 miles.
5.4 Example 5 Solving Problem Involving Bearing (Second Method) (cont.)
It takes 1.8 hours at 350 mph to fly from A to B, so
90°
34°
56°
630 mi
98°
The distance from B to C is about 350 miles.
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51. 5.4 Example 6 Using Trigonometry to Measure a Distance (page 536)
A method that surveyors use to determine a small distance d between two points P and Q is called the subtense bar method. The subtense bar with length b is centered at Q and situated perpendicular to the line of sight between P and Q.
(a) Find d with θ = 2°41′38″ and b = cm.
From the figure, we have
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52. 5.4 Example 6 Using Trigonometry to Measure a Distance (cont.)
Convert θ to decimal degrees:
Five significant digits
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53. 5.4 Example 6 Using Trigonometry to Measure a Distance (page 536)
(b) Angle θ usually cannot be measured more accurately than to the nearest 1″. How much change would there be in the value of d if θ were measured 1″ larger?
Since θ is 1″ larger, θ = 2°41′39″ ≈
The difference is
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54. 5.4 Example 7 Solving a Problem Involving Angles of Elevation (page 536)
Marla needs to find the height of a building. From a given point on the ground, she finds that the angle of elevation to the top of the building is 74.2°. She then walks back 35 feet. From the second point, the angle of elevation to the top of the building is 51.8°. Find the height of the building.
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55. 5.4 Example 7 Solving a Problem Involving Angles of Elevation (cont.)
There are two unknowns, the distance from the base of the building, x, and the height of the building, h.
In triangle ABC
In triangle BCD
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56. 5.4 Example 7 Solving a Problem Involving Angles of Elevation (cont.)
Set the two expressions for h equal and solve for x.
Since h = x tan 74.2°, substitute the expression for x to find h.
The building is about 69 feet tall.
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57. 5.4 Example 7 Solving a Problem Involving Angles of Elevation (cont.)
Graphing Calculator Solution
Superimpose coordinate axes on the figure with D at the origin.
The coordinates of A are (35, 0).
The tangent of the angle between the x-axis and the graph of a line with equation y = mx + b is the slope of the line. For line DB, m = tan 51.8º.
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58. 5.4 Example 7 Solving a Problem Involving Angles of Elevation (cont.)
Since b = 0, the equation of line DB is
The equation of line AB is
Use the coordinates of A and the point-slope form to find the equation of AB:
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59. 5.4 Example 7 Solving a Problem Involving Angles of Elevation (cont.)
Graph y1 and y2, then find the point of intersection. The y-coordinate gives the height, h.
The building is about 69 feet tall.
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