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, c) are called angle-side opposite pairs. Second, as mentioned earlier, we will strive to solve for quantities using the original data given in the problem whenever possible. While this is not always the easiest or fastest way to proceed, it 1as well as the measure of said angle 2as well as the length of said side 11.2 The Law of Sines 897 minimizes the chances of propagated error.3 Third, since many of the applications which require solving triangles ‘in the wild’ rely on degree measure, we shall adopt this convention for the time being.4 The Pythagorean Theorem along with Theorems 10.4 and 10.10 allow us to easily handle any given right triangle problem, but what if the triangle isn’t a right triangle? In certain cases, we can use the Law of Sines to help. Theorem 11.2. The Law of Sines: Given a triangle with angle-side opposite pairs (α, a), (β, b) and (γ, c), the following ratios hold or, equivalently, sin(α) a = sin(β) b = sin(γ) c a sin(α) = b sin(β) = c sin(γ) The proof of the Law of Sines can be broken into three cases. For our ﬁrst case, consider the triangle ABC below, all of whose angles are acute, with angle-side opposite pairs (α, a), (β, b) and (γ, c). If we drop an altitude from vertex B, we divide the triangle into two right triangles: ABQ and BCQ. If we call the length of the altitude h (for height), we get from Theorem 10.4 that sin(α) = h a so that h = c sin(α) = a sin(γ). After some rearrangement of the last equation, we get sin(α). If we drop an altitude from vertex A, we can proceed as above using the triangles ABQ and ACQ to get sin(β) B, completing the proof for this case. c and sin(γ) = h a = sin(γ) b = sin(γ For our next case consider the triangle ABC below with obtuse angle α. Extending an altitude from vertex A gives two right triangles, as in the previous case: ABQ and ACQ. Proceeding as before, we get h = b sin(γ
) and h = c sin(β) so that sin(β). b = sin(γ 3Your Science teachers should thank us for this. 4Don’t worry! Radians will be back before you know it! 898 Applications of Trigonometry Dropping an altitude from vertex B also generates two right triangles, ABQ and BCQ. We know that sin(α) = h c so that h = c sin(α). Since α = 180◦ − α, sin(α) = sin(α), so in fact, we have h = c sin(α). Proceeding to BCQ, we get sin(γ) = h a so h = a sin(γ). Putting this together with the previous equation, we get sin(γ), and we are ﬁnished with this case. c = sin(α The remaining case is when ABC is a right triangle. In this case, the Law of Sines reduces to the formulas given in Theorem 10.4 and is left to the reader. In order to use the Law of Sines to solve a triangle, we need at least one angle-side opposite pair. The next example showcases some of the power, and the pitfalls, of the Law of Sines. Example 11.2.2. Solve the following triangles. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle. 1. α = 120◦, a = 7 units, β = 45◦ 2. α = 85◦, β = 30◦, c = 5.25 units 3. α = 30◦, a = 1 units, c = 4 units 4. α = 30◦, a = 2 units, c = 4 units 5. α = 30◦, a = 3 units, c = 4 units 6. α = 30◦, a = 4 units, c = 4 units Solution. b sin(45◦) = 1. Knowing an angle-side opposite pair, namely α and a, we may proceed in using the Law of √ 6 Sines. Since β = 45◦, we use 3 ≈ 5.72 units. Now that we have two angle-side pairs, it is time to ﬁnd the third. To ﬁnd γ, we use the fact that the sum of the measures of the angles in a triangle is 180◦. Hence, γ =
180◦ − 120◦ − 45◦ = 15◦. To ﬁnd c, we have no choice but to used the derived value γ = 15◦, yet we can minimize the propagation of error here by using the given angle-side opposite pair (α, a). The Law of Sines gives us sin(120◦) so b = 7 sin(45◦) sin(120◦) so that c = 7 sin(15◦) sin(120◦) ≈ 2.09 units.5 sin(120◦) = 7 c sin(15◦) = 7 7 2. In this example, we are not immediately given an angle-side opposite pair, but as we have the measures of α and β, we can solve for γ since γ = 180◦ − 85◦ − 30◦ = 65◦. As in the previous example, we are forced to use a derived value in our computations since the only 5The exact value of sin(15◦) could be found using the diﬀerence identity for sine or a half-angle formula, but that becomes unnecessarily messy for the discussion at hand. Thus “exact” here means 7 sin(15◦) sin(120◦). 11.2 The Law of Sines 899 angle-side pair available is (γ, c). The Law of Sines gives rearrangement, we get a = 5.25 sin(85◦) which yields sin(65◦). After the usual sin(65◦) ≈ 5.77 units. To ﬁnd b we use the angle-side pair (γ, c) sin(65◦) hence b = 5.25 sin(30◦) sin(65◦) ≈ 2.90 units. sin(30◦) = 5.25 b a sin(85◦) = 5.25 β = 45◦ a = 7 c ≈ 2.09 α = 120◦ γ = 15◦ b ≈ 5.72 Triangle for number 1 β = 30◦ c = 5.25 a ≈ 5.77 α = 85◦ γ = 65◦ b ≈ 2.90 Triangle for number 2 1 4 = sin(30◦) 3. Since we are given (α, a) and c
with α = 30◦, or sin(γ) = 4 sin(30◦) case, we have sin(γ) 4 = sin(30◦) = 2 3 3 900 Applications of Trigonometry 3 3 : γ = arcsin 2 we must have that 0◦ < γ < 150◦. There are two angles γ that fall in this range and have radians ≈ 138.19◦. At radians ≈ 41.81◦ and γ = π − arcsin 2 sin(γ) = 2 3 this point, we pause to see if it makes sense that we actually have two viable cases to consider. As we have discussed, both candidates for γ are ‘compatible’ with the given angle-side pair (α, a) = (30◦, 3) in that both choices for γ can ﬁt in a triangle with α and both have a sine of 2 3. The only other given piece of information is that c = 4 units. Since c > a, it must be true that γ, which is opposite c, has greater measure than α which is opposite a. In both cases, γ > α, so both candidates for γ are compatible with this last piece of given information as radians ≈ 41.81◦, we well. Thus have two triangles on our hands. In the case γ = arcsin 2 3 ﬁnd6 β ≈ 180◦ − 30◦ − 41.81◦ = 108.19◦. Using the Law of Sines with the angle-side opposite radians pair (α, a) and β, we ﬁnd b ≈ 3 sin(108.19◦) ≈ 138.19◦, we repeat the exact same steps and ﬁnd β ≈ 11.81◦ and b ≈ 1.23 units.7 Both triangles are drawn below. sin(30◦) ≈ 5.70 units. In the case γ = π − arcsin 2 3 β ≈ 11.81◦ c = 4 β ≈ 108.19◦ a = 3 α = 30◦ γ ≈ 41.81◦ α = 30◦ c = 4 a = 3 γ ≈ 138.19◦ b ≈ 5.70 b ≈ 1.23 6. For this
last problem, we repeat the usual Law of Sines routine to ﬁnd that sin(γ) so that sin(γ) = 1 2. Since γ must inhabit a triangle with α = 30◦, we must have 0◦ < γ < 150◦. Since the measure of γ must be strictly less than 150◦, there is just one angle which satisﬁes both required conditions, namely γ = 30◦. So β = 180◦ − 30◦ − 30◦ = 120◦ and, using the √ Law of Sines one last time, b = 4 sin(120◦) 3 ≈ 6.93 units. 4 = sin(30◦) 4 sin(30◦) = 4 c = 4 β = 120◦ a = 4 α = 30◦ γ = 30◦ b ≈ 6.93 Some remarks about Example 11.2.2 are in order. We ﬁrst note that if we are given the measures of two of the angles in a triangle, say α and β, the measure of the third angle γ is uniquely 6 radians, γ = arcsin 2 6To ﬁnd an exact expression for β, we convert everything back to radians: α = 30◦ = π 3 radians and 180◦ = π radians. Hence, β = π − π 7An exact answer for β in this case is β = arcsin 2 6 − arcsin 2 − π 3 6 − arcsin 2 = 5π 6 radians ≈ 11.81◦. 3 3 radians ≈ 108.19◦. 11.2 The Law of Sines 901 determined using the equation γ = 180◦ − α − β. Knowing the measures of all three angles of a triangle completely determines its shape. If in addition we are given the length of one of the sides of the triangle, we can then use the Law of Sines to ﬁnd the lengths of the remaining two sides to determine the size of the triangle. Such is the case in numbers 1 and 2 above. In number 1, the given side is adjacent to just one of the angles – this is called the ‘Angle-Angle-Side’ (AAS) case.8 In number 2, the given side is adjacent to both angles which means we are in
the so-called ‘Angle-Side-Angle’ (ASA) case. If, on the other hand, we are given the measure of just one of the angles in the triangle along with the length of two sides, only one of which is adjacent to the given angle, we are in the ‘Angle-Side-Side’ (ASS) case.9 In number 3, the length of the one given side a was too short to even form a triangle; in number 4, the length of a was just long enough to form a right triangle; in 5, a was long enough, but not too long, so that two triangles were possible; and in number 6, side a was long enough to form a triangle but too long to swing back and form two. These four cases exemplify all of the possibilities in the Angle-Side-Side case which are summarized in the following theorem. Theorem 11.3. Suppose (α, a) and (γ, c) are intended to be angle-side pairs in a triangle where α, a and c are given. Let h = c sin(α) If a < h, then no triangle exists which satisﬁes the given criteria. If a = h, then γ = 90◦ so exactly one (right) triangle exists which satisﬁes the criteria. If h < a < c, then two distinct triangles exist which satisfy the given criteria. If a ≥ c, then γ is acute and exactly one triangle exists which satisﬁes the given criteria Theorem 11.3 is proved on a case-by-case basis. If a < h, then a < c sin(α). If a triangle were c = sin(α) to exist, the Law of Sines would have sin(γ) a = 1, which is impossible. In the ﬁgure below, we see geometrically why this is the case. so that sin(γ) = c sin(α) > a a a a h = c sin(α sin(α) a < h, no triangle a = h, γ = 90◦ Simply put, if a < h the side a is too short to connect to form a triangle. This means if a ≥ h, we are always guaranteed to have at least one triangle, and the remaining parts of the theorem 8If this sounds familiar, it should. From high school Ge
ometry, we know there are four congruence conditions for triangles: Angle-Angle-Side (AAS), Angle-Side-Angle (ASA), Side-Angle-Side (SAS) and Side-Side-Side (SSS). If we are given information about a triangle that meets one of these four criteria, then we are guaranteed that exactly one triangle exists which satisﬁes the given criteria. 9In more reputable books, this is called the ‘Side-Side-Angle’ or SSA case. 902 Applications of Trigonometry c a a = a a = sin(γ) < 1 which means there are two solutions to sin(γ) = c sin(α) tell us what kind and how many triangles to expect in each case. If a = h, then a = c sin(α) and so that sin(γ) = c sin(α) the Law of Sines gives sin(α) a = 1. Here, γ = 90◦ as required. Moving along, now suppose h < a < c. As before, the Law of Sines10 gives sin(γ) = c sin(α). Since h < a, c sin(α) < a or c sin(α) : an acute angle which we’ll call γ0, and its supplement, 180◦ − γ0. We need to argue that each of these angles ‘ﬁt’ into a triangle with α. Since (α, a) and (γ0, c) are angle-side opposite pairs, the assumption c > a in this case gives us γ0 > α. Since γ0 is acute, we must have that α is acute as well. This means one triangle can contain both α and γ0, giving us one of the triangles promised in the theorem. If we manipulate the inequality γ0 > α a bit, we have 180◦ −γ0 < 180◦ −α which gives (180◦ − γ0) + α < 180◦. This proves a triangle can contain both of the angles α and (180◦ − γ0), giving us the second triangle predicted in the theorem. To prove the last case in the theorem, we assume a ≥ c. Then α ≥ γ, which forces γ to be an acute angle. Hence, we get only one triangle in this case,
completing the proof. a a c a a h α γ0 γ0 h < a < c, two triangles c α h a γ a ≥ c, one triangle One last comment before we use the Law of Sines to solve an application problem. In the AngleSide-Side case, if you are given an obtuse angle to begin with then it is impossible to have the two triangle case. Think about this before reading further. Example 11.2.3. Sasquatch Island lies oﬀ the coast of Ippizuti Lake. Two sightings, taken 5 miles apart, are made to the island. The angle between the shore and the island at the ﬁrst observation point is 30◦ and at the second point the angle is 45◦. Assuming a straight coastline, ﬁnd the distance from the second observation point to the island. What point on the shore is closest to the island? How far is the island from this point? Solution. We sketch the problem below with the ﬁrst observation point labeled as P and the second as Q. In order to use the Law of Sines to ﬁnd the distance d from Q to the island, we ﬁrst need to ﬁnd the measure of β which is the angle opposite the side of length 5 miles. To that end, we note that the angles γ and 45◦ are supplemental, so that γ = 180◦ − 45◦ = 135◦. We can now 5 ﬁnd β = 180◦ − 30◦ − γ = 180◦ − 30◦ − 135◦ = 15◦. By the Law of Sines, we have sin(15◦) which gives d = 5 sin(30◦) sin(15◦) ≈ 9.66 miles. Next, to ﬁnd the point on the coast closest to the island, which we’ve labeled as C, we need to ﬁnd the perpendicular distance from the island to the coast.11 d sin(30◦) = 10Remember, we have already argued that a triangle exists in this case! 11Do you see why C must lie to the right of Q? 11.2 The Law of Sines 903 Let x denote the distance from the second observation point Q to the point C and let y denote the distance from C to the island. Using The
orem 10.4, we get sin (45◦) = y d. After some rearranging, we ﬁnd y = d sin (45◦) ≈ 9.66 ≈ 6.83 miles. Hence, the island is approximately 6.83 miles from the coast. To ﬁnd the distance from Q to C, we note that β = 180◦ − 90◦ − 45◦ = 45◦ so by symmetry,12 we get x = y ≈ 6.83 miles. Hence, the point on the shore closest to the island is approximately 6.83 miles down the coast from the second observation point. √ 2 2 Sasquatch Island Sasquatch Island β β d ≈ 9.66 miles d ≈ 9.66 miles y miles γ Q 30◦ 5 miles P 45◦ Shoreline 45◦ Q C x miles We close this section with a new formula to compute the area enclosed by a triangle. Its proof uses the same cases and diagrams as the proof of the Law of Sines and is left as an exercise. Theorem 11.4. Suppose (α, a), (β, b) and (γ, c) are the angle-side opposite pairs of a triangle. Then the area A enclosed by the triangle is given by A = 1 2 bc sin(α) = 1 2 ac sin(β) = 1 2 ab sin(γ) Example 11.2.4. Find the area of the triangle in Example 11.2.2 number 1. Solution. From our work in Example 11.2.2 number 1, we have all three angles and all three sides to work with. However, to minimize propagated error, we choose A = 1 2 ac sin(β) from Theorem 11.4 because it uses the most pieces of given information. We are given a = 7 and β = 45◦, and we calculated c = 7 sin(15◦) sin (45◦) =≈ 5.18 square units. The reader is encouraged to check this answer against the results obtained using the other formulas in Theorem 11.4. sin(120◦). Using these values, we ﬁnd A = 1 7 sin(15◦) sin(120◦) 2 (7) 12Or by Theorem 10.4 again... 904 Applications of Trigonometry 11.2.1 Exercises
In Exercises 1 - 20, solve for the remaining side(s) and angle(s) if possible. As in the text, (α, a), (β, b) and (γ, c) are angle-side opposite pairs. 1. α = 13◦, β = 17◦, a = 5 2. α = 73.2◦, β = 54.1◦, a = 117 3. α = 95◦, β = 85◦, a = 33.33 4. α = 95◦, β = 62◦, a = 33.33 5. α = 117◦, a = 35, b = 42 6. α = 117◦, a = 45, b = 42 7. α = 68.7◦, a = 88, b = 92 8. α = 42◦, a = 17, b = 23.5 9. α = 68.7◦, a = 70, b = 90 10. α = 30◦, a = 7, b = 14 11. α = 42◦, a = 39, b = 23.5 12. γ = 53◦, α = 53◦, c = 28.01 13. α = 6◦, a = 57, b = 100 14. γ = 74.6◦, c = 3, a = 3.05 15. β = 102◦, b = 16.75, c = 13 16. β = 102◦, b = 16.75, c = 18 17. β = 102◦, γ = 35◦, b = 16.75 18. β = 29.13◦, γ = 83.95◦, b = 314.15 19. γ = 120◦, β = 61◦, c = 4 20. α = 50◦, a = 25, b = 12.5 21. Find the area of the triangles given in Exercises 1, 12 and 20 above. (Another Classic Application: Grade of a Road) The grade of a road is much like the pitch of a roof (See Example 10.6.6) in that it expresses the ratio of rise/run. In the case of a road, this ratio is always positive because it is measured going uphill and it is usually given as a percentage. For example, a road which rises 7 feet for every 100
feet of (horizontal) forward progress is said to have a 7% grade. However, if we want to apply any Trigonometry to a story problem involving roads going uphill or downhill, we need to view the grade as an angle with respect to the horizontal. In Exercises 22 - 24, we ﬁrst have you change road grades into angles and then use the Law of Sines in an application. 22. Using a right triangle with a horizontal leg of length 100 and vertical leg with length 7, show that a 7% grade means that the road (hypotenuse) makes about a 4◦ angle with the horizontal. (It will not be exactly 4◦, but it’s pretty close.) 23. What grade is given by a 9.65◦ angle made by the road and the horizontal?13 13I have friends who live in Paciﬁca, CA and their road is actually this steep. It’s not a nice road to drive. 11.2 The Law of Sines 905 24. Along a long, straight stretch of mountain road with a 7% grade, you see a tall tree standing perfectly plumb alongside the road.14 From a point 500 feet downhill from the tree, the angle of inclination from the road to the top of the tree is 6◦. Use the Law of Sines to ﬁnd the height of the tree. (Hint: First show that the tree makes a 94◦ angle with the road.) (Another Classic Application: Bearings) In the next several exercises we introduce and work with the navigation tool known as bearings. Simply put, a bearing is the direction you are heading according to a compass. The classic nomenclature for bearings, however, is not given as an angle in standard position, so we must ﬁrst understand the notation. A bearing is given as an acute angle of rotation (to the east or to the west) away from the north-south (up and down) line of a compass rose. For example, N40◦E (read “40◦ east of north”) is a bearing which is rotated clockwise 40◦ from due north. If we imagine standing at the origin in the Cartesian Plane, this bearing would have us heading into Quadrant I along the terminal side of θ = 50◦. Similarly, S50◦W would point into Quadrant III along the terminal
side of θ = 220◦ because we started out pointing due south (along θ = 270◦) and rotated clockwise 50◦ back to 220◦. Counter-clockwise rotations would be found in the bearings N60◦W (which is on the terminal side of θ = 150◦) and S27◦E (which lies along the terminal side of θ = 297◦). These four bearings are drawn in the plane below. N N40◦E N60◦W 60◦ 40◦ W E 50◦ 27◦ S50◦W S27◦E S The cardinal directions north, south, east and west are usually not given as bearings in the fashion described above, but rather, one just refers to them as ‘due north’, ‘due south’, ‘due east’ and ‘due west’, respectively, and it is assumed that you know which quadrantal angle goes with each cardinal direction. (Hint: Look at the diagram above.) 25. Find the angle θ in standard position with 0◦ ≤ θ < 360◦ which corresponds to each of the bearings given below. (a) due west (b) S83◦E (c) N5.5◦E (d) due south 14The word ‘plumb’ here means that the tree is perpendicular to the horizontal. 906 Applications of Trigonometry (e) N31.25◦W (f) S72◦4112W15 (g) N45◦E (h) S45◦W 26. The Colonel spots a campﬁre at a of bearing N42◦E from his current position. Sarge, who is positioned 3000 feet due east of the Colonel, reckons the bearing to the ﬁre to be N20◦W from his current position. Determine the distance from the campﬁre to each man, rounded to the nearest foot. 27. A hiker starts walking due west from Sasquatch Point and gets to the Chupacabra Trailhead before she realizes that she hasn’t reset her pedometer. From the Chupacabra Trailhead she hikes for 5 miles along a bearing of N53◦W which brings her to the Muﬃn Ridge Observatory. From there, she knows a
bearing of S65◦E will take her straight back to Sasquatch Point. How far will she have to walk to get from the Muﬃn Ridge Observatory to Sasquach Point? What is the distance between Sasquatch Point and the Chupacabra Trailhead? 28. The captain of the SS Bigfoot sees a signal ﬂare at a bearing of N15◦E from her current location. From his position, the captain of the HMS Sasquatch ﬁnds the signal ﬂare to be at a bearing of N75◦W. If the SS Bigfoot is 5 miles from the HMS Sasquatch and the bearing from the SS Bigfoot to the HMS Sasquatch is N50◦E, ﬁnd the distances from the ﬂare to each vessel, rounded to the nearest tenth of a mile. 29. Carl spies a potential Sasquatch nest at a bearing of N10◦E and radios Jeﬀ, who is at a bearing of N50◦E from Carl’s position. From Jeﬀ’s position, the nest is at a bearing of S70◦W. If Jeﬀ and Carl are 500 feet apart, how far is Jeﬀ from the Sasquatch nest? Round your answer to the nearest foot. 30. A hiker determines the bearing to a lodge from her current position is S40◦W. She proceeds to hike 2 miles at a bearing of S20◦E at which point she determines the bearing to the lodge is S75◦W. How far is she from the lodge at this point? Round your answer to the nearest hundredth of a mile. 31. A watchtower spots a ship oﬀ shore at a bearing of N70◦E. A second tower, which is 50 miles from the ﬁrst at a bearing of S80◦E from the ﬁrst tower, determines the bearing to the ship to be N25◦W. How far is the boat from the second tower? Round your answer to the nearest tenth of a mile. 32. Skippy and Sally decide to hunt UFOs. One night, they position themselves 2 miles apart on an abandoned stretch of desert runway. An hour into their investigation, Skippy spies a UFO hovering over a spot on the runway directly between him and Sally. He records the angle of inclination from
the ground to the craft to be 75◦ and radios Sally immediately to ﬁnd the angle of inclination from her position to the craft is 50◦. How high oﬀ the ground is the UFO at this point? Round your answer to the nearest foot. (Recall: 1 mile is 5280 feet.) 15See Example 10.1.1 in Section 10.1 for a review of the DMS system. 11.2 The Law of Sines 907 33. The angle of depression from an observer in an apartment complex to a gargoyle on the building next door is 55◦. From a point ﬁve stories below the original observer, the angle of inclination to the gargoyle is 20◦. Find the distance from each observer to the gargoyle and the distance from the gargoyle to the apartment complex. Round your answers to the nearest foot. (Use the rule of thumb that one story of a building is 9 feet.) 34. Prove that the Law of Sines holds when ABC is a right triangle. 35. Discuss with your classmates why knowing only the three angles of a triangle is not enough to determine any of the sides. 36. Discuss with your classmates why the Law of Sines cannot be used to ﬁnd the angles in the triangle when only the three sides are given. Also discuss what happens if only two sides and the angle between them are given. (Said another way, explain why the Law of Sines cannot be used in the SSS and SAS cases.) 37. Given α = 30◦ and b = 10, choose four diﬀerent values for a so that (a) the information yields no triangle (b) the information yields exactly one right triangle (c) the information yields two distinct triangles (d) the information yields exactly one obtuse triangle Explain why you cannot choose a in such a way as to have α = 30◦, b = 10 and your choice of a yield only one triangle where that unique triangle has three acute angles. 38. Use the cases and diagrams in the proof of the Law of Sines (Theorem 11.2) to prove the area formulas given in Theorem 11.4. Why do those formulas yield square units when four quantities are being multiplied together? 908 Applications of Trigonometry 11.2.2 Answers 1. 3. 5. 7. α = 13◦ β = 17◦ a = 5 γ = 150◦ b �
� 6.50 c ≈ 11.11 Information does not produce a triangle Information does not produce a triangle α = 68.7◦ β ≈ 76.9◦ γ ≈ 34.4◦ c ≈ 53.36 a = 88 α = 68.7◦ β ≈ 103.1◦ γ ≈ 8.2◦ c ≈ 13.47 a = 88 b = 92 b = 92 9. Information does not produce a triangle 11. 13. 15. 17. 19. α = 42◦ β ≈ 23.78◦ γ ≈ 114.22◦ a = 39 c ≈ 53.15 b = 23.5 α = 6◦ β ≈ 169.43◦ γ ≈ 4.57◦ c ≈ 43.45 a = 57 b = 100 α = 6◦ β ≈ 10.57◦ γ ≈ 163.43◦ c ≈ 155.51 a = 57 b = 100 α ≈ 28.61◦ β = 102◦ a ≈ 8.20 b = 16.75 c = 13 γ ≈ 49.39◦ α = 43◦ γ = 35◦ β = 102◦ a ≈ 11.68 b = 16.75 c ≈ 9.82 Information does not produce a triangle 2. 4. 6. 8. 10. 12. 14. 16. 18. 20. α = 73.2◦ β = 54.1◦ γ = 52.7◦ c ≈ 97.22 b ≈ 99.00 a = 117 β = 62◦ α = 95◦ a = 33.33 b ≈ 29.54 c ≈ 13.07 γ = 23◦ α = 117◦ β ≈ 56.3◦ γ ≈ 6.7◦ c ≈ 5.89 b = 42 a = 45 b = 23.5 α = 42◦ β ≈ 67.66◦ γ ≈ 70.34◦ c ≈ 23.93 a = 17 α = 42◦ β ≈ 112.34◦ γ ≈ 25.66◦ c ≈ 11.00 a = 17 b = 23.5 α = 30◦ β = 90◦ γ = 60◦ √ 3 b = 14 a =
7 c = 7 α = 53◦ β = 74◦ a = 28.01 b ≈ 33.71 c = 28.01 γ = 53◦ α ≈ 78.59◦ β ≈ 26.81◦ γ = 74.6◦ b ≈ 1.40 a = 3.05 α ≈ 101.41◦ β ≈ 3.99◦ γ = 74.6◦ b ≈ 0.217 a = 3.05 c = 3 c = 3 Information does not produce a triangle α = 66.92◦ β = 29.13◦ γ = 83.95◦ c ≈ 641.75 a ≈ 593.69 b = 314.15 α = 50◦ β ≈ 22.52◦ γ ≈ 107.48◦ a = 25 c ≈ 31.13 b = 12.5 21. The area of the triangle from Exercise 1 is about 8.1 square units. The area of the triangle from Exercise 12 is about 377.1 square units. The area of the triangle from Exercise 20 is about 149 square units. ≈ 0.699 radians, which is equivalent to 4.004◦ 22. arctan 7 100 23. About 17% 24. About 53 feet 11.2 The Law of Sines 909 25. (a) θ = 180◦ (b) θ = 353◦ (c) θ = 84.5◦ (d) θ = 270◦ (e) θ = 121.25◦ (f) θ = 197◦1848 (g) θ = 45◦ (h) θ = 225◦ 26. The Colonel is about 3193 feet from the campﬁre. Sarge is about 2525 feet to the campﬁre. 27. The distance from the Muﬃn Ridge Observatory to Sasquach Point is about 7.12 miles. The distance from Sasquatch Point to the Chupacabra Trailhead is about 2.46 miles. 28. The SS Bigfoot is about 4.1 miles from the ﬂare. The HMS Sasquatch is about 2.9 miles from the ﬂare. 29. Jeﬀ is about 371 feet from the nest. 30. She is about 3.02 miles from the
lodge 31. The boat is about 25.1 miles from the second tower. 32. The UFO is hovering about 9539 feet above the ground. 33. The gargoyle is about 44 feet from the observer on the upper ﬂoor. The gargoyle is about 27 feet from the observer on the lower ﬂoor. The gargoyle is about 25 feet from the other building. 910 Applications of Trigonometry 11.3 The Law of Cosines In Section 11.2, we developed the Law of Sines (Theorem 11.2) to enable us to solve triangles in the ‘Angle-Angle-Side’ (AAS), the ‘Angle-Side-Angle’ (ASA) and the ambiguous ‘Angle-Side-Side’ (ASS) cases. In this section, we develop the Law of Cosines which handles solving triangles in the ‘Side-Angle-Side’ (SAS) and ‘Side-Side-Side’ (SSS) cases.1 We state and prove the theorem below. Theorem 11.5. Law of Cosines: Given a triangle with angle-side opposite pairs (α, a), (β, b) and (γ, c), the following equations hold a2 = b2 + c2 − 2bc cos(α) b2 = a2 + c2 − 2ac cos(β) c2 = a2 + b2 − 2ab cos(γ) or, solving for the cosine in each equation, we have cos(α) = b2 + c2 − a2 2bc cos(β) = a2 + c2 − b2 2ac cos(γ) = a2 + b2 − c2 2ab To prove the theorem, we consider a generic triangle with the vertex of angle α at the origin with side b positioned along the positive x-axis. B = (c cos(α), c sin(α)) c α a A = (0, 0) b C = (b, 0) From this set-up, we immediately ﬁnd that the coordinates of A and C are A(0, 0) and C(b, 0). From Theorem 10.3, we know that since the point B(x, y) lies on a circle of radius c, the coordinates 1Here, ‘Side-Angle-Side’ means that we are
given two sides and the ‘included’ angle - that is, the given angle is adjacent to both of the given sides. 11.3 The Law of Cosines 911 of B are B(x, y) = B(c cos(α), c sin(α)). (This would be true even if α were an obtuse or right angle so although we have drawn the case when α is acute, the following computations hold for any angle α drawn in standard position where 0 < α < 180◦.) We note that the distance between the points B and C is none other than the length of side a. Using the distance formula, Equation 1.1, we get a = (c cos(α) − b)2 + (c sin(α) − 0)2 2 (c cos(α) − b)2 + c2 sin2(α) a2 = a2 = (c cos(α) − b)2 + c2 sin2(α) a2 = c2 cos2(α) − 2bc cos(α) + b2 + c2 sin2(α) a2 = c2 cos2(α) + sin2(α) + b2 − 2bc cos(α) a2 = c2(1) + b2 − 2bc cos(α) a2 = c2 + b2 − 2bc cos(α) Since cos2(α) + sin2(α) = 1 The remaining formulas given in Theorem 11.5 can be shown by simply reorienting the triangle to place a diﬀerent vertex at the origin. We leave these details to the reader. What’s important about a and α in the above proof is that (α, a) is an angle-side opposite pair and b and c are the sides adjacent to α – the same can be said of any other angle-side opposite pair in the triangle. Notice that the proof of the Law of Cosines relies on the distance formula which has its roots in the Pythagorean Theorem. That being said, the Law of Cosines can be thought of as a generalization of the Pythagorean Theorem. If we have a triangle in which γ = 90◦, then cos(γ) = cos (90◦) = 0 so we get the familiar relationship c2 = a2 + b2. What this means is that in the larger mathematical sense
, the Law of Cosines and the Pythagorean Theorem amount to pretty much the same thing.2 Example 11.3.1. Solve the following triangles. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle. 1. β = 50◦, a = 7 units, c = 2 units 2. a = 4 units, b = 7 units, c = 5 units Solution. 1. We are given the lengths of two sides, a = 7 and c = 2, and the measure of the included angle, β = 50◦. With no angle-side opposite pair to use, we apply the Law of Cosines. We get b2 = 72 + 22 − 2(7)(2) cos (50◦) which yields b = In order to determine the measures of the remaining angles α and γ, we are forced to used the derived value for b. There are two ways to proceed at this point. We could use the Law of Cosines again, or, since we have the angle-side opposite pair (β, b) we could use the Law of Sines. The advantage to using the Law of Cosines over the Law of Sines in cases like this is that unlike the sine function, the cosine function distinguishes between acute and obtuse angles. The cosine of an acute is positive, whereas the cosine of an obtuse angle is negative. Since the sine of both acute and obtuse angles are positive, the sine of an angle alone is not 53 − 28 cos (50◦) ≈ 5.92 units. 2This shouldn’t come as too much of a shock. All of the theorems in Trigonometry can ultimately be traced back to the deﬁnition of the circular functions along with the distance formula and hence, the Pythagorean Theorem. 912 Applications of Trigonometry enough to determine if the angle in question is acute or obtuse. Since both authors of the textbook prefer the Law of Cosines, we proceed with this method ﬁrst. When using the Law of Cosines, it’s always best to ﬁnd the measure of the largest unknown angle ﬁrst, since this will give us the obtuse angle of the triangle if there is one. Since the largest angle is opposite the longest side, we choose to ﬁnd α ﬁ
rst. To that end, we use the formula cos(α) = b2+c2−a2 53 − 28 cos (50◦) and c = 2. We get3 and substitute a = 7, b = 2bc cos(α) = 2 − 7 cos (50◦) 53 − 28 cos (50◦) Since α is an angle in a triangle, we know the radian measure of α must lie between 0 and π radians. This matches the range of the arccosine function, so we have α = arccos 2 − 7 cos (50◦) 53 − 28 cos (50◦) radians ≈ 114.99◦ At this point, we could ﬁnd γ using γ = 180◦ − α − β ≈ 180◦ − 114.99◦ − 50◦ = 15.01◦, that is if we trust our approximation for α. To minimize propagation of error, however, we could use the Law of Cosines again,4 in this case using cos(γ) = a2+b2−c2. Plugging in a = 7, b = 53 − 28 cos (50◦) and c = 2, we get γ = arccos sketch the triangle below. √ 7−2 cos(50◦) 53−28 cos(50◦) 2ab radians ≈ 15.01◦. We β = 50◦ a = 7 c = 2 α ≈ 114.99◦ γ ≈ 15.01◦ b ≈ 5.92 As we mentioned earlier, once we’ve determined b it is possible to use the Law of Sines to ﬁnd the remaining angles. Here, however, we must proceed with caution as we are in the ambiguous (ASS) case. It is advisable to ﬁrst ﬁnd the smallest of the unknown angles, since we are guaranteed it will be acute.5 In this case, we would ﬁnd γ since the side opposite γ is smaller than the side opposite the other unknown angle, α. Using the angle-side opposite pair (β, b), we get sin(γ). The usual calculations produces γ ≈ 15.01◦ and α = 180◦ − β − γ ≈ 180◦ − 50◦ − 15.01
◦ = 114.99◦. sin(50◦) 53−28 cos(50◦) 2 = √ 2. Since all three sides and no angles are given, we are forced to use the Law of Cosines. Following our discussion in the previous problem, we ﬁnd β ﬁrst, since it is opposite the longest side, radians ≈ 101.54◦. As in b. We get cos(β) = a2+c2−b2 5, so we get β = arccos − 1 = − 1 2ac 5 3after simplifying... 4Your instructor will let you know which procedure to use. It all boils down to how much you trust your calculator. 5There can only be one obtuse angle in the triangle, and if there is one, it must be the largest. 11.3 The Law of Cosines 913 the previous problem, now that we have obtained an angle-side opposite pair (β, b), we could proceed using the Law of Sines. The Law of Cosines, however, oﬀers us a rare opportunity to ﬁnd the remaining angles using only the data given to us in the statement of the problem. Using this, we get γ = arccos 5 7 radians ≈ 44.42◦ and α = arccos 29 35 radians ≈ 34.05◦. β ≈ 101.54◦ c = 5 a = 4 α ≈ 34.05◦ γ ≈ 44.42◦ b = 7 We note that, depending on how many decimal places are carried through successive calculations, and depending on which approach is used to solve the problem, the approximate answers you obtain may diﬀer slightly from those the authors obtain in the Examples and the Exercises. A great example of this is number 2 in Example 11.3.1, where the approximate values we record for the measures of the angles sum to 180.01◦, which is geometrically impossible. Next, we have an application of the Law of Cosines. Example 11.3.2. A researcher wishes to determine the width of a vernal pond as drawn below. From a point P, he ﬁnds the distance to the eastern-most point of the pond to be 950 feet, while the distance to the western-most point of the pond from
P is 1000 feet. If the angle between the two lines of sight is 60◦, ﬁnd the width of the pond. 1000 feet 950 feet 60◦ P Solution. We are given the lengths of two sides and the measure of an included angle, so we may apply the Law of Cosines to ﬁnd the length of the missing side opposite the given angle. Calling this length w (for width), we get w2 = 9502 + 10002 − 2(950)(1000) cos (60◦) = 952500 from which we get w = 952500 ≈ 976 feet. √ 914 Applications of Trigonometry In Section 11.2, we used the proof of the Law of Sines to develop Theorem 11.4 as an alternate formula for the area enclosed by a triangle. In this section, we use the Law of Cosines to derive another such formula - Heron’s Formula. Theorem 11.6. Heron’s Formula: Suppose a, b and c denote the lengths of the three sides of a triangle. Let s be the semiperimeter of the triangle, that is, let s = 1 2 (a + b + c). Then the area A enclosed by the triangle is given by A = s(s − a)(s − b)(s − c) We prove Theorem 11.6 using Theorem 11.4. Using the convention that the angle γ is opposite the side c, we have A = 1 2 ab sin(γ) from Theorem 11.4. In order to simplify computations, we start by manipulating the expression for A2. A2 = 2 ab sin(γ) 1 2 a2b2 sin2(γ) = = 1 4 a2b2 4 1 − cos2(γ) since sin2(γ) = 1 − cos2(γ). The Law of Cosines tells us cos(γ) = a2+b2−c2 2ab, so substituting this into our equation for A2 gives A2 = = = = = = = = 1 − 1 − 2 1 − cos2(γ) a2 + b2 − c2 2ab a2 + b2 − c22 4a2b2 4a2b2 − a2 + b2 − c22 4a2b2 a2b2 4 a2b2 4 a2b2 4 a2b
2 4 4a2b2 − a2 + b2 − c22 16 (2ab)2 − a2 + b2 − c22 16 2ab − a2 + b2 − c2 2ab + a2 + b2 − c2 16 c2 − a2 + 2ab − b2 a2 + 2ab + b2 − c2 16 diﬀerence of squares. 11.3 The Law of Cosines 915 A2 = = = = = c2 − a2 − 2ab + b2 a2 + 2ab + b2 − c2 16 c2 − (a − b)2 (a + b)2 − c2 16 (c − (a − b))(c + (a − b))((a + b) − c)((a + b) + c) 16 perfect square trinomials. diﬀerence of squares. (b + c − a)(a + c − b)(a + b − c)(a + b + c) 16 (a + c − b) 2 (b + c − a) 2 (a + b − c) 2 · · · (a + b + c) 2 At this stage, we recognize the last factor as the semiperimeter, s = 1 complete the proof, we note that 2 (a + b + c) = a+b+c 2. To (s − a − 2a 2 = b + c − a 2 Similarly, we ﬁnd (s − b) = a+c−b 2 and (s − c) = a+b−c 2. Hence, we get A2 = (b + c − a) 2 · (a + c − b) 2 · (a + b − c) 2 · (a + b + c) 2 = (s − a)(s − b)(s − c)s so that A = s(s − a)(s − b)(s − c) as required. We close with an example of Heron’s Formula. Example 11.3.3. Find the area enclosed of the triangle in Example 11.3.1 number 2. Solution. We are given a = 4, b = 7 and c = 5. Using these values, we ﬁnd s = 1 2 (4 + 7 + 5) = 8, (s − a) = 8 − 4 = 4,
(s − b) = 8 − 7 = 1 and (s − c) = 8 − 5 = 3. Using Heron’s Formula, we get √ A = s(s − a)(s − b)(s − c) = 6 ≈ 9.80 square units. (8)(4)(1)(3) = 96 = 4 √ 916 Applications of Trigonometry 11.3.1 Exercises In Exercises 1 - 10, use the Law of Cosines to ﬁnd the remaining side(s) and angle(s) if possible. 1. a = 7, b = 12, γ = 59.3◦ 2. α = 104◦, b = 25, c = 37 3. a = 153, β = 8.2◦, c = 153 4. a = 3, b = 4, γ = 90◦ 5. α = 120◦, b = 3, c = 4 6. a = 7, b = 10, c = 13 7. a = 1, b = 2, c = 5 8. a = 300, b = 302, c = 48 9. a = 5, b = 5, c = 5 10. a = 5, b = 12, ; c = 13 In Exercises 11 - 16, solve for the remaining side(s) and angle(s), if possible, using any appropriate technique. 11. a = 18, α = 63◦, b = 20 12. a = 37, b = 45, c = 26 13. a = 16, α = 63◦, b = 20 14. a = 22, α = 63◦, b = 20 15. α = 42◦, b = 117, c = 88 16. β = 7◦, γ = 170◦, c = 98.6 17. Find the area of the triangles given in Exercises 6, 8 and 10 above. 18. The hour hand on my antique Seth Thomas schoolhouse clock in 4 inches long and the minute hand is 5.5 inches long. Find the distance between the ends of the hands when the clock reads four o’clock. Round your answer to the nearest hundredth of an inch. 19. A geologist wants to measure the diameter of a crater. From her camp, it is 4 miles to the northern-most point of the crater and 2 miles to the southern-
most point. If the angle between the two lines of sight is 117◦, what is the diameter of the crater? Round your answer to the nearest hundredth of a mile. 20. From the Pedimaxus International Airport a tour helicopter can ﬂy to Cliﬀs of Insanity Point by following a bearing of N8.2◦E for 192 miles and it can ﬂy to Bigfoot Falls by following a bearing of S68.5◦E for 207 miles.6 Find the distance between Cliﬀs of Insanity Point and Bigfoot Falls. Round your answer to the nearest mile. 21. Cliﬀs of Insanity Point and Bigfoot Falls from Exericse 20 above both lie on a straight stretch of the Great Sasquatch Canyon. What bearing would the tour helicopter need to follow to go directly from Bigfoot Falls to Cliﬀs of Insanity Point? Round your angle to the nearest tenth of a degree. 6Please refer to Page 905 in Section 11.2 for an introduction to bearings. 11.3 The Law of Cosines 917 22. A naturalist sets oﬀ on a hike from a lodge on a bearing of S80◦W. After 1.5 miles, she changes her bearing to S17◦W and continues hiking for 3 miles. Find her distance from the lodge at this point. Round your answer to the nearest hundredth of a mile. What bearing should she follow to return to the lodge? Round your angle to the nearest degree. 23. The HMS Sasquatch leaves port on a bearing of N23◦E and travels for 5 miles. It then changes course and follows a heading of S41◦E for 2 miles. How far is it from port? Round your answer to the nearest hundredth of a mile. What is its bearing to port? Round your angle to the nearest degree. 24. The SS Bigfoot leaves a harbor bound for Nessie Island which is 300 miles away at a bearing of N32◦E. A storm moves in and after 100 miles, the captain of the Bigfoot ﬁnds he has drifted oﬀ course. If his bearing to the harbor is now S70◦W, how far is the SS Bigfoot from Nessie Island? Round your answer to the nearest hundredth of a mile. What course should the captain set to head to the island? Round your angle to the
nearest tenth of a degree. 25. From a point 300 feet above level ground in a ﬁretower, a ranger spots two ﬁres in the Yeti National Forest. The angle of depression7 made by the line of sight from the ranger to the ﬁrst ﬁre is 2.5◦ and the angle of depression made by line of sight from the ranger to the second ﬁre is 1.3◦. The angle formed by the two lines of sight is 117◦. Find the distance between the two ﬁres. Round your answer to the nearest foot. (Hint: In order to use the 117◦ angle between the lines of sight, you will ﬁrst need to use right angle Trigonometry to ﬁnd the lengths of the lines of sight. This will give you a Side-Angle-Side case in which to apply the Law of Cosines.) ﬁre 117◦ ﬁre ﬁretower 26. If you apply the Law of Cosines to the ambiguous Angle-Side-Side (ASS) case, the result is a quadratic equation whose variable is that of the missing side. If the equation has no positive real zeros then the information given does not yield a triangle. If the equation has only one positive real zero then exactly one triangle is formed and if the equation has two distinct positive real zeros then two distinct triangles are formed. Apply the Law of Cosines to Exercises 11, 13 and 14 above in order to demonstrate this result. 27. Discuss with your classmates why Heron’s Formula yields an area in square units even though four lengths are being multiplied together. 7See Exercise 78 in Section 10.3 for the deﬁnition of this angle. 918 Applications of Trigonometry 11.3.2 Answers 1. 3. 5. 7. 9. 11. 13. 15. α ≈ 35.54◦ β ≈ 85.16◦ γ = 59.3◦ c ≈ 10.36 a = 7 b = 12 α ≈ 85.90◦ β = 8.2◦ a = 153 b ≈ 21.88 c = 153 γ ≈ 85.90◦ α = 120◦ β ≈ 25.28◦ γ ≈ 34.72◦ a = c = 4 b = 3 37
√ Information does not produce a triangle α = 60◦ β = 60◦ γ = 60 = 20 α = 63◦ β ≈ 98.11◦ γ ≈ 18.89◦ a = 18 α = 63◦ β ≈ 81.89◦ γ ≈ 35.11◦ c ≈ 11.62 a = 18 c ≈ 6.54 b = 20 Information does not produce a triangle α = 42◦ a ≈ 78.30 b = 117 β ≈ 89.23◦ γ ≈ 48.77◦ c = 88 α = 104◦ a ≈ 49.41 b = 25 β ≈ 29.40◦ γ ≈ 46.60◦ c = 37 α ≈ 36.87◦ β ≈ 53.13◦ γ = 90 ≈ 32.31◦ β ≈ 49.58◦ γ ≈ 98.21◦ a = 7 c = 13 b = 10 α ≈ 83.05◦ β ≈ 87.81◦ γ ≈ 9.14◦ b = 302 a = 300 c = 48 α ≈ 22.62◦ β ≈ 67.38◦ γ = 90◦ c = 13 b = 12 a = 5 α ≈ 55.30◦ β ≈ 89.40◦ γ ≈ 35.30◦ a = 37 c = 26 b = 45 α = 63◦ β ≈ 54.1◦ γ ≈ 62.9◦ c ≈ 21.98 a = 22 b = 20 α ≈ 3◦ γ = 170◦ β = 7◦ a ≈ 29.72 b ≈ 69.2 c = 98.6 √ 2. 4. 6. 8. 10. 12. 14. 16. √ √ 17. The area of the triangle given in Exercise 6 is The area of the triangle given in Exercise 8 is The area of the triangle given in Exercise 10 is exactly 30 square units. 1200 = 20 51764375 ≈ 7194.75 square units. 3 ≈ 34.64 square units. 18. The distance between the ends of the hands at four o’clock is about 8.26 inches. 19. The diameter of the crater is about 5.22 miles. 20. About 313 miles 21.
N31.8◦W 22. She is about 3.92 miles from the lodge and her bearing to the lodge is N37◦E. 23. It is about 4.50 miles from port and its heading to port is S47◦W. 24. It is about 229.61 miles from the island and the captain should set a course of N16.4◦E to reach the island. 25. The ﬁres are about 17456 feet apart. (Try to avoid rounding errors.) 11.4 Polar Coordinates 919 11.4 Polar Coordinates In Section 1.1, we introduced the Cartesian coordinates of a point in the plane as a means of assigning ordered pairs of numbers to points in the plane. We deﬁned the Cartesian coordinate plane using two number lines – one horizontal and one vertical – which intersect at right angles at a point we called the ‘origin’. To plot a point, say P (−3, 4), we start at the origin, travel horizontally to the left 3 units, then up 4 units. Alternatively, we could start at the origin, travel up 4 units, then to the left 3 units and arrive at the same location. For the most part, the ‘motions’ of the Cartesian system (over and up) describe a rectangle, and most points can be thought of as the corner diagonally across the rectangle from the origin.1 For this reason, the Cartesian coordinates of a point are often called ‘rectangular’ coordinates. In this section, we introduce a new system for assigning coordinates to points in the plane – polar coordinates. We start with an origin point, called the pole, and a ray called the polar axis. We then locate a point P using two coordinates, (r, θ), where r represents a directed distance from the pole2 and θ is a measure of rotation from the polar axis. Roughly speaking, the polar coordinates (r, θ) of a point measure ‘how far out’ the point is from the pole (that’s r), and ‘how far to rotate’ from the polar axis, (that’s θ). y P (−3, 4) P (r, θ) 3 2 1 r θ −4 −3 −2 −1 −1 1 2 3 4 x Pole r Polar Axis −2 −3 −4 For example, if we wished to
plot the point P with polar coordinates 4, 5π 6 move out along the polar axis 4 units, then rotate 5π, we’d start at the pole, 6 radians counter-clockwise. P 4, 5π 6 r = 4 Pole θ = 5π 6 Pole Pole We may also visualize this process by thinking of the rotation ﬁrst.3 To plot P 4, 5π 6 we rotate 5π this way, 6 counter-clockwise from the polar axis, then move outwards from the pole 4 units. 1Excluding, of course, the points in which one or both coordinates are 0. 2We will explain more about this momentarily. 3As with anything in Mathematics, the more ways you have to look at something, the better. The authors encourage the reader to take time to think about both approaches to plotting points given in polar coordinates. 920 Applications of Trigonometry Essentially we are locating a point on the terminal side of 5π 6 which is 4 units away from the pole. θ = 5π 6 Pole θ = 5π 6 Pole P 4, 5π 6 Pole If r < 0, we begin by moving in the opposite direction on the polar axis from the pole. For example, to plot Q −3.5, π 4 we have r = −3.5 Pole θ = π 4 Pole Pole Q −3.5, π 4 If we interpret the angle ﬁrst, we rotate π Here we are locating a point 3.5 units away from the pole on the terminal side of 5π 4 radians, then move back through the pole 3.5 units. 4, not π 4. θ = π 4 Pole θ = π 4 Pole Pole Q −3.5, π 4 As you may have guessed, θ < 0 means the rotation away from the polar axis is clockwise instead of counter-clockwise. Hence, to plot R 3.5, − 3π 4 we have the following. r = 3.5 Pole Pole θ = − 3π 4 Pole R 3.5, − 3π 4 From an ‘angles ﬁrst’ approach, we rotate − 3π R is the point on the terminal side of θ = − 3π 4 then move out 3.5 units from the pole. We see that 4 which is 3.5 units from the pole. Pole θ = −
3π 4 Pole θ = − 3π 4 Pole R 3.5, − 3π 4 11.4 Polar Coordinates 921 The points Q and R above are, in fact, the same point despite the fact that their polar coordinate representations are diﬀerent. Unlike Cartesian coordinates where (a, b) and (c, d) represent the same point if and only if a = c and b = d, a point can be represented by inﬁnitely many polar coordinate pairs. We explore this notion more in the following example. Example 11.4.1. For each point in polar coordinates given below plot the point and then give two additional expressions for the point, one of which has r > 0 and the other with r < 0. 4. P −3, − π 4 3. P 117, − 5π 2 2. P −4, 7π 6 1. P (2, 240◦) Solution. 1. Whether we move 2 units along the polar axis and then rotate 240◦ or rotate 240◦ then move out 2 units from the pole, we plot P (2, 240◦) below. θ = 240◦ Pole Pole P (2, 240◦) We now set about ﬁnding alternate descriptions (r, θ) for the point P. Since P is 2 units from the pole, r = ±2. Next, we choose angles θ for each of the r values. The given representation for P is (2, 240◦) so the angle θ we choose for the r = 2 case must be coterminal with 240◦. (Can you see why?) One such angle is θ = −120◦ so one answer for this case is (2, −120◦). For the case r = −2, we visualize our rotation starting 2 units to the left of the pole. From this position, we need only to rotate θ = 60◦ to arrive at location coterminal with 240◦. Hence, our answer here is (−2, 60◦). We check our answers by plotting them. Pole θ = −120◦ θ = 60◦ Pole P (2, −120◦) P (−2, 60◦) 2. We plot −4, 7π 6 6 radians. Since r = −4 < 0, we ﬁnd our point lies
4 units from the pole on the terminal side of π 6. by ﬁrst moving 4 units to the left of the pole and then rotating 7π P −4, 7π 6 Pole Pole θ = 7π 6 922 Applications of Trigonometry To ﬁnd alternate descriptions for P, we note that the distance from P to the pole is 4 units, so any representation (r, θ) for P must have r = ±4. As we noted above, P lies on the terminal as one of our answers. To ﬁnd a diﬀerent side of π representation for P with r = −4, we may choose any angle coterminal with the angle in the as our second answer. original representation of P −4, 7π 6 6, so this, coupled with r = 4, gives us 4, π. We pick − 5π 6 and get −4, − 5π 6 6 P 4, π 6 θ = π 6 Pole θ = − 5π 6 Pole P −4, − 5π 6, we move along the polar axis 117 units from the pole and rotate 3. To plot P 117, − 5π 2 clockwise 5π 2 radians as illustrated below. Pole θ = − 5π 2 Pole P 117, − 5π 2 Since P is 117 units from the pole, any representation (r, θ) for P satisﬁes r = ±117. For the r = 117 case, we can take θ to be any angle coterminal with − 5π 2. In this case, we choose as one answer. For the r = −117 case, we visualize moving left 117 θ = 3π units from the pole and then rotating through an angle θ to reach P. We ﬁnd that θ = π 2 satisﬁes this requirement, so our second answer is −117, π 2 2, and get 117, 3π. 2 Pole θ = 3π 2 Pole θ = π 2 P 117, 3π 2 P −117, π 2 11.4 Polar Coordinates 923 4. We move three units to the left of the pole and follow up with a clockwise rotation of π 4 radians to plot P −3, − π 4. We see that P lies on the terminal side of 3π 4. P
−3, − π 4 θ = − π 4 Pole Pole. To Since P lies on the terminal side of 3π ﬁnd a diﬀerent representation for P with r = −3, we may choose any angle coterminal with − π 4, one alternative representation for P is 3, 3π 4 for our ﬁnal answer −3, 7π 4. We choose θ = 7π. 4 4 P 3, 3π 4 P −3, 7π 4 θ = 3π 4 Pole θ = 7π 4 Pole Now that we have had some practice with plotting points in polar coordinates, it should come as no surprise that any given point expressed in polar coordinates has inﬁnitely many other representations in polar coordinates. The following result characterizes when two sets of polar coordinates determine the same point in the plane. It could be considered as a deﬁnition or a theorem, depending on your point of view. We state it as a property of the polar coordinate system. Equivalent Representations of Points in Polar Coordinates Suppose (r, θ) and (r, θ) are polar coordinates where r = 0, r = 0 and the angles are measured in radians. Then (r, θ) and (r, θ) determine the same point P if and only if one of the following is true: r = r and θ = θ + 2πk for some integer k r = −r and θ = θ + (2k + 1)π for some integer k All polar coordinates of the form (0, θ) represent the pole regardless of the value of θ. The key to understanding this result, and indeed the whole polar coordinate system, is to keep in mind that (r, θ) means (directed distance from pole, angle of rotation). If r = 0, then no matter how much rotation is performed, the point never leaves the pole. Thus (0, θ) is the pole for all 924 Applications of Trigonometry values of θ. Now let’s assume that neither r nor r is zero. If (r, θ) and (r, θ) determine the same point P then the (non-zero) distance from P to the pole in each case must be the same. Since this distance is controlled by the ﬁrst coordinate, we have that
either r = r or r = −r. If r = r, then when plotting (r, θ) and (r, θ), the angles θ and θ have the same initial side. Hence, if (r, θ) and (r, θ) determine the same point, we must have that θ is coterminal with θ. We know that this means θ = θ + 2πk for some integer k, as required. If, on the other hand, r = −r, then when plotting (r, θ) and (r, θ), the initial side of θ is rotated π radians away from the initial side of θ. In this case, θ must be coterminal with π + θ. Hence, θ = π + θ + 2πk which we rewrite as θ = θ + (2k + 1)π for some integer k. Conversely, if r = r and θ = θ + 2πk for some integer k, then the points P (r, θ) and P (r, θ) lie the same (directed) distance from the pole on the terminal sides of coterminal angles, and hence are the same point. Now suppose r = −r and θ = θ + (2k + 1)π for some integer k. To plot P, we ﬁrst move a directed distance r from the pole; to plot P, our ﬁrst step is to move the same distance from the pole as P, but in the opposite direction. At this intermediate stage, we have two points equidistant from the pole rotated exactly π radians apart. Since θ = θ + (2k + 1)π = (θ + π) + 2πk for some integer k, we see that θ is coterminal to (θ + π) and it is this extra π radians of rotation which aligns the points P and P. Next, we marry the polar coordinate system with the Cartesian (rectangular) coordinate system. To do so, we identify the pole and polar axis in the polar system to the origin and positive x-axis, respectively, in the rectangular system. We get the following result. Theorem 11.7. Conversion Between Rectangular and Polar Coordinates: Suppose P
is represented in rectangular coordinates as (x, y) and in polar coordinates as (r, θ). Then x = r cos(θ) and y = r sin(θ) x2 + y2 = r2 and tan(θ) = y x (provided x = 0) In the case r > 0, Theorem 11.7 is an immediate consequence of Theorem 10.3 along with the quotient identity tan(θ) = sin(θ) If r < 0, then we know an alternate representation for (r, θ) cos(θ). is (−r, θ + π). Since cos(θ + π) = − cos(θ) and sin(θ + π) = − sin(θ), applying the theorem to (−r, θ + π) gives x = (−r) cos(θ + π) = (−r)(− cos(θ)) = r cos(θ) and y = (−r) sin(θ + π) = (−r)(− sin(θ)) = r sin(θ). Moreover, x2 + y2 = (−r)2 = r2, and y x = tan(θ + π) = tan(θ), so the theorem is true in this case, too. The remaining case is r = 0, in which case (r, θ) = (0, θ) is the pole. Since the pole is identiﬁed with the origin (0, 0) in rectangular coordinates, the theorem in this case amounts to checking ‘0 = 0.’ The following example puts Theorem 11.7 to good use. Example 11.4.2. Convert each point in rectangular coordinates given below into polar coordinates with r ≥ 0 and 0 ≤ θ < 2π. Use exact values if possible and round any approximate values to two decimal places. Check your answer by converting them back to rectangular coordinates. 1. P 2, −2 √ 3 2. Q(−3, −3) 3. R(0, −3) 4. S(−3, 4) 11.4 Polar Coordinates 925 Solution. 1. Even though we are not explicitly told to do so, we can avoid many common mistakes by taking 3 shows that the time to plot the points before we do any calculations. Plotting P
2, −2 √ it lies in Quadrant IV. With x = 2 and y = −2 = 4 + 12 = 16 so r = ±4. Since we are asked for r ≥ 0, we choose r = 4. To ﬁnd θ, we have that tan(θ) = y 3, and since P lies in Quadrant IV, we know θ is a Quadrant IV angle. We are asked to have 0 ≤ θ < 2π,. To check, we convert (r, θ) = 4, 5π 3. Hence, our answer is 4, 5π so we choose θ = 5π 3 back to rectangular coordinates and we ﬁnd x = r cos(θ) = 4 cos 5π = 2 and 3 y = r sin(θ) = 4 sin 5π 3 3. This tells us θ has a reference angle of π 3, we get r2 = x2 + y2 = (2)2 + −, as required. x = −2 = 4 = −2 32 √ √ 18 = ±3 2. The point Q(−3, −3) lies in Quadrant III. Using x = y = −3, we get r2 = (−3)2 + (−3)2 = 18 2. We ﬁnd 4. Since Q lies in Quadrant III, 4, which satisﬁes the requirement that 0 ≤ θ < 2π. Our ﬁnal answer is. To check, we ﬁnd x = r cos(θ) = (3 = −3 so r = ± tan(θ) = −3 we choose θ = 5π (r, θ) = 3 −3 = 1, which means θ has a reference angle of π 2. Since we are asked for r ≥ 0, we choose r = 3 = (3 2) √ √ √ √ − √ 2) cos 5π 4 2, 5π 4 2 2 and y = r sin(θ) = (3 √ 2) sin 5π 4 √ = (3 − √ 2 2 2) = −3, so we are done. y y θ = 5π 3 x θ = 5π 4 x P Q P has rectangular coordinates (2, −2 P has polar coordinates 4, 5π 3 �
� 3) Q has rectangular coordinates (−3, −3) Q has polar coordinates 3 √ 2, 5π 4 3. The point R(0, −3) lies along the negative y-axis. While we could go through the usual computations4 to ﬁnd the polar form of R, in this case we can ﬁnd the polar coordinates of R using the deﬁnition. Since the pole is identiﬁed with the origin, we can easily tell the point R is 3 units from the pole, which means in the polar representation (r, θ) of R we know r = ±3. Since we require r ≥ 0, we choose r = 3. Concerning θ, the angle θ = 3π 2 satisﬁes 0 ≤ θ < 2π 4Since x = 0, we would have to determine θ geometrically. 926 Applications of Trigonometry with its terminal side along the negative y-axis, so our answer is 3, 3π 2 x = r cos(θ) = 3 cos 3π 2 = (3)(0) = 0 and y = r sin(θ) = 3 sin 3π 2 = 3(−1) = −3.. To check, we note −3 = − 4 4. The point S(−3, 4) lies in Quadrant II. With x = −3 and y = 4, we get r2 = (−3)2 +(4)2 = 25 so r = ±5. As usual, we choose r = 5 ≥ 0 and proceed to determine θ. We have tan(θ) = y x = 4 3, and since this isn’t the tangent of one the common angles, we resort to using the arctangent function. Since θ lies in Quadrant II and must satisfy 0 ≤ θ < 2π, we choose θ = π − arctan 4 ≈ (5, 2.21). To radians. Hence, our answer is (r, θ) = 5, π − arctan 4 3 3 check our answers requires a bit of tenacity since we need to simplify expressions of the form: cos π − arctan 4. These are good review exercises and are hence 3 left to the reader. We ﬁnd cos π − arctan 4 5, so
that 3 x = r cos(θ) = (5) − 3 5 = − 3 = −3 and y = r sin(θ) = (5) 4 5 = 4 which conﬁrms our answer. and sin π − arctan 4 3 5 and sin π − arctan 4 = 4 3 y y S θ = 3π 2 x R θ = π − arctan 4 3 x R has rectangular coordinates (0, −3) R has polar coordinates 3, 3π 2 S has rectangular coordinates (−3, 4) S has polar coordinates 5, π − arctan 4 3 Now that we’ve had practice converting representations of points between the rectangular and polar coordinate systems, we now set about converting equations from one system to another. Just as we’ve used equations in x and y to represent relations in rectangular coordinates, equations in the variables r and θ represent relations in polar coordinates. We convert equations between the two systems using Theorem 11.7 as the next example illustrates. Example 11.4.3. 1. Convert each equation in rectangular coordinates into an equation in polar coordinates. (a) (x − 3)2 + y2 = 9 (b) y = −x (c) y = x2 2. Convert each equation in polar coordinates into an equation in rectangular coordinates. (a) r = −3 (b) θ = 4π 3 (c) r = 1 − cos(θ) 11.4 Polar Coordinates 927 Solution. 1. One strategy to convert an equation from rectangular to polar coordinates is to replace every occurrence of x with r cos(θ) and every occurrence of y with r sin(θ) and use identities to simplify. This is the technique we employ below. (a) We start by substituting x = r cos(θ) and y = sin(θ) into (x−3)2+y2 = 9 and simplifying. With no real direction in which to proceed, we follow our mathematical instincts and see where they take us.5 (r cos(θ) − 3)2 + (r sin(θ))2 = 9 r2 cos2(θ) − 6r cos(θ) + 9 + r2 sin2(θ) = 9 r2 cos2(θ) + sin2(θ) − 6r cos(
θ) = 0 Subtract 9 from both sides. Since cos2(θ) + sin2(θ) = 1 Factor. r2 − 6r cos(θ) = 0 r(r − 6 cos(θ)) = 0 We get r = 0 or r = 6 cos(θ). From Section 7.2 we know the equation (x − 3)2 + y2 = 9 describes a circle, and since r = 0 describes just a point (namely the pole/origin), we choose r = 6 cos(θ) for our ﬁnal answer.6 (b) Substituting x = r cos(θ) and y = r sin(θ) into y = −x gives r sin(θ) = −r cos(θ). Rearranging, we get r cos(θ) + r sin(θ) = 0 or r(cos(θ) + sin(θ)) = 0. This gives r = 0 or cos(θ) + sin(θ) = 0. Solving the latter equation for θ, we get θ = − π 4 + πk for integers k. As we did in the previous example, we take a step back and think geometrically. We know y = −x describes a line through the origin. As before, r = 0 describes the origin, but nothing else. Consider the equation θ = − π 4. In this equation, the variable r is free,7 meaning it can assume any and all values including r = 0. If we imagine plotting points (r, − π 4 ) for all conceivable values of r (positive, negative and zero), we are essentially drawing the line containing the terminal side of θ = − π 4 which is none other than y = −x. Hence, we can take as our ﬁnal answer θ = − π 4 here.8 (c) We substitute x = r cos(θ) and y = r sin(θ) into y = x2 and get r sin(θ) = (r cos(θ))2, or r2 cos2(θ) − r sin(θ) = 0. Factoring, we get r(r cos2(θ) − sin(θ)) = 0 so that either r = 0 or r cos2
(θ) = sin(θ). We can solve the latter equation for r by dividing both sides of the equation by cos2(θ), but as a general rule, we never divide through by a quantity that may be 0. In this particular case, we are safe since if cos2(θ) = 0, then cos(θ) = 0, and for the equation r cos2(θ) = sin(θ) to hold, then sin(θ) would also have to be 0. Since there are no angles with both cos(θ) = 0 and sin(θ) = 0, we are not losing any 5Experience is the mother of all instinct, and necessity is the mother of invention. Study this example and see what techniques are employed, then try your best to get your answers in the homework to match Jeﬀ’s. 2 into r = 6 cos(θ), we recover the point r = 0, so we aren’t losing anything 6Note that when we substitute θ = π by disregarding r = 0. 7See Section 8.1. 8We could take it to be any of θ = − π 4 + πk for integers k. 928 Applications of Trigonometry information by dividing both sides of r cos2(θ) = sin(θ) by cos2(θ). Doing so, we get r = sin(θ) cos2(θ), or r = sec(θ) tan(θ). As before, the r = 0 case is recovered in the solution r = sec(θ) tan(θ) (let θ = 0), so we state the latter as our ﬁnal answer. 2. As a general rule, converting equations from polar to rectangular coordinates isn’t as straight forward as the reverse process. We could solve r2 = x2 + y2 for r to get r = ± x2 + y2 + πk for and solving tan(θ) = y integers k. Neither of these expressions for r and θ are especially user-friendly, so we opt for a second strategy – rearrange the given polar equation so that the expressions r2 = x2 + y2, r cos(θ) = x, r sin(θ) = y and/or tan(θ) = y x requires the
arctangent function to get θ = arctan y x present themselves. x (a) Starting with r = −3, we can square both sides to get r2 = (−3)2 or r2 = 9. We may now substitute r2 = x2 + y2 to get the equation x2 + y2 = 9. As we have seen,9 squaring an equation does not, in general, produce an equivalent equation. The concern here is that the equation r2 = 9 might be satisﬁed by more points than r = −3. On the surface, this appears to be the case since r2 = 9 is equivalent to r = ±3, not just r = −3. However, any point with polar coordinates (3, θ) can be represented as (−3, θ + π), which means any point (r, θ) whose polar coordinates satisfy the relation r = ±3 has an equivalent10 representation which satisﬁes r = −3. (b) We take the tangent of both sides the equation θ = 4π Since tan(θ) = y if, geometrically, the equations θ = 4π 3 and y = x The same argument presented in number 1b applies equally well here so we are done. 3 to get tan(θ) = tan 4π 3. 3. Of course, we pause a moment to wonder 3 generate the same set of points.11 x, we get y 3 or c) Once again, we need to manipulate r = 1 − cos(θ) a bit before using the conversion formulas given in Theorem 11.7. We could square both sides of this equation like we did in part 2a above to obtain an r2 on the left hand side, but that does nothing helpful for the right hand side. Instead, we multiply both sides by r to obtain r2 = r − r cos(θ). We now have an r2 and an r cos(θ) in the equation, which we can easily handle, but we also have another r to deal with. Rewriting the equation as r = r2 + r cos(θ) and squaring both sides yields r2 = r2 + r cos(θ)2. Substituting r2 = x2 + y2 and r cos(θ) = x gives x2 + y2 = x2 +
y2 + x2. Once again, we have performed some 9Exercise 5.3.1 in Section 5.3, for instance... 10Here, ‘equivalent’ means they represent the same point in the plane. As ordered pairs, (3, 0) and (−3, π) are diﬀerent, but when interpreted as polar coordinates, they correspond to the same point in the plane. Mathematically speaking, relations are sets of ordered pairs, so the equations r2 = 9 and r = −3 represent diﬀerent relations since they correspond to diﬀerent sets of ordered pairs. Since polar coordinates were deﬁned geometrically to describe the location of points in the plane, however, we concern ourselves only with ensuring that the sets of points in the plane generated by two equations are the same. This was not an issue, by the way, when we ﬁrst deﬁned relations as sets of points in the plane in Section 1.2. Back then, a point in the plane was identiﬁed with a unique ordered pair given by its Cartesian coordinates. 11In addition to taking the tangent of both sides of an equation (There are inﬁnitely many solutions to tan(θ) = √ √ 3 is only one of them!), we also went from y x = 3, in which x cannot be 0, to y = x √ 3, 3 in which we assume and θ = 4π x can be 0. 11.4 Polar Coordinates 929 algebraic maneuvers which may have altered the set of points described by the original equation. First, we multiplied both sides by r. This means that now r = 0 is a viable solution to the equation. In the original equation, r = 1 − cos(θ), we see that θ = 0 gives r = 0, so the multiplication by r doesn’t introduce any new points. The squaring of both sides of this equation is also a reason to pause. Are there points with coordinates (r, θ) which satisfy r2 = r2 + r cos(θ)2 but do not satisfy r = r2 + r cos(θ)? Suppose (r, θ) satisﬁes r2 = r2 + r cos(θ)2. Then r = ± (r)2 +
r cos(θ). If we have that r = (r)2+r cos(θ), we are done. What if r = − (r)2 + r cos(θ) = −(r)2−r cos(θ)? We claim that the coordinates (−r, θ + π), which determine the same point as (r, θ), satisfy r = r2 + r cos(θ). We substitute r = −r and θ = θ + π into r = r2 + r cos(θ) to see if we get a true statement. −r − −(r)2 − r cos(θ) (r)2 + r cos(θ) (r)2 + r cos(θ)? = (−r)2 + (−r cos(θ + π))? = (r)2 − r cos(θ + π)? = (r)2 − r(− cos(θ)) = (r)2 + r cos(θ) Since r = −(r)2 − r cos(θ) Since cos(θ + π) = − cos(θ) Since both sides worked out to be equal, (−r, θ + π) satisﬁes r = r2 + r cos(θ) which means that any point (r, θ) which satisﬁes r2 = r2 + r cos(θ)2 has a representation which satisﬁes r = r2 + r cos(θ), and we are done. In practice, much of the pedantic veriﬁcation of the equivalence of equations in Example 11.4.3 is left unsaid. Indeed, in most textbooks, squaring equations like r = −3 to arrive at r2 = 9 happens without a second thought. Your instructor will ultimately decide how much, if any, justiﬁcation is warranted. If you take anything away from Example 11.4.3, it should be that relatively nice things in rectangular coordinates, such as y = x2, can turn ugly in polar coordinates, and vice-versa. In the next section, we devote our attention to graphing equations like the ones given in Example 11.4.3 number 2 on the Cartesian coordinate plane without converting back to rectangular coordinates
. If nothing else, number 2c above shows the price we pay if we insist on always converting to back to the more familiar rectangular coordinate system. 930 Applications of Trigonometry 11.4.1 Exercises In Exercises 1 - 16, plot the point given in polar coordinates and then give three diﬀerent expressions (c) r > 0 and θ ≥ 2π for the point such that (a) r < 0 and 0 ≤ θ ≤ 2π, (b) r > 0 and θ ≤ 0 2, 1. π 3 5. 12, − 7π 6 9. (−20, 3π) 2. 5, 7π 4 6. 3, − 10. −4, 5π 4 5π 4 3. 1 3, 3π 2 √ 7. 2 2, −π 11. −1, 2π 3 13. −3, − 11π 6 14. −2.5, − π 4 15. √ − 5, − 4π 3 4. 8. 5 2 7 2, 5π 6, − 13π 6 −3, 12. π 2 16. (−π, −π) In Exercises 17 - 36, convert the point from polar coordinates into rectangular coordinates. 18. 2, π 3 22. −4, 5π 6 7π 6 19. 11, − 23. 9, 7π 2 20. (−20, 3π) 24. −5, − 9π 4 26. (−117, 117π) 27. (6, arctan(2)) 28. (10, arctan(3)) 17. 5, 7π 4 21. 3 5, π 2 25. 42, 13π 6 29. −3, arctan 31. 33. 2, π − arctan −1, π + arctan 4 3 1 2 3 4 34. 2 3 35. (π, arctan(π)) 36. 13, arctan 30. 5, arctan − 4 3 32. − 1 2, π − arctan (5), π + arctan 2 √ 2 12 5 In Exercises 37 - 56, convert the point from rectangular coordinates into polar coordinates with r ≥ 0 and 0 ≤ θ < 2π. 37. (0, 5) 41. (−3, 0) 38. (3, √
3) 39. (7, −7) √ 42. − √ 2 2, 43. −4, −4 √ 3 √ 3) 40. (−3, − √ 44. 3 4, − 1 4 11.4 Polar Coordinates 931 45. − √ 3 3 10 3 10, − √ 46. − √ 5, − 5 47. (6, 8) √ √ 5) 5, 2 48. ( 49. (−8, 1) √ 50. (−2 √ 10, 6 10) 53. (24, −7) 54. (12, −9) 51. (−5, −12) √ 2 4 √ 6 4, 55. 52. − √ 5 15 √ 5 2 15, − 56. − √ 65 5 2, √ 65 5 In Exercises 57 - 76, convert the equation from rectangular coordinates into polar coordinates. Solve for r in all but #60 through #63. In Exercises 60 - 63, you need to solve for θ 57. x = 6 61. y = −x 58. x = −3 √ 3 62. y = x 59. y = 7 60. y = 0 63. y = 2x 64. x2 + y2 = 25 65. x2 + y2 = 117 66. y = 4x − 19 67. x = 3y + 1 68. y = −3x2 69. 4x = y2 70. x2 + y2 − 2y = 0 71. x2 − 4x + y2 = 0 72. x2 + y2 = x 73. y2 = 7y − x2 75. x2 + (y − 3)2 = 9 74. (x + 2)2 + y2 = 4 76. 4x2 + 4 y − 2 1 2 = 1 In Exercises 77 - 96, convert the equation from polar coordinates into rectangular coordinates. 77. r = 7 81. θ = 2π 3 78. r = −3 82. θ = π 79. r = 83. θ = √ 2 3π 2 80. θ = π 4 84. r = 4 cos(θ) 85. 5r = cos(θ) 86. r = 3 sin(θ) 87. r = −2 sin(θ) 88.
r = 7 sec(θ) 89. 12r = csc(θ) 90. r = −2 sec(θ) √ 91. r = − 5 csc(θ) 92. r = 2 sec(θ) tan(θ) 93. r = − csc(θ) cot(θ) 94. r2 = sin(2θ) 95. r = 1 − 2 cos(θ) 96. r = 1 + sin(θ) 97. Convert the origin (0, 0) into polar coordinates in four diﬀerent ways. 98. With the help of your classmates, use the Law of Cosines to develop a formula for the distance between two points in polar coordinates. 932 Applications of Trigonometry 11.4.2 Answers 1. 2,, π 3 2, − 5π 3 −2,, 4π 3 7π 3 2, 2. 5, 7π 4 π 4,, −5, 5, 3π 4 15π 4 5, − 3. 1 3 1 3 4. 5 2 5 2,,, 3π,, 5π 6 7π 7π 2,, 11π 6 17π 6 y y 2 1 1 −1 1 2 x −1 1 2 3 x −1 −2 −3 −2 −1 −1 1 2 3 x −2 −3 11.4 Polar Coordinates 933 5. 12, − 12, −, −12,, 12, 7π 6 19π 6 11π 6 17π 6 6. 3, − 3, −, −3,, 3, 5π 4 13π 4 7π 4 11π 4 √ √ 7. 2 2 √ 2, −π, −2 √ 2, −3π, 2 2, 0 2, 3π 8. 7 2 7 2, −, − 13π 23π 6 5π 6 y 6 3 −12 −9 −6 −3 x y 3 2 1 −3 −2 −1 1 2 3 x −1 −2 −3 3 2 1 y −3 −2 −1 1 2 3 x −1 −2 −3 4 3 2 1 y −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 934 Applications of Trigonometry 9. (−20, 3π), (−20, π) (20
, −2π), (20, 4π) 10. −4, 4, − 5π 4 7π 4 11. −1, 1, − 2π 3 π 3 13π 4,, −4, 4, 9π 4 8π 3,, 1, −1, 11π 3 12. −3, 3, −,, π 2 π 2 5π 2 −3, 3, 7π 2 10 x 20 −20 −10 y y 1 −1 4 3 2 1 −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 2 1 y −2 −1 1 2 x −1 −2 y 3 2 1 −3 −2 −1 −1 1 2 3 x −2 −3 11.4 Polar Coordinates 935 13. −3, − 11π 6, 3, − 5π 6 π 6 −3, 19π 6, 3,,, 14. −2.5, − 2.5, − π 4 5π 4 −2.5, 2.5, 7π 4 11π 4 √ 15. − 5, − √ 5, − π 3 4π 3,, √ √ − 2π 3 5, 5, 11π 3 16. (−π, −π), (−π, π) (π, −2π), (π, 2π3 −2 −1 −1 −2 −3 2 1 −2 −1 −1 −2 2 1 −2 −1 −1 −2 3 2 1 −3 −2 −1 1 2 3 x −1 −2 −3 936 Applications of Trigonometry √ 5 2 2, − √ 2 5 2 17. 18. 1, √ 3 19. − √ 11 2 3, 11 2 22. 2 √ 3, −2 23. (0, −9) 20. (20, 0) 24 28. √ 10, 3 √ 10 √ 26 52 √ 5 26 52, − 32. 36. (5, 12) √ 2 3, 40. 44. 1 2, 11π 6 7π 6 4 3 48. (5, arctan (2)) 26. (117, 0) 30. (3, −4 34. √ 2 3, 38. 42. 2, √ 46. 3π 4 10, �
� 6 5π 4 √ 5 6 5 5 √ 12 5, 27. 31 35. π√ 1+π2, √ π2 1+π2 7π 4 √ 7 2, 39. 43. 8, 4π 3 47. 10, arctan 21. 0, 3 5 √ 25. 21 3, 21 29. − 9 5, − 12 5 33. 4 5, 3 5 37. 5, π 2 41. (3, π) 4π 3, 3 5 √ 51. 13, π + arctan 25, 2π − arctan 1 8 12 5 7 24 45. 49. 53. 55. 65, π − arctan 50. (20, π − arctan(3)), π + arctan (2) 52. 1 3 54. 15, 2π − arctan 3 4 √ 2 2, π 3 56. √ 13, π − arctan(2) 57. r = 6 sec(θ) 58. r = −3 sec(θ) 59. r = 7 csc(θ) 60. θ = 0 61. θ = 3π 4 √ 65. r = 117 62. θ = π 3 63. θ = arctan(2) 64. r = 5 66. r = 19 4 cos(θ)−sin(θ) 67. x = 1 cos(θ)−3 sin(θ) 68. r = − sec(θ) tan(θ) 3 69. r = 4 csc(θ) cot(θ) 70. r = 2 sin(θ) 71. r = 4 cos(θ) 72. r = cos(θ) 11.4 Polar Coordinates 937 73. r = 7 sin(θ) 74. r = −4 cos(θ) 75. r = 6 sin(θ) 76. r = sin(θ) 77. x2 + y2 = 49 78. x2 + y2 = 9 79. x2 + y2 = 2 80. y = x √ 81. y = − 3x 83. x = 0 82. y = 0 84. x2 + y2 = 4x or (x − 2)2 + y2
= 4 85. 5x2 + 5y2 = x or x − 2 1 10 + y2 = 1 100 86. x2 + y2 = 3y or x2 + y − 2 3 2 = 9 4 87. x2 + y2 = −2y or x2 + (y + 1)2 = 1 88. x = 7 89. y = 1 12 √ 91. y = − 5 93. y2 = −x 90. x = −2 92. x2 = 2y 94. x2 + y22 = 2xy 95. x2 + 2x + y22 = x2 + y2 96. x2 + y2 + y2 97. Any point of the form (0, θ) will work, e.g. (0, π), (0, −117), 0, = x2 + y2 23π 4 and (0, 0). 938 Applications of Trigonometry 11.5 Graphs of Polar Equations In this section, we discuss how to graph equations in polar coordinates on the rectangular coordinate plane. Since any given point in the plane has inﬁnitely many diﬀerent representations in polar coordinates, our ‘Fundamental Graphing Principle’ in this section is not as clean as it was for graphs of rectangular equations on page 23. We state it below for completeness. The Fundamental Graphing Principle for Polar Equations The graph of an equation in polar coordinates is the set of points which satisfy the equation. That is, a point P (r, θ) is on the graph of an equation if and only if there is a representation of P, say (r, θ), such that r and θ satisfy the equation. Our ﬁrst example focuses on some of the more structurally simple polar equations. Example 11.5.1. Graph the following polar equations. 1. r = 4 2. r = −3 √ 2 3. θ = 5π 4 4. θ = − 3π 2 Solution. In each of these equations, only one of the variables r and θ is present making the other variable free.1 This makes these graphs easier to visualize than others. 1. In the equation r = 4, θ is free. The graph of this equation is, therefore, all points which have a polar coordinate representation (4, θ), for any choice of θ. Graph
ically this translates into tracing out all of the points 4 units away from the origin. This is exactly the deﬁnition of circle, centered at the origin, with a radius of 44 x 4 In r = 4, θ is free The graph of r = 4 −4 2. Once again we have θ being free in the equation r = −3 2, θ) gives us a circle of radius 3 form (−3 √ √ 2 centered at the origin. 2. Plotting all of the points of the √ 1See the discussion in Example 11.4.3 number 2a. 11.5 Graphs of Polar Equations 939 4 x 4 In r = −3 √ 2, θ is free −4 The graph of r = −3 √ 2 3. In the equation θ = 5π. 4, r is free, so we plot all of the points with polar representation r, 5π What we ﬁnd is that we are tracing out the line which contains the terminal side of θ = 5π 4 when plotted in standard position. 4 y r < 0 θ = 5π 4 y 4 r = 0 x −4 x 4 r > 0 In θ = 5π 4, r is free The graph of θ = 5π 4 −4 4. As in the previous example, the variable r is free in the equation θ = − 3π 2. Plotting r, − 3π 2 for various values of r shows us that we are tracing out the y-axis. 940 Applications of Trigonometry y r > 0 r = 0 θ = − 3π 2 y 4 x −4 x 4 r < 0 −4 In θ = − 3π 2, r is free The graph of θ = − 3π 2 Hopefully, our experience in Example 11.5.1 makes the following result clear. Theorem 11.8. Graphs of Constant r and θ: Suppose a and α are constants, a = 0. The graph of the polar equation r = a on the Cartesian plane is a circle centered at the origin of radius \|a\|. The graph of the polar equation θ = α on the Cartesian plane is the line containing the terminal side of α when plotted in standard position. Suppose we wish to graph r = 6 cos(θ). A reasonable way to start is to treat θ as the independent variable, r
as the dependent variable, evaluate r = f (θ) at some ‘friendly’ values of θ and plot the resulting points.2 We generate the table below. θ 0 π 4 π 2 3π 4 π 5π 4 3π 2 7π 4 2π r = 6 cos(θ) 6 √ 3 2 3 √ √ (r, θ) (6, 0) 2, π 4 0, π 2 2, 3π 4 (−6, π) √ 2, 5π 4 0, 3π 2 √ 2, 7π 4 (6, 2π) 0 2 −3 √ −3 −6 √ −3 √ 3 2 6 2 −3 0 3 y 3 −3 3 x 6 2For a review of these concepts and this process, see Sections 1.4 and 1.6. 11.5 Graphs of Polar Equations 941 Despite having nine ordered pairs, we get only four distinct points on the graph. For this reason, we employ a slightly diﬀerent strategy. We graph one cycle of r = 6 cos(θ) on the θr-plane3 and use it to help graph the equation on the xy-plane. We see that as θ ranges from 0 to π 2, r ranges from 6 to 0. In the xy-plane, this means that the curve starts 6 units from the origin on the positive x-axis (θ = 0) and gradually returns to the origin by the time the curve reaches the y-axis (θ = π 2 ). The arrows drawn in the ﬁgure below are meant to help you visualize this process. In the θr-plane, the arrows are drawn from the θ-axis to the curve r = 6 cos(θ). In the xy-plane, each of these arrows starts at the origin and is rotated through the corresponding angle θ, in accordance with how we plot polar coordinates. It is a less-precise way to generate the graph than computing the actual function values, but it is markedly faster. r 6 3 −3 −6 π 2 π 3π 2 2π θ y θ runs from 0 to π 2 x Next, we repeat the process as θ ranges from π 2 to π. Here, the r values are all negative. This means that in
the xy-plane, instead of graphing in Quadrant II, we graph in Quadrant IV, with all of the angle rotations starting from the negative x-axis. r 6 3 −3 −6 y θ runs from π 2 to π π 2 π 3π 2 2π θ x r < 0 so we plot here As θ ranges from π to 3π 2, the r values are still negative, which means the graph is traced out in Quadrant I instead of Quadrant III. Since the \|r\| for these values of θ match the r values for θ in 3The graph looks exactly like y = 6 cos(x) in the xy-plane, and for good reason. At this stage, we are just graphing the relationship between r and θ before we interpret them as polar coordinates (r, θ) on the xy-plane. 942 Applications of Trigonometry, we have that the curve begins to retrace itself at this point. Proceeding further, we ﬁnd 2 ≤ θ ≤ 2π, we retrace the portion of the curve in Quadrant IV that we ﬁrst traced 2 ≤ θ ≤ π. The reader is invited to verify that plotting any range of θ outside the interval 0, π 2 that when 3π out as π [0, π] results in retracting some portion of the curve.4 We present the ﬁnal graph below. r 6 3 −3 −6 π 2 π θ y 3 −3 3 x 6 r = 6 cos(θ) in the θr-plane r = 6 cos(θ) in the xy-plane Example 11.5.2. Graph the following polar equations. 1. r = 4 − 2 sin(θ) 2. r = 2 + 4 cos(θ) 3. r = 5 sin(2θ) 4. r2 = 16 cos(2θ) Solution. 1. We ﬁrst plot the fundamental cycle of r = 4 − 2 sin(θ) on the θr-axes. To help us visualize what is going on graphically, we divide up [0, 2π] into the usual four subintervals 0, π 2, π, 2 π, 3π 2, r decreases from 2 4 to 2.
This means that the curve in the xy-plane starts 4 units from the origin on the positive x-axis and gradually pulls in towards the origin as it moves towards the positive y-axis. 2, 2π, and proceed as we did above. As θ ranges from 0 to π and 3π, π r 6 4 2 π 2 π 3π 2 2π θ y θ runs from 0 to π 2 x 4The graph of r = 6 cos(θ) looks suspiciously like a circle, for good reason. See number 1a in Example 11.4.3. 11.5 Graphs of Polar Equations 943 Next, as θ runs from π oﬀ, we gradually pull the graph away from the origin until we reach the negative x-axis. 2 to π, we see that r increases from 2 to 4. Picking up where we left θ runs from π 2 to 3π 2 2π θ, we see that r increases from 4 to 6. On the xy-plane, the curve Over the interval π, 3π 2 sweeps out away from the origin as it travels from the negative x-axis to the negative y-axis 3π 2 2π θ θ runs from π to 3π 2 Finally, as θ takes on values from 3π xy-plane pulls in from the negative y-axis to ﬁnish where we started. 2 to 2π, r decreases from 6 back to 4. The graph on the r 6 4 2 y x π 2 π 3π 2 2π θ θ runs from 3π 2 to 2π We leave it to the reader to verify that plotting points corresponding to values of θ outside the interval [0, 2π] results in retracing portions of the curve, so we are ﬁnished. 944 Applications of Trigonometry sin(θ) in the θr-plane 3π 2 2π y 2 −4 x 4 −6 r = 4 − 2 sin(θ) in the xy-plane. 2. The ﬁrst thing to note when graphing r = 2 + 4 cos(θ) on the θr-plane over the interval [0, 2π] is that the graph crosses through the θ-axis. This corresponds to the graph of the curve passing through the origin
in the xy-plane, and our ﬁrst task is to determine when this happens. Setting r = 0 we get 2 + 4 cos(θ) = 0, or cos(θ) = − 1 2. Solving for θ in [0, 2π] gives θ = 2π 3. Since these values of θ are important geometrically, we break the interval [0, 2π] into six subintervals: 0, π 2, 2π. As 3, π, π, 4π 2 θ ranges from 0 to π 2, r decreases from 6 to 2. Plotting this on the xy-plane, we start 6 units out from the origin on the positive x-axis and slowly pull in towards the positive y-axis. 3 and θ = 4π and 3π, 2π, 4π 3, 3π, π 2, 2π 3 3 2 r 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ y θ runs from 0 to π 2 x On the interval π will eventually cross through) the origin. Not only do we reach the origin when θ = 2π theorem from Calculus5 states that the curve hugs the line θ = 2π, r decreases from 2 to 0, which means the graph is heading into (and 3, a 3 as it approaches the origin. 2, 2π 3 5The ‘tangents at the pole’ theorem from second semester Calculus. 11.5 Graphs of Polar Equations 945 r 6 4 2 −2 y θ = 2π 3 2π 3 4π 3 π 2 π 3π 2 2π θ x 3, π, r ranges from 0 to −2. Since r ≤ 0, the curve passes through the On the interval 2π origin in the xy-plane, following the line θ = 2π 3 and continues upwards through Quadrant IV towards the positive x-axis.6 Since \|r\| is increasing from 0 to 2, the curve pulls away from the origin to ﬁnish at a point on the positive x-axis. r 6 4 2 −2 y θ = 2π 3 2π 3 4π 3 π 2 π 3π 2 2π θ x Next, as θ progresses from π to 4
π graph in the ﬁrst quadrant, heading into the origin along the line θ = 4π 3. 3, r ranges from −2 to 0. Since r ≤ 0, we continue our 6Recall that one way to visualize plotting polar coordinates (r, θ) with r < 0 is to start the rotation from the left 3 and π radians from the negative x-axis in side of the pole - in this case, the negative x-axis. Rotating between 2π this case determines the region between the line θ = 2π 3 and the x-axis in Quadrant IV. 946 Applications of Trigonometry r 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ y θ = 4π 3 x On the interval 4π line θ = 4π 3, 3π 2, r returns to positive values and increases from 0 to 2. We hug the 3 as we move through the origin and head towards the negative y-axis. r 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ y θ = 4π 3 x As we round out the interval, we ﬁnd that as θ runs through 3π to 6, and we end up back where we started, 6 units from the origin on the positive x-axis. 2 to 2π, r increases from 2 out y r 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ x θ runs from 3π 2 to 2π 11.5 Graphs of Polar Equations 947 Again, we invite the reader to show that plotting the curve for values of θ outside [0, 2π] results in retracing a portion of the curve already traced. Our ﬁnal graph is below. r 6 4 2 −2 2π 3 4π 3 π 2 π 3π 2 2π θ y θ = 2π 3 2 θ = 4π 3 −2 2 6 x r = 2 + 4 cos(θ) in the θr-plane r = 2 + 4 cos(θ) in the xy-plane 3. As usual, we start by graphing a fundamental cycle of r = 5 sin(2θ) in the θr-plane, which in this case, occurs as
θ ranges from 0 to π. We partition our interval into subintervals to help us with the graphing, namely 0, π, π 4, r 4 increases from 0 to 5. This means that the graph of r = 5 sin(2θ) in the xy-plane starts at the origin and gradually sweeps out so it is 5 units away from the origin on the line θ = π 4. 4, π. As θ ranges from 0 to π and 3π, π 2, 3π 3π 4 π θ x r 5 −5 Next, we see that r decreases from 5 to 0 as θ runs through π heading negative as θ crosses π as the curve heads to the origin. 2. Hence, we draw the curve hugging the line θ = π, and furthermore, r is 2 (the y-axis) 4, π 2 948 Applications of Trigonometry y π 4 π 2 3π 4 π θ x r 5 −5 As θ runs from π pulls away from the negative y-axis into Quadrant IV. 2 to 3π 4, r becomes negative and ranges from 0 to −5. Since r ≤ 0, the curve y π 4 π 2 3π 4 π θ x r 5 −5 For 3π 4 ≤ θ ≤ π, r increases from −5 to 0, so the curve pulls back to the origin. y π 4 π 2 3π 4 π θ x r 5 −5 11.5 Graphs of Polar Equations 949 Even though we have ﬁnished with one complete cycle of r = 5 sin(2θ), if we continue plotting beyond θ = π, we ﬁnd that the curve continues into the third quadrant! Below we present a graph of a second cycle of r = 5 sin(2θ) which continues on from the ﬁrst. The boxed labels on the θ-axis correspond to the portions with matching labels on the curve in the xy-plane. r 5 π 1 5π 4 2 3π 2 3 7π 4 4 2π θ 4 1 −5 We have the ﬁnal graph below 3π 4 π 5π 4 3π 2 7π 4 2π θ −5 x 5 −5 −5 r = 5
sin(2θ) in the θr-plane r = 5 sin(2θ) in the xy-plane 4. Graphing r2 = 16 cos(2θ) is complicated by the r2, so we solve to get r = ± 16 cos(2θ) = ±4 cos(2θ). How do we sketch such a curve? First oﬀ, we sketch a fundamental period cos(2θ) is of r = cos(2θ) which we have dotted in the ﬁgure below. When cos(2θ) < 0,. On the intervals which remain, undeﬁned, so we don’t have any values on the interval π cos(2θ) ranges from 0 to 1 as well.7 From cos(2θ) ranges from 0 to 1, inclusive. Hence, cos(2θ) ranges continuously from 0 to ±4, respectively. Below we this, we know r = ±4 cos(2θ) on the θr plane and use them to sketch the graph both r = 4 corresponding pieces of the curve r2 = 16 cos(2θ) in the xy-plane. As we have seen in earlier cos(2θ) and r = −4 4, 3π 4 7Owing to the relationship between y = x and y = √ x over [0, 1], we also know cos(2θ) ≥ cos(2θ) wherever the former is deﬁned. 950 Applications of Trigonometry examples, the lines θ = π serve as guides for us to draw the curve as is passes through the origin. 4, which are the zeros of the functions r = ±4 4 and θ = 3π cos(2θ), θ = 3π 4 π 4 π 2 3π 4 4 3 π θ r = 4 cos(2θ) and r = −4cos(2θ) As we plot points corresponding to values of θ outside of the interval [0, π], we ﬁnd ourselves retracing parts of the curve,8 so our ﬁnal answer is below. r 4 θ = 3π 3π 4 π θ −4 x 4 −4 r = ±4 cos(2θ) in the θ
r-plane −4 r2 = 16 cos(2θ) in the xy-plane A few remarks are in order. First, there is no relation, in general, between the period of the function f (θ) and the length of the interval required to sketch the complete graph of r = f (θ) in the xyplane. As we saw on page 941, despite the fact that the period of f (θ) = 6 cos(θ) is 2π, we sketched the complete graph of r = 6 cos(θ) in the xy-plane just using the values of θ as θ ranged from 0 to π. In Example 11.5.2, number 3, the period of f (θ) = 5 sin(2θ) is π, but in order to obtain the complete graph of r = 5 sin(2θ), we needed to run θ from 0 to 2π. While many of the ‘common’ polar graphs can be grouped into families,9 the authors truly feel that taking the time to work through each graph in the manner presented here is the best way to not only understand the polar 8In this case, we could have generated the entire graph by using just the plot r = 4cos(2θ), but graphed over the interval [0, 2π] in the θr-plane. We leave the details to the reader. 9Numbers 1 and 2 in Example 11.5.2 are examples of ‘lima¸cons,’ number 3 is an example of a ‘polar rose,’ and number 4 is the famous ‘Lemniscate of Bernoulli.’ 11.5 Graphs of Polar Equations 951 coordinate system, but also prepare you for what is needed in Calculus. Second, the symmetry seen in the examples is also a common occurrence when graphing polar equations. In addition to the usual kinds of symmetry discussed up to this point in the text (symmetry about each axis and the origin), it is possible to talk about rotational symmetry. We leave the discussion of symmetry to the Exercises. In our next example, we are given the task of ﬁnding the intersection points of polar curves. According to the Fundamental Graphing Principle for Polar Equations on page 938, in order for a point P to be on the graph of a
polar equation, it must have a representation P (r, θ) which satisﬁes the equation. What complicates matters in polar coordinates is that any given point has inﬁnitely many representations. As a result, if a point P is on the graph of two diﬀerent polar equations, it is entirely possible that the representation P (r, θ) which satisﬁes one of the equations does not satisfy the other equation. Here, more than ever, we need to rely on the Geometry as much as the Algebra to ﬁnd our solutions. Example 11.5.3. Find the points of intersection of the graphs of the following polar equations. 1. r = 2 sin(θ) and r = 2 − 2 sin(θ) 2. r = 2 and r = 3 cos(θ) 3. r = 3 and r = 6 cos(2θ) Solution. 4. r = 3 sin θ 2 and r = 3 cos θ 2 1. Following the procedure in Example 11.5.2, we graph r = 2 sin(θ) and ﬁnd it to be a circle centered at the point with rectangular coordinates (0, 1) with a radius of 1. The graph of r = 2 − 2 sin(θ) is a special kind of lima¸con called a ‘cardioid.’10 y 2 −2 2 x −4 r = 2 − 2 sin(θ) and r = 2 sin(θ) It appears as if there are three intersection points: one in the ﬁrst quadrant, one in the second quadrant, and the origin. Our next task is to ﬁnd polar representations of these points. In 10Presumably, the name is derived from its resemblance to a stylized human heart. 952 Applications of Trigonometry 2. From this, we get θ = π 6 into r = 2 sin(θ), we get r = 2 sin π order for a point P to be on the graph of r = 2 sin(θ), it must have a representation P (r, θ) which satisﬁes r = 2 sin(θ). If P is also on the graph of r = 2−2 sin(θ), then P has a (possibly diﬀerent) representation
P (r, θ) which satisﬁes r = 2 sin(θ). We ﬁrst try to see if we can ﬁnd any points which have a single representation P (r, θ) that satisﬁes both r = 2 sin(θ) and r = 2 − 2 sin(θ). Assuming such a pair (r, θ) exists, then equating11 the expressions for r gives 2 sin(θ) = 2 − 2 sin(θ) or sin(θ) = 1 6 + 2πk = 1, which for integers k. Plugging θ = π is also the value we obtain when we substitute it into r = 2 − 2 sin(θ). Hence, 1, π is one 6 representation for the point of intersection in the ﬁrst quadrant. For the point of intersection, so this is in the second quadrant, we try θ = 5π our answer here. What about the origin? We know from Section 11.4 that the pole may be represented as (0, θ) for any angle θ. On the graph of r = 2 sin(θ), we start at the origin when θ = 0 and return to it at θ = π, and as the reader can verify, we are at the origin exactly when θ = πk for integers k. On the curve r = 2 − 2 sin(θ), however, we reach the origin when θ = π 2 + 2πk for integers k. There is no integer value of k for which πk = π 2 + 2πk which means while the origin is on both graphs, the point is never reached simultaneously. In any case, we have determined the three points of intersection to be 1, π 6 6. Both equations give us the point 1, 5π 2, and more generally, when θ = π 6 + 2πk or θ = 5π = 2 1 2 and the origin., 1, 5π 6 6 6 2. As before, we make a quick sketch of r = 2 and r = 3 cos(θ) to get feel for the number and location of the intersection points. The graph of r = 2 is a circle, centered at the origin, with a radius of 2. The graph of r = 3 cos(θ
) is also a circle - but this one is centered at the point with rectangular coordinates 3 2, 0 and has a radius of 3 2. y 2 −2 2 3 x −2 r = 2 and r = 3 cos(θ) We have two intersection points to ﬁnd, one in Quadrant I and one in Quadrant IV. Proceeding as above, we ﬁrst determine if any of the intersection points P have a representation (r, θ) which satisﬁes both r = 2 and r = 3 cos(θ). Equating these two expressions for r, we get cos(θ) = 2 3. To solve this equation, we need the arccosine function. We get 11We are really using the technique of substitution to solve the system of equations r = 2 sin(θ) r = 2 − 2 sin(θ) 11.5 Graphs of Polar Equations 953 θ = arccos 2 + 2πk or θ = 2π − arccos 2 3 3 2, arccos 2 as one representation for our answer in Quadrant I, and 2, 2π − arccos 2 3 3 as one representation for our answer in Quadrant IV. The reader is encouraged to check these results algebraically and geometrically. + 2πk for integers k. From these solutions, we get 3. Proceeding as above, we ﬁrst graph r = 3 and r = 6 cos(2θ) to get an idea of how many intersection points to expect and where they lie. The graph of r = 3 is a circle centered at the origin with a radius of 3 and the graph of r = 6 cos(2θ) is another four-leafed rose.12 y 6 3 −6 −3 3 x 6 −3 −6 r = 3 and r = 6 cos(2θ), 3, 5π 6, 3, 7π 6 and 3, 11π 6 6 + πk or θ = 5π 2. Solving, we get θ = π It appears as if there are eight points of intersection - two in each quadrant. We ﬁrst look to see if there any points P (r, θ) with a representation that satisﬁes both r = 3 and r = 6 cos(2θ). For these points, 6
cos(2θ) = 3 or cos(2θ) = 1 6 + πk for integers k. Out of all of these solutions, we obtain just four distinct points represented by 3, π. To determine the coordinates of the remaining four 6 points, we have to consider how the representations of the points of intersection can diﬀer. We know from Section 11.4 that if (r, θ) and (r, θ) represent the same point and r = 0, then either r = r or r = −r. If r = r, then θ = θ+2πk, so one possibility is that an intersection point P has a representation (r, θ) which satisﬁes r = 3 and another representation (r, θ+2πk) for some integer, k which satisﬁes r = 6 cos(2θ). At this point,13 if we replace every occurrence of θ in the equation r = 6 cos(2θ) with (θ+2πk) and then see if, by equating the resulting expressions for r, we get any more solutions for θ. Since cos(2(θ + 2πk)) = cos(2θ + 4πk) = cos(2θ) for every integer k, however, the equation r = 6 cos(2(θ + 2πk)) reduces to the same equation we had before, r = 6 cos(2θ), which means we get no additional solutions. Moving on to the case where r = −r, we have that θ = θ + (2k + 1)π for integers k. We look to see if we can ﬁnd points P which have a representation (r, θ) that satisﬁes r = 3 and another, 12See Example 11.5.2 number 3. 13The authors have chosen to replace θ with θ + 2πk in the equation r = 6 cos(2θ) for illustration purposes only. We could have just as easily chosen to do this substitution in the equation r = 3. Since there is no θ in r = 3, however, this case would reduce to the previous case instantly. The reader is encouraged to follow this latter procedure in the interests of eﬃciency. 954 Applications of Trigonometry
(−r, θ + (2k + 1)π), that satisﬁes r = 6 cos(2θ). To do this, we substitute14 (−r) for r and (θ + (2k + 1)π) for θ in the equation r = 6 cos(2θ) and get −r = 6 cos(2(θ + (2k + 1)π)). Since cos(2(θ + (2k + 1)π)) = cos(2θ + (2k + 1)(2π)) = cos(2θ) for all integers k, the equation −r = 6 cos(2(θ + (2k + 1)π)) reduces to −r = 6 cos(2θ), or r = −6 cos(2θ). Coupling this equation with r = 3 gives −6 cos(2θ) = 3 or cos(2θ) = − 1 3 + πk. From these solutions, we obtain15 the remaining four intersection points with representations −3, π 3, which we can readily check graphically. 3 + πk or θ = 2π 2. We get θ = π and −3, 5π 3, −3, 2π 3, −3, 4π 3. Using the techniques 4. As usual, we begin by graphing r = 3 sin θ 2 presented in Example 11.5.2, we ﬁnd that we need to plot both functions as θ ranges from 0 to 4π to obtain the complete graph. To our surprise and/or delight, it appears as if these two equations describe the same curve! and r = 3 cos θ 2 y 3 −3 3 x −3 r = 3 sin θ 2 and r = 3 cos θ 2 appear to determine the same curve in the xy-plane and r = 3 cos θ 2 To verify this incredible claim,16 we need to show that, in fact, the graphs of these two equations intersect at all points on the plane. Suppose P has a representation (r, θ) which satisﬁes both r = 3 sin θ. Equating these two expressions for r gives 2. While normally we discourage dividing by a variable = 3 cos θ the equation 3 sin θ 2 2 expression (in case it could be 0), we note here
that if 3 cos θ = 0, then for our equation 2 = 0 as well. Since no angles have both cosine and sine equal to zero, to hold, 3 sin θ 2 to get we are safe to divide both sides of the equation 3 sin θ 2 tan θ 2 + 2πk for integers k. From these solutions, however, we 2 = 1 which gives θ = π by 3 cos θ 2 = 3 cos θ 2 14Again, we could have easily chosen to substitute these into r = 3 which would give −r = 3, or r = −3. 15We obtain these representations by substituting the values for θ into r = 6 cos(2θ), once again, for illustration purposes. Again, in the interests of eﬃciency, we could ‘plug’ these values for θ into r = 3 (where there is no θ) and represents the same point get the list of points: 3, π as −3, π, we still get the same set of solutions.. While it is not true that 3, π and 3, 5π, 3, 4π, 3, 2π 3 3 3 3 3 16A quick sketch of r = 3 sin θ and r = 3 cos θ in the θr-plane will convince you that, viewed as functions of r, 3 2 2 these are two diﬀerent animals. 11.5 Graphs of Polar Equations 955 √ 3 2 2, π 2 2 + πk = 3 cos θ 2 [θ + 2πk] = 3 cos θ get only one intersection point which can be represented by. We now investigate other representations for the intersection points. Suppose P is an intersection point with a representation (r, θ) which satisﬁes r = 3 sin θ and the same point P has a diﬀerent 2. Substituting representation (r, θ + 2πk) for some integer k which satisﬁes r = 3 cos θ 2 into the latter, we get r = 3 cos 1 2 + πk. Using the sum formula for, since sin (πk) = ±3 cos θ cosine, we expand 3 cos θ 2 sin(πk) = 0 for all integers k, and cos (πk) = ±1
for all integers k. If k is an even integer, we get the same equation r = 3 cos θ as before. If k is odd, we get r = −3 cos θ. This 2 2 = −1. Solving, latter expression for r leads to the equation 3 sin θ 2 √ we get θ = − π 2, − π. Next, we assume P has a representation (r, θ) which satisﬁes r = 3 sin θ and a representation 2 for some integer k. Substituting (−r) for (−r, θ + (2k + 1)π) which satisﬁes r = 3 cos θ 2 r and (θ + (2k + 1)π) in for θ into r = 3 cos θ 2 [θ + (2k + 1)π]. Once gives −r = 3 cos 1 2 again, we use the sum formula for cosine to get 2 + 2πk for integers k, which gives the intersection point cos(πk) − 3 sin θ 2, or tan θ 2 3 = −3 cos θ 2 2 2 2 cos 1 2 [θ + (2k + 1)π] = cos 2 θ 2 + (2k+1)π cos (2k+1)π 2 = cos θ 2 = ± sin θ 2 − sin θ 2 sin (2k+1)π 2 = 0 and sin (2k+1)π (2k+1)π where the last equality is true since cos 2 2 2 [θ + (2k + 1)π] can be rewritten as r = ±3 sin θ Hence, −r = 3 cos 1 (2k+1)π = 1, and the equation −r = 3 cos 1 = sin π then sin 2 2 reduces to −r = −3 sin θ, or r = 3 sin θ 2 2 What this means is that if a polar representation (r, θ) for the point P satisﬁes r = 3 sin( θ then the representation (−r, θ + π) for P automatically satisﬁes r = 3 cos θ 2 equations r = 3 sin( θ = ±1 for integers k.. If we choose k = 0, 2 [θ + (
2k + 1)π] in this case which is the other equation under consideration! 2 ),. Hence the 2 ) determine the same set of points in the plane. 2 ) and r = 3 cos( θ 2 Our work in Example 11.5.3 justiﬁes the following. Guidelines for Finding Points of Intersection of Graphs of Polar Equations To ﬁnd the points of intersection of the graphs of two polar equations E1 and E2: Sketch the graphs of E1 and E2. Check to see if the curves intersect at the origin (pole). Solve for pairs (r, θ) which satisfy both E1 and E2. Substitute (θ + 2πk) for θ in either one of E1 or E2 (but not both) and solve for pairs (r, θ) which satisfy both equations. Keep in mind that k is an integer. Substitute (−r) for r and (θ + (2k + 1)π) for θ in either one of E1 or E2 (but not both) and solve for pairs (r, θ) which satisfy both equations. Keep in mind that k is an integer. 956 Applications of Trigonometry Our last example ties together graphing and points of intersection to describe regions in the plane. Example 11.5.4. Sketch the region in the xy-plane described by the following sets. 1. (r, θ) \| 0 ≤ r ≤ 5 sin(2θ), 0 ≤ θ ≤ π 2 2. (r, θ) \| 3 ≤ r ≤ 6 cos(2θ), 0 ≤ θ ≤ π 6 3. (r, θ) \| 2 + 4 cos(θ) ≤ r ≤ 0, 2π 4. (r, θ) \| 0 ≤ r ≤ 2 sin(θ), 0 ≤ θ ≤ π 6 3 ≤ θ ≤ 4π 3 ∪ (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), π 6 ≤ θ ≤ π 2 Solution. Our ﬁrst step in these problems is to sketch the graphs of the polar equations involved to get a sense of the geometric situation. Since all of the equations in this example are found in either Example 11.5.2 or Example 11.5.3, most of the work is done for
us. 1. We know from Example 11.5.2 number 3 that the graph of r = 5 sin(2θ) is a rose. Moreover, we know from our work there that as 0 ≤ θ ≤ π 2, we are tracing out the ‘leaf’ of the rose which lies in the ﬁrst quadrant. The inequality 0 ≤ r ≤ 5 sin(2θ) means we want to capture all of the points between the origin (r = 0) and the curve r = 5 sin(2θ) as θ runs through 0, π 2. Hence, the region we seek is the leaf itself. r 5 −5 π 4 π 2 3π 4 π θ y x (r, θ) \| 0 ≤ r ≤ 5 sin(2θ), 0 ≤ θ ≤ π 2 2. We know from Example 11.5.3 number 3 that r = 3 and r = 6 cos(2θ) intersect at θ = π 6, so the region that is being described here is the set of points whose directed distance r from the origin is at least 3 but no more than 6 cos(2θ) as θ runs from 0 to π 6. In other words, we are looking at the points outside or on the circle (since r ≥ 3) but inside or on the rose (since r ≤ 6 cos(2θ)). We shade the region below and r = 6 cos(2θ) (r, θ) \| 3 ≤ r ≤ 6 cos(2θ), 0 ≤ θ ≤ π 6 11.5 Graphs of Polar Equations 957 3. From Example 11.5.2 number 2, we know that the graph of r = 2 + 4 cos(θ) is a lima¸con whose ‘inner loop’ is traced out as θ runs through the given values 2π 3. Since the values r takes on in this interval are non-positive, the inequality 2 + 4 cos(θ) ≤ r ≤ 0 makes sense, and we are looking for all of the points between the pole r = 0 and the lima¸con as θ ranges over the interval 2π r. In other words, we shade in the inner loop of the lima¸con. 3 to 4π 3, 4π 3 6 4 2 −2 2
π 3 4π 3 π 2 π 3π 2 2π θ y θ = 2π 3 θ = 4π 3 x (r, θ) \| 2 + 4 cos(θ) ≤ r ≤ 0, 2π 3 ≤ θ ≤ 4π 3 4. We have two regions described here connected with the union symbol ‘∪.’ We shade each in turn and ﬁnd our ﬁnal answer by combining the two. In Example 11.5.3, number 1, we found that the curves r = 2 sin(θ) and r = 2 − 2 sin(θ) intersect when θ = π 6. Hence, for the, we are shading the region between the origin ﬁrst region, (r, θ) \| 0 ≤ r ≤ 2 sin(θ), 0 ≤ θ ≤ π 6 (r = 0) out to the circle (r = 2 sin(θ)) as θ ranges from 0 to π 6, which is the angle of intersection of the two curves. For the second region, (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), π, θ picks up 6 ≤ θ ≤ π where it left oﬀ at π 2. In this case, however, we are shading from the origin (r = 0) out to the cardioid r = 2 − 2 sin(θ) which pulls into the origin at θ = π 2. Putting these two regions together gives us our ﬁnal answer. 6 and continues to sin(θ) and r = 2 sin(θ) (r, θ) \| 0 ≤ r ≤ 2 sin(θ), 0 ≤ θ ≤ π 6 (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), π ∪ 6 ≤ θ ≤ π 2 958 Applications of Trigonometry 11.5.1 Exercises In Exercises 1 - 20, plot the graph of the polar equation by hand. Carefully label your graphs. 1. Circle: r = 6 sin(θ) 2. Circle: r = 2 cos(θ) 3. Rose: r = 2 sin(2θ) 4. Rose: r = 4 cos(2θ) 5. Rose: r =
5 sin(3θ) 6. Rose: r = cos(5θ) 7. Rose: r = sin(4θ) 8. Rose: r = 3 cos(4θ) 9. Cardioid: r = 3 − 3 cos(θ) 10. Cardioid: r = 5 + 5 sin(θ) 11. Cardioid: r = 2 + 2 cos(θ) 12. Cardioid: r = 1 − sin(θ) 13. Lima¸con: r = 1 − 2 cos(θ) 14. Lima¸con: r = 1 − 2 sin(θ) 15. Lima¸con: r = 2 √ 3 + 4 cos(θ) 16. Lima¸con: r = 3 − 5 cos(θ) 17. Lima¸con: r = 3 − 5 sin(θ) 18. Lima¸con: r = 2 + 7 sin(θ) 19. Lemniscate: r2 = sin(2θ) 20. Lemniscate: r2 = 4 cos(2θ) In Exercises 21 - 30, ﬁnd the exact polar coordinates of the points of intersection of graphs of the polar equations. Remember to check for intersection at the pole (origin). 21. r = 3 cos(θ) and r = 1 + cos(θ) 22. r = 1 + sin(θ) and r = 1 − cos(θ) 23. r = 1 − 2 sin(θ) and r = 2 24. r = 1 − 2 cos(θ) and r = 1 25. r = 2 cos(θ) and r = 2 27. r2 = 4 cos(2θ) and r = 2 √ 3 sin(θ) √ 26. r = 3 cos(θ) and r = sin(θ) 28. r2 = 2 sin(2θ) and r = 1 29. r = 4 cos(2θ) and r = 2 30. r = 2 sin(2θ) and r = 1 In Exercises 31 - 40, sketch the region in the xy-plane described by the given set. 31. {(r, θ) \| 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π} 32. {
(r, θ) \| 0 ≤ r ≤ 4 sin(θ), 0 ≤ θ ≤ π} 33. (r, θ) \| 0 ≤ r ≤ 3 cos(θ), − π 2 ≤ θ ≤ π 2 34. (r, θ) \| 0 ≤ r ≤ 2 sin(2θ), 0 ≤ θ ≤ π 2 11.5 Graphs of Polar Equations 959 35. (r, θ) \| 0 ≤ r ≤ 4 cos(2θ), − π 4 ≤ θ ≤ π 37. (r, θ) \| 1 + cos(θ) ≤ r ≤ 3 cos(θ), − π 4 3 ≤ θ ≤ π 3 36. (r, θ) \| 1 ≤ r ≤ 1 − 2 cos(θ), π 2 ≤ θ ≤ 3π 2 38. (r, θ) \| 1 ≤ r ≤ 2 sin(2θ), 13π √ 12 ≤ θ ≤ 17π 12 39. (r, θ) \| 0 ≤ r ≤ 2 ∪ (r, θ) \| 0 ≤ r ≤ 2 cos(θ), π 3 sin(θ), 0 ≤ θ ≤ π 6 ∪ (r, θ) \| 0 ≤ r ≤ 1, π 40. (r, θ) \| 0 ≤ r ≤ 2 sin(2θ), 0 ≤ θ ≤ π 12 12 ≤ In Exercises 41 - 50, use set-builder notation to describe the polar region. Assume that the region contains its bounding curves. 41. The region inside the circle r = 5. 42. The region inside the circle r = 5 which lies in Quadrant III. 43. The region inside the left half of the circle r = 6 sin(θ). 44. The region inside the circle r = 4 cos(θ) which lies in Quadrant IV. 45. The region inside the top half of the cardioid r = 3 − 3 cos(θ) 46. The region inside the cardioid r = 2 − 2 sin(θ) which lies in Quadrants I and IV. 47. The inside of the petal of the rose r = 3 cos(4θ) which lies on the positive x-axis 48. The region inside the circle r = 5 but outside the circle r
= 3. 49. The region which lies inside of the circle r = 3 cos(θ) but outside of the circle r = sin(θ) 50. The region in Quadrant I which lies inside both the circle r = 3 as well as the rose r = 6 sin(2θ) While the authors truly believe that graphing polar curves by hand is fundamental to your understanding of the polar coordinate system, we would be derelict in our duties if we totally ignored the graphing calculator. Indeed, there are some important polar curves which are simply too difﬁcult to graph by hand and that makes the calculator an important tool for your further studies in Mathematics, Science and Engineering. We now give a brief demonstration of how to use the graphing calculator to plot polar curves. The ﬁrst thing you must do is switch the MODE of your calculator to POL, which stands for “polar”. 960 Applications of Trigonometry This changes the “Y=” menu as seen above in the middle. Let’s plot the polar rose given by r = 3 cos(4θ) from Exercise 8 above. We type the function into the “r=” menu as seen above on the right. We need to set the viewing window so that the curve displays properly, but when we look at the WINDOW menu, we ﬁnd three extra lines. In order for the calculator to be able to plot r = 3 cos(4θ) in the xy-plane, we need to tell it not only the dimensions which x and y will assume, but we also what values of θ to use. From our previous work, we know that we need 0 ≤ θ ≤ 2π, so we enter the data you see above. (I’ll say more about the θ-step in just a moment.) Hitting GRAPH yields the curve below on the left which doesn’t look quite right. The issue here is that the calculator screen is 96 pixels wide but only 64 pixels tall. To get a true geometric perspective, we need to hit ZOOM SQUARE (seen below in the middle) to produce a more accurate graph which we present below on the right. In function mode, the calculator automatically divided the interval [Xmin, Xmax] into 96 equal subintervals. In polar mode, however, we must specify how to split up the interval [�
�min, θmax] using the θstep. For most graphs, a θstep of 0.1 is ﬁne. If you make it too small then the calculator takes a long time to graph. It you make it too big, you get chunky garbage like this. You will need to experiment with the settings in order to get a nice graph. Exercises 51 - 60 give you some curves to graph using your calculator. Notice that some of them have explicit bounds on θ and others do not. 11.5 Graphs of Polar Equations 961 51. r = θ, 0 ≤ θ ≤ 12π 52. r = ln(θ), 1 ≤ θ ≤ 12π 53. r = e.1θ, 0 ≤ θ ≤ 12π 54. r = θ3 − θ, −1.2 ≤ θ ≤ 1.2 55. r = sin(5θ) − 3 cos(θ) 57. r = arctan(θ), −π ≤ θ ≤ π 59. r = 1 2 − cos(θ) 56. r = sin3 θ 2 + cos2 θ 3 58. r = 1 1 − cos(θ) 60. r = 1 2 − 3 cos(θ) 61. How many petals does the polar rose r = sin(2θ) have? What about r = sin(3θ), r = sin(4θ) and r = sin(5θ)? With the help of your classmates, make a conjecture as to how many petals the polar rose r = sin(nθ) has for any natural number n. Replace sine with cosine and repeat the investigation. How many petals does r = cos(nθ) have for each natural number n? Looking back through the graphs in the section, it’s clear that many polar curves enjoy various forms of symmetry. However, classifying symmetry for polar curves is not as straight-forward as it was for equations back on page 26. In Exercises 62 - 64, we have you and your classmates explore some of the more basic forms of symmetry seen in common polar curves. 62. Show that if f is even17 then the graph of r = f (θ) is symmetric about the x-axis. (a) Show that f (θ) = 2 + 4
cos(θ) is even and verify that the graph of r = 2 + 4 cos(θ) is indeed symmetric about the x-axis. (See Example 11.5.2 number 2.) (b) Show that f (θ) = 3 sin θ 2 is not even, yet the graph of r = 3 sin θ 2 is symmetric about the x-axis. (See Example 11.5.3 number 4.) 63. Show that if f is odd18 then the graph of r = f (θ) is symmetric about the origin. (a) Show that f (θ) = 5 sin(2θ) is odd and verify that the graph of r = 5 sin(2θ) is indeed symmetric about the origin. (See Example 11.5.2 number 3.) (b) Show that f (θ) = 3 cos θ 2 is not odd, yet the graph of r = 3 cos θ 2 is symmetric about the origin. (See Example 11.5.3 number 4.) 64. Show that if f (π − θ) = f (θ) for all θ in the domain of f then the graph of r = f (θ) is symmetric about the y-axis. (a) For f (θ) = 4 − 2 sin(θ), show that f (π − θ) = f (θ) and the graph of r = 4 − 2 sin(θ) is symmetric about the y-axis, as required. (See Example 11.5.2 number 1.) 17Recall that this means f (−θ) = f (θ) for θ in the domain of f. 18Recall that this means f (−θ) = −f (θ) for θ in the domain of f. 962 Applications of Trigonometry (b) For f (θ) = 5 sin(2θ), show that f π − π 4 = f π 4, yet the graph of r = 5 sin(2θ) is symmetric about the y-axis. (See Example 11.5.2 number 3.) In Section 1.7, we discussed transformations of graphs. classmates explore transformations of polar graphs. In Exercise 65 we have you and your 65. For Exercises 65a and 65b below, let f (�
�) = cos(θ) and g(θ) = 2 − sin(θ). (a) Using your graphing calculator, compare the graph of r = f (θ) to each of the graphs of. Repeat this process for g(θ). In general, how do you think the graph of r = f (θ + α) compares with the graph of r = f (θ)? and r = f θ − 3π 4, r = f θ + 3π 4 (b) Using your graphing calculator, compare the graph of r = f (θ) to each of the graphs of r = 2f (θ), r = 1 2 f (θ), r = −f (θ) and r = −3f (θ). Repeat this process for g(θ). In general, how do you think the graph of r = k · f (θ) compares with the graph of r = f (θ)? (Does it matter if k > 0 or k < 0?) 66. In light of Exercises 62 - 64, how would the graph of r = f (−θ) compare with the graph of r = f (θ) for a generic function f? What about the graphs of r = −f (θ) and r = f (θ)? What about r = f (θ) and r = f (π − θ)? Test out your conjectures using a variety of polar functions found in this section with the help of a graphing utility. 67. With the help of your classmates, research cardioid microphones. 68. Back in Section 1.2, in the paragraph before Exercise 53, we gave you this link to a fascinating list of curves. Some of these curves have polar representations which we invite you and your classmates to research. 11.5 Graphs of Polar Equations 963 11.5.2 Answers 1. Circle: r = 6 sin(θ) y −6 6 −6 3. Rose: r = 2 sin(2θ) y −2 2 −2 2. Circle: r = 2 cos(θ) y 2 6 x −2 2 x −2 4. Rose: r = 4 cos(2θ) y 4 θ = 3π 4 θ = π 4 2 x −4 4 x 5. Rose: r = 5 sin(3θ) y
5 θ = 2π 3 θ = π 3 −5 5 x −4 6. Rose: r = cos(5θ) y 1 θ = 7π 10 θ = 3π 10 θ = 9π 10 −1 θ = π 10 1 x −5 −1 964 Applications of Trigonometry 7. Rose: r = sin(4θ) y 1 θ = 3π 4 θ = π 4 −1 1 x 8. Rose: r = 3 cos(4θ) y θ = 5π 8 3 θ = 3π 8 θ = 7π 8 −3 θ = π 8 3 x −1 −3 9. Cardioid: r = 3 − 3 cos(θ) 10. Cardioid: r = 5 + 5 sin(θ) y 6 3 y 10 5 −6 −3 3 6 x −10 −5 5 10 x −3 −6 −5 −10 11. Cardioid: r = 2 + 2 cos(θ) 12. Cardioid: r = 1 − sin(θ) y 4 2 y 2 1 −4 −2 2 4 x −2 −1 1 2 x −2 −4 −1 −2 11.5 Graphs of Polar Equations 965 13. Lima¸con: r = 1 − 2 cos(θ) 14. Lima¸con: r = 1 − 2 sin(θ = 5π 6 −3 −1 1 3 x −3 −1 1 3 x −1 −3 θ = 5π 3 15. Lima¸con: r = 2 √ 3 + 4 cos(θ) y √ 2 3 + 4 θ = 5π 6 √ 2 3 √ −2 3 − 4 θ = 7π 6 √ −2 3 √ −1 −3 16. Lima¸con: r = 3 − 5 cos(θ) y 8 3 θ = arccos 3 5 −8 −2 8 x −3 −8 θ = 2π − arccos 3 5 17. Lima¸con: r = 3 − 5 sin(θ) 18. Lima¸con: r = 2 + 7 sin(θ) y 8 θ = π − arcsin 3 5 θ = arcsin 3 5 y 9 5 −8 −3 3
−2 −8 8 x −9 θ = π + arcsin −2 2 7 2 9 θ = 2π − arcsin 2 7 x −9 966 Applications of Trigonometry 19. Lemniscate: r2 = sin(2θ) 20. Lemniscate: r2 = 4 cos(2θ) y 1 y 2 θ = 3π 4 θ = π 4 −1 1 x −2 2 x −1 −2 3 2, π 3, 3 2, 5π 3, pole √ 2 + 2, √ 2 − 2 2, 7π 4 2, 3π 4, pole 21. r = 3 cos(θ) and r = 1 + cos(θ) y 3 2 1 −3 −2 −1 1 2 3 −1 −2 −3 x 22. r = 1 + sin(θ) and r = 1 − cos(θ) y 2 1 −2 −1 1 2 x −1 −2 11.5 Graphs of Polar Equations 967 2, 7π 6, 2, 11π 6 1,, π 2 1, 3π 2, (−1, 0) √ 3, π 6, pole 23. r = 1 − 2 sin(θ) and r = 2 y 3 1 −3 −1 1 3 x −1 −3 24. r = 1 − 2 cos(θ) and r = 1 y 3 1 −3 −1 1 3 x −1 −3 25. r = 2 cos(θ) and r = 2 √ 3 sin(θ) y 4 3 2 1 −3 −2 −1 1 2 3 x −1 −2 −3 −4 968 Applications of Trigonometry √ 3 10 10, arctan(3), pole √ 2,, π 6 √, 5π 6 2, √, 7π 6 2, √ 2, 11π 6 1,, π 12 1,, 5π 12 1, 13π 12, 1, 17π 12 26. r = 3 cos(θ) and r = sin(θ) y 3 2 1 −3 −2 −1 1 2 3 x −1 −2 −3 27. r2 = 4 cos(2θ) and r = √ 2 y 2 −2 2 x −2 28. r
2 = 2 sin(2θ) and r = 1 y √ 2 1 −1 11.5 Graphs of Polar Equations 969 29. r = 4 cos(2θ) and r = 2 y 4 2, 2,, π 6 11π 6, 7π 6 2, 2, 5π 6, −2,, π 3, −2,, 2π 3 −2, 4π 3, −2, 5π 3 −4 4 x −4 30. r = 2 sin(2θ) and r = 1 y 2 −2 2 x −2, π 12, 17π 12 1, 1, 1, −1,, 19π 12, 5π 12 1,, 7π 12 −1, −1, 23π 12, 13π 12 −1,, 11π 12 970 Applications of Trigonometry 31. {(r, θ) \| 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π} 32. {(r, θ) \| 0 ≤ r ≤ 4 sin(θ), 0 ≤ θ ≤ π3 −2 −1 1 2 3 −1 −2 −3 x x −4 −3 −2 −1 1 2 3 4 −1 −2 −3 −4 33. (r, θ) \| 0 ≤ r ≤ 3 cos(θ), − π 2 ≤ θ ≤ π 2 34. (r, θ) \| 0 ≤ r ≤ 2 sin(2θ), 3 −2 −1 1 2 3 −1 −2 −3 x −2 2 x −2 35. (r, θ) \| 0 ≤ r ≤ 4 cos(2θ), − π 4 ≤ θ ≤ π 4 36. (r, θ) \| 1 ≤ r ≤ 1 − 2 cos(θ), π 2 ≤ θ ≤ 3π 2 y 4 y 3 1 −4 4 x −3 −1 1 3 x −4 −1 −3 11.5 Graphs of Polar Equations 971 37. (r, θ) \| 1 + cos(θ) ≤ r ≤ 3 cos(θ), − 3 −2 −1 1 2 3 −1 −2 −3 x 38. (r, θ) \| 1 ≤ r ≤ y 2 sin(2θ), 13π 12 ≤ θ ≤ 17π 12 �
� − 2 −1 √ 2 1 −r, θ) \| 0 ≤ r ≤ 2 cos(θ), π 6 ≤ θ ≤ π 2 39. (r, θ) \| 0 ≤ r ≤ 2 3 sin(θ), 3 −2 −1 1 2 3 x −1 −2 −3 −4 972 Applications of Trigonometry ∪ (r, θ) \| 0 ≤ r ≤ 1, π 12 ≤ θ ≤ π 4 40. (r, θ) \| 0 ≤ r ≤ 2 sin(2θ), 0 ≤ θ ≤ π 12 y −2 2 −2 2 x 41. {(r, θ) \| 0 ≤ r ≤ 5, 0 ≤ θ ≤ 2π} 42. (r, θ) \| 0 ≤ r ≤ 5, π ≤ θ ≤ 3π 2 43. (r, θ) \| 0 ≤ r ≤ 6 sin(θ), π 44. (r, θ) \| 4 cos(θ) ≤ r ≤ 0 45. {(r, θ) \| 0 ≤ r ≤ 3 − 3 cos(θ), 0 ≤ θ ≤ π} 46. (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), 0 ≤ θ ≤ π 2 or (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), 3π 47. (r, θ) \| 0 ≤ r ≤ 3 cos(4θ), 0 ≤ θ ≤ π 8 or (r, θ) \| 0 ≤ r ≤ 3 cos(4θ), − ≤ 5π 2 ∪ (r, θ) \| 0 ≤ r ≤ 2 − 2 sin(θ), 3π 2 ≤ θ ≤ 2π ∪ (r, θ) \| 0 ≤ r ≤ 3 cos(4θ), 15π 8 ≤ θ ≤ 2π 48. {(r, θ) \| 3 ≤ r ≤ 5, 0 ≤ θ ≤ 2π} 49. (r, θ) \| 0 ≤ r ≤ 3 cos(θ), − π 50. (r, θ) \| 0 ≤ r ≤ 6 sin(2θ), 0 ≤ θ ≤ π 12 12 ≤ θ ≤ π (r, θ) \| 0 ≤
r ≤ 6 sin(2θ), 5π 2 2 ≤ θ ≤ 0 ∪ {(r, θ) \| sin(θ) ≤ r ≤ 3 cos(θ), 0 ≤ θ ≤ arctan(3)} ∪ (r, θ) \| 0 ≤ r ≤ 3, π 12 ≤ θ ≤ 5π 12 ∪ 11.6 Hooked on Conics Again 973 11.6 Hooked on Conics Again In this section, we revisit our friends the Conic Sections which we began studying in Chapter 7. Our ﬁrst task is to formalize the notion of rotating axes so this subsection is actually a follow-up to Example 8.3.3 in Section 8.3. In that example, we saw that the graph of y = 2 x is actually a hyperbola. More speciﬁcally, it is the hyperbola obtained by rotating the graph of x2 − y2 = 4 counter-clockwise through a 45◦ angle. Armed with polar coordinates, we can generalize the process of rotating axes as shown below. 11.6.1 Rotation of Axes Consider the x- and y-axes below along with the dashed x- and y-axes obtained by rotating the xand y-axes counter-clockwise through an angle θ and consider the point P (x, y). The coordinates (x, y) are rectangular coordinates and are based on the x- and y-axes. Suppose we wished to ﬁnd rectangular coordinates based on the x- and y-axes. That is, we wish to determine P (x, y). While this seems like a formidable challenge, it is nearly trivial if we use polar coordinates. Consider the angle φ whose initial side is the positive x-axis and whose terminal side contains the point P. y y P (x, y) = P (x, y) x θ φ θ x We relate P (x, y) and P (x, y) by converting them to polar coordinates. Converting P (x, y) to polar coordinates with r > 0 yields x = r cos(θ + φ) and y = r sin(θ + φ). To convert the point P (x, y) into polar coordinates, we ﬁrst match the polar axis with the positive x-axis, choose
the same r > 0 (since the origin is the same in both systems) and get x = r cos(φ) and y = r sin(φ). Using the sum formulas for sine and cosine, we have x = r cos(θ + φ) = r cos(θ) cos(φ) − r sin(θ) sin(φ) = (r cos(φ)) cos(θ) − (r sin(φ)) sin(θ) = x cos(θ) − y sin(θ) Sum formula for cosine Since x = r cos(φ) and y = r sin(φ) 974 Applications of Trigonometry Similarly, using the sum formula for sine we get y = x sin(θ) + y cos(θ). These equations enable us to easily convert points with xy-coordinates back into xy-coordinates. They also enable us to easily convert equations in the variables x and y into equations in the variables in terms of x and y.1 If we want equations which enable us to convert points with xy-coordinates into xy-coordinates, we need to solve the system x cos(θ) − y sin(θ) = x x sin(θ) + y cos(θ) = y for x and y. Perhaps the cleanest way2 to solve this system is to write it as a matrix equation. Using the machinery developed in Section 8.4, we write the above system as the matrix equation AX = X where A = cos(θ) − sin(θ) cos(θ) sin(θ Since det(A) = (cos(θ))(cos(θ)) − (− sin(θ))(sin(θ)) = cos2(θ) + sin2(θ) = 1, the determinant of A is not zero so A is invertible and X = A−1X. Using the formula given in Equation 8.2 with det(A) = 1, we ﬁnd so that A−1 = cos(θ) − sin(θ) sin(θ) cos(θ) X = A−1X x y x y = = cos(θ) − sin(θ) sin(θ) cos(θ) x y
x cos(θ) + y sin(θ) −x sin(θ) + y cos(θ) From which we get x = x cos(θ) + y sin(θ) and y = −x sin(θ) + y cos(θ). To summarize, Theorem 11.9. Rotation of Axes: Suppose the positive x and y axes are rotated counterclockwise through an angle θ to produce the axes x and y, respectively. Then the coordinates P (x, y) and P (x, y) are related by the following systems of equations x = x cos(θ) − y sin(θ) y = x sin(θ) + y cos(θ) and x = x cos(θ) + y sin(θ) y = −x sin(θ) + y cos(θ) We put the formulas in Theorem 11.9 to good use in the following example. 1Sound familiar? In Section 11.4, the equations x = r cos(θ) and y = r sin(θ) make it easy to convert points from polar coordinates into rectangular coordinates, and they make it easy to convert equations from rectangular coordinates into polar coordinates. 2We could, of course, interchange the roles of x and x, y and y and replace φ with −φ to get x and y in terms of x and y, but that seems like cheating. The matrix A introduced here is revisited in the Exercises. 11.6 Hooked on Conics Again 975 Example 11.6.1. Suppose the x- and y- axes are both rotated counter-clockwise through the angle θ = π 3 to produce the x- and y- axes, respectively. 1. Let P (x, y) = (2, −4) and ﬁnd P (x, y). Check your answer algebraically and graphically. 2. Convert the equation 21x2 + 10xy √ 3 + 31y2 = 144 to an equation in x and y and graph. Solution. √ 3, Theorem 11.9 gives x = x cos(θ) + y sin(θ) = 2 cos π √ 1. If P (x, y) = (2, −4) then x = 2 and y = −4. Using these values for x and y
along with which simpliﬁes 3. Similarly, y = −x sin(θ) + y cos(θ) = (−2) sin π + (−4) cos π which √ 3 3 3. To check our answer 3. Hence P (x, y to x = 1 − 2 √ gives y = − algebraically, we use the formulas in Theorem 11.9 to convert P (x, y) = 1 − 2 back into x and y coordinates. We get + (−4) sin π 3 3 − 2 = −2 − 3, −2 − 3, −2 − √ √ 3 x = x cos(θ) − y sin(θ) √ − (−2 − 3) cos ) sin π 3 = ( Similarly, using y = x sin(θ) + y cos(θ), we obtain y = −4 as required. To check our answer graphically, we sketch in the x-axis and y-axis to see if the new coordinates P (x, y) = 3 ≈ (−2.46, −3.73) seem reasonable. Our graph is below. 1 − 2 3, −x, y) = (2, −4) P (x, y) ≈ (−2.46, −3.73) 2. To convert the equation 21x2 +10xy we substitute x = x cos π 3 − y sin π 3 √ 3+31y2 = 144 to an equation in the variables x and y, = x 2 + y and y = x sin π 3 + y cos 976 Applications of Trigonometry and simplify. While this is by no means a trivial task, it is nothing more than a hefty dose of Beginning Algebra. We will not go through the entire computation, but rather, the reader should take the time to do it. Start by verifying that x2 = (x)2 4 − √ xy 2 3 + 3(y)2 4, √ (x)2 4 3 − xy 2 − xy = √ 3, y2 = 3(x)2 4 + √ xy 2 3 + (y)2 4 To our surprise and delight, the equation 21x2 + 10xy 3 + 31y2 = 144 in xy-coordinates 4 + (y)2 reduces to 36(x)2 + 16
(y)2 = 144, or (x)2 9 = 1 in xy-coordinates. The latter is an ellipse centered at (0, 0) with vertices along the y-axis with (xy-coordinates) (0, ±3) and whose minor axis has endpoints with (xy-coordinates) (±2, 0). We graph it below. (y) 21x2 + 10xy √ 3 + 31y2 = 144 √ The elimination of the troublesome ‘xy’ term from the equation 21x2 + 10xy 3 + 31y2 = 144 in Example 11.6.1 number 2 allowed us to graph the equation by hand using what we learned in Chapter 7. It is natural to wonder if we can always do this. That is, given an equation of the form Ax2 +Bxy +Cy2 +Dx+Ey +F = 0, with B = 0, is there an angle θ so that if we rotate the x and yaxes counter-clockwise through that angle θ, the equation in the rotated variables x and y contains no xy term? To explore this conjecture, we make the usual substitutions x = x cos(θ) − y sin(θ) and y = x sin(θ) + y cos(θ) into the equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 and set the coeﬃcient of the xy term equal to 0. Terms containing xy in this expression will come from the ﬁrst three terms of the equation: Ax2, Bxy and Cy2. We leave it to the reader to verify that x2 = (x)2 cos2(θ) − 2xy cos(θ) sin(θ) + (y)2 sin(θ) xy = (x)2 cos(θ) sin(θ) + xy cos2(θ) − sin2(θ) − (y)2 cos(θ) sin(θ) y2 = (x)2 sin2(θ) + 2xy cos(θ) sin(θ) + (y)2 cos2(θ) 11.6 Hooked on Conics Again 977 The contribution to the xy-term from Ax2 is −2A cos(θ
) sin(θ), from Bxy it is B cos2(θ) − sin2(θ), and from Cy2 it is 2C cos(θ) sin(θ). Equating the xy-term to 0, we get −2A cos(θ) sin(θ) + B cos2(θ) − sin2(θ) + 2C cos(θ) sin(θ) = 0 −A sin(2θ) + B cos(2θ) + C sin(2θ) = 0 Double Angle Identities From this, we get B cos(2θ) = (A − C) sin(2θ), and our goal is to solve for θ in terms of the coeﬃcients A, B and C. Since we are assuming B = 0, we can divide both sides of this equation by B. To solve for θ we would like to divide both sides of the equation by sin(2θ), provided of course that we have assurances that sin(2θ) = 0. If sin(2θ) = 0, then we would have B cos(2θ) = 0, and since B = 0, this would force cos(2θ) = 0. Since no angle θ can have both sin(2θ) = 0 and cos(2θ) = 0, we can safely assume3 sin(2θ) = 0. We get cos(2θ) B. We have just proved the following theorem. B, or cot(2θ) = A−C sin(2θ) = A−C Theorem 11.10. The equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 with B = 0 can be transformed into an equation in variables x and y without any xy terms by rotating the xand y- axes counter-clockwise through an angle θ which satisﬁes cot(2θ) = A−C B. We put Theorem 11.10 to good use in the following example. Example 11.6.2. Graph the following equations. 1. 5x2 + 26xy + 5y2 − 16x √ 2 + 16y √ 2 − 104 = 0 2. 16x2 + 24xy + 9y2 + 15x
− 20y = 0 Solution. 1. Since the equation 5x2 + 26xy + 5y2 − 16x 2 − 104 = 0 is already given to us in the form required by Theorem 11.10, we identify A = 5, B = 26 and C = 5 so that cot(2θ) = A−C 26 = 0. This means cot(2θ) = 0 which gives θ = π 2 k for integers k. √ √ 2 + y 2 2 We choose θ = π 2. The reader should verify that 4 so that our rotation equations are x = x 4 + π and y = x B = 5−5 2 − y 2 + 16y √ √ 2 2 2 √ √ x2 = (x)2 2 − xy + (y)2 2, xy = (x)2 2 − (y)2 2, y2 = (x)2 2 + xy + (y)2 2 Making the other substitutions, we get that 5x2 + 26xy + 5y2 − 16x 2 − 104 = 0 2 + 16y reduces to 18(x)2 − 8(y)2 + 32y − 104 = 0, or (x)2 9 = 1. The latter is the equation of a hyperbola centered at the xy-coordinates (0, 2) opening in the x direction with vertices (±2, 2) (in xy-coordinates) and asymptotes y = ± 3 4 − (y−2)2 2 x + 2. We graph it below. √ √ 3The reader is invited to think about the case sin(2θ) = 0 geometrically. What happens to the axes in this case? 978 Applications of Trigonometry 24 2. From 16x2 + 24xy + 9y2 + 15x − 20y = 0, we get A = 16, B = 24 and C = 9 so that cot(2θ) = 7 24. Since this isn’t one of the values of the common angles, we will need to use inverse functions. Ultimately, we need to ﬁnd cos(θ) and sin(θ), which means we have two If we use the arccotangent function immediately, after the usual calculations we options.. To get cos(�
�) and sin(θ) from this, we would need to use half angle 2 arccot 7 get θ = 1 identities. Alternatively, we can start with cot(2θ) = 7 24, use a double angle identity, and then go after cos(θ) and sin(θ). We adopt the second approach. From cot(2θ) = 7 24, we have tan(2θ) = 24 7, which 7. Using the double angle identity for tangent, we have gives 24 tan2(θ) + 14 tan(θ) − 24 = 0. Factoring, we get 2(3 tan(θ) + 4)(4 tan(θ) − 3) = 0 which gives tan(θ) = − 4 4. While either of these values of tan(θ) satisﬁes the equation. To 4, since this produces an acute angle,4 θ = arctan 3 cot(2θ) = 7. ﬁnd the rotation equations, we need cos(θ) = cos arctan 3 4 Using the techniques developed in Section 10.6 we get cos(θ) = 4 5. Our rotation 5 + 4y 5 − 3y equations are x = x cos(θ) − y sin(θ) = 4x 5. As usual, we now substitute these quantities into 16x2 + 24xy + 9y2 + 15x − 20y = 0 and simplify. As a ﬁrst step, the reader can verify 3 or tan(θ) = 3 24, we choose tan(θ) = 3 5 and y = x sin(θ) + y cos(θ) = 3x and sin(θ) = sin arctan 3 4 5 and sin(θ) = 3 1−tan2(θ) = 24 2 tan(θ) 4 x2 = 16(x)2 25 − 24xy 25 + 9(y)2 25, xy = 12(x)2 25 + 7xy 25 − 12(y)2 25, y2 = 9(x)2 25 + 24xy 25 + 16(y)2 25 Once the dust settles, we get 25(x)2 − 25y = 0, or y = (x)2, whose graph is a parabola opening
along the positive y-axis with vertex (0, 0). We graph this equation below = arctan 3 4 x 5x2 + 26xy + 5y2 − 16x √ 2 + 16y √ 2 − 104 = 0 16x2 + 24xy + 9y2 + 15x − 20y = 0 4As usual, there are inﬁnitely many solutions to tan(θ) = 3. The 4 reader is encouraged to think about why there is always at least one acute answer to cot(2θ) = A−C B and what this means geometrically in terms of what we are trying to accomplish by rotating the axes. The reader is also encouraged to keep a sharp lookout for the angles which satisfy tan(θ) = − 4 4. We choose the acute angle θ = arctan 3 3 in our ﬁnal graph. (Hint: 3 = −1.) − 4 3 4 11.6 Hooked on Conics Again 979 We note that even though the coeﬃcients of x2 and y2 were both positive numbers in parts 1 and 2 of Example 11.6.2, the graph in part 1 turned out to be a hyperbola and the graph in part 2 worked out to be a parabola. Whereas in Chapter 7, we could easily pick out which conic section we were dealing with based on the presence (or absence) of quadratic terms and their coeﬃcients, Example 11.6.2 demonstrates that all bets are oﬀ when it comes to conics with an xy term which require rotation of axes to put them into a more standard form. Nevertheless, it is possible to determine which conic section we have by looking at a special, familiar combination of the coeﬃcients of the quadratic terms. We have the following theorem. Theorem 11.11. Suppose the equation Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 describes a non-degenerate conic section.a If B2 − 4AC > 0 then the graph of the equation is a hyperbola. If B2 − 4AC = 0 then the graph of the equation is a parabola. If B2 − 4AC < 0 then the graph of the equation is an ellipse or circle. aRecall that this means its graph is either a
circle, parabola, ellipse or hyperbola. See page 497. As you may expect, the quantity B2 −4AC mentioned in Theorem 11.11 is called the discriminant of the conic section. While we will not attempt to explain the deep Mathematics which produces this ‘coincidence’, we will at least work through the proof of Theorem 11.11 mechanically to show that it is true.5 First note that if the coeﬃcient B = 0 in the equation Ax2 +Bxy +Cy2 +Dx+Ey +F = 0, Theorem 11.11 reduces to the result presented in Exercise 34 in Section 7.5, so we proceed here under the assumption that B = 0. We rotate the xy-axes counter-clockwise through an angle θ which satisﬁes cot(2θ) = A−C to produce an equation with no xy-term in accordance with B Theorem 11.10: A(x)2 + C(y)2 + Dx + Ey + F = 0. In this form, we can invoke Exercise 34 in Section 7.5 once more using the product AC. Our goal is to ﬁnd the product AC in terms of the coeﬃcients A, B and C in the original equation. To that end, we make the usual substitutions x = x cos(θ) − y sin(θ) y = x sin(θ) + y cos(θ) into Ax2 + Bxy + Cy2 + Dx + Ey + F = 0. We leave it to the reader to show that, after gathering like terms, the coeﬃcient A on (x)2 and the coeﬃcient C on (y)2 are A = A cos2(θ) + B cos(θ) sin(θ) + C sin2(θ) C = A sin2(θ) − B cos(θ) sin(θ) + C cos2(θ) In order to make use of the condition cot(2θ) = A−C the power reduction formulas. After some regrouping, we get B, we rewrite our formulas for A and C using 2A = [(A + C) + (A − C) cos(2θ)] +
B sin(2θ) 2C = [(A + C) − (A − C) cos(2θ)] − B sin(2θ) Next, we try to make sense of the product (2A)(2C) = {[(A + C) + (A − C) cos(2θ)] + B sin(2θ)} {[(A + C) − (A − C) cos(2θ)] − B sin(2θ)} 5We hope that someday you get to see why this works the way it does. 980 Applications of Trigonometry We break this product into pieces. First, we use the diﬀerence of squares to multiply the ‘ﬁrst’ quantities in each factor to get [(A + C) + (A − C) cos(2θ)] [(A + C) − (A − C) cos(2θ)] = (A + C)2 − (A − C)2 cos2(2θ) Next, we add the product of the ‘outer’ and ‘inner’ quantities in each factor to get −B sin(2θ) [(A + C) + (A − C) cos(2θ)] +B sin(2θ) [(A + C) − (A − C) cos(2θ)] = −2B(A − C) cos(2θ) sin(2θ) The product of the ‘last’ quantity in each factor is (B sin(2θ))(−B sin(2θ)) = −B2 sin2(2θ). Putting all of this together yields 4AC = (A + C)2 − (A − C)2 cos2(2θ) − 2B(A − C) cos(2θ) sin(2θ) − B2 sin2(2θ) From cot(2θ) = A−C sin(2θ) = A−C twice along with the Pythagorean Identity cos2(2θ) = 1 − sin2(2θ) to get B, we get cos(2θ) B, or (A−C) sin(2θ) = B cos(2θ). We use this substitution 4AC = (A + C
)2 − (A − C)2 cos2(2θ) − 2B(A − C) cos(2θ) sin(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 1 − sin2(2θ) − 2B cos(2θ)B cos(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 + (A − C)2 sin2(2θ) − 2B2 cos2(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 + [(A − C) sin(2θ)]2 − 2B2 cos2(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 + [B cos(2θ)]2 − 2B2 cos2(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 + B2 cos2(2θ) − 2B2 cos2(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 − B2 cos2(2θ) − B2 sin2(2θ) = (A + C)2 − (A − C)2 − B2 cos2(2θ) + sin2(2θ) = (A + C)2 − (A − C)2 − B2 = A2 + 2AC + C2 − A2 − 2AC + C2 − B2 = 4AC − B2 Hence, B2 − 4AC = −4AC, so the quantity B2 − 4AC has the opposite sign of AC. The result now follows by applying Exercise 34 in Section 7.5. Example 11.6.3. Use Theorem 11.11 to classify the graphs of the following non-degenerate conics. 1. 21x2 + 10xy √ 3 + 31y2 = 144 2. 5x2 + 26xy + 5y2 − 16x √ 2 + 16y √ 2 − 104 = 0 3. 16x2
+ 24xy + 9y2 + 15x − 20y = 0 Solution. This is a straightforward application of Theorem 11.11. 11.6 Hooked on Conics Again 981 1. We have A = 21, B = 10 3)2 − 4(21)(31) = −2304 < 0. Theorem 11.11 predicts the graph is an ellipse, which checks with our work from Example 11.6.1 number 2. 3 and C = 31 so B2 − 4AC = (10 √ √ 2. Here, A = 5, B = 26 and C = 5, so B2 − 4AC = 262 − 4(5)(5) = 576 > 0. Theorem 11.11 classiﬁes the graph as a hyperbola, which matches our answer to Example 11.6.2 number 1. 3. Finally, we have A = 16, B = 24 and C = 9 which gives 242 − 4(16)(9) = 0. Theorem 11.11 tells us that the graph is a parabola, matching our result from Example 11.6.2 number 2. 11.6.2 The Polar Form of Conics In this subsection, we start from scratch to reintroduce the conic sections from a more uniﬁed perspective. We have our ‘new’ deﬁnition below. Deﬁnition 11.1. Given a ﬁxed line L, a point F not on L, and a positive number e, a conic section is the set of all points P such that the distance from P to F the distance from P to L = e The line L is called the directrix of the conic section, the point F is called a focus of the conic section, and the constant e is called the eccentricity of the conic section. We have seen the notions of focus and directrix before in the deﬁnition of a parabola, Deﬁnition 7.3. There, a parabola is deﬁned as the set of points equidistant from the focus and directrix, giving an eccentricity e = 1 according to Deﬁnition 11.1. We have also seen the concept of eccentricity before. It was introduced for ellipses in Deﬁnition 7.5 in Section 7
.4, and later extended to hyperbolas in Exercise 31 in Section 7.5. There, e was also deﬁned as a ratio of distances, though in these cases the distances involved were measurements from the center to a focus and from the center to a vertex. One way to reconcile the ‘old’ ideas of focus, directrix and eccentricity with the ‘new’ ones presented in Deﬁnition 11.1 is to derive equations for the conic sections using Deﬁnition 11.1 and compare these parameters with what we know from Chapter 7. We begin by assuming the conic section has eccentricity e, a focus F at the origin and that the directrix is the vertical line x = −d as in the ﬁgure below. y d r cos(θ) P (r, θ) r θ x = −d O = F x 982 Applications of Trigonometry Using a polar coordinate representation P (r, θ) for a point on the conic with r > 0, we get e = the distance from P to F the distance from P to L = r d + r cos(θ) so that r = e(d + r cos(θ)). Solving this equation for r, yields r = ed 1 − e cos(θ) At this point, we convert the equation r = e(d + r cos(θ)) back into a rectangular equation in the variables x and y. If e > 0, but e = 1, the usual conversion process outlined in Section 11.4 gives6 1 − e22 e2d2 x − e2d 1 − e2 2 1 − e2 e2d2 + y2 = 1 1−e2, so the major axis has length 2ed 1−e2 and the minor axis has length 2ed√ e2d 1−e2, 0 We leave it to the reader to show if 0 < e < 1, this is the equation of an ellipse centered at with major axis along the x-axis. Using the notation from Section 7.4, we have a2 = e2d2 (1−e2)2 and b2 = e2d2 1−e2. Moreover, we ﬁnd that one focus is (0, 0) and working through the formula given in Deﬁnition 7.
5 gives the eccentricity 1−e2, 0 to be e, as required. If e > 1, then the equation generates a hyperbola with center whose transverse axis lies along the x-axis. Since such hyperbolas have the form (x−h)2 b2 = 1, we need to take the opposite reciprocal of the coeﬃcient of y2 to ﬁnd b2. We get7 a2 = e2d2 (e2−1)2 and b2 = − e2d2 e2−1 and the conjugate axis has length 2ed√. e2−1 Additionally, we verify that one focus is at (0, 0), and the formula given in Exercise 31 in Section ed 1−e cos(θ) reduces to 7.5 gives the eccentricity is e in this case as well. If e = 1, the equation r =. This is a parabola with vertex r = − d 2, the focus is (0, 0), the focal diameter is 2d and the directrix is x = −d, as required. Hence, we have shown that in all cases, our ‘new’ understanding of ‘conic section’, ‘focus’, ‘eccentricity’ and ‘directrix’ as presented in Deﬁnition 11.1 correspond with the ‘old’ deﬁnitions given in Chapter 7. 1−cos(θ) which gives the rectangular equation y2 = 2d x + d 2, 0 opening to the right. In the language of Section 7.3, 4p = 2d so p = d e2−1, so the transverse axis has length 2ed a2 − y2 (1−e2)2 = e2d2 1−e2 = e2d2 e2d d 2 ed Before we summarize our ﬁndings, we note that in order to arrive at our general equation of a conic 1−e cos(θ), we assumed that the directrix was the line x = −d for d > 0. We could have just as r = easily chosen the directrix to be x = d, y = −d or y = d. As the reader can verify, in these cases 1+e sin(θ), respectively. The key thing to
1−e sin(θ) and r = we obtain the forms r = remember is that in any of these cases, the directrix is always perpendicular to the major axis of an ellipse and it is always perpendicular to the transverse axis of the hyperbola. For parabolas, knowing the focus is (0, 0) and the directrix also tells us which way the parabola opens. We have established the following theorem. 1+e cos(θ), r = ed ed ed 6Turn r = e(d + r cos(θ)) into r = e(d + x) and square both sides to get r2 = e2(d + x)2. Replace r2 with x2 + y2, expand (d + x)2, combine like terms, complete the square on x and clean things up. 7Since e > 1 in this case, 1 − e2 < 0. Hence, we rewrite 1 − e22 = e2 − 12 to help simplify things later on. 11.6 Hooked on Conics Again 983 Theorem 11.12. Suppose e and d are positive numbers. Then the graph of r = ed 1−e cos(θ) is the graph of a conic section with directrix x = −d. the graph of r = ed 1+e cos(θ) is the graph of a conic section with directrix x = d. the graph of r = ed 1−e sin(θ) is the graph of a conic section with directrix y = −d. the graph of r = ed 1+e sin(θ) is the graph of a conic section with directrix y = d. In each case above, (0, 0) is a focus of the conic and the number e is the eccentricity of the conic. If 0 < e < 1, the graph is an ellipse whose major axis has length 2ed 1−e2 and whose minor axis has length 2ed√ 1−e2 If e = 1, the graph is a parabola whose focal diameter is 2d. If e > 1, the graph is a hyperbola whose transverse axis has length 2ed e2−1 and whose conjugate axis has length 2ed√ e2−1. We test out Theorem 11.12 in the next example. Example 11.6.4. Sketch the
graphs of the following equations. 1. r = 4 1 − sin(θ) Solution. 2. r = 12 3 − cos(θ) 3. r = 6 1 + 2 sin(θ) 4 1. From r = 1−sin(θ), we ﬁrst note e = 1 which means we have a parabola on our hands. Since ed = 4, we have d = 4 and considering the form of the equation, this puts the directrix at y = −4. Since the focus is at (0, 0), we know that the vertex is located at the point (in rectangular coordinates) (0, −2) and must open upwards. With d = 4, we have a focal diameter of 2d = 8, so the parabola contains the points (±4, 0). We graph r = 1−sin(θ) below. 4 12 2. We ﬁrst rewrite r = 4 1−(1/3) cos(θ). 3−cos(θ) in the form found in Theorem 11.12, namely r = Since e = 1 3 satisﬁes 0 < e < 1, we know that the graph of this equation is an ellipse. Since ed = 4, we have d = 12 and, based on the form of the equation, the directrix is x = −12. This means that the ellipse has its major axis along the x-axis. We can ﬁnd the vertices of the ellipse by ﬁnding the points of the ellipse which lie on the x-axis. We ﬁnd r(0) = 6 and r(π) = 3 which correspond to the rectangular points (−3, 0) and (6, 0), so these are our 2, 0.8 vertices. The center of the ellipse is the midpoint of the vertices, which in this case is 3 2, 0 and this allows us to ﬁnd the We know one focus is (0, 0), which is 3 1−e2 = (2)(4) 2 from the center 3 8As a quick check, we have from Theorem 11.12 the major axis should have length 2ed 1−(1/3)2 = 9. 984 Applications of Trigonometry other focus (3, 0), even though we are not
asked to do so. Finally, we know from Theorem 2ed√ 3 which means the endpoints 1−e2 = 11.12 that the length of the minor axis is of the minor axis are 3 1−(1/3)2 = 6 4√ 2. We now have everything we need to graph r = 12 √ √ 3−cos(θ). 2, ±3 3 2 1 −4 −3 −2 −1 −1 1 2 3 4 −2 −3 y = −4 r = 4 1−sin(θ) y 4 3 2 1 −3 −2 −1 −1 1 2 3 4 5 6 x −2 −3 −4 x = −12 r = 12 3−cos(θ) 3. From r = 6 1+2 sin(θ) we get e = 2 > 1 so the graph is a hyperbola. Since ed = 6, we get d = 3, and from the form of the equation, we know the directrix is y = 3. This means the transverse axis of the hyperbola lies along the y-axis, so we can ﬁnd the vertices by looking where the hyperbola intersects the y-axis. We ﬁnd r π = −6. These two 2 points correspond to the rectangular points (0, 2) and (0, 6) which puts the center of the hyperbola at (0, 4). Since one focus is at (0, 0), which is 4 units away from the center, we know the other focus is at (0, 8). According to Theorem 11.12, the conjugate axis has a length of 3. Putting this together with the location of the vertices, we get that √ 3 the asymptotes of the hyperbola have slopes ± 2 3. Since the center of the hyperbola √ 2 = 2 and r 3π 2 = (2)(6) √ 22−1 = ± 2ed√ = 4 e2−1 √ 3 is (0, 4), the asymptotes are y = ± √ 3 3 x + 4. We graph the hyperbola below5 −4 −3 −2 −+2 sin(θ) 11.6 Hooked on Conics Again 985 In light of Section 11.6.1, the reader may wonder what the rotated form of
the conic sections would look like in polar form. We know from Exercise 65 in Section 11.5 that replacing θ with (θ − φ) in an expression r = f (θ) rotates the graph of r = f (θ) counter-clockwise by an angle φ. For instance, to graph r = 1−sin(θ), which we obtained in Example 11.6.4 number 1, counter-clockwise by π all we need to do is rotate the graph of r = 4 1−sin(θ− π 4 ) 4 radians, as shown below4 −3 −2 −1 −1 −2 −3 r = 4 1−sin(θ− π 4 ) Using rotations, we can greatly simplify the form of the conic sections presented in Theorem 11.12, since any three of the forms given there can be obtained from the fourth by rotating through some multiple of π 2. Since rotations do not aﬀect lengths, all of the formulas for lengths Theorem 11.12 remain intact. In the theorem below, we also generalize our formula for conic sections to include circles centered at the origin by extending the concept of eccentricity to include e = 0. We conclude this section with the statement of the following theorem. Theorem 11.13. Given constants > 0, e ≥ 0 and φ, the graph of the equation r = 1 − e cos(θ − φ) is a conic section with eccentricity e and one focus at (0, 0). If e = 0, the graph is a circle centered at (0, 0) with radius. If e = 0, then the conic has a focus at (0, 0) and the directrix contains the point with polar coordinates (−d, φ) where d = e. minor axis has length 2ed√ 1−e2 – If 0 < e < 1, the graph is an ellipse whose major axis has length 2ed 1−e2 and whose – If e = 1, the graph is a parabola whose focal diameter is 2d. – If e > 1, the graph is a hyperbola whose transverse axis has length 2ed e2−1 and whose conjugate axis has length 2ed√ e2−1. 986 Applications of Trigonometry 11.6.3 Exercises Graph the following
equations. 1. x2 + 2xy + y2 − x √ √ 2 − 6 = 0 2. 7x2 − 4xy √ 3 + 3y2 − 2x − 2y √ 3 − 5 = 0 3. 5x2 + 6xy + 5y2 − 4 2x + 4 √ 2y = 0 2 + y √ 5. 13x2 − 34xy √ 3 + 47y2 − 64 = 0 7. x2 − 4xy + 4y2 − 2x √ 5 − y √ 5 = 0 Graph the following equations. 9. r = 11. r = 2 1 − cos(θ) 3 2 − cos(θ) 13. r = 4 1 + 3 cos(θ) 15. r = 2 1 + sin(θ − π 3 ) √ √ 4. x2 + 2 6. x2 − 2 3xy + 3y2 + 2 √ 3x − 2y − 16 = 0 3xy − y2 + 8 = 0 8. 8x2 + 12xy + 17y2 − 20 = 0 10. r = 12. r = 3 2 + sin(θ) 2 1 + sin(θ) 14. r = 2 1 − 2 sin(θ) 16. r = 6 3 − cos θ + π 4 The matrix A(θ) = cos(θ) − sin(θ) cos(θ) sin(θ) is called a rotation matrix. We’ve seen this matrix most recently in the proof of used in the proof of Theorem 11.9. 17. Show the matrix from Example 8.3.3 in Section 8.3 is none other than A π 4. 18. Discuss with your classmates how to use A(θ) to rotate points in the plane. 19. Using the even / odd identities for cosine and sine, show A(θ)−1 = A(−θ). Interpret this geometrically. 11.6 Hooked on Conics Again 987 11.6.4 Answers 1. x2 + 2xy + y2 − becomes (x)2 = −(y − 3) after rotating counter-clockwise through x2 + 2xy + y2 − x √ 2 + y √ 3. 5x2 + 6xy + 5y2
− 4 2x + 4 √ 2 − 6 = 0 √ 2y = 0 becomes (x)2 + (y+2)2 counter-clockwise through θ = π 4. 4 = 1 after rotating. 7x2 − 4xy 3 + 3y2 − 2x − 2y 3 − 5 = 0 9 + (y)2 = 1 after rotating becomes (x−2)2 counter-clockwise through √ 7x2 − 4xy 3 + 3y2 − 2x − 2y √ 3 − 5 = 0 √ 4. x2 + 2 3xy + 3y2 + 2 √ 3x − 2y − 16 = 0 becomes(x)2 = y + 4 after rotating counter-clockwise through 5x2 + 6xy + 5y2 − 4 √ 2x + 4 √ 2y = 0 √ x2 + 2 3xy + 3y2 + 2 √ 3x − 2y − 16 = 0 988 Applications of Trigonometry √ 5. 13x2 − 34xy 3 + 47y2 − 64 = 0 becomes (y)2 − (x)2 counter-clockwise through θ = π 6. 16 = 1 after rotating √ 6. x2 − 2 3xy − y2 + 8 = 0 4 − (y)2 becomes (x)2 counter-clockwise through θ = π 3 4 = 1 after rotating 13x2 − 34xy √ 3 + 47y2 − 64 = 0 √ x2 − 2 3xy − y2 + 8 = 0 7. x2 − 4xy + 4y2 − 2x √ 5 − y √ 5 = 0 becomes (y)2 = x after rotating counter-clockwise through θ = arctan 1 2 y. y 8. 8x2 + 12xy + 17y2 − 20 = 0 becomes (x)2 + (y)2 4 = 1 after rotating counter-clockwise through θ = arctan(2) y x y x θ = arctan 1 2 x θ = arctan(2) x x2 − 4xy + 4y2 − 2x √ 5 − y √ 5 = 0 8x2 + 12xy + 17y2 − 20 = 0 11.6 Hooked on Conics Again 989 2 9. r = 1−
cos(θ) is a parabola directrix x = −2, vertex (−1, 0) focus (0, 0), focal diameter 4 10. r = 3 2+sin(θ) = 3 2 1+ 1 2 sin(θ) is an ellipse directrix y = 3, vertices (0, 1), (0, −3) center (0, −2), foci (0, 0), (0, −2) minor axis length 4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 −4 −3 −2 −1 1 2 3 4 x −1 −2 −4 11. r = 3 2−cos(θ) = 3 2 1− 1 2 cos(θ) is an ellipse directrix x = −3, vertices (−1, 0), (3, 0) center (1, 0), foci (0, 0), (2, 0) minor axis length 2 √ 3 y 4 3 2 1 −4 −3 −2 −1 1 2 3 4 x −1 −2 −4 12. r = 2 1+sin(θ) is a parabola directrix y = 2, vertex (0, 1) focus (0, 0), focal diameter 4 y 4 3 2 1 −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 990 Applications of Trigonometry 13. r = 4 1+3 cos(θ) is a hyperbola directrix x = 4 center 3 conjugate axis length 2 3, vertices (1, 0), (2, 0) √ 2 2, 0, foci (0, 0), (3, 0) 14. r = 2 1−2 sin(θ) is a hyperbola directrix y = −1, vertices 0, − 2 3 center 0, − 4, foci (0, 0), 0, − 8 3 3 √ conjugate axis length 2 3 3, (0, −24 −3 −2 −1 1 2 3 4 x −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 15. r = 2 3 ) is 1+sin(θ− π 2 the parabola r = 1+sin(θ) rotated through 1 −2
−3 −4 is the ellipse 16. r = 6 3−cos(θ+ π 4 ) 6 3−cos(θ) = r = 3 cos(θ) rotated through φ = − π 4 14 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 φ = − π 4 x 11.7 Polar Form of Complex Numbers 991 11.7 Polar Form of Complex Numbers √ In this section, we return to our study of complex numbers which were ﬁrst introduced in Section 3.4. Recall that a complex number is a number of the form z = a + bi where a and b are real −1. The number a is called the real part of numbers and i is the imaginary unit deﬁned by i = z, denoted Re(z), while the real number b is called the imaginary part of z, denoted Im(z). From Intermediate Algebra, we know that if z = a + bi = c + di where a, b, c and d are real numbers, then a = c and b = d, which means Re(z) and Im(z) are well-deﬁned.1 To start oﬀ this section, we associate each complex number z = a + bi with the point (a, b) on the coordinate plane. In this case, the x-axis is relabeled as the real axis, which corresponds to the real number line as usual, and the y-axis is relabeled as the imaginary axis, which is demarcated in increments of the imaginary unit i. The plane determined by these two axes is called the complex plane. Imaginary Axis (3, 0) ←→ z = 3 0 1 2 3 4 Real Axis (−4, 2) ←→ z = −4 + 2i 4i 3i 2i i −4 −3 −2 −1 −i −2i −3i (0, −3) ←→ z = −3i −4i The Complex Plane Since the ordered pair (a, b) gives the rectangular coordinates associated with the complex number z = a + bi, the expression z = a + bi is called the rectangular form of z. Of course, we could just as easily associate z with a pair of polar coordinates (r, θ). Although it is not as straightforward as the de�

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