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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.... |
) 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 ... |
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... |
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 se... |
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◦ ... |
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 t... |
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 t... |
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 t... |
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... |
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 = ... |
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 th... |
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 t... |
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... |
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 syste... |
� 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 = ... |
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 ... |
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 Ap... |
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 ho... |
, 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.... |
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 w... |
◦ = 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 =... |
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 l... |
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) ... |
(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,... |
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 Bigf... |
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... |
√ 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... |
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... |
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-cl... |
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 b... |
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 hav... |
−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 ... |
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... |
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 a... |
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 answe... |
� 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... |
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 coordin... |
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 ans... |
(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) wou... |
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 he... |
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... |
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... |
. 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 ... |
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 Exe... |
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 devel... |
, −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... |
� 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 5... |
= 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... |
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 (−... |
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 ... |
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 Qu... |
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 c... |
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... |
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 th... |
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 ran... |
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 ... |
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 ... |
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, ... |
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 fo... |
) 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... |
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 + (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 ... |
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 14Agai... |
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 represen... |
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 intersectio... |
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 ... |
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 si... |
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 co... |
(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... |
= 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 pol... |
�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 gr... |
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... |
�) = 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 calculato... |
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 si... |
−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... |
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 A... |
� − 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 ... |
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 not... |
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... |
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... |
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, ... |
(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 ... |
) 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... |
− 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 ve... |
�) 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... |
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... |
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... |
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 int... |
)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 − 2B... |
+ 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 √... |
.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 ... |
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 ... |
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 parab... |
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... |
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... |
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 a... |
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 ... |
− 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 = ... |
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 −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.... |
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