Chapter 10 Additional Topics in Trigonometry 1
Right Triangle applications Law of Sines and Cosines Parametric equations Polar coordinates Curves in polar coordinates Summary 2
Chapter 10.1 Right Triangle Applications 3
Finding Sides We could always find sides if we could find the sin and cosine However, there were limited angles for which we could find those values Now we are going to use a calculator to find trig functions for arbitrary angles 4
Example We have a right triangle with sides 3, 4, and 5 m. Find the sine and cosine of the angle opposite the shortest side 5
Solution We have a right triangle with sides 3, 4, and 5 m. Find the sine and cosine of the angle opposite the shortest side The shortest side is 3, we have sin a = 3/5, need to find a Arcsine a = 0.64 or 36.9 deg 6
Example The three sides of a right triangle are 5, 12, and 13 m. find the angle opposite the shortest side 7
Solution The three sides of a right triangle are 5, 12, and 13 m. find the angle opposite the shortest side Sin x = 5/13 = =0.38, 8
Angle of Depression and Elevation Angle of Elevation Horizon Horizon Angle of Depression Tower, e.g. 9
Example Joe is standing on top of a mountain that he knows to be 1000 feet higher than the lake below him. The lake is 500 feet from the base of the mountain. What is the angle of depression of his sightline to the lake? 10
Solution Arctan(1/2) = 0.46 Angle = - 0.46 = 2.68 radians Horizon Angle of Depression Arctan(1/2) Mountain, 1000 ft Lake 500 ft 11
Chapter 10.2 Laws of sines and cosines 12
Derivation of the Law of Sines b C h a B c A Sin A = h/b, Sin B = h/a so b Sin A = a Sin B or Sin A/Sin B = a/b Equivalently, Sin A/a = Sin B/b Similarly, Sin C/Sin A = c/a, Sin C/c = Sin A/a, etc. Holds if know ASA, AAS, or SSA, S = Side, A = angle 13
Example Requires calculator: Side b = 385 m Angle B = 47 deg Angle A = 108 deg Find angle C, side a, side b 14
Solution Requires calculator: Side b = 385 m Angle B = 47 deg Angle A = 108 deg Find angle C, side a, side b Sin B / b = sin A / a, a = b sin A/sin B = 500 Angle C = 180 47 108 = 25 c = b sin c/sin b = 222 15
Example Angle A = 20.4 deg Angle B = 63.4 deg Side c = 12.9 m Find the rest of the sides 16
Solution Angle A = 20.4 deg Angle B = 63.4 deg Side c = 12.9 m Find the rest of the sides Angle C = 180 20.4 63.4 = 96.2 a = c sina/sin C = 4.5 m b = c sin B/sin C = 11.6 m 17
SAS B A C If fix side a, Angle A, side c, then angle C and B and side b can vary; SAS does not give a unique solution and may have no solution 18
Example b = 100, c = 60, C = 28 deg A sin B/b = sin C/c C B 2 B 1 sin B 1 = sin 28 (5/3) B 1 = 51.5 deg B 1 + B 2 = 180, so B 2 = 128.5 deg Therefore A 1 = 100.5 deg. A 2 = 23.5 deg A can be 125.7, or 51.0 19
Example b = 4 2, c = 8, C = 45 deg 20
Solution b = 4 2, c = 8, C = 45 deg Sin B = sin (45) (4 2)/8 = ½ B can be 30 deg or b can be 150 deg Since C = 45 deg, B cannot be 150 deg Now we can find a and b A = 180 45 30 = 105 a = 8 sin 45 / sin 105 = 10.9 21
Can this be a triangle? A = 34 deg B = 73 deg C = 52 deg a = 14 b = 22 c = 18 22
Solution A = 34 deg B = 73 deg C = 52 deg a = 14 b = 22 c = 18 14 = 22 sin(34)/sin(73) = 12.9 NO 23
Can this be a triangle? A = 42 deg B = 57 deg C = 81 deg a = 7 b = 9 c = 22 24
Solution A = 42 deg B = 57 deg C = 81 deg a = 7 b = 9 c = 22 7 + 9 = 16 < 22, not a triangle. 25
Triangles for? A = 38 deg a = 432 b = 382 26
Solution A = 38 deg a = 432 b = 382 432 = 382 sin (B)/ sin (38) Or sin B = (382/432) sin 38 = 0.7; B = 32.9 deg or 147 deg Angle C = 180 38 147 = -5 deg ; not posible Or Angle C = 180 38 32.7 = 109.1 deg Is one triangle 27
sin B/5.2 = sin 65/4.9 One or Two Triangles 28
Solution sin B/5.2 = sin 65/4.9 Sin B = 0.96, B = 74 deg or 105 deg C = 180 74 65 = 41 deg Or C = 180 74 105 = 1 deg Can be 2 triangles 29
Sin A /a = Sin B /b = Sin C /c Law of Sines 30
Law of Cosines Derivation C A b cos A = x/b or sin A = = y/b y a x = b cos A or y = b sin A B c x a 2 = (x-c) 2 + y 2 = (b cos A c) 2 + b 2 sin 2 A = b 2 cos 2 A 2bc cos A - c 2 + b 2 sin 2 A = b 2 (cos 2 A + sin 2 A) + c 2-2bc 2 cos A = b 2 + c 2 2bc cos A 31
a 2 = b 2 + c 2 2bc cos A Law of Cosines (SAS) b 2 = a 2 + c 2 2ac cos B c 2 = a 2 + b 2 2ab cos C Use when know two sides and an angle or three sides 32
Example B = 95 deg a = 16 c = 7 Find b: b 2 = a 2 + c 2 2ac cos B = 16x16 + 7x7 2x16 x 7 x cos(95) = 256 + 49-272 cos (95) = 324.523 b = 18.01 33
Example Continued B = 95 deg a = 16 c = 7 b = 18 Sin A/a = Sin B /b = 0.055 A = 62.3 deg Sin C = 22.6 deg 34
SSS Triangles a = 15 b = 25 c = 28 c 2 = a 2 + b 2-2ab cos C Cos C = -(c 2 - a 2 - b 2 )/(2ab) = -(784 225 625)/750 = 0.088 C = 85 deg Similarly A = 32.3 deg. B = 180-323.3-85 = 62.7 35
Review Law of Sines: ASA, SAA, Sometimes SSA - ambiguous Law of Cosines: SSS, SAS How about AAA? 36
Review Law of Sines: ASA, SAA, Sometimes SSA - ambiguous If know two angles, know all; If know only one angle, possibly two solutions Law of Cosines: SSS, SAS Need to know 3 of 4; 3 sides and one angle How about AAA? Similar triangle; need a side to determine lengths 37
Area of a Triangle We know area = ½ base x height; C a However, there are many variations on this. b h B In this triangle, h = b sin A = ½ bc sin A, similarly = ½ ab sin C = ½ ac sin B A c The area is one-half the product of two sides times the sine of the angle between them. 38
Example Find the area: a = 16.2 cm b = 25.6 cm C = 28.3 deg 39
Solution Find the area: a = 16.2 cm b = 25.6 cm C = 28.3 deg Drop perpendicular from Angle A to side b Area is ½ ab sin C = 171.7 cm 40
Another Area Approach Requires two angles and one side: Since area = ½ bc sin A, b = 2 Area /(c sina) We have, from Area = ½ ac sin B that a = 2 Area / (csin B) So, 2 Area = 2Area / (c sin B) x 2 Area/(c sin A) x sin C Rearranging: c 2 sin A sin B = 2 Area sin C Or, Area = c 2 sin A sin B /(2 sin C) This holds for any combo of sides and angles. 41
Example a= 34.5 ft B = 87.9 deg C = 29.3 deg 42
a= 34.5 ft B = 87.9 deg C = 29.3 deg Area = a 2 sin C sin B /(2 sin A) Angle A = 180 87.9 29.3 = 62.8 Area = 34.5 2 sin 29.3 sin 87.9 / sin 62.8 = 654.5 sqft 1190 0.48 1 / 0.89 = 43
a 2 + b 2 2ab cos C = c 2 Yet another approach So a 2 + b 2 c 2 = 2ab cos C (a 2 + b 2 c 2 )/2ab = cos C Now Area = ½ ab sin C = ½ ab sqrt(1 cos 2 C) = ½ ab sqrt{ 1- [(a 2 + b 2 c 2 )/2ab] 2 } This says that we can find the area by only knowing the sides! 44
Chapter 10.5 Parametric Equations 45
Example of Parametric Equations Let x = 2t and y = t2 2, t 0 The points x and y give us a location, a point, at any time t Let t = 1, x = 2 and y = ½, P is at (2, ½) The definitions of x and y are parametric equations, and t is the parameter Often t is time, and the equations constrain an object s motion 46
Working with Parametric Equations Let x = 2t and y = t2 2, t 0 We often want to eliminate t so have equations in x and y Let t = x/2, then y = t 2 /2 = (x/2) 2 /2 = x 2 /4, a parabola 47
Example x = 6 cos t, y = 3 sin t, 0 t 2π Let us see if we can eliminate the parameter t x/6 = cos t, y/3 = sin t sin 2 t + cos 2 t = 1, so (x/6) 2 + (y/3) 2 = 1 This is the equation of an ellipse The parametric equations for an ellipse are x = a cos t and y = b sin t 48
In each of these plots, the arrow is the direction of increasing t 49
A Circle and an Ellipse A circle is x 2 + y 2 = r 2 You can also translate a circle: (x x1) 2 + (y y1) 2 = r 2 An ellipse just changes the x/y ratio: (x/a) 2 + (y/b) 2 = r 2 50
Example Let x = 1 + 2t and y = 2 + 4t, 0 t Find an equation for x and y for the path traced out 51
Solution x 1 t = (x-1)/2, so y = 2 + 4 ( 2 y = 2x is just a line ) = 2 + 2x 2 = 2x 52
Example Let x = 1+ t, y = t 2 53
Solution Let x = 1+ t, y = t 2 t = x-1, so y = (x-1) 2 54
Example x = 2 sin t, y = 3 cos t 55
Example x = 2 sin t, y = 3 cos t Sin t = x/2, cos t = y/3; cos 2 t + sin 2 t = 1 = x/2 + y/3 56
x = cos 3 t, y = sin 3 t, 0 t 2 π Write in terms of x and y Example 57
Solution x= cos 3 t, so cos t = 3 x, cos 2 t = 3 x 2, and similarly for y Now, cos 2 t + sin 2 t = 1, so 3 x 2 + 3 y 2 = 1 58
Example x = -4 + 3tan t y = 7 2 sec t 59
Solution x = -4 + 3tan t y = 7 2 sec t Tan t = (x+4)/3 Sec t = -(y-7)/2 sec 2 t - tan 2 t = 1 = (y-7) 2 /4- (x+4) 2 = 1 60
Solution x = 2cos t + 2t, y = 2sint sin 2t Sin 2t = 2 cos t sin t = (y-2 sin t)(x 2t)/2 3 2 1 0-2 -1 0 1 2 3 4-1 -2-3 61
x = 4 cos 2t, y = 6 sin 2t More Examples X = 2cos t, y = 2 sin t x = 3t + 2, y = 3t -2 X = 2 tan t, y = 2 cos 2 t 62
Section 10.6 Polar Coordinates 63
Introduction Points are represented in terms of an angle and a radius: P(r,θ) The angle is measured from the positive x axis A negative value is a long the same line, but in the opposite direction P(r, ) r P(-r, ) 64
Examples Plot the following given in polar notation: (3, 2 /3) (-5, - /4) (0, ) 65
cos = x/r so x = r cos sin = x/r so x = r sin x 2 + y 2 = r 2 tan = y/x Equivalencies 66
To Convert from Rectangular to Polar Given x, y, r = x 2 + y 2 = arctan(y/x) 67
Rectangular to Polar x = r cos, y = r sin ; usually we want r 0, 0 < 2 Convert to Polar: (-1, -1) (1, 3) (-2, 0) 68
Solution Convert to Polar: (-1, -1) becomes (1, 5 /4) (1, 3) becomes (2, /3 ) (-2, 0) becomes (2, ) (why and not zero?) 69
Polar to Rectangular Convert to rectangular coordinates P = (5, 5 /6) r = 2 cos r = 4 70
P = (5, 5 /6); (5 3/2, 5/2) Solution r = 2 cos ; x = r cos, so cos = x/r. Multiply both sides by r x 2 + y 2 = 2x r = 4 ; x 2 + y 2 = 4 71
Convert to Rectangular r 2 = sin 2 72
Solution r 2 = sin 2 sin 2 = 2sin cos = (2/ r 2 ) x y So (x 2 + y 2 ) 2 = 2xy 73
r = cos + 2sin Converting an Equation 74
Solution r = cos + 2sin Multiply both sides by r r 2 = rcos + 2rsin x 2 + y 2 = x + 2y 75
Distances Distance between two points in polar coordinates, (r1, 1) and (r2, 2) : d = r12 + r22 2r1r2cos ( 2 1) See next page for derivation 76
Distance Derivation The distance between two points (x1, y1) and (x2 y2) in rectangular coordinates is (x1 x2) 2 +(y1 y2) 2 Let r1 = x1 2 + y1 2 and r2 = x2 2 + y2 2 and x1 = r1 cos 1, etc. Then x1 x2 = r1 cos 1 r2 cos 2 and y1 y 2 = r1 sin 1 r2 sin 2 Squaring each, we have: (x1 x2) 2 = r1 2 cos 2 1-2r1r2 cos 1 cos 2 + r2 2 cos 2 2 (y1 y2) 2 = r1 2 sin 2 1-2r1r2 sin 1 sin 2 + r2 2 sin 2 2 Adding gives r1 2 + r2 2-2r1r2cos 1 cos 2 sin 1 sin 2 = r1 2 + r2 2-2r1r2cos ( 2-1) 77
An example Find the distance between (2, 5 /6) and (4, /6) 78
Solution Find the distance between (2, 5 /6) and (4, /6) r1 2 + r2 2-2r1r2cos ( 1-2) = 4 + 16 16 cos(5 /6 - /6) = 20 16 cos (-2 /3) = 20 16(-1/2) = 28 Note: the order of the angles is immaterial since cos x = cos (-x) 79
Circles A circle centered at the origin has the equation r = a, where a is the radius A circle of radius a, centered at (r1, 1) has the equation: r 2 + r1 2 2r r1 cos( - 1) = a 2 80
Polar Equation of a Line? If a line goes through the origin, then we just have = c, where c is a constant For an arbitrary line, cos( - ) = d/r or d = r cos( - ), where d is the distance from the origin to the line and (d, ) is the foot of the perpendicular to the line P(r, ) d x axis 81
Chapter 10.7 Curves in Polar Coordinates 82
Graphing Just as with rectangular coordinates, create a table of r,. A compass rose of angles (e.g., the unit circle) might help 83
Graph r = / Example; Spiral of Archimedes r 0 0 /4 1/4 /2 1/2 3 /4 3/4 1 5 /4 5/4 3 /2 3/2 7 /4 7/4 2 2 9 /4 9/4 5 /2 5/2 11 /4 11/4 3 12 13 /4 13/4 7 /2 7/2 15 /4 15/4 4 4 3 2 1 0-4 -2 0 2 4 6-1 -2-3 -4 From r and you might just find rcos and rsin and plot x and y 84
Symmetry 1. If in the equation substituting - for changes nothing, then the graph is symmetric about the x axis 2. If in the equation substituting - for and r for r changes nothing, then the graph is symmetric about the y axis 3. If in the equation substituting - for changes nothing, then the graph is symmetric about the y axis 4. If in the equation substituting r for r changes nothing, then the graph is symmetric about the origin HOWEVER: A graph may be symmetric without these holding 85
Examples r 2 = 4cos 2 1. 4cos 2 = 4cos (-2 ), symmetric about x 2. 4cos 2 = 4cos (-2 ), and (-r) 2 = r 2, symmetric about y 3. cos ( - ) = cos, symmetric about y 4. Interchanging r and r no change, symmetric abut origin 86
Which Goes with Which? 1) r = 3 + 3 sin θ 3) r = 3 + 3 cos θ 2) r = 3-3 cos θ 4) r = 3 3 sin θ A C B D 87
Solution 1: D 2: B 3: A 4: C 88