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Radian Measure : r Radian Measure r 1 radian Radian Measure : Radian Measure There are 2π radians in a full rotation – once around the circle There are 360° in a full rotation To convert from degrees to radians or radians to degrees, use the proportion 2π = 360° π = 180° Examples : Examples Find the radian measure equivalent of 210°. = 135° 180° = π Length of Arc : r Length of Arc l θ θ must be in radians Fraction of circle Length of arc Circumference = 2πr Area of Sector : r Area of Sector Fraction of circle Area of sector Area of circle = π r 2 θ θ must be in radians Slide 9: r θ θ must be in radians Page 6 of tables Page 7 of tables Examples : Examples l = 2·5 8 l = rθ = 20 cm l 2·5 8 cm A circle has radius length 8 cm. An angle of 2.5 radians is subtended by an arc. Find the length of the arc. Slide 11: (i) Find the length of the minor arc pq. (ii) Find the area of the minor sector opq. p q o 10 cm 0·8 rad p q o 12 cm l = rθ = 10(0·8) = 8 cm l = rθ Q1. Q2. Slide 12: Q3. The bend on a running track is a semi-circle of radius A runner, on the track, runs a distance of 20 metres on the bend. The angles through which the runner has run is A. Find to three significant figures, the measure of A in radians. = 0·6283.. = 0·628 radians l = rθ Slide 13: Q4. A bicycle chain passes around two circular cogged wheels. Their radii are 9 cm and 2·5 cm. If the larger wheel turns through 100 radians, through how many radians will the smaller one turn? 100 radians l = rθ l = 9 100 = 900 cm 900 = 2·5θ θ = 360 radians Slide 14: The diagram shows a sector (solid line) circumscribed by a circle (dashed line). k k 60º r r 30º (i) Find the radius of the circle in terms of k. 2006 Paper 2 Q4 (c) Slide 15: The diagram shows a sector (solid line) circumscribed by a circle (dashed line). k k r r (ii) Show that the circle encloses an area which is double that of the sector. 2006 Paper 2 Q4 (c) Area of circle = π r2 Area of sector Twice area of sector The Unit Circle : The Unit Circle Imagine a circle on the co-ordinate plane, with its center at the origin, and a radius of 1. Choose a point on the circle somewhere in the first quadrant. The Unit Circle : The Unit Circle Connect the origin to the point, and from that point drop a perpendicular to the x-axis. This creates a right triangle with hypotenuse of 1. 1 The Unit Circle : The Unit Circle The length of sides of the triangle are the x and y co-ordinates of the chosen point. Applying the definitions of the trigonometric ratios to this triangle gives x y 1 θ The Unit Circle : The co-ordinates of the chosen point are the cosine and sine of the angle . This provides a way to define functions sin and cos for all real numbers . The other trigonometric functions can be defined from these. The Unit Circle Trigonometric Functions : Trigonometric Functions x y 1 θ cosecant secant cotan Around the Circle : Around the Circle As that point moves around the unit circle into the second, third and fourth quadrants, the new definitions of the trigonometric functions still hold. Reference Angles : Reference Angles The angles whose terminal sides fall in the 2nd, 3rd, and 4th quadrants will have values of sine, cosine and other trig functions which are identical (except for sign) to the values of angles in 1st quadrant. The acute angle which produces the same values is called the reference angle. Second Quadrant : Second Quadrant Original angle θ Reference angle For an angle , in the second quadrant, the reference angle is In the second quadrant, sin is positive cos is negative tan is negative Third Quadrant : Third Quadrant Original angle θ Reference angle For an angle , in the third quadrant, the reference angle is – In the third quadrant, sin is negative cos is negative tan is positive Fourth Quadrant : Fourth Quadrant Original angle θ Reference angle For an angle , in the fourth quadrant, the reference angle is 2 In the fourth quadrant, sin is negative cos is positive tan is negative All Students Take Care : All Students Take Care All Students Take Care Use the phrase “All Students Take Care” to remember the signs of the trigometric functions in the different quadrants. Examples : Examples Find sin240° in surd form. – Draw the angle on the unit circle – In the 3rd quadrant sine is negative – Find the angle to nearest x-axis 60º Page 9 of tables Examples : Examples cosθ = – 0·5. Find the two possible values of θ, where 0º ≤ θ ≤ 360°. S A T C 60º cosA = 0·5 cos is negative in two quadrants 2nd 3rd 180º + 60º 240º 180º – 60º 120º Solving Triangles : ADJ OPP HYP Solving Triangles Slide 30: Trigonometric ratios in surd form 1 1 1 2 30º 45º 45º 60º Page 9 of tables Cosine Rule : c a b Cosine Rule C A B a2 = b2 + c2 – 2bccosA b a c b2 = a2 + c2 – 2accosB c2 = a2 + b2 – 2abcosC Page 9 of tables Cosine Rule : By Pythagoras’ Theorem a2 = (c – x)2 + h2 a2 = c2 – 2cx + x2 + h2 a2 = b2 + c2 – 2cx Cosine Rule c a b a2 = b2 + c2 – 2bccosA c b a x c – x b2 = x2 + h2 A h Cosine Rule : The Cosine Rule can be used to find a third side of a triangle if you have the other two sides and the angle between them. Cosine Rule Included angle Examples : Find the unknown side in the triangle below: l 5 m 12 m 43o Identify sides a,b,c and angle Ao a = l b = 5 c = 12 A = 43º Write down the Cosine Rule Substitute values and find a2 a2 = 52 + 122 – 2 5 12 cos 43o a2 = 81·28 Take square root of both sides a = 9·02 m Examples a2 = b2 + c2 – 2bccosA a2 = 25 + 144 – 120(0·731) Slide 35: 137o 17·5 cm 12·2 cm a2 = 12·22 + 17·52 – ( 2 12·2 17·5 cos 137o ) a2 = 148·84 + 306·25 – ( 427 – 0·731 ) a2 = 455·09 + 312·137 a2 = 767·227 a = 27·7 cm Find the unknown side in the triangle below: Examples a2 = b2 + c2 – 2bccosA a = ? b = 12·2 c = 17·5 A = 137º Slide 36: 20o x 6 62 = 102 + x2 – (2 10 x cos 20o) 36 = 100 + x2 – 20x( 0·9397) 0 = x2 – 18·79x + 64 Find the two possible values for the unknown side. Examples 10 a = 6 b = 10 c = x A = 20º a2 = b2 + c2 – 2bccosA Slide 37: 20o x 6 0 = x2 – 18·79x + 64 Find the two possible values for the unknown side. Examples 10 a = 6 b = 10 c = x A = 20º a = 1 b = –18·79 c = 64 Slide 38: Find the length of the unknown side in the triangles below: L = 47.5cm M = 5·05 m G = 12.4cm Sine Rule : c a b Sine Rule C A B b a c Page 9 of tables Slide 40: ______ ______ (i) Find dc , correct to the nearest cm. Multiply both sides by sin 50º dc sin 48º In the triangle abc, d is a point on [bc]. bd = 5 cm, ac = 7cm, bca = 82º and cad = 50º . 48º = sin 50º 7 = 7·215… = 7 cm Angle sum of ∆ is 180º 7 Sine Rule __ Slide 41: (ii) Find ab , correct to the nearest cm. In the triangle abc, d is a point on [bc]. bd = 5 cm, ac = 7cm, bca = 82º and cad = 50º . 5 d 50º 7 + = 12 12 Cosine Rule: a2 = b2 + c2 – 2bc cosA ab 2 = 122 + 72 – 2(12)(7)cos82º = 144 + 49 – 168cos82º = 169·6189… = 13·02… = 13 cm ab = 169·6189 a c b 7 82º Area of triangle : c a b Area of triangle C A B b a c Page 6 of tables Slide 43: Calculate the area of the triangle pqr, giving your answer correct to one decimal place. In the triangle pqr, | pq | = 10 cm, | pr | = 12 cm and |pqr | = 42. 10 cm p r 42 12 cm q 33·9 = 58·192… = 58·2 cm2 104·1 Angle sum of ∆ = 180 |rpq | = 180 – 42 – 33.9 Must be the included angle Identities involving Cosine Rule : Identities involving Cosine Rule Using the usual notation for a triangle, prove that c(bcosA – acosB) = b2 – a2 Slide 45: Using the usual notation for a triangle, prove that c(bcosA – acosB) = b2 – a2 Identities involving Cosine Rule cos2θ + sin2θ = 1 : cos2θ + sin2θ = 1 θ (x, y) r y x By Pythagoras’ Theorem x 2 + y 2 = r 2 Divide both sides by r 3D Trigonometry : 3D Trigonometry Slide 48: r s h p q β α p, q and r are points on level ground, [sr] is a vertical flagpole of height h. The angles of elevation of the top of the flagpole from p and q are α and β, respectively. (i) If | α | = 60º and | β | = 30º, express | pr | and | qr | in terms of h. 30º 60º Slide 49: ADJ OPP s h r p q 30º 60º 60º Slide 50: s h r p q OPP ADJ 30º 60º Slide 51: r s h p q a2 = b2 + c2 – 2bccosA Pythagoras’ Theorem 30º 60º Slide 52: r p q a2 = b2 + c2 – 2bccosA Slide 53: The great pyramid at Giza in Egypt has a square base and four triangular faces. The base of the pyramid is of side 230 metres and the pyramid is 146 metres high. The top of the pyramid is directly above the centre of the base. (i) Calculate the length of one of the slanted edges, correct to the nearest metre. 230 m 230 m Pythagoras’ theorem x = 105800 2 325·269.. x 2 x = 162·6 = 162·6 162·6 146 2006 Paper 2 Q5 (b) Slide 54: The great pyramid at Giza in Egypt has a square base and four triangular faces. The base of the pyramid is of side 230 metres and the pyramid is 146 metres high. The top of the pyramid is directly above the centre of the base. (i) Calculate the length of one of the slanted edges, correct to the nearest metre. 146 m 162·6 m Pythagoras’ theorem l = 47754·76 2 l = 218·528.. l = 219 m 162·6 146 2006 Paper 2 Q5 (b) Slide 55: (ii) Calculate, correct to two significant numbers, the total area of the four triangular faces of the pyramid (assuming they are smooth flat surfaces) 219 m 230 m = 34736 2 = 186·375.. h = 186·4 m h 115 m Pythagoras’ theorem = 21436 m2 2006 Paper 2 Q5 (b) Slide 56: (ii) Calculate, correct to two significant numbers, the total area of the four triangular faces of the pyramid (assuming they are smooth flat surfaces) 219 m = 34736 = 186·375.. h = 186·4 m h 115 m Pythagoras’ theorem Total area = 21436 4 = 85744 m2 = 86000 m2 2006 Paper 2 Q5 (b) Slide 57: θ 2θ 3x x q pqrs is a vertical wall of height h on level ground. p is a point on the ground in front of the wall. The angles of elevation of r from p is θ and the angle of elevation of s from p is 2θ. | pq | = 3| pt |. Find θ. p s r h t 2005 Paper 2 Q5 (c) Slide 58: θ 2θ 3x x q pqrs is a vertical wall of height h on level ground. p is a point on the ground in front of the wall. The angles of elevation of r from p is θ and the angle of elevation of s from p is 2θ. | pq | = 3| pt |. Find θ. p s r h t 2005 Paper 2 Q5 (c) Slide 59: tan θ x θ 2θ 3x x q p s r h t = tan 2θ x 3 Let t = tan θ 2005 Paper 2 Q5 (c) Slide 60: a c d (i) Find | bac | to the nearest degree. 4.D Question 4 b 5 m 5 2 4 A abc is an isosceles triangle on a horizontal plane, such that |ab| = |ac| = 5 and |bc| = 4. m is the midpoint of [bc]. Slide 61: abc is an isosceles triangle on a horizontal plane, such that |ab| = |ac| = 5 and |bc| = 4. m is the midpoint of [bc]. a c d b 5 m 5 2 (ii) A vertical pole [ad] is erected at a such that |ad | = 2, find |amd | to the nearest degree. 2 2 21 am = amd 4.D Question 4 Trigonometric Identities : Trigonometric Identities An identity is an equation which is true for all values of the variable. There are many trig identities that are useful in changing the appearance of an expression. The most important ones should be committed to memory. Trigonometric Identities : Trigonometric Identities Reciprocal Identities Quotient Identities Trigonometric Identities : Trigonometric Identities Pythagorean Identities The fundamental Pythagorean identity Divide by sin2 x Divide by cos2 x Solving Trig Equations : Solving Trig Equations To solve trigonometric equations: If there is more than one trigonometric function, use identities to simplify Let a variable represent the remaining function Solve the equation for this new variable Reinsert the trigonometric function Determine the argument which will produce the desired value Slide 66: (1 – cos2A) (i) Using cos 2A = cos2A – sin2A, or otherwise, prove 2005 Paper 2 Q4 (b) cos 2A = cos2A – sin2A cos 2A = cos2A – 1 + cos2A cos 2A = 2cos2A – 1 1 + Slide 67: = cos x (ii) Hence, or otherwise, solve the equation 1 + cos 2x = cos x, where 0º ≤ x ≤ 360º. 2005 Paper 2 Q4 (b) 1 + cos 2x 2cos2x – cos x = 0 2cos2x cos x(2cos x – 1) = 0 From (i) cos x = 0 Slide 68: Expand Collect like terms Rearrange Factorise Slide 69: Replace t with sin x Trigonometric Formulas : Trigonometric Formulas Page 9 of tables Slide 71: Trigonometric Formulas Slide 72: Replace B with – B Slide 73: Replace B with A Slide 75: 360° 90° 180° 270° y = sin x –1 1 Graphs of the Trig Functions : Graphs of the Trig Functions 2π π 1 –1 y = sin x Fluctuates from 0 to a high of 1, down to –1, and back to 0, in a space of 2. –2π –π Graphs of the Trig Functions : Graphs of the Trig Functions 2π π 1 –1 –2π –π Slide 78: y = tan x Inverse Trig Functions : Inverse Trig Functions Definition: For all one-to-one functions, the inverse function is the set of ordered pairs obtained by interchanging the first and second elements of each pair in the original function. Notation: If f is a given function, then f -1 denotes the inverse of f. Slide 80: Graph of y = x 2 f (2) = 4 and f (–2) = 4 so what is an inverse function supposed to do with 4? By definition, a function cannot generate two different outputs for the same input, so this function does not have an inverse. Horizontal line test Slide 81: Graph of y = x 2 By taking only one half of the graph: x ≥ 0 , the graph now passes the horizontal line test and we do have an inverse. Note how each graph reflects across the line y = x onto its inverse. y = sin x : 2π π 1 –1 –2π –π y = sin x The sine function does not pass the horizontal line test To overcome this, the value for θ is restricted to – 90º ≤ θ ≤ 90º These are known as the “Principal Values” Slide 83: y = sin x Slide 84: -1 1 -1 1 x y y = sin–1 x Slide 85: The thing to remember is that for the trig function the input is the angle and the output is the ratio, but for the inverse trig function the input is the ratio and the output is the angle. y = cos x : 2π π 1 –1 –2π –π y = cos x The restriction used for y = sin x in not suitable Domain [0, π] and the range is [–1, 1] Slide 87: y = cos–1 x -1 1 p/6 p/3 p/2 2p/3 5p/6 p x y Slide 88: y = tan x y = tan–1x y = tan–1 x Slide 89: The table below will summarize the parameters we have so far. Remember, the angle is the input for a trig function and the ratio is the output. For the inverse trig functions the ratio is the input and the angle is the output. When x < 0, y = cos–1x will be in 2nd quadrant When x < 0, y = sin–1x will be in 4th quadrant When x < 0, y = tan–1x will be in 4th quadrant Slide 90: The graphs give you the big picture concerning the behavior of the inverse trig functions. Calculators are helpful with calculations. But special triangles can be very helpful with respect to the basics. Slide 92: Negative inputs for the cos–1 can be a little tricky. Negative inputs for the cos–1 generate angles in the 2nd Quadrant so we have to use 60 degrees as a reference angle in the 2nd Quadrant. Slide 93: θ Slide 96: 2006 Paper 2 Q5 (a) (i) Copy and complete the table below for f : x → tan–1x, giving the values for f (x) in terms of π. (ii) Draw the graph of y = f (x) in the domain –2 ≤ x ≤ 2, scaling the y-axis in terms of π. 0 You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.