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STUDY KNOWHOW PROGRAM STUDY AND LEARNING CENTRE Trigonometry B Dec 2016 Contents Pythagoras’ Theorem ............................................................................................................................................................... 3 Right Triangle Trigonometry................................................................................................................................................ 5 Sine Rule........................................................................................................................................................................................ 9 Cosine Rule ................................................................................................................................................................................ 13 Angular measurement - Radians...................................................................................................................................... 15 Circular Functions .................................................................................................................................................................. 17 Trigonometric Equations .................................................................................................................................................... 20 Graphs of Sine and Cosine Functions ............................................................................................................................. 21 B Dec 2016 2 PYTHAGORAS’ THEOREM Right-angled triangles A right-angled triangle has an angle of 90⁰. In a right-angled triangle the side opposite the right angle is called the hypotenuse. It is also the longest side. hypotenuse Right angle (90o) Pythagoras’ Theorem Pythagoras’ theorem states: In a right angled triangle the square of the length of the hypotenuse (h) is equal to the sum of the squares of the other two sides. h b h2 = a2 + b2 a Pythagoras’ Theorem can be used to find a side length of a right angled triangle when you know the other two side lengths. Examples: Finding the hypotenuse Find the length of the hypotenuse (h) in the following triangle. Use Pythagoras theorem h2 = a2 + b2 with a = 6cm. and b = 8 cm: h2 = a2 + b2 h2 = 62 + 82 h2 = 100 h = 10 cm h 6 [Note: The correct units can be put in at the end] 8 Finding a shorter side Find the value of x in the triangle below. x 2.7 4.2 B Dec 2016 h2 = a2 + b2 4.22 = 2.72 + x2 4.22 - 2.72 = x2 10.35 = x2 x = 3.22 3 Pythagorean triples In some right-angled triangles all three sides have integer values. These three values form a Pythagorean triple. Some examples are triangles with sides: (3,4,5), (5,12,13), (7,24 25) and (8,15,17) Check! Multiples of these, such as (6,8,10) and (9,12,15) are also Pythagorean triples. Exercises Find the missing sides in the following. (1) (2) a 9mm h 6.5 20.2 12mm (3) (4) 7cm 4.8 cm a c Answers: (1) 15 mm. B Dec 2016 14 cm 6.2 cm (2) 19.13 (3) 7.84 cm. (4) 12.12 cm. 4 RIGHT TRIANGLE TRIGONOMETRY Trigonometry is a branch of mathematics involving the study of triangles, and has applications in fields such as engineering, surveying, navigation, optics, and electronics. The ability to use and manipulate trigonometric functions is necessary in other branches of mathematics, including calculus, vectors and complex numbers. Right-angled Triangles In a right-angled triangle the three sides are given special names. The side opposite the right angle is called the hypotenuse (h) – this is always the longest side of the triangle. The other two sides are named in relation to another known angle (or an unknown angle under consideration). If this angle is known or under consideration this side is called the opposite side because it is opposite the angle h This side is called the adjacent side because it is adjacent to or near the angle Trigonometric Ratios In a right-angled triangle the following ratios are defined for a given angle θ sine θ = θ cosine θ = θ tangent θ = θ These ratios are abbreviated to sin θ, cos θ, and tan θ respectively. A useful memory aid is SOH CAH TOA: Sin=Opp/Hyp Cos=Adj/Hyp Tan=Opp/Adj These ratios can be used to find unknown sides and angles in right-angled triangles. Examples Evaluating ratios In the right-angled triangle below evaluate sin θ, cos θ, and tan θ. sin θ = = = 0.8 5 4 cosθ = = = 0.6 tan θ = = = 1.33 3 B Dec 2016 5 Finding angles. Find the value of the angle in the triangle below Method 1. 2. 3. 4. 19.7 13.4 Determine which ratio to use. Write the relevant equation. Substitute values from given information. Solve the equation for the unknown In this triangle we know two sides and need to find the angle . The known sides are the opposite side and the hypotenuse. The ratio that relates the opposite side and the hypotenuse is the sine ratio. sin θ = sin θ = sin θ = 0 6082 θ = sin-1(0.6082) [Using sin-1 function on the calculator] θ = 42.9⁰ Finding side lengths Find the value of the indicated unknown side length in each of the following right-angled triangles. (a) bc b 27o 42 cm In this problem we know an angle, and the adjacent side. The side to be determined is the opposite side. The ratio that relates these two sides is the tangent ratio. tan θ = tan 27⁰ = b = 42 × tan27⁰ b = 21.4cm In this problem we know an angle, and the adjacent side. The unknown side is the hypotenuse. The ratio that relates these two sides is the cosine ratio. x (b) 35o 7 cos θ = cos 35⁰ = x = 7 × cos 35⁰ x = 8.55 B Dec 2016 See Exercise 2 6 Special angles and exact values There are some special angles for which the trigonometric functions have exact values rather than decimal approximations. Applying the rules for sine, cosine and tangent to the triangles below, exact values for the sine, cosine and tangent of the angles 30⁰, 45⁰ and 60⁰ can be found. Exercises Exercise 1 Using the right-angled triangle below find: (a) sin θ, (b) tan θ, (c) cos α, (d) tan α (Hint: Use Pythagoras theorem to find the hypotenuse) Exercise 2 Find the value of the indicated unknown (side length or angle) in each of the following diagrams. (a) (b) a 62o 47o 14cm 4.71 mm a B Dec 2016 7 (c) (d) 4.8 cm x 6.2 cm z 6.5 20.2 (e) (f) 500 a b 34 27o 42 (g) In a right-angled triangle sin = 0.55 and the hypotenuse is 21 mm. Find the length of each of the other two sides. Answers Exercise 1 (a) 12/13 = 0.9231 (b) 12/5 = 2.4 Exercise 2 (a) = 6.6 cm (f) b = 47.1 B Dec 2016 (c) 12/13 = 0.9231 (b) a = 3.4 mm (c) z = 7.8, = 37 7 (g) 11.6mm. and 17.5mm. (d) 5/12 = 0.4167 (d) θ =18.8, x = 19.1 8 (e) a = 44.4 SINE RULE The sine rule can be used to find angles and sides in any triangle (not just a right-angled triangle) when given: (i) One side and any two angles OR (ii) Two sides and an angle opposite one of the given sides. In the triangle ABC below: angles A,B,C, are the angles at the vertices A,B,C respectively a,b,c are the side lengths opposite the angles A,B,C respectively. B a c C b A The sine rule states: a b c = = or sin A sin B sin C sin A sin B sin C = = a b c or Examples 1. In triangle PQR find: Q a) side length p b) side length q p P 70° 30° q R a) Side length p: Use the sine rule in the form p q r = = sin P sin Q sin R The relevant part of the formula is p r = sin P sin R p 15 = sin 70° sin 30° p= 15 x sin 70° sin 30° p = 28 2 cm b) Side length q: Angle Q is found using the fact that the sum of the three interior angles of a triangle add to 180°. Q = 180° (70° + 30°) = 80° B Dec 2016 9 = From the sine rule 28 2 q = sin 70° sin 80° q= 28 2 x sin 80° sin 70° q = 29 6 cm 2. In triangle ABC find: B a) angle C b) angle A c) side length a. 126° a C A b = 20 m a) Angle C: sin A sin B sin C = = Use the sine rule in the form a b c sin B sin C = b c The relevant part of the formula is sin126° sin C = 20 12 sin C = 12 𝑥 sin 126° 20 sin ⬚ C = 0 485 C = sin− ( 0 485) C = 29° b) Angle A: A = 180° (126° + 29°) A = 25° c) Side length a: Use the sine rule in the form a b c = = sin A sin B sin C The relevant part of the formula is a b = sin A sin B a 20 = sin 25° sin 126° a= 20 x sin 25° sin 126° a = 10 4 m B Dec 2016 10 3. Given a triangle ABC with angle A = 30°, adjacent side = 15 cm and opposite side = 8 cm as shown below, find angle C. In this case there are two possible solutions. This is called the ambiguous case of the sine rule. B 30° A C1 C2 Use the sine rule in the form sin A sin C = a c sin 30° sin C = 8 12 12 x sin 30° 8 sin C = 0.75 sin C = C = sin− ( 0 75) C = 48.6° Another solution to 3 is ABC2 C = sin− (0 75) This solution gives the triangle ABC1 = 131.4° = (180° 48.6°). This solution gives triangle Note: In any non-right-angled triangle, where two sides and the non-included angle are given check for the ambiguous case if: the angle is acute and the length of the side adjacent to the angle is greater than the length of the side opposite the angle. Exercises Exercise 1 For the following triangles find the unknown sides. B a) 65° c=? Q b) 10 r=? C A 30° P b=? t=? c) S 130° 27 r=? 15° T 11 38 45° q=? R B Dec 2016 85° R Exercise 2 For the following triangles find all unknown angles and sides. a) b) M n= ? 21 80° 16 5 N 8.5 L c) B A 35° b=? B 6.4 A a=? 65° 4.7 C Answers. Exercise 1 a) b = 18.1, c =19.9. b) q = 53.5, r = 41.2. c) t = 9.1, r = 20.2. Exercise 2 a) L = 35.4° , N = 64.6°, n = 7.8. b) Ambiguous case: C = 131.2°, B = 13.8°, b = 6.7 or C = 48.8, B = 96.2, b = 27.7 c) B = 41.7°, A = 73.3°, a = 6.8 B Dec 2016 12 C COSINE RULE The Cosine Rule can be used to solve non-right triangles Cosine Rule a2 = b2 + c2 – 2bc cos α b2 = a2 + c2 – 2ac cos β c2 = a2 + b2 – 2ab cos Υ The angles α, β and γ are respectively opposite the sides a, b, and c. N.B. The side on the left hand side of the equation is opposite the angle listed at the end of the Equation: a2 = b2 + c2 – 2bc cos α Use the Cosine Rule when you are given two sides and the angle between them three sides Examples 1. Find the value of a in this triangle a2 = b2 + c2 – 2bc cos α a a2 = 122 + 152 – 2×12×15 cos 83o a2 = 144 + 225 – 360 × cos 83o b =12 830 c = 15 a2 = 369 – 43.87 a2 = 325.13 a = 18.03 2. Find the size of angle β in this triangle: b2 = a2 + c2 – 2ac cos β 112 = 52 + 72 – 2×5×7×cos β 121 = 25 + 49 – 70 cos β b =11 a=5 ° 47 = – 70 cos β cos β = β = 132⁰11’ B Dec 2016 13 c=7 Exercise 1. Use the sine OR cosine rule to find the pronumeral shown: a) b) a 6.6 23.1 840 19.6 2.3 1610 b c) d) 0.93 13.6 780 350 8.2 1.25 a c 2. Find θ a) b) 15.6 7.21 α0 9.2 β0 9.99 10.1 Answers 1 a) 28.7 2 a) 34..2o B Dec 2016 b) 4.38 c) 8.33 4.93 d) 1.05 b) 42.9o 14 ANGULAR MEASUREMENT - RADIANS Definition of a radian Though angles have commonly been measured in degrees they may also be measured in units known as radians. One radian is the angle created by bending the radius length around the arc of a circle. Converting between radians and degrees Because the circumference of a circle is given by the formula C = 2πr, we know 2π radians (2πc ) is a complete rotation and the same as 360 degrees. Similarly half a rotation or 180 degrees = radians (180⁰ = πc). Angles that represent fractional parts of a circle can be expressed in terms of π: Angle in Degrees Angle in Radians π 2 π 4 π 3 π 6 90 45 60 30 270 3 × 90 = 3 × = For other angles rearranging πc = 180⁰ gives: Examples 1. Convert 60⁰ to radians. 1c = 𝑂 𝜋 and 1𝑂 = 2. Convert 240⁰ to radians. 1 = 1 = 60 = 60 × 240 = 240 × 60 = B Dec 2016 𝜋𝐶 = 1.05C 240 = 15 3. Convert 4. ‘Convert 6 5c to degrees radians to degrees ° 1c = = 4 × 1c = ° ° 6.5C = 6.5 × ° 6.5C = 372.4⁰ = 45⁰ Note: The symbol for radian, c , is often omitted. Exercise 1. Convert the following degrees to radians a) 30⁰ d) 450⁰ b) 270⁰ e) 135⁰ c) 20⁰ f) 57.3⁰ 2. Convert the following radians to degrees a) b) d) 3 5π e) c) π f) 1 radian Answers 1. a) b) c) d) e) f) 1 radian 2. a) 90⁰ d) 630⁰ b) 225⁰ e) 180⁰ c) 330⁰ f) 57.3⁰ B Dec 2016 16 CIRCULAR FUNCTIONS The trigonometric ratios that have been defined in right-angled triangles can be extended to angles greater than 90⁰ by considering angles as rotations within a unit circle. The centre of the unit circle is at the point (0,0) and it has a radius of one unit: Angles are considered as rotations from the positive x-axis. Angles greater than 180⁰ and negative angles can also be defined in terms of the unit circle. Anticlockwise rotations are considered positive and clockwise rotations are negative: An angle of 135o or An angle of -75⁰ c If P(x,y) is any point on the circle, and An angle of 210⁰ or – 150⁰ is the angle POQ in the triangle POQ as shown: y P(x, y) y = sin ( ) 1 O x Q x = cos( ) B Dec 2016 17 Then from the trigonometric properties of a right triangle cos( ) = = = x, sin( ) = cos( ) = x, = = y, tan( ) = sin ( ) = y, = tan( ) = Examples P (0.45, 0.89) 63o (i) (ii) (iii) (iv) sin 0 = 0 cos 180 = -1 sin (-90) = -1 sin 63o = 0.89 (v) tan(-180) = (vi) tan 63o = (vii) cos (-297o) = 0.45 =0 − = 1.98 Cos, sin and tan values can also be calculated using a scientific calculator. [Hint: make sure your calculator is in degrees or radian mode accordingly]. Exact values Using the table of exact values and the symmetry of the unit circle it is also possible to find the exact values for multiples of 30⁰, 45⁰, 60⁰. can be constructed: 30⁰ = Sin Cos 1 2 √ √3 2 √ 1 Tan 45⁰ = 60⁰ = = √ √ = √ 1 2 1 √3 √3 Example Evaluate sin 330⁰ Plotting 330⁰ on a unit circle shows that sin 330⁰ is closely related to sin 30⁰. The y-coordinates differ only by sign because the distances from the x-axis are the same. 1 sin 30⁰ = (from the table) and sin 330⁰ = 2 . 1) Evaluate cos 150⁰ B Dec 2016 18 Plotting 150⁰ on a unit circle shows that cos 150⁰ is closely related to cos 30⁰. The x-coordinates differ only by sign because the distances from the y-axis are the same. √3 cos 30⁰ = √ (from the table) and cos 150⁰ = 3) Evaluate tan Plotting on a unit circle shows that tan 2 . is closely related to tan . Tan values of diametrically opposite angles are the same. tan = √3 (from the table) and tan = √3 . Note: By drawing a sketch diagram it is always possible to use the symmetry of the unit circle and the values in the table to find the exact value of any multiple of 30⁰, 45⁰, 60⁰. The only difference will be a change of sign and the following diagram may assist in determining whether a negative sign will be required. Rotating anticlockwise from the first quadrant: All trigonometric functions are +ve in the first quadrant Sin is +ve for angles in the second quadrant Tan is +ve for angles in the third quadrant Cos is +ve for angles in the fourth quadrant It can be helpful to use a mnemonic such as All Stations To Camberwell to remember these properties of the trigonometric functions Exercise 1) What are the coordinates for points on the unit circle that make the following angles with the positive x-axis? a) 30⁰ d) 270⁰ 2) Find the exact value for: b) 125⁰ e) -180⁰ c) -60⁰ f) 720⁰ a) sin 330 b) cos 210 c) sin( 30 ) d)cos 90 e) tan 300 f) cos 180 g) sin( 120 ) h) cos 315 Answers 1. 2. a) (0.87,0.5) a) -0.5 B Dec 2016 b) (-0.56,0.82) b) √ c) (0.5,- 0.87) d) (0,-1) e) (– 1,0) f) (1,0) c) -0.5 d) 0 e) √3 f) -1 19 TRIGONOMETRIC EQUATIONS Often the value of the trigonometric function is given and the corresponding angle(s), within a given domain, are required. Examples 1) Given sin = 0.3 find all values of in the domain 0 360⁰ The diagram shows that there are two angles in the given domain (one complete positive rotation) that will have a sin value of 0.3. Using the calculator to find the first value: sin = 0.3 = sin-1 (0.3) = 17.46⁰ Using symmetry the angle in the 2nd. quadrant is 180 – 17.46 = 162.54⁰ The solutions are 2) Solve = 0.5 over the domain -2 = 17.46⁰, 162.54⁰ 2 The diagram shows that there will be four angles in the given domain (one complete positive rotation and one complete negative rotation) that will have a cos value of 0.5. Using the table of exact values to find the first value: cos = 0.5 = cos-1 (0.5) = Use symmetry to find the other angles. The solutions are 3) Solve tan 2x = -3 over the domain -90 = , , 180⁰ First, the domain must be adjusted to suit the angle 2x in the equation: -90 x 180⁰ ⟹ -180 2x 360⁰ Using the calculator to find the first value: tan 2 = -3 2 = tan-1 (-3) 2 = -71.57⁰ Tan values of diametrically opposite angles are the same and the diagram shows how to use symmetry to find other angles in the domain 180 2x 360⁰: 2x = -71.57⁰, 108.43⁰, 288.43⁰ Finally we divide by 2 to find solutions for x: x = -35.78⁰, 54.22⁰, 144.22⁰ Exercises a) If sin = 0.25 find c) If cos = 0.4 find e) If tan = -1.5 find Answers a) 14.5⁰ , 165.5⁰ d) 113.6⁰ , - 113.6⁰ B Dec 2016 for 0⁰ for -180⁰ for 0⁰ 180⁰ b) If tan = 0.8 find ⁰ 360 d) If cos = -0.4 find 360⁰ f) If cos = -0.3 find b) 38.7⁰ , 218.7⁰ e) 123.7⁰ , 303.7⁰ for 0⁰ for -180⁰ for 0⁰ c) 66.4⁰ , -66.4⁰ f) 107.5⁰ , 252.5⁰ 20 360⁰ 360⁰ 360⁰ GRAPHS OF SINE AND COSINE FUNCTIONS The functions y = sin and y = cos have a domain of R and a range of [-1, 1]. The graphs of both functions have an amplitude of 1 and a period of 2π radians (repeats every 2π units). y = sin 𝑥 y = cos y 1.5 1 0.5 -8 -6 -4 -2 2 4 x 6 -0.5 -1 -1.5 [Remember 3.142 so 2 6.284] Change of amplitude and period The graphs of both y = sin and y = cos Examples 1) y = 3 sin have amplitude | | and period 2) y = 3cos 2 y y 4 4 3 2 2 1 -6 -4 -2 2 4 6 -6 x -1 -4 -2 2 4 6 x -2 -2 -4 -3 y = 3 cos 2 𝑥 has: amplitude 3 (|𝑎| = 3) 2π period = π (n = 2) y = 3 sin 𝑥 has: amplitude 3 (|𝑎| = 3) period 2π (n = 1) Vertical translation The graphs of y = sin + k, y = cos up k units for k > 0 and down k units for k < 0. are the graphs of y = sin y = cos The graphs of y = sin 𝑥 and y = 2 + sin 𝑥 are shown B Dec 2016 21 translated Horizontal translation Replacing x with (x ) shifts the graphs of y = sin Replacing x with (x+ ) shifts the graphs of y = sin Examples 1) y = ( and y = cos and y = cos 2) y = ) ( y -6 -4 y 1.5 1 1 0.5 0.5 2 4 6 8 x -6 -4 -2 2 -0.5 -0.5 -1 -1 -1.5 -1.5 (4 First change y = 3 the graph is clear. y 4 4 6 8 x ) ) to the form y = 3 (4 y = 3 𝑠𝑖𝑛 4( 𝑥 𝜋 4( ) so that the horizontal translation of ) 3 𝜋 1 1 2 3 𝜋 The period is = The amplitude is 3. There is a horizontal shift right of [NB: Dotted graph is y = 3 𝑠𝑖𝑛 𝑥] 2 -1 units to the right. units to the left. ) 1.5 -2 3) y = 3 -2 horizontally horizontally x 𝜋 . -1 -2 -3 Reflection Changing the sign of x – axis. B Dec 2016 in the equations y = sin or y = cos 22 results in a reflection about the Example y= 3 2 y y = 3cos 2𝑥 4 3 The graph of y = 3cos 2𝑥 is a reflection of y = 3cos 2𝑥 (dotted) in the x-axis 2 1 -3 -2 The period is -1 1 2 3 x 4 𝜋 = 𝜋 The amplitude is 3. -1 -2 -3 Exercises 1.Sketch the graph of the following functions for one complete cycle stating the amplitude and period: (a) y = 2cos (b) y = 2sin 3 (c) y = sin 2 (d) y = 3cos (e) y = -2sin 3 2. Sketch the graph of the following functions for one complete cycle stating the amplitude and period. (a) y = 2sin( ) (b) 3cos(x ) 3. Sketch the graph of the following functions for one complete cycle stating the amplitude and period. (a) y = 2sin(3 ) (b) 3cos(4x 2 ) (c) y = 2sin(2x + ) Answers 1(a) Amplitude = 2. Period = 2 1(b) Amplitude = 2. Period = 2/3 y y 2 2 1 1 1(c) Amplitude = 0.5. Period = y 0.5 1 2 3 4 5 x 6 0.5 -1 -1 -2 -2 1(d) 1 1.5 2 x 0.5 -0.5 1(e) Amplitude = 3. Period = 4 y Amplitude = 2. Period = 2/3 y 3 2 2 1 1 2 4 6 8 10 12 x -1 0.5 1 1.5 2 x -2 -1 -3 -2 B Dec 2016 23 1 1.5 2 2.5 3 x 2(a) Amplitude = 2. Period = 2 2(b) Amplitude = 1. Period = 2 y 1.5 y 1 2 0.5 1 1 1 2 3 4 5 6 x 2 3 4 5 x -0.5 -1 -1 -1.5 -2 3(a) Amplitude = 2. Period = 2/3 3(b) 3(c) Amplitude = 3. Period = /2 Amplitude = 2. Period = y y y 2 3 2 2 1 1 1 0.5 1 1.5 2 x 0.5 1 1.5 x 0.5 -1 -1 -1 -2 -2 B Dec 2016 -2 -3 24 1 1.5 2 x