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Asia Pacific University of Technology and Innovation Trigonometry Definition Trigonometry deals with the relationship between the sides and angles of the triangle. Examples of things that we can measure in a triangle are the lengths of the sides, the angles and the area of the triangle and so forth. Triangles are used in the construction of houses, skyscrapers, bridges or any structure where stability is desired. This branch of Mathematics is used to solve problems in Surveying, Geography, Astronomy and Navigation etc., Definition on angle An angle is when two rays (think of a ray as “half” of a line) have their end point in common. The two rays make up the “sides” of the angle, called the initial and terminal side. A picture of an angle is shown below. Terminal Initial Pythagorean Theorem Given a right angled triangle with sides of length a, b and c as shown below (c being the longest side, which is also called the hypotenuse) then a2 + b2 = c2 . c a b Example: Use the Pythagorean Theorem to find the missing side of the right angled triangle shown below. 3 c 7 Solution: In this triangle we are given the lengths of the “legs” (i.e. the sides joining the right angle) and we are missing the hypotenuse, or c. So 32 + 72= c2 or c2 = 58 or c = √58 = 7.616 Relation between radians and degrees We have two ways to measure angles. In degrees a full revolution corresponds to 360° while in radians a full revolution corresponds to 2π rads. we have that 360° = 2π rads. This can be rearranged to give the following useful (if not quite correct) relationship, 180 rad 1 rad 180 This gives a way to convert from radians to degrees or from degrees to radians. Examples 1. 2. Convert 240° to radians Convert 3π/8 radians to degrees. Solution: 1. 2. 240° = 240°. = 4 rad 3 180 3 3 180 . 67.5 8 8 Review questions 1. Convert 150° into radian 1 radian int o deg ree 4 Answers 2. 1. 2. 150° = 150°. 180 = 5 rad 6 1 180 1 7 180 14195 4 4 22 Quadrants Let X OX and YOY be two lines at right angles to each other in the figure. We call X OX and YOY as x-axis and y-axis respectively. Clearly these axes divide the entire plane into FOUR equal parts, called Quadrants. Note: Angle is measured anti-clockwise from positive x-axis. Memory Aid – “ALL Silver Tea Cups” I II III IV - Positive The trigonometric functions:\ The six trigonometric functions are sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec) and cosecant (csc). They are defined in terms of ratios in the following way. Note: Sine is the Opposite over the Hypotenuse, Cosine is the Adjacent over the Hypotenuse and Tangent is the Opposite over the Adjacent. Opposite side b Hypotenuse a Adjacent side c Cos B = Hypotenuse a Opposite side b Tan B = Adjacent side c Sin B = Hypotenuse a Opposite side b Hypotenuse a Sec B = Adjacent side c Adjacent side c Cot B = Opposite side b Cosec B = Note: Sin = Sine Cos = Cosine Tan = Tangent Cosec [or] csc = cosecant Sec = secant Cot = cotangent When θ is an acute angle, the definition of these ratios are given as above. Example If (2, 3) is a point on the terminal side of θ, find all the six trigonometric ratios. Solution: x = 2, y = 3, r = x 2 y 2 4 9 13 y 3 sin r 13 x 2 cos r 13 y 3 tan x 2 r 13 cos ec y 3 r x x cot y sec 13 2 2 3 Review question 7 1. If tan , find cos and cos ec . 25 8 2. If cos , find tan 10 Answers x 8 1. cos x 8, r 10, r = 252 72 625 49 674 25.96 26 r 10 x 25 cos r 26 r 26 cos ec y 7 y 6 2. tan x 8 Sine Rule Sine Rule can be used with any triangle as long as we have been given enough information. a b c = = SinA SinB SinC Example 1 Find the length of a in the following triangle. Solution: Given information in the following table. Side a=? b=? c = 10 Angle A = 41° B=? C = 34° Matching up the sides and angles, we get a c sin A sin C a 10 sin 41 sin 34 Rearranging to solve for a, 10 0.656 10 sin 41 a 11.74m a 0.559 sin 34 Note: b and B are unknown in the above example, so don’t choose the ratio. Example 2 In triangle ABC, A = 35 017 , C = 45 013 and b = 42.1 cm, Find B, a and c using Sine Law. Solution: Given information in the following table. Side Angle A = 35 017 a=? B ? [1800 35017 45013] 9930 b = 42.1 C = 45 013 c=? a 42.1 0 sin 35 17 sin 99 030 42.1 a sin 35 017 = 24.65 o sin 99 30 42.1 c 0 sin 99 30 sin 45013 42.1 c sin 45 013 = 30.3 sin 99 o 30 Example 3 Triangle PQR has PQ = 7.73 cm, QR = 13 cm, and PRQ 35 . Calculate the two possible values of RPQ . Solution: 7.73 13 sin 35 sin P 13 sin 35 sin P 0.9646 7.73 P = sin 1 0.9646 74.71or105.29 Review questions 1. In a triangle ABC, A = 40°, B = 52°, a = 9 cm, find C and b. (Note: c is the side opposite to the angle C) 2. In triangle ABC, A = 56 0 , C = 73 0 and a = 11 cm, find B, b and c. Answers 1. C = 180° - 52° - 40° = 88° b 9 0 sin 52 sin 40 0 9 b sin 52 0 = 11.03 cm 0 sin 40 2. B = 180 – 56 – 73 = 51 11 b sin 56 sin 51 11 sin 51 10.31 cm sin 56 11 c sin 56 sin 73 11 c sin 73 12.69 cm sin 56 b Cosine Rule Cosine Rule can be used with any triangle as long as we have been given enough information. The Cosine rule can be used to find: 1. 2. An unknown side when two sides of the triangle and the included angle are given (SAS) An unknown angle when 3 sides are given (SSS). Finding an unknown side 2 2 2 2 2 2 2 2 2 a = b + c – 2bcCosA b = a + c – 2acCosB c = a + b – 2abCosC Alternate version Finding an unknown angle b2 c 2 a 2 2bc 2 a c2 b2 CosB 2ac 2 a b2 c2 CosC 2ab CosA Example Angle A in the following triangle is 45o, length b is 2 units and length c is 3 units. Find angles B and C and the length a using cosine rule. Solution: Given information in the following table. Side Angle a=? A = 45° b=2 B=? c=3 C=? Note: If an unknown side to find when two sides of the triangle and the included angle are given (SAS), the following rule is to be used. 2 2 2 a = b + c – 2bcCosA a2 = 22 + 32 -2 (2)(3) cos (45o) a2=4 + 9 -12 cos (45o) a = 2.1 To work out the angle B, we can rewrite the cosine rule. b2=a2 + c2- 2ac cos B 4=4.5+9-12.6*cosB 4 – 4.5 – 9 = - 12.6 cos B 9.5 cos B 12.6 Cos B = 0.75 B = 410 24 41 a2 c2 b2 Note : CosB can also be used . 2ac Note that A + B + C = 180° So C = 180° - 45° - 41°= 94° Review questions 1. In triangle ABC, AB = 7 cm, AC = 13 cm, If its area = 30 cm2, calculate length of BC 2. For the diagram below, find the length of a. Answers 1. BC 2 AC 2 AB 2 2 AB( AC) cos A 7 2 132 2(13)(7) cos 41.250 = 218 – 136.8422 = 81.1578 BC = 81.1578 = 9.009 1 1 (b)(c) sin A 30 (7)(13)(sin A) 2 2 0 A 41.25 Note: Area = 2. a 212 32 2 2(21)(32) cos 40 0 20.87 Graph of trigonometric functions Formula for drawing the trig functions Amplitude = A = A Period = 360 B If we sketch the graph of y = a sin (x), a > 1 stretches the graph in the y-axis direction a < 1 compresses the graph in the y-axis direction Negative value of a flips graph in the x-axis Here a = A B defines how many times it repeats itself in 360° Example 1 Draw the graph of y = 3 sin x A B 60 (91)(sin A) sin A 60 91 Solution: Amplitude = A = 3 3 Period = x y 360 = 360° 1 -2π 0 3 2 3 -π 0 0 2 -3 0 2 3 π 0 3 2 -3 2π 0 Note: As this curve has a period 2π, we can extend this curve in both directions along the x-axis for as many cycles as we please. Example 2 Draw the graph of y = - cos 3x Solution: Amplitude = A = 1 1 Period = x y 360 = 120° 3 0 -1 30° 0 60° 1 90° 0 120° -1 Note: The "minus" sign tells us that the graph is upside down Review question Draw the graph of y = 5 cos x 2 Answer Amplitude = A = 5 5 Period = x y 360 = 720° 1 2 0 180° 5 0 360° -5 540° 0 720° 5 Note: you are reminded to draw using the table of values. Area for any triangle 1 Area = ab sin C 2 1 Area = bc sin A 2 1 Area = ac sin B 2 Example Find the area of triangle ABC which has AB = 12 cm, BC = 14 cm and ABC = 35o. Solution: 12 14 sin 35 0 Area = = 48.18 cm2 2 Review questions 1. In ABC , A 57 0 , AB 8.5cm and AC 6.3cm. Calculate the area of ABC. 2. ABC is a triangle such that AB = 5 cm and AC = 10 cm and A 120 o . Find the area of the triangle ABC. Answers 1 1. Area of ABC bc sin A 2 1 (6.3)(8.5) sin 57 0 22.46cm 2 2 1 2. Area of the triangle ABC = (5)(10) sin 120 o 2 21.65cm 2` Table of Fundamental trigonometric Identities Reciprocal identities Pythogorean Identities Quotient Identities sin u 1 Sin u = Sin2u + Cos2u = 1 Tan u csc u cos u 1 cos u Cos u = Sin2u = 1 - Cos2u Cot u sin u sec u 1 Tan u = Cos2u = 1- Sin2u cot u 1 Csc u = 1 + tan2u = Sec2u sin u 1 Sec u = 1 + cot2u = Csc2u cos u 1 Cot u = tan u Co-Function Identities Even Identities Sin u cos u Sin(-u) = - sin u 2 Cos u sin u Cos(-u) = cos u 2 Tan u cot u Tan(-u) = – tan u 2 Csc u sec u Csc(-u) = - csc u 2 Sec u Cscu Sec(-u) = sec u 2 Cot u tan u Cot(-u) = - cot u 2 Example 1 Show that sec2 x + csc2 x = sec2x csc2 x Solution: The problem means that we are to write the left-hand side, and then show, through substitutions and algebra, that we can transform it to look like the right hand side. We begin: 1 1 = Sec2x + csc2x Using Reciprocal identities 2 cos x sin 2 x sin 2 x cos 2 x cos 2 x sin 2 x 1 2 cos x sin 2 x 1 1 2 cos x sin 2 x = Sec2x + csc2x =RHS on adding the fractions Using Pythagorean identities Using Reciprocal identities Example 2 Show that 1 1 2 sec 2 1 sin 1 sin Solution: LHS = 1 1 1 sin 1 sin on adding the fractions 1 sin 1 sin 1 sin 2 1 sin 1 sin 2 2 by Pythogorean Identities 2 2 1 sin 1 sin cos 2 2 2 sec 2 RHS by reciporcal identities. 2 cos Note: LHS = Left hand side, RHS = Right hand side Review questions 1. 2. Prove cos x tan x sin x sec x Prove the trigonometric identity sin 2 cos 2 1 2 cos 2 . Answers 1. 2. sin x cos 2 x sin 2 x cos x tan x sin x cos x . sin x sec x cos x cos x sin 2 cos 2 (1 cos 2 ) cos 2 1 2 cos 2 Solving trigonometric equations A trigonometric equation is any equation that contains a trigonometric function. Example 1 Solve for x in the following equation for 0 x 2 2 sin x 1 0 Solution: 2 sin x 1 sin x 1 2 sin x 0.5 x = sin 1 0.5 30 Since the value is positive, the angle should be in QI and QII, hence 5 x and x 6 6 Note: sin 30° = 0.5 and also sin 150° = 0.5 So, the angle should be in QI and QII The answers can be checked numerically and graphically. Example 2 Solve for x in the following equation for 0 x 2 x (i) cos = 0.866 2 (ii) tan 2x 3.732 Solution: x (i) cos 0.866 2 Hence, reference angle = 300 x 2 Since 0 0 x360 0 ,0 0 180 0 Thus, (ii) x is in 1st quadrant only. 2 x x 60 0 30 0 2 tan 2 x 3.732, tan 75 0 3.732 Hence, reference angle = 750 0 0 0 0 Since 0 x360 ,0 2 x720 Thus, 2x is in 1st quadrant and 3rd quadrant 2 x 180 0 750 ,360 0 750 ,360 0 (180 0 750 ),360 0 (360 0 750 ) x 52.50 ,142.50 ,232.50 ,322.50 Review questions 1. 2. Solve the following trigonometric equations cos 0.807 for 0 360 . Solve the equation 2 cos ec 1 6 sin for 0 o 360 o . Answers cos 0.807 1. cos 36.20 0.807 2. Hence the reference angle 36.20 Since cos is a negative value, is in quadrant II or III. Therefore, =180o-36.20o, 180o+36.20o = 143.80o, 216.20o 2 cos ec 1 6 sin 2 1 6 sin sin 6 sin 2 sin 2 0 (3 sin 2)( 2 sin 1) 0 2 1 sin or sin 3 2 o o 221.81 , 318.9 , 30o, 150o Points to remember: 2 Pythagorean theorem Relation between radians and degrees Six trigonometric functions Sine Rule Cosine Rule Graph of trigonometric functions a b a a Sec B = c d a Cb( x) Cc( x) = dx = SinA SinB SinC Sin B = a2 = b2 + c2 – 2bc sin A b2 = c2 + a2 – 2ca sin B c2 = a2 + b2 – 2ab sin C Amplitude = A Period = Area for any triangle Trigonometric Identities Solving trigonometric equation Formula/Concept + b = c2 π radian = 180° 1 radian = 57.3° [ approximately] c b a ; Cos B = ;Tan B = ;Cosec B = a c b c ; Cot B = b 2 360 B 1 ab sin C 2 1 Area = bc sin A 2 1 Area = ac sin B 2 Refer the table given Refer the following table. Use Identity when necessary. Area = For plotting the graphs, the values of y 0, , corresponding to x. 6 4 3 5 , , 2 3 , 2 3 5 7 5 4 , , , , , , , , 3 2 3 7 11 , , . 4 6 4 6 6 4 3