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Trigonometry and Discrete Math Midterm Exam 2017 Practice Your preparation for the midterm exam should include reviewing your old tests and quizzes, going over your notes and projects, and completing this review. Good luck! Trigonometry and Discrete Math Midterm Exam Topics 2016-17 Unit 1: Trigonometry Chapter 1- Trigonometry and Right Triangles 1.1 Basic Trig- Sine, Cosine, Tangent ratios of triangles SOH, CAH, TOA Solve for missing sides Solve for missing angles (inverse) Angle of elevation and angle of depression word problems using sine, cosine, or tangent 1.2 Reciprocal Trig Functions- Secant, Cosecant, Cotangent ratios of triangles sinβ‘(π΄) sinβ‘(π©) sinβ‘(πΆ) 1.3 Law of Sines π = π = π Law of Cosines π2 = π 2 + π 2 β 2ππ β cos(π΄) π+π+π 1.4 Area of Oblique Triangles- Herons Formula (SSS) π΄πππ = βπ (π β π)(π β π)(π β π)β‘where π = 2 1 SAS Method π΄πππ = 2 ππ β sin(π΄) Chapter 2- Trig in the Unit Circle 180 2.1 Convert Radians to Degrees- multiply by π π Convert Degrees to Radians- multiply by 180 Coterminal angles- add or subtract 360β° or 2Ο Reference Angles- always refer to the x-axis 2.2 Unit Circle- STUDY YOUR UNIT CIRCLE! (and how to find sine, cosine, tangent values) 2.4 Graphing Sine and Cosine, with transformations Period = 2Ο/b Starting/ ending points = set ( ) to 0 and to 2Ο Make sure all graphs include 5 points along the x-axis and 3 points on the y-axis 2.5 Graphing tangent functions, with transformations Tangent- period = Ο/b Intercepts = set ( ) to 0 and to Ο Asymptotes = halfway between asymptotes 2.6 Graphing Cotangent functions, with transformations Asymptotes = set ( ) to 0 and to Ο Intercepts = halfway between asymptotes 2.7 Graphing Secant and Cosecant functions Graph just like a sine or cosine, but add in additional parabola pieces between asymptotes Chapter 3- Trig Identities 3.1 Verify that a trig identity is correct by using mathematical operations and trig identities Remember to work on the more complicated side 3.2 Solving Trig equations Remember to refer to the unit circle at the last step Sine and Cosine answers β must add +2nΟ Tangent answers β must add +nΟ Ferris Wheel Design Project 1. What was the real world problem? 2. What examples of trig did you use and what did you use it for? 3. Connection to math modeling and trigonometric functions What you will be given on the exam: Law of Sines Formula Law of Cosines Formula Heronβs Formula (Area) SAS Area Formula The Unit Circle 51) The leaning tower of Pisa is tilted at an angle of 96β° from the ground. An observer standing 100m from the base of the tower measures the angle of elevation to be 30β°. Find the height of the tower using either law of sines or law of cosines. 52) After the hurricane, the small tree in the yard was leaning. To keep it from falling, a 6 foot strap was nailed into the ground 4 feet from the base of the tree. The strap was nailed to the tree 3.5 feet above the ground. What is the angle that the tree is leaning? Use the law of sines or the law of cosines to solve. 53) The area of a triangular room is 140 feet2. The length of side AB is 18 feet and angle B is 75β°. Find the length of side BC. 54) A triangular field is surveyed. The length of one side of the field measured 365 meters and another side was 267 meters. The third side measured 308 meters. Find the area of the farmerβs field. Verify the following trigonometric expressions using all identities. π πππ π πππ π πππ πππ π 55) 1βπππ π + 1+πππ π = 2ππ ππ 56)β‘π‘πππ + πππ‘π = πππ π + π πππ 57) π‘ππ2 π(1 β π ππ2 π) = π ππ2 π 58)β‘πππ‘ππ ππππ πππ = 1 59) 1 1βπ πππ 1 + 1+π πππ = 2π ππ 2 π 60)β‘ππ ππ β πππ ππππ‘π = π πππ Solve the following trigonometric equations for the radian measure using the unit circle. 61) 7πππ‘π₯ β β3 = 4πππ‘π₯ 62) 2 = 4πππ 2 π₯ + 1 51) Set Up: π ππ30 π₯ = π ππ54 100 (Solve for 3rd angle first, then use law of sines) x = 61.8 m 52) Set Up: (6)2 = (3.5)2 + (4)2 β 2(3.5)(4)πππ π ΞΈ = 106.3β° 1 53) Set Up: 150 = 2 (18)π(π ππ75) Length of a = 16 feet 54) Semi perimeter = 470 Set Up: π΄ = β470(470 β 365)(470 β 308)(470 β 267) Area = 40,285.5 meters 2 (1+πππ π)π πππ π πππ(1βπππ π) π πππ+π ππππππ π+π πππβπ ππππππ π 55) (1+πππ π)1βπππ π + 1+πππ π(1βπππ π) β (πππ‘π)π πππ πππ π(π‘πππ) 56) (πππ‘π)π‘πππ + πππ‘π(π‘πππ) β 1βπππ π+πππ πβπππ 2 π πππ π π πππ πππ π π πππ β + β π πππ 1 1 πππ π πππ‘ππ‘πππ β πππ π+π πππ π ππ2 π 57) π‘ππ2 π(1 β π ππ2 π)β‘ β π‘ππ2 π(πππ 2 π) β πππ 2 π β 58) πππ‘ππ ππππ πππ β (1+π πππ)1 πππ π π πππ β 1 πππ π β π πππ 1 1 πππ 2 π 1 2π πππ 2π πππ β πππ π + π πππ β π ππ2 π β1 1(1βπ πππ) 1+π πππ+1βπ πππ 2 2 59) (1+π πππ)1βπ πππ + 1+π πππ(1βπ πππ) β 1βπ πππ+π πππβπ ππ2 π β 1βπ ππ2 π β πππ 2 π β 2π ππ 2 π 1 60) ππ ππ β πππ ππππ‘π β π πππ β π πππ π πππ π 1 61) π‘πππ₯ = β3 β π₯ = 3 + ππ, π₯ = 1 π 1 β π πππ β π πππ β 4π 3 62) πππ π₯ = ± 2 β π₯ = 3 + 2ππ, π₯ = πππ 2 π π πππ β 1βπππ 2 π π πππ β π ππ2 π π πππ β π πππ + ππ 2π 3 + 2ππ, π₯ = 4π 3 2 β 1βπππ 2 π β π ππ2 π β π πππ β 2ππ ππ + 2ππ, π₯ = 5π 3 + 2ππ
