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
1.3
Law of Sines sin⁡(𝐴) 𝑎
Law of Cosines 𝑎 2
= sin⁡(𝑩)
= 𝑏 2 𝑏
=
+ 𝑐 2 sin⁡(𝐶) 𝑐
− 2𝑏𝑐 ⋅ cos(𝐴)
1.4
Area of Oblique Triangles- Herons Formula (SSS) 𝐴𝑟𝑒𝑎 = √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)⁡ where 𝑠 = 𝑎+𝑏+𝑐
2
SAS Method 𝐴𝑟𝑒𝑎 =
1
2 𝑏𝑐 ⋅ sin(𝐴)
Chapter 2- Trig in the Unit Circle
2.1
Convert Radians to Degrees- multiply by
180 𝜋 𝜋
Convert Degrees to Radians- multiply by
180
Coterminal angles- add or subtract 360 ⁰ or 2π
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 = 2π/b
Starting/ ending points = set ( ) to 0 and to 2π
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 = π/b and
Intercepts = set ( ) to 0 and to π
Asymptotes = halfway between asymptotes
2.6
Graphing Cotangent functions, with transformations
Asymptotes = set ( ) to 0 and to π
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
Simplify expressions by using mathematical operations and trig identities
3.2
Verify that a trig identity is correct- remember to work on the more complicated side
3.3
Solving Trig equations
Remember to refer to the unit circle at the last step
Sine and Cosine answers – must add +2nπ
Tangent answers – must add +nπ
Unit 2: Mathematical Modeling
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

What you will be given on the exam:
Law of Sines Formula
Law of Cosines Formula
Heron’s Formula (Area)
SAS Area Formula
The Unit Circle

47) The leaning tower of Pisa is tilted at an angle of 96 ⁰ from the ground. An observer standing 100m from the base of the tower measures the angle of elevation to be 30 ⁰ . Find the height of the tower using either law of sines or law of cosines.
48) 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.
49) The area of a triangular room is 140 feet
2
. The length of side AB is 18 feet and angle B is 75 ⁰ . Find the length of side BC.
50) 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 farmer’s field.
Simplify the following trigonometric expressions using all identities. 𝑠𝑖𝑛𝜃
51)
1−𝑐𝑜𝑠𝜃
+ 𝑠𝑖𝑛𝜃
1+𝑐𝑜𝑠𝜃
52) ⁡ 𝑠𝑖𝑛𝜃 𝑡𝑎𝑛𝜃
+ 𝑐𝑜𝑠𝜃 𝑐𝑜𝑡𝜃
Verify the following trigonometric equations using all trig identities.
53) 𝑡𝑎𝑛 2 𝜃(1 − 𝑠𝑖𝑛 2 𝜃) = 𝑠𝑖𝑛 2 𝜃 54) ⁡𝑐𝑜𝑡𝜃𝑠𝑒𝑐𝜃𝑠𝑖𝑛𝜃 = 1
1
55)
1−𝑠𝑖𝑛𝜃
1
+
1+𝑠𝑖𝑛𝜃
= 2𝑠𝑒𝑐 2 𝜃 56) ⁡𝑐𝑠𝑐𝜃 − 𝑐𝑜𝑠𝜃𝑐𝑜𝑡𝜃 = 𝑠𝑖𝑛𝜃
Solve the following trigonometric equations for the radian measure using the unit circle.
57) 7𝑐𝑜𝑡𝑥 − √3 = 4𝑐𝑜𝑡𝑥 58) 2 = 4𝑐𝑜𝑠 2 𝑥 + 1

47) Set Up: 𝑠𝑖𝑛30 𝑥
= 𝑠𝑖𝑛54
(Solve for 3
100 rd angle first, then use law of sines) x = 61.8 m
48) Set Up: (6) 2 = (3.5) 2 + (4) 2 − 2(3.5)(4)𝑐𝑜𝑠𝜃
θ = 106.3
⁰
49) Set Up: 150 =
1
(18)𝑎(𝑠𝑖𝑛75)
2
Length of a = 16 feet
50) Semi perimeter = 470
Set Up: 𝐴 = √470(470 − 365)(470 − 308)(470 − 267)
Area = 40,285.5 meters
2
(1+𝑐𝑜𝑠𝜃)𝑠𝑖𝑛𝜃
51)
(1+𝑐𝑜𝑠𝜃)1−𝑐𝑜𝑠𝜃
(𝑐𝑜𝑡𝜃)𝑠𝑖𝑛𝜃
52)
(𝑐𝑜𝑡𝜃)𝑡𝑎𝑛𝜃
+
+ 𝑠𝑖𝑛𝜃(1−𝑐𝑜𝑠𝜃)
1+𝑐𝑜𝑠𝜃(1−𝑐𝑜𝑠𝜃)
→ 𝑠𝑖𝑛𝜃+𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃+𝑠𝑖𝑛𝜃−𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃
1−𝑐𝑜𝑠𝜃+𝑐𝑜𝑠𝜃−𝑐𝑜𝑠 2 𝜃
2𝑠𝑖𝑛𝜃
→
1−𝑐𝑜𝑠 2 𝜃
→
2𝑠𝑖𝑛𝜃 𝑠𝑖𝑛 2 𝜃
2
→ 𝑠𝑖𝑛𝜃
→ 2𝑐𝑠𝑐𝜃 𝑐𝑜𝑠𝜃(𝑡𝑎𝑛𝜃) 𝑐𝑜𝑡𝜃(𝑡𝑎𝑛𝜃)
→ 𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃
∙ 𝑠𝑖𝑛𝜃
+
1 𝑐𝑜𝑠𝜃
∙
1 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 𝑐𝑜𝑡𝜃𝑡𝑎𝑛𝜃
→ 𝑐𝑜𝑠𝜃+𝑠𝑖𝑛𝜃
1
→ 𝑐𝑜𝑠𝜃 + 𝑠𝑖𝑛𝜃
53) 𝑡𝑎𝑛 2 𝜃(1 − 𝑠𝑖𝑛 2 𝜃)⁡ → 𝑡𝑎𝑛 2 𝜃(𝑐𝑜𝑠 2 𝑠𝑖𝑛
2 𝜃) → 𝑐𝑜𝑠 𝜃
2 𝜃
∙ 𝑐𝑜𝑠
2 𝜃
→ 𝑠𝑖𝑛
1
2 𝜃
54) 𝑐𝑜𝑡𝜃𝑠𝑒𝑐𝜃𝑠𝑖𝑛𝜃 → 𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃
(1+𝑠𝑖𝑛𝜃)1
55)
(1+𝑠𝑖𝑛𝜃)1−𝑠𝑖𝑛𝜃
+
∙
1 𝑐𝑜𝑠𝜃
1(1−𝑠𝑖𝑛𝜃)
∙ 𝑠𝑖𝑛𝜃
1
1+𝑠𝑖𝑛𝜃(1−𝑠𝑖𝑛𝜃)
→
→ 1
1+𝑠𝑖𝑛𝜃+1−𝑠𝑖𝑛𝜃
1−𝑠𝑖𝑛𝜃+𝑠𝑖𝑛𝜃−𝑠𝑖𝑛 2 𝜃
2
→
1−𝑠𝑖𝑛 2 𝜃
2
→ 𝑐𝑜𝑠 2 𝜃
→ 2𝑠𝑒𝑐 2 𝜃
1
56) 𝑐𝑠𝑐𝜃 − 𝑐𝑜𝑠𝜃𝑐𝑜𝑡𝜃 → 𝑠𝑖𝑛𝜃 𝜋
57) 𝑡𝑎𝑛𝑥 = √3 → 𝑥 =
3
− 𝑐𝑜𝑠𝜃
1
+ 𝑛𝜋, 𝑥 =
58) 𝑐𝑜𝑠𝑥 = ±
1
2
→ 𝑥 = 𝜋
3
+ 2𝑛𝜋, 𝑥 =
∙
4𝜋
3 𝑐𝑜𝑠𝜃 𝑠𝑖𝑛𝜃
1
→ 𝑠𝑖𝑛𝜃
2𝜋
3
+ 𝑛𝜋
−
+ 2𝑛𝜋, 𝑥 = 𝑐𝑜𝑠
2 𝑠𝑖𝑛𝜃 𝜃
→
1−𝑐𝑜𝑠 𝑠𝑖𝑛𝜃
2 𝜃
→ 𝑠𝑖𝑛
2 𝑠𝑖𝑛𝜃 𝜃
→ 𝑠𝑖𝑛𝜃
4𝜋
3
+ 2𝑛𝜋, 𝑥 =
5𝜋
3
+ 2𝑛𝜋