Trigonometry 2
15th November 2016
Welcome!
• Trigonometry 2
• Question Focused Approach
• 2 hours - 10mins break at 7pm
• Any questions? – Just Ask!!
• Material: https://web.actuaries.ie/students/tutorials
Today’s Topics
1.
2.
3.
4.
5.
Exam Tips
Recap on Trigonometry Part 1
The 4 Quadrants
Trigonometry Identities
Past exam questions
Trigonometry Exam Tips:
• Draw a diagram every time.
• Label all diagrams.
• Check the mode on your calculator – radians or degrees.
• Know how and when to round your answer.
Recap: Important Formulae
1.
2.
3.
4.
5.
6.
7.
8.
Formulae for sin, cos and tan
Pythagoras Theorem
Cosine Rule
Sine Rule
Area of Triangle
Length of Arc
Area of section of circle
Formula for radians to degrees
Recap: Important Formulae
1. Formulae for sin, cos and tan
𝑜𝑝𝑝
ℎ𝑦𝑝
𝑎𝑑𝑗
𝑜𝑝𝑝
sin 𝜃 =
; cos 𝜃 =
; tan 𝜃 =
ℎ𝑦𝑝
𝑎𝑑𝑗
Pythagoras Theorem ℎ𝑦𝑝2 = 𝑜𝑝𝑝2 + 𝑎𝑑𝑗 2
2.
3. Cosine Rule
4. Sine Rule
a 2  b 2  c 2  2bcCosA
sin𝐴
𝑎
=
5. Area of Triangle =
sin𝐵 sin𝐶
=
𝑏
𝑐
1
𝑎𝑏 sin 𝐶
2
or
𝑎
sin𝐴
=
𝑏
sin𝐵
=
𝑐
sin𝐶
Recap: Important Formulae
6. Length of Arc 𝑙 = 𝑟𝜃
1
2
7. Area of section of circle Area = 𝑟 2 𝜃
8. Formula for radians to degrees 𝜋 𝑟𝑎𝑑𝑖𝑎𝑛𝑠 = 180°
Log tables – Pages 13, 14, 15 and 16
The 4 Quadrants
90°
θ° = x°
θ° = 180° - x°
2nd Quadrant
1st
Quadrant
180°
3rd
Quadrant
4th
0°
360°
Quadrant
θ° = 180° + x°
θ° = 360° - x°
270°
Finding the solutions for θ°
90°
θ° = x°
θ° = 180° - x°
Sin +
All +
180°
0°
360°
Cos +
Tan +
θ° = 180° + x°
θ° = 360° - x°
270°
Solving for θ° - The 4 Quadrants
x = the reference angle
1st Quadrant
θ° = x°
2nd Quadrant
θ° = 180 ° - x°
3rd Quadrant
θ° = 180 ° + x°
4th Quadrant
θ° = 360 ° - x°
Finding the solutions for θ°
90°
θ° = x°
θ° = 180° - x°
2nd Quadrant
1st
Quadrant
45°
180°
3rd
Quadrant
4th
0°
360°
Quadrant
θ° = 180° + x°
θ° = 360° - x°
270°
Finding the solutions for θ°
90°
θ° = x°
θ° = 180° - x°
2nd Quadrant
1st
Quadrant
120°
180°
3rd
Quadrant
4th
0°
360°
Quadrant
θ° = 180° + x°
θ° = 360° - x°
270°
Finding the solutions for θ°
90°
θ° = x°
θ° = 180° - x°
2nd Quadrant
1st
Quadrant
210°
180°
3rd
Quadrant
4th
0°
360°
Quadrant
θ° = 180° + x°
θ° = 360° - x°
270°
Finding the solutions for θ°
90°
θ° = x°
θ° = 180° - x°
2nd Quadrant
1st
Quadrant
180°
3rd
4th
Quadrant 320°
Quadrant
θ° = 180° + x°
0°
360°
θ° = 360° - x°
270°
The angles 30ᴼ,45ᴼ,60ᴼ and 90ᴼ
Finding the solutions for θ°
Question:
cos(θ°) = ½
Find the solutions for θ° (given 0≤θ°≤360°).
Warning: There is more than one answer!
Finding the solutions for θ°
Step 1: Find the reference angle x°
cos(θ°) = ½
reference angle = x°
= the angle whose cos is ½
a.k.a. cos-1[½]
= 60°
Finding the solutions for θ°
Step 2: What quadrants could θ° be in?
cos(θ°) is positive in the 1st and 4th Quadrants so we will have 2
answers from these Quadrants.
Finding the solutions for θ°
Step 3: Find the solutions using the formulae:
•
•
•
•
1st Quadrant: θ° = x°
2nd Quadrant: θ° = 180 ° - x°
3rd Quadrant: θ° = 180 ° + x°
4th Quadrant: θ° = 360 ° - x°
1st Quadrant
4th Quadrant
θ° = x°
θ° = 60°
θ° = 360° - x°
= 360° - 60°=300°
Solutions: θ° = {60°,300°}
Check on calculator!
Finding the solutions for θ°
Find the solutions for θ° for the following:
1. cos(θ°) =
2. sin(θ°) =
3
2
1
2
3. Tan(θ°) = -1
where 0≤θ°≤360°.
You may use a calculator!
Step 1
Find the reference angle
Step 2
What quadrants could θ° be in?
Step 3
Find the solutions
Finding the solutions for θ°
Find the solutions for θ° for the following:
1. cos(θ°) =
3
2
=>Reference angle = 30°
– Cos is positive in the 1st and 4th quadrants
– 1st quadrant = 30°
– 4th quadrant = 360° - 30° = 330°
Finding the solutions for θ°
Find the solutions for θ° for the following:
2. sin(θ°) =
1
2
=> reference angle 45°
1st quadrant: 45°
2nd quadrant: 180° – 45° = 135°
3. tan(θ°) = -1=> reference angle 45°
2nd quadrant: 180° - 45° = 135°
4th quadrant: 360° - 45° = 315°
Trigonometry Identities Question
• Hint! Check page 14 of log tables for sin2x
• Take out the common term and put the rest in brackets
• Let it equal to zero
Trigonometry Identities Question - Solution
•
•
•
•
•
•
•
sin2x + sinx = 0
2sinxcosx + sinx = 0
sinx(2cosx + 1) = 0
sinx = 0
or
sinx = 0
Reference angle = 0°
x = 0°,180°,360°
2cosx + 1 = 0
Trigonometry Identities Question – Solution
•
•
•
•
•
•
•
sin2x + sinx = 0
sinx = 0
or
2cosx + 1 = 0
cosx = -1/2
=> reference angle = 60°
Cos is negative in the 2nd and 3rd quadrants
2nd quadrant: 180° - 60° = 120°
3nd quadrant: 180° + 60° = 240°
x= 0°, 120°, 180°, 240 ° or 360°
Sine Rule
Page 16
Sine Rule
• Do not need to have a right angle
• Do need to have at least one matching pair (angle and
opposite side) and one more angle/side
Cosine Rule
What is it used for?
1) If given a triangle and the lengths of the three sides but no angle.
or
2) If given two sides and an angle
The cosine rule states that for any triangle:
a² = b² + c² - 2bc cosA
Area of a triangle
1
𝐴𝑟𝑒𝑎 = 𝑎𝑏𝑆𝑖𝑛𝐶
2
Or
𝐴𝑟𝑒𝑎 =
1
𝑏𝑐𝑆𝑖𝑛𝐴
2
Or
𝐴𝑟𝑒𝑎 =
1
𝑎𝑐𝑆𝑖𝑛𝐵
2
Question 1 (LC 2014)
Question 1 - Solution
120 m
72.15°
150 m
134 m
Question 2 (LC 2013)
(iii)
Question 2 Solutions
• Sin-1 (0.756) = 49.11°
• Sin is positive in the
1st and 2nd quadrants
• 180° – 49° = 131°
5 cm
3 cm
3 cm
27°
Question 2 Solutions
(iii)
• Use 49° as 49°< 90°
• 27° is the angle at Y
• All angles must sum to 180° so the remaining angle must equal 104°
Question 3 (LC 2015)
(a) diagram
(a) Joan is playing golf. She is 150 m from the centre of a circular green of diameter
30 m. The diagram shows the range of directions in which Joan can hit the ball so that
it could land on the green. Find α, the measure of the angle of this range of directions.
Give your answer, in degrees, correct to one decimal place.
Question 3 Solutions
sin−1 0.1 = 5.739°
Question 3 Solution
• Part (b) alternative solution
Question 4 (LC 2016)
Question 4 Solutions
Question 4 Solutions
Question 4 Solutions
Question 4 (continued) (LC 2016)
Question 4(b) Solution
Next Tutorial
• 22nd November
• Same location
• 6 – 8pm
• Geometry 2
• Tutorial material:
https://web.actuaries.ie/students/tutorials