trig

1. TRIGONOMETRY
HIGHER LEVEL

2. REAL LIFE APPLICATIONS
For Finding Distances
Between Celestial Bodies.
For Calculating the Height
and Steepness of
Mountains.
For Ocean Navigation.
Construction Industry
Sports
Calculating Tides
WHY STUDY TRIGONOMETRY ?

3. DESIGN OF
BEAUTIFUL
THINGS…
R.I.P

4. IT SAID PUT IT IN THE OVEN AT 130 DEGREES
MISCONCEPTIONS Terms confuse students.
What do they mean?
Relevance of Topic to their lives.
Make it relevant
Applications and connections
poorly dealt with in text books
Explanation of why
- Graphs, Equations & Identities

5. WHAT ARE THE ORIGINS OF THE TERMS?
HOW MANY OF YOU WERE TAUGHT THIS IN SCHOOL?
HOW MANY OF YOU WONDERED OR FOUND
THE TERMS DAUNTING AT JC LEVEL ?

6. ORIGINS OF
TRIGONOMETRY
• Trigonometry: Origins from
the Greek word ‘Trigonon’
for Triangle.
HYPOTENUSE: is Greek for ‘stretched under’
Greeks saw Hypotenuse as being stretched
over the legs of a right angles triangle
like a bow and arrow.
SINE comes from the Latin word ‘Sinus’
Which means bend, cavity or bosom of
a piece of clothing. Through several Arabic
and Sanskrit translations it came to mean
bowstring. The term originally applied to the line segment MB.
The ratio of the ‘sine’ MB to the radius, OA, is the SINE of angle AOB.

7. TANGENT comes from the Latin word ‘tangens’
Which means ‘to touch’. It was originally applied to the
line segment AD – the segment of the tangent to the
circle at A which is ‘cut off’ by the extension of radius OB.
The ratio of the ‘tangent’ AD to the radius, OA, is the
TANGENT of angle AOB.
D
A
B
O
COSINE was originally written CO-SINE which means
‘complementi of Sinus’ - the sine of the complementary angle.
Hence, the COSINE of angle AOB is the sine of the
complementary angle.
SECANT comes from the Latin word ‘Secans’ meaning ‘to
cut’. It was originally applied to the
line segment OD, the line that cuts off the tangent.
The ratio of the ‘secant’ OD to the radius, OA, is the
SECANT of angle AOB. Cotangent and Cosecant can be
derived using the same complementary idea as cosine.

8. PROBLEM AREAS ON THE COURSE
• Unit Circle – Rote learning the C.A.S.T / A.S.T.C quadrants
without really knowing what they are about
• Equations – Recognising all the possible questions and
understanding the connection with Trig graphs
• Identities – Why we use them. They seem very irrelevant to
students.
• 3D Trig – Visualisation is a big issue with some students so it
is worth spending time on it.

9. CONNECTIONS WITH OTHER LC TOPICS
COMPLEX
NUMBERS
GEOMETRY
CALCULUS
SLOPES & AREAS
NETS & AREAS
COORDINATE
GEOMETRY
– LINE AND CIRCLE
FUNCTION
GRAPHS
PERCENTAGE
ERRORS
THIS CAN SAVE A LOT OF TIME WHEN PLANNING THE COURSE

10. REVISION OF JUNIOR CERTIFICATE
KNOWLEDGE
• JC ONLY DEALS WITH RIGHT ANGLED TRIANGLES
• THIS IMPLIES ANGLES LESS THAN 90 DEGREES
ADJACENT
OPPOSITE
HYPOTENUSE
OPPOSITE
ADJACENT
HYPOTENUSE
A
B

11. RATIOS OF LENGTHS OF SIDES
HYPOTENUSE OPPOSITE
ADJACENT
A
sin 𝐴 =
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝐬𝐢𝐧 𝑨 =
𝒐
𝒉
cos 𝐴 =
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡
𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝐜𝐨𝐬 𝑨 =
𝒂
𝒉
tan 𝐴 =
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡
𝐭𝐚𝐧 𝑨 =
𝒐
𝒂

12. CARE IS NEEDED WHEN IDENTIFYING RELEVANT SIDES
HYPOTENUSE
HYPOTENUSE

13. PYTHAGORAS’ THEOREM
THIS IS THE RELATIONSHIP BETWEEN THE LENGTH OF SIDES OF A RIGHT
ANGLED TRIANGLE.
We need 2 sides to find the third side
𝒂𝟐
+ 𝒃𝟐
= 𝑯𝟐
H
4
3
𝟑𝟐 + 𝟒𝟐 = 𝑯𝟐
9 +𝟏𝟔 = 𝑯𝟐
EXAMPLE:
A
𝑯𝟐 = 𝟐𝟓
H = 𝟓

14. PYTHAGOREAN TRIPLES
THESE ARE WHOLE NUMBERS THAT SATISFY PYTHAGORAS’ THEOREM.
3, 4, 5
This triangle multiplied by a scale factor of 2 gives 6, 8, 10
5 4
3
A
is known as a Pythagorean triple
multiplied by a scale factor of 3 gives
3, 4, 5 9, 12, 15
And so on.
The same will apply to multiples of the Pythagorean triples 5, 12, 13 and 7, 24, 25

15. RECALL: RATIOS OF LENGTHS OF SIDES ARE FIXED FOR
A GIVEN ANGLE
15
12
9
A
sin 𝐴 =
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝐬𝐢𝐧 𝑨 =
𝒐
𝒉
cos 𝐴 =
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡
𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
𝐜𝐨𝐬 𝑨 =
𝒂
𝒉
tan 𝐴 =
𝑂𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝐴𝑑𝑗𝑎𝑐𝑒𝑛𝑡
𝐭𝐚𝐧 𝑨 =
𝒐
𝒂
3
4
5
10
6
8
𝑺𝑶 𝒕𝒂𝒏 𝑨 =
𝟒
𝟑
=
𝟖
𝟔
=
𝟏𝟐
𝟗

16. INTRODUCING HIGHER LEVEL IDEAS…
There are opportunities to introduce Higher Level Trig in TY.
The following topics can be done using JC trigonometry theory
but are excellent for getting the students better at visualizing.

17. 3D TRIGONOMETRY
CLASS ACTIVITY TO REINFORCE
RATIOS FROM JC
AND APPLICATION
MAKE 3D MODELS:
1) GIVE STUDENTS DIFFERENT DIMENSIONS
TO TEST RATIOS
2) USEFUL TO KEEP IN FOLDER AS A REFERENCE

18. TRIG
PUZZLES
WHAT WOULD BE THE KEY
PROBLEMS FOR STUDENTS HERE?
ARE THERE ANY SHORTCUTS
TO LOOK OUT FOR?

19. GEOGEBRA 3D

20. Radian measure
• Take the time to introduce radians
• I radian is the measure of the angle
subtended by an arc of the same length
as the radius.
https://www.ck12.org/trigonometry/radian-measure/lesson/Radian-Measure-TRIG/

21. New Formula to introduce: Length of an Arc
• Think of 1 radian as
• approximately 60 degrees
What you use at JC: What you use at LC:
𝑳 = 𝒓𝜽

22. Area of a sector:
A=
𝟏
𝟐
𝒓𝟐𝜽
Practise these 2 formula well
using this as an opportunity
to convert from radians to degrees
and visa versa.

23. EARLY TRIGONOMETRY: ARCS & SECTORS
Around 250 BC Eratosthenes of Cyrene used geometry
to estimate the size of the Earth. He noticed that at
mid-day on the summer solstice, the sun was almost
exactly overhead at Syene (present-day Aswan in
Egypt), because it shone straight down a vertical well.
On the same day of the year, the shadow of a tall
column indicated that the sun’s position at Alexandria
(northern coast of Egypt) was one fiftieth of a full circle
away from the vertical ( i.e 𝜽 ).
The Greeks knew that the Earth is spherical, and
Alexandria was almost due north of Syene, so the
geometry of a circular section of the sphere enabled
them to calculate the radius of the Earth.

24. Eratosthenes knew that camel trains took 50 days to
get from Alexandria to Syene, and they travelled a
distance of 100 ‘Stadia’ each day. We do not know the
exact length of a Stadium but one estimate found is
that 1 Stadium = 157 metres
Using this information find the radius of the Earth.
See how accurate they were in 250 BC by comparing
your answer to modern day figure for the Earth’s
radius. Keeping in mind that most people in Europe
still believed the world was flat until the Spanish
expedition organised by Magellan proved otherwise
in 1522 !
https://www.facebook.com/businessinsider/videos/10154023449809071/

25. TRIGONOMETRY
THE UNIT CIRCLE

26. The Unit Circle: A Circle with Centre (0,0) and Radius 1 unit
How do we find the coordinates of the
point 𝑃(𝑥, 𝑦) ?
Using our trigonometry ratios for
right angled triangles:
∴ 𝑷(𝒄𝒐𝒔 𝑨 , 𝒔𝒊𝒏 𝑨)
These are the coordinates for al points P on the circle
𝒄𝒐𝒔 𝑨 =
𝒙
𝟏
𝒔𝒊𝒏 𝑨 =
𝒚
𝟏
𝒙 = 𝒄𝒐𝒔 𝑨 𝒚 = 𝒔𝒊𝒏 𝑨

27. Applications of the unit circle for trigonometry
… your first Trig proof
Can you state the equation of this circle?
Hint: it must describe all points on the circle.
Equation: 𝒙𝟐
+ 𝒚𝟐
= 𝟏
Important relationship between Cos A and Sin A:
cos 𝐴 2
+ sin 𝐴 2
= 1
𝑐𝑜𝑠2𝐴 + 𝑠𝑖𝑛2𝐴 = 1

28. … Connection between Coordinate Geometry
and Trigonometry
(+, +)
(−, +)
(−, −) (+, −)
For all angles:
0 < 𝐴 < 90° 𝑆𝑖𝑛 𝐴 + & 𝐶𝑜𝑠 𝐴 +
90 < 𝐴 < 180° 𝑆𝑖𝑛 𝐴 +
270 < 𝐴 < 360° 𝐶𝑜𝑠𝐴 +

29. …Now we can find ratios for angles Greater than 90°
𝑆𝑖𝑛 𝐵 = 𝑆𝑖𝑛 𝐴
𝐵 = 180 − 𝐴
∴ 𝑆𝑖𝑛 𝐴 = 𝑆𝑖𝑛(180 − 𝐴)
Angles 𝟗𝟎 < 𝑩 < 𝟏𝟖𝟎

30. …Consider angles greater than 180°
What can we say about the 3 green triangles?
What is the same about the coordinates
and what is different?
𝑆𝑖𝑛 𝐴 =
Cos 𝐵 =
Cos 𝐵 =
𝑆𝑖𝑛 𝜃 =
𝑆𝑖𝑛 𝐵
− 𝐶𝑜𝑠𝐴
𝐶𝑜𝑠 𝜃
−𝑆𝑖𝑛 𝐵

31. The diagram shows a triangle ABC together with its
incircle. By considering the sum of the areas of the
triangles PBC, PCA and PAB , or otherwise,
calculate the radius, r cm, of the circle.
THINK ABOUT
EXTENSION PROBLEMS AND
MAKING CONNECTIONS
WITH OTHER TOPICS:
LINKS WITH GEOMETRY

32. TRIGONOMETRIC FUNCTION GRAPHS
𝒇 𝜽 = 𝐬𝐢𝐧 𝜽
𝒇 𝜽 = 𝒂 𝐬𝐢𝐧 𝒃𝜽 + 𝒄
𝒇 𝜽 = 𝟐𝐬𝐢𝐧 𝟐𝜽 + 𝟏
𝒇 𝜽 = 𝟐𝒄𝒐𝒔𝟑𝜽
𝒇 𝜽 = 𝒂 𝐬𝐢𝐧 𝒃𝜽 + 𝒄
𝒇 𝜽 = 𝒕𝒂𝒏 𝜽

33. APPLICATIONS OF TRIG FUNCTIONS IN REAL LIFE?
SHM
MOTION OF
SPRINGS
TIDES
SOUND WAVES /
MUSICAL INSTRUMENTS/
HEARING FUNCTION
PENDULUM CLOCK
CAR SHOCK
ABSORBERS
BUNGEE JUMPING
/ DIVING BOARDS

34. TRIG GRAPHS

35. EXPLAIN THE TERMS
The RANGE of the function is the interval from the lowest to the highest value: = [−1, 1]
The PERIOD is the angle turned in one complete cycle, starting at any position. Hence
the term PERIODIC graph: = 2𝜋
When given the graph the period can be found using
2𝜋
𝑏
The AMPLITUDE is the distance from the centre of oscillation to the
maximum/ minimum: = 1
The CENTRAL line about which the function oscillates: = 𝑇ℎ𝑒 𝑥 𝑎𝑥𝑖𝑠
From the above we can determine that 𝒂 = 𝟏, 𝒃 = 𝟏, 𝒄 = 𝟎

36. DETERMINING THE CONSTANTS
THESE IDENTIFY DIFFERENT GRAPHS
CAN YOU IDENTIFY:
THE RANGE
THE AMPLITUDE
THE PERIOD
The RANGE: = [−𝟐, 𝟐]
a = c =
b =
2 1 0
The PERIOD: = 𝟐𝝅
The AMPLITUDE: = 𝟐
HENCE, THE EQUATION IS:
𝒇 𝜽 = 𝟐 𝒔𝒊𝒏 𝜽
EXAMPLE 1

37. EXAMPLE 2
CAN YOU IDENTIFY:
THE RANGE
THE AMPLITUDE
THE PERIOD
The RANGE: = [𝟎, 𝟐]
a = c =
b =
1 1 1
The PERIOD: = 𝟐𝝅
The AMPLITUDE: = 𝟏
HENCE, THE EQUATION IS:
𝒇 𝜽 = 𝒔𝒊𝒏 𝜽 + 𝟏

38. EXAMPLE 3
CAN YOU IDENTIFY:
THE RANGE
THE AMPLITUDE
THE PERIOD
The RANGE: = [−𝟐, 𝟐]
a = c =
b =
2 2 0
The PERIOD: = 𝝅
The AMPLITUDE: = 𝟐
HENCE, THE EQUATION IS:
𝒇 𝜽 = 𝟐 𝒔𝒊𝒏 𝟐𝜽

39. CONSOLIDATE USING A REAL LIFE
APPLICATION:
SHM – TIDES EXAMPLE
At 1200 noon, the depth of water in a harbor is at its lowest and
shows a depth of 3 metres. At 530 pm that evening, the depth
is at its greatest at 13 metres.
What is the function graph to represent this tide.
Extension Q to think about
A yacht needs a depth of 10 metres to sail out of the harbor.
What is the earliest time that afternoon the yacht can sail?

40. DRAW THE FUNCTION GRAPH:
WHAT DO WE KNOW?
MIN DEPTH = 3m at 12 noon
MAX DEPTH = 13m at 530 pm
CAN YOU FIND THE REQUIRED CONSTANTS
a, b AND c ?
C = mid way between low and high tide
c = 6
The time between low and high tide
is 5.5 hours which is half a period
so the period is 11 hours.
b =
𝟏𝟏
𝟐𝝅
So let
2𝜋
𝑏
= 11
Amplitude is 5 metres
a = 5
𝑃𝑒𝑟𝑖𝑜𝑑 =
2𝜋
𝑏
Start by plotting the axis with c =6,
then the low and high tide and using the period of 11 Hours work backwards to sketch.
NB Finding b is not necessary to sketch graph : the period is the key