#### Transcript Slide 1

```Chapter 5
An Introduction to
Trigonometric Functions
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5.1 Angle Measure, Special Triangles and Special Angles
Learning Objectives
In Section 5.1 you will learn how to:
 A. Use the vocabulary associated with a study of angles and triangles
 B. Find fixed ratios of the sides of special triangles
 C. Use radians for angle measure and compute circular arc length and area using
 D. Convert between degrees and radians for nonstandard angles
 E. Solve applications involving angular velocity and linear velocity using radians
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5.1
Angle Measure, Special Triangles and Special
Angles
Vocabulary
An angle of
meaasuring
90° is a right
angle.
An angle of measuring 180° is
called a straight angle.
A half line or all points extended
from a single point in a single
direction
The joining of two rays at
a common endpoint.
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We know that the measure of angle is degree and 1° =
1
360
of a full rotation.
Going a little deeper, are there units smaller or finer than degree?
Yes, they are called minutes and seconds.
• Each degree can be divided into 60 smaller parts called minutes (1° = 60′ )
• Each minutes can be divided into 60 smaller parts called seconds (1′ = 60′′)
Therefore, we also have the following converting ratios:
1
1′ = of a degree
1′′ =
60
1
3600
of a degree
Notation: sixty-one degree, eighteen minutes, and forty-five seconds
61° 18′ 45′′
Application
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Some exercises:
1) Determine the measure of each angle described:
a. The complement of a 57° angle: 33° angle (why?)
b. The supplement of a 132° angle: 48° angle (why?)
c. 𝜃 and 39° angle are complement. What is the measure of 𝜃? 𝜃 = 59° (why?)
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5.1
Angle Measure, Special Triangles and Special
Angles
1’ (1 minute) =
1
60
 11 
 11 
a. 61 18   45

60
3600
 60 
 3600
 18   45 
 61 


 60   3600
 61  0.3  0.0125
 61 .3125
1
of a degree and 1”(1 second) = 3600 of a degree
b. 142.2075o  142o  0.2075o
o
 142  0.207560'
 142o 12.45'
o
 142 12'0.4560"
o
 142 12' 27"
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5.1
Angle Measure, Special Triangles and Special
Angles
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Similar triangles:
Two triangles are similar if their corresponding angles are equal and/or their
corresponding sides are proportional.
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5.1
Angle Measure, Special Triangles and Special
Angles
5 .5 H

8
44
5.5 ft
8 ft
H
44 ft
8H  242
H  30.25 ft
H  30 ft 3 in
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5.1
Angle Measure, Special Triangles and Special
Angles
45o
45o
5 2
5
5
5
2
45o
45o
5
5
2
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5.1
Angle Measure, Special Triangles and Special
Angles
60o
8
4
30o
4 3
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Angle measure can be thought of the amount of rotation from a fixed ray called
the initial side to a rotated ray called the terminal side. (this way of thinking
allows us to have the concept of positive and negative angles depend on the
direction of the rotation).
Counterclockwise rotations give positive angle.
Clockwise rotations give negative angle.
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Coterminal angles: share the same initial and terminal sides
Coterminal angles will always differ by multiples of 360°. In other words, for any
integer 𝑘, angles 𝜃 and 𝜃 + 360𝑘 will be coterminal.
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5.1
Angle Measure, Special Triangles and Special
Angles
360o
If k is an integer then 60o + 360k is coterminal
If k= 1 then 60o + 360(1) = 420o
If k= 2 then 60o + 360(2) = 780o
If k= -1 then 60o + 360(-1) = -300o
If k= -2 then 60o + 360(-2) = -660o
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Sometimes, it might be helpful to imagine the angle on the xy-coordinate plane (we
refer that the angle is in standard position). In that case, we may put the vertex of
the angle at the origin.
A few vocabularies:
• A central circle is a circle on the xy-plane with its center at the origin.
• A central angle is an angle whose vertex is at the center of the circle.
• For a central angle 𝜃 intersecting the circle at points B and C, we say the
circular arc BC, denoted 𝐵𝐶, subtends ∠𝐵𝐴𝐶
The letter s is usually used to represent the arc length. If we define one radian (1
rad) to be the measure of an angle subtended by an arc equal in length to the
radius, then 𝜃 = 1 𝑟𝑎𝑑 when 𝑠 = 𝑟.
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5.1
Angle Measure, Special Triangles and Special
Angles
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5.1
Angle Measure, Special Triangles and Special
Angles
 = 3.5
r = 10 cm
s=r
s = 10 cm *3.5
s = 35 cm
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1-17
Using the central angle 𝜃 measured in radians, we develop a formula for the area of a
circular sector using a proportion.
Recall: the circumference of the circle is 𝐶 = 2𝜋𝑟. This implies that the radius, or an arc
length of measurement r, can be wrapped around the circumference of the circle 2𝜋 ≈ 6.28
times.
This shows the radian measure of a full 360° rotation is 2𝜋.
2𝜋 𝑟𝑎𝑑 = 360°
The ratio of the area of a sector to total area = the ratio of the subtended angle to 2𝜋.
Using 𝒜 to represent the area of the sector, we have
𝒜
𝜃 𝒜 2𝜋 =𝜃𝜋𝑟 2
1 2
=
𝒜
=
𝑟 𝜃
𝜋𝑟 2 2𝜋
2
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5.1
Angle Measure, Special Triangles and Special
Angles

r
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5.1
Angle Measure, Special Triangles and Special
Angles

3π
=4
r = 72ft
1 2
A r 
2
1
2 3
 (72)
2
4
15552 

8
 1944 ft 2
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5.1
Angle Measure, Special Triangles and Special
Angles
deg 

180 o
    5 rad
75 
o 
 180  12
o
 deg m ultiplyby
180o

  180o  180o

 
 7.5o
24   
24
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5.1
Angle Measure, Special Triangles and Special
Angles
  78o  51o
 
o
 27 
o 
180


3

20
s  r
 3 
s  3960

 20 
s  594 mi
 1866mi
M
Q

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r
1-22
5.1
Angle Measure, Special Triangles and Special
Angles
The angular velocity of an object is the amount of rotation per unit time.
The linear velocity of an object is defined as a change of position or distance traveled
per unit time.
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5.1
Angle Measure, Special Triangles and Special
Angles

r = 13 in
300rev 3002  600


1 min
1 min
1 min
600 24,504 .4 in
V  13 in 

1 min
1 min
24,504.4 in  60 min  1 ft  1 m i 



  23.2m ph
V
1 min  1 hr  12in  5280ft 
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5.1
Angle Measure, Special Triangles and Special
Angles
Linear Speed
Angular Speed
5.1
Angle Measure, Special Triangles and Special
Angles
A child is spinning a rock at the end of a 3-foot rope at the rate of
160 revolutions per minute (rpm). Find the linear speed of the
rock when it is released.
160 rev 160  2 320



1 min
1 min
1 min
 320 
V  3 ft 
  3016ft per min
 1 min 
V  34.3 mi per hr
5.2 Unit Circles and the Trigonometry of Real Numbers
1-28
Circle: a set of points on the plane that are a fixed distance called the radius from a
fixed point called the center.
The equation of a circle centered at origin, (0,0), and the radius r is given by
𝑥 2 + 𝑦 2 = 𝑟 2 (𝑤ℎ𝑦? )
If the central circle has radius equal 1 (𝑥 2 + 𝑦 2 = 1), then it is known as the unit
circle.
Observe that (0,1), (1,0), (0,-1), (-1,0) are four particular points on the unit circles.
They are known as quadrantal points.
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Some exercises:
1
1. Find a point on the unit circle given 𝑦 = with (x,y) in QII.
2
𝑥2 + 𝑦2 = 1
2
1
𝑥2 +
=1
2
1
𝑥2 + = 1
4
3
𝑥2 =
4
3
𝑥=±
2
We want the point in QII so we pick 𝑥 = −
3 1
,
2 2
3
,
2
and the point is −
3 1
,
2 2
.
be in? QI
FACT: If (𝑎, 𝑏) is on the unit circle, then (−𝑎, 𝑏), (𝑎, −𝑏), (−𝑎, −𝑏) are also on the unit circle
because a circle is symmetric to both axes and the origins.
So −
3 1
,
2 2
3 1
,
2 2
−
3
5
2. Name the quadrant containing − , −
4
5
3
1
,−
2
2
3
1
,−
2
2
are on the unit circle.
and verify it’s on the unit circle. Then use symmetry to
find three other points on the circle. (Do at home and turn it in)
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1-31
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From the example of 𝑡 =
3𝜋
,
2
it gives rise to the concept of domain.
By definition, cosine (cos) and sine (sin) have no restrictions for its domain.
For functions with x in the denominator (sec and tan), we want to avoid odd
𝜋
multiples of because the x-coordinate of the related point is zero.
2
𝜋
2
𝑡 ≠ + 𝜋𝑘 for all integers k
For functions with y in the denominator (csc and cot), we want to avoid multiples
of 𝜋 because the y-coordinate of the related point is zero.
𝑡 ≠ 𝜋𝑘 for all integers k
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In the previous example, we were able to determine the values of the trig functions
even though t is unknown. In many cases, we need to find the value of t.
3
For instance, what is the value of t given cos 𝑡 = −
with t in QII?
2
Exercises of this type fall into two categories: (1) you recognize the given number
1
2
as one of the special values: ±{0, ,
2 3 3
, , ,
2 2 3
3, 1}; or (2) you don’t.
In the first case, you can often name the real number t after a careful consideration
of the related quadrant and required sign.
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5.3 Graph of Since and Cosine; Cosecant and Secant Functions
Learning Objectives
In Section 5.3 you will learn how to:
 A. Graph f(t)=sin t using special values and symmetry
 B. Graph f(t)=cos t using special values and symmetry
 C. Graph sine and cosine functions with various amplitudes and periods
 D. Investigate graphs of the reciprocal functions f(t)=csc(Bt) and f(t)=sec(Bt)
 E. Write the equation for a given graph
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1
2
3
Noting that = 0.5, ≈ 0.71, ≈ 0.87, we plot these points and connect them with a smooth
2
2
2
curve to graph 𝑦 = sin 𝑡 in the interval [0,2𝜋].
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1-44
What happens if we expand the table from 2𝜋 to 4𝜋?
The function values begin to repeat. One way of verifying that it is true is by using the concept of
13𝜋
𝜋
𝜋
9𝜋
reference angle. For instance, sin
= sin
since the reference angle 𝜃𝑟 = ; sin
=
sin
𝜋
4
since 𝜃𝑟 =
𝜋
,
4
6
6
6
4
and so on.
Hence sine function is an example of what is called a periodic function. (cycle through a set pattern of
values).
Definition: a function f is said to be periodic if there is a positive number P such that 𝑓(𝑡 + 𝑃) =
𝑓(𝑡) for all t in the domain. The smallest number P for which this occurs is called the period of f.
For the sine function we have sin 𝑡 = sin(𝑡 + 2𝜋), as in sin
sin
𝜋
4
13𝜋
6
= sin
𝜋
6
+ 2𝜋 and sin
9𝜋
4
=
+ 2𝜋 . So we conclude that the sine function is periodic with period 𝑃 = 2𝜋.
FACT: In general, for all real number t, sin 𝑡 = sin(𝑡 + 2𝜋𝑘) for all integers k.
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Observation: both the graph and the unit circle confirm that the range of 𝑦 = sin 𝑡 is −1,1 and that
𝑦 = sin 𝑡 is an odd function.
𝜋
Recall: a function f is odd if 𝑓 −𝑡 = −𝑓(𝑡). A quick verification for the sine function: sin − =
− sin
𝜋
2
2
.
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1-46
a. 𝑓
𝜋
4
𝜋
4
+ 8𝜋 = sin
𝜋
+ 8𝜋 = sin
𝜋
𝜋
6
b. 𝑓 − 6 = sin − 6 = − sin
c. 𝑓
17𝜋
2
= sin
17𝜋
2
= sin
𝜋
2
𝜋
4
+ 4 2𝜋
=−
1
2
= sin
𝜋
4
=
2
2
(IV)
(II)
+ 8𝜋 = sin
𝜋
2
+ 4 2𝜋
= sin
𝜋
2
= 1 (I)
d. 𝑓 21𝜋 = sin 21𝜋 = sin 𝜋 + 20𝜋 = sin 𝜋 = 0 (V)
e. 𝑓
11𝜋
2
= sin
11𝜋
2
= sin
3𝜋
2
+ 4𝜋 = sin
3𝜋
2
+ 2 2𝜋
= sin
3𝜋
2
= −1 (III)
FACT: In general, for all real number t,
sin 𝑡 = sin(𝑡 + 2𝜋𝑘) for all integers k.
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1-47
cos t
3
2
1
cos t
1
1
2
3
2
0
1
2
0
−
1
2
−
2
2
−
3
2
-1
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1-48
1
2
There is another way to get the values for cosine: we know that ± , ±
3
2
, ±
3
1
,±
2
2
, ±
2
2
,±
2
2
are all points on the unit circle since they satisfy 𝑥 2 + 𝑦 2 = 1. Since cos 𝑡 = 𝑥 and sin 𝑡 = 𝑦, the
1
equation cos2 𝑡 + sin2 𝑡 = 1 can be obtained by direct substitution. This means if sin 𝑡 = ± then
cos 𝑡 = ±
3
2
2
and vice versa, with the sign taken from the appropriate quadrant.
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1-49
The function is decreasing for t in 0, 𝜋 , and increasing for t in (𝜋, 2𝜋).
Similar to sine function, cosine is also periodic with a period of 2𝜋.
And we note that cosine is an even function, meaning cos(𝑡) = cos(−𝑡) for all t in the domain.
𝜋
Recall: a function f is even if 𝑓 −𝑡 = 𝑓(𝑡). A quick verification for the cosine function: cos − =
cos
𝜋
2
2
= 0.
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1-50
Observing the graph of the functions 𝑦 = sin 𝑥 and 𝑦 = cos 𝑥 again:
They are oscillating back and forth, and one question to consider is how far is the distance between the
maximum and minimum value. Now half of this distance is what called an amplitude. Another way of
thinking of amplitude is the height of the function.
With the basic sine and cosine function, it is easy to see that the amplitude is 1. However, how can we tell
the amplitude of a more complicated sine and cosine functions?
ANSWER: if 𝑦 = 𝐴𝑠𝑖𝑛(𝐵𝑥) or 𝑦 = 𝐴𝑐𝑜𝑠(𝐵𝑥), the amplitude can be found by computing 𝐴 .
Quick examples:
• 𝑦 = 4𝑠𝑖𝑛(2𝑥) has an amplitude of 𝐴 = 4 = 4
• 𝑦 = −5𝑐𝑜𝑠(3𝑥) has an amplitude of 𝐴 = −5 = 5
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1-51
Graphing the basic 𝑠𝑖𝑛 𝑥 function by using its characteristics:
Amplitude
|1|=1
Period
2𝜋
Starting point 𝑥 = 0
End point
𝑥 = 0 + 2𝜋 = 2𝜋
Mid-point
𝑥=
0+2𝜋
2
=𝜋
Q1
𝑥=
0+𝜋
2
𝜋
2
Q3
𝑥=
𝜋+2𝜋
2
=
=
Notice that we are focusing in the
interval [0,2𝜋]. If we were to aim our
focus to the graph of this function on
𝜋
a different interval, say [− , 3𝜋],
2
how would the graph look like?
3𝜋
2
The starting point, end point and mid-point are
the zeroes of the sine function while Q1 and Q3
represent the x-coordinates where the maximum
and minimum values occur.
52
Let us look at a variation of 3𝑠𝑖𝑛 𝑥 and see how different it is from the basic one.
𝑦 = 3 sin 𝑥
𝑦 = sin 𝑥
We see that the amplitude affects the height of the graph. It alters the maximum and minimum
values. These values still occur at the same x-values as the basic function, but the maximum and
minimum values are no longer 1 and -1. They depend on the value of the amplitude.
53
Continue with a different variation of graph of 𝑠𝑖𝑛 2𝑥 and see how it differs from the basic one.
𝑦 = sin 𝑥
𝑦 = sin
𝑠𝑖𝑛3𝑥
2𝑥
1
𝑦 = sin 𝑥
2
What did we notice? It seems like multiply the input x by 2 means each cycle will complete
twice as fast.
What if I multiply x by 3?
1
What if I multiply x by ?
2
54
We conclude to ourselves that changing the input of the sin 𝑥 function will affect its graph
horizontally. More importantly, it changes the period and the x-values where the graph
intersects the x-axis (remember these are called the zeroes of the function).
To find the period P of 𝑦 = 𝐴𝑠𝑖𝑛(𝐵𝑥) and 𝑦 = 𝐴𝑐𝑜𝑠(𝐵𝑥), we do
2𝜋
𝑃=
𝐵
1
2
Example: Sketch the graph of 𝑦 = sin(4𝑥)
1
2
2𝜋
4
1
Amplitude
𝐴=
Period
𝑃=
Starting point
𝑥=0
End point
𝑥 = 0+2 =
Mid-point
0+ 2
Q1
Q3
=2
=
𝜋
2
𝜋
𝜋
𝑥=
𝑥=
𝑥=
2
𝜋
4
0+
2
𝜋 𝜋
+
4 2
2
𝜋
2
=
𝜋
4
=
𝜋
8
=
3𝜋
8
55
Observe that once we correctly graph one cycle of the function, extending the graph in
either direction is not a hard matter because the cycle repeats itself.
56
Another variation that we may see are horizontal and vertical shifting in the function
𝑠𝑖𝑛 𝑥. This is relatively straightforward
𝑦 = sin 𝑥 ± 𝐶 ± 𝐷
means that the function sin 𝑥 shift either left or right C units and up or down D unit
(depend on the signs in front of C and D).
Example: Sketch the graph of 𝑦 = sin 𝑥 +
𝜋
2
−1
𝑦 = sin 𝑥
𝜋
𝑦 = sin 𝑥 +
2
𝑦 = sin 𝑥 +
𝜋
−1
2
57
One more transformation we might see is the graph of 𝑦 = − sin 𝑥. This is a vertical reflection
of the original function 𝑦 = sin 𝑥. Its graph will look like this
Notice: sin(−𝑥) = − sin 𝑥 since the sine function is an odd function.
58
Example: Sketch the graph of 𝑦 = 3 sin
1
𝑥
2
+
𝜋
2
Solution:
Amplitude: 𝐴 = 3 = 3
Period: 𝑃 =
2𝜋
1
2
= 4𝜋
Phase shift:
1
𝑥
2
1
𝑥
2
+
𝜋
2
=0
=−
𝜋
2
𝑥 = −𝜋
(this is the starting point)
Vertical shift: None
End point: 𝑥 = −𝜋 + 4𝜋 = 3𝜋
Mid-point: 𝑥 =
𝑄1 =
𝑄3 =
−𝜋+3𝜋
2
=𝜋
Amplitude
3
Period
4𝜋
Starting point
𝑥 = −𝜋
End point
𝑥 = 3𝜋
Mid-point
𝑥=𝜋
Q1
𝑥=0
Q3
𝑥 = 2𝜋
−𝜋+𝜋
=0
2
𝜋+3𝜋
= 2𝜋
2
59
60
Example: Sketch the graph of
𝜋
𝑦 = −3 cos 𝑥 −
4
 Amplitude: 𝐴 = 3 = 3
2𝜋
 Period: 𝑃 = = 2𝜋
1
𝜋
 Phase shift: 𝑥 − = 0 ⟹ 𝑥 =
4
starting point)
 Vertical shift: none
𝜋
9𝜋
 End point: 𝑥 = + 2𝜋 =
4
 Mid-point: 𝑥 =
 Q1: 𝑥 =
 Q2: 𝑥 =
𝜋 5𝜋
+
4 4
2
=
5𝜋 9𝜋
+
4
4
2
𝜋 9𝜋
+
4 4
2
𝜋
4
(this is my
4
=
5𝜋
4
3𝜋
=
4
7𝜋
4
Notice that for cosine function, Q1 and Q2
become the zeroes of the function while the
starting point, end point and mid-point become
where the minimum/maximum occur.
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1-61
Once we know how to graph 𝑦 = 𝐴 sin(𝐵𝑥) and 𝑦 = 𝐴 cos(𝐵𝑥), graphing 𝑦 = 𝐴 csc(𝐵𝑥) and 𝑦 =
𝐴 sec(𝐵𝑥) can be done easily. Why?
Cosecant and secant function are the reciprocal of sine and cosine function:
1
1
csc 𝑥 =
sec 𝑥 =
sin 𝑥
cos 𝑥
One of the crucial thing worth noticing is that cosecant and secant will have vertical asymptote at every
point where sine and cosine equal to 0, respectively.
Example: graph the function 𝑦 = csc 𝑥 for 𝑥 ∈ [0,4𝜋]
We follow the same procedure as before to graph 𝑦 = sin 𝑥 first
Notice that sin 𝑥 = 0 at 𝑥 = 0, 𝜋, 2𝜋, 3𝜋, 4𝜋. So these are the
places where the vertical asymptotes appear.
𝜋
5𝜋
𝜋
5𝜋
Also, sin 2 = sin 2 = 1 implies csc 2 = csc 2 = 1.
sin
3𝜋
2
= sin
7𝜋
2
= −1 implies csc
3𝜋
2
= csc
7𝜋
2
= −1.
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𝑦 = sec 𝑥
𝑦 = cos 𝑥
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So far we have been dealing with given a function then graph that function. Can we do this conversely,
meaning given the graph, can we write the equation that it represents?
In this section, we look at the simple cases by examining the amplitude and the period of sine and cosine
function.
Example: given the following graph, write the equation of the function
By looking at the behavior of the graph, this is the graph of some function 𝑦 = 𝐴 sin(𝐵𝑥). Since the
3
3
amplitude of this graph is , 𝐴 = .
The period 𝑃 =
8𝜋
.
3
4
Now 𝑃 =
2𝜋
𝐵
3
4
4
so to find B, we do 𝐵 =
Therefore, the function is 𝑦 = sin
3
𝑥
4
2𝜋
𝑃
=
2𝜋
8𝜋
3
=
3
4
.
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1-64
Exercises: do at home and turn it in my next Thursday, January 30 2013 before the exam 1 starts.
1.
2. Graph the following functions. Make sure to include these information: amplitude, period, starting
point, end point, mid-point, Q1, Q3
𝜋
a) 𝑦 = − sin 𝑥 −
b) 𝑦 = cos 2𝑥 +
c) 𝑦 = −2 csc
4
𝜋
2
1
𝑥
3
(Just graph one period of the function correctly)
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1-65
5.4
5.4 Graphs of Tangent and Cotangent Functions
Learning Objectives
In Section 5.4 you will learn how to:
 A. Graph y = tan t using asymptotes, zeroes, and the ratio (sin t)/(cos t)
 B. Graph y = cot t using asymptotes, zeroes, and the ratio (cos t)/(sin t)
 C. Identify and discuss important characteristics of y = tan t and y = cot t
 D. Graph y = A tan Bt and y = A cot Bt with various values of A and B
 E. Solve applications of y = tan t and y = cot t
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5.4
0
1
3
1
3
UND − 3
-1
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−
1
3
0
1-67
5.4
EXAMPLE 1 CONTINUED
Vertical Asymptotes
A few good points to know:
tan 0 = 0 ⟹ 0,0 is on the graph
𝜋
𝜋
tan = 1 ⟹ , 1 is on the graph
Period
𝜋
𝜋
𝜋
2
2
Usually starts at − to for one period
4
4
𝜋
tan(− ) = −1 ⟹ − , −1 is on the graph
4
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4
1-68
5.4
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EXAMPLE 2 CONTINUED
Vertical Asymptotes
𝜋
,1
4
𝜋
,0
2
3𝜋
, −1
4
Period
Usually starts at 0 to 𝜋 for one period
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5.4
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5.4
What is the period
of 𝑓 𝑡 = tan(𝑡)?
3𝜋
4
−𝜋4
𝜋
4
5𝜋
4
𝜋 7𝜋 13𝜋
,
6 6 6
5𝜋
and − differ?
6
How do ,
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5.4
𝑃𝑒𝑟𝑖𝑜𝑑
𝜋
𝜋
2𝜋
P= − −
=
3
3
3
2𝜋 𝜋
P=
=
3
𝐵
2𝜋 𝜋
=
3
𝐵
3𝜋
3 𝜋3 2 = 𝐴
𝐵=
−2
=
𝐴
tan
𝑦 = 𝐴 tan 𝑡
2𝜋
2 22
3
𝐵=
3𝜋
2
𝒚
−2 = 𝐴 tan
3
4
𝑦 = 𝐴 tan 𝑡
2
−2 = 𝐴(−1)
𝟑
= 𝟐 𝒕𝒂𝒏
𝒕
𝟐
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1-77
5.5 Transformations and Applications of Trigonometric Graphs
Learning Objectives
In Section 5.5 you will learn how to:
 A. Apply vertical translations in context
 B. Apply horizontal translations in context
 C. Solve applications involving harmonic motion
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Sinusoidal pattern
•
•
A pattern that can be modeled by a sine (or cosine) function.
One particular example is outdoor temperatures
𝑦 = 𝐴𝑠𝑖𝑛(𝐵𝑡)
𝜋
•
We take 𝑡 = 0 corresponds to 12:00 noon. Note that 𝐴 = 15 and 𝑃 = 24 ⟹ 𝐵 =
•
•
If a vertical shift is applied, then the function would have the form 𝑦 = 𝐴𝑠𝑖𝑛 𝐵𝑡 + D.
D can be thought of as the average value. If we let M be the maximum value and m be the minimum value,
we have
M−m
2
𝐴+𝐷 =
𝐷−𝐴 =
M−m
2
M+m
2
+
−
M+m
2
M−m
2
=
=
2𝑀
2
2m
2
=𝐴
M+m
2
.
12
=𝐷
= 𝑀 (amplitude + average value = maximum value)
= m (average value – amplitude = minimum value)
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1-79
Note: 𝑃 = 24 and 𝐵 =
𝜋
as stated before. Hence,
𝜋
𝑦 = 𝐴 sin
𝑡 +𝐷
12
We know 𝑀 = 85 and 𝑚 = 61, thus we can compute
the amplitude A and the average value D
M − m 85 − 61
𝐴=
=
= 12
2
2
M + m 85 + 61
𝐷=
=
= 73
2
2
𝜋
Therefore, 𝑦 = 12 sin
𝑡 + 73 is the sinusoidal
12
12
equation model for the daily temperature.
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1-80
𝜋
2𝜋
5
5
Since 𝐵 = , we can compute the period 𝑃 = 𝜋 = 10.
This means the population of this species rises and falls
over a 10-yr cycle.
amplitude + average value = maximum value
So 𝑀 = 1200 + 9000 = 10,200
average value – amplitude = minimum value
So 𝑚 = 9000 – 1200 = 7800
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1-81
Horizontal Translation: 𝑦 = 𝐴 sin(𝐵𝑡 + 𝐶) + 𝐷
𝜋
𝑃 𝑡 = 1200 cos
𝑡 − 2.5 + 9000
5
𝜋
𝜋
= 1200 cos 𝑡 − + 9000
5
2
So both graph represents the same function. The only
difference is that one of them is in shifted form and the
other is in standard form.
Standard form: 𝑦 = 𝐴 sin(𝐵𝑡 ± 𝐶) + 𝐷
𝐶
Shifted form: 𝑦 = 𝐴 sin 𝐵 𝑡 ±
+𝐷
𝐵
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1-82
Amplitude: 𝐴 = 2.5 = 2.5
2𝜋
2𝜋
Period: 𝑃 =
= 𝜋 =8
𝐵
4
Average value: 𝐷 = 6
Horizontal shift: we write the equation in shifted form: 𝑦 = 2.5 sin
For the endpoints of the primary interval, we solve 0 ≤
𝜋
4
𝜋
4
𝑡+3
+ 6. So it shifts 3 units to the left.
𝑡 + 3 < 2𝜋, which gives −3 ≤ 𝑡 < 5.
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1-84
Simple Harmonic Motion
•
The periodic motion of springs, tides, sound exhibit what is known as harmonic motion, which can be
modeled using sinusoidal functions.
•
For object in harmonic motion, the input variable t is always a time unit. Amplitude and period are defined
as before.
Frequency: the number of cycles it completes per unit time. Frequency can be computed as
1
𝐵
𝑓= =
𝑃 2𝜋
•
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1-85
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1-86
5.6 The Trigonometry of Right Triangles
Learning Objectives
In Section 5.6 you will learn how to:
 A. Find values of the six trigonometric functions from their ratio definitions
 B. Solve a right triangle given one angle and one side
 C. Solve a right triangle given two sides
 D. Use cofunctions and complements to write equivalent expressions
 E. Solve applications involving angles of elevation and depression
 F. Solve general applications of right triangles
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1-87
Trig functions using right
circle
SOH CAH TOA
Using Pythagorean theorem, we can figure out the length of
𝑎𝑑𝑗 =
72 − 42 = 33
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1-89
SOLVING RIGHT TRIANGLES GIVEN ONE ANGLE AND ONE SIDE
𝜽
𝟎
𝝅
(𝟑𝟎°)
𝟔
𝝅
(𝟒𝟓°)
𝟒
𝝅
(𝟔𝟎°)
𝟑
𝝅
(𝟗𝟎°)
𝟐
sin 𝜃
0
1
2
1
3
2
3
2
1
2
1
cos 𝜃
tan 𝜃
0
2
2
2
2
1
3
undefined
csc 𝜃
undefined
2 3
=
3
3
1
1
3
=
3
3
2
2
2
s𝑒𝑐 𝜃
1
cot 𝜃
undefined
2 3
3
3
= 2
2
1
0
2
2
undefined
3
3
0
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1-90
To solve for a right triangle means to find the measure of all three
angles and all three sides
To solve for c, we do
sin 30° =
𝑐=
17.9
𝑐
17.9
17.9
=
= 35.8
1
sin 30°
2
To solve for b, we can apply the Pythagorean theorem to get 𝑏 ≈ 31
Since ∠𝐴 and ∠𝐵 are complements, that means 𝐵 = 60°.
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1-91
Tips for solving right triangle:
1. Angles must sum up to 180°
2. Pythagorean Theorem
3. Special triangles
4. SOH CAH TOA
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1-92
Using calculator can help to find the value of trig
functions of angles that are less familiar.
We see that 𝐵 = 58° because angles A and B are
complements.
To solve for b, we do
tan 32° =
𝑏=
24
𝑏
24
24
≈
≈ 38.41
tan 32°
0.624869
Knowing b now, we can apply to Pythagorean
theorem or use another trig ratio to figure out c.
Here I will use a trig ratio:
24
sin 32° =
𝑐
𝑐=
24
≈ 45.29
sin 32°
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1-93
SOLVING RIGHT TRIANGLES GIVEN TWO SIDES
In some case, you are given the value of the trig function at an angle but not the angle
itself. (i.e. sin 𝜃 = 0.7604, what is 𝜃?)
An easy case: sin 𝜃 =
2
,
2
what is 𝜃?
From the table, it is obvious that 𝜃 = 45°. Rigorously speaking, 𝜃 = sin−1
2
2
= 45°
NOTATION: 𝜃 = sin−1 𝑥 or 𝜃 = arcsin 𝑥
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From the example above, we see that
𝛼 + 𝛽 = 90° ⟹ 𝛽 = 90° − 𝛼
𝑎
Now, sin 𝛼 = 𝑐 ; cos 𝛽 =
𝑎
𝑐
So sin 𝛼 = cos 𝛽 = cos(90° − 𝛼)
In words: the sine of an angle is equal to
cosine of its complement.
Therefore, sine and cosine are called
cofunctions.
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QUICK REVIEW
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1101
What does it mean to solve a right triangle?
A: finding all the angles’ measurement and the length of each side.
Cofunctions:
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1102
APPLICATIONS USING ANGLE OF ELEVATION/DEPRESSION
Terminology:
An angle of elevation is defined to be the acute angle form by a horizontal line of
orientation (parallel to level ground) and a line of sight. An angle of depression is
likewise defined but involves a line of sight that is below the horizontal line of orientation.
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1103
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1104
Singh is
in there
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1105
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1106
5.7 Trigonometry and the Coordinate Plane
Learning Objectives
In Section 5.7 you will learn how to:
 A. Define the trigonometric functions using the coordinates of a point
 B. Using reference angles to evaluate the trig functions for any angle
 C. Solve applications using the trig functions of any angle
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1-107
Given a 30-60-90 triangle with hypotenuse of 10, we can use the ratio in the 30-60-90
triangle to figure out the length of the two legs: 5 and 5 3. If we place it in QI with the
30° angle at the origin and the longer leg along the x-axis, we get
sin 30° =
𝑦
5
1
=
=
𝑟 10 2
𝑥 5 3
3
cos 30° = =
=
𝑟
10
2
5
3
tan 30° =
=
=
3
5 3
3
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1
1108
Therefore, we can define the six trig functions in terms of 𝑥, 𝑦, and 𝑟, where 𝑟 =
𝑥2 + 𝑦2
The slope of the line coincident with the
hypotenuse is 5
5
=
3
3
.
3
Since the line goes
through the origin, using the point-slope
formula, we get the equation of that line
3
𝑦=
𝑥
3
Notice: no matter what point we pick on that
line, it will be at the 60° vertex of a right
triangle if we draw a perpendicular line from
that point to the x-axis.
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1109
For convenience, we pick 𝑥 = 6 (you can pick any value x
you want). Hence, 𝑦 = 2 3, which leads to the point (6,2 3)
being on the line and at the vertex of a 60° triangle.
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𝑦
5
2
sin 𝜃 = =
=
𝑟 5 2
2
𝑟
csc 𝜃 = = 2
𝑦
𝑥
5
2
cos 𝜃 = = −
=−
𝑟
2
5 2
𝑟
sec 𝜃 = = − 2
𝑥
𝑦
5
tan 𝜃 = =
= −1
𝑥 −5
𝑥
cot 𝜃 = = −1
𝑦
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REFERENCE ANGLE REVISIT
To find reference angle, it is not enough to just plus/minus 360° from the original angle.
One technique to find reference angle:
1. Plus/minus 360° to make your original angle is within 0° to 360° (if necessary)
2. Draw out the new angle to see which quadrant it is in
3. Use the above to help figure out the reference angle
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Angle 315° is in QIV. So to find the reference angle in this case, we compute: 𝜃𝑟 =
360° − 315° = 45°
b. Angle 150° is in QII. So to find the reference angle in this case, we compute: 𝜃𝑟 =
180° − 150° = 30°
c. Angle −121° is not within 0° to 360°. So I’m going to add 360° until the new angle
is within 0° to 360° for the first time:
−121° + 360° = 239°
Since 0° ≤ 239° ≤ 360° and it is in QIII, hence the reference angle 𝜃𝑟 = 239° − 180° =
59°
d. Once again, 425° is not within 0° to 360°. So in this case, I’m going to subtract 360°
until the new angle is within 0° to 360° for the first time:
425° − 360° = 65°
Since 0° ≤ 65° ≤ 360° and it is in QI, hence the reference angle 𝜃𝑟 = 65°
a.
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The terminal side is in QIV where x is positive
and y is negative. With 𝜃 = 45°, we have
2
sin 315° = −
2
2
cos 315° =
2
tan 315° = −1
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1-116
sin 𝜃 is positive in QI and QII, while cos 𝜃 is negative in QII and QIII. Both
conditions are satisfied in QII only.
𝑦
Since sin 𝜃 = , we have that 𝑦 = 5 and 𝑟 = 13. By using Pythagorean theorem,
𝑟
we have 𝑥 = 132 − 52 = 144 = ±12. To be in QII, we pick 𝑥 = −12
Hence, cos 𝜃 = −
12
13
Moreover, csc 𝜃 =
and tan 𝜃 =
13
,
5
sec 𝜃 = −
13
−12
13
,
12
and cot 𝜃 = −
12
5
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1-118
```
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