Lesson 4.2 Powerpoint - peacock

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Transcript Lesson 4.2 Powerpoint - peacock

4.2
Trigonometric Functions:
The Unit Circle
Objectives
 Identify a unit circle and describe its relationship
to real numbers.
 Evaluate trigonometric functions using the unit
circle.
 Use domain and period to evaluate sine and
cosine functions, and use a calculator to
evaluate trigonometric functions.
2
The Unit Circle
3
The Unit Circle
The two historical perspectives of trigonometry incorporate
different methods for introducing the trigonometric
functions.
One such perspective follows and is based on the unit
circle.
Consider the unit circle given by
x2 + y2 = 1
Unit circle
as shown at the right.
4
The Unit Circle
Imagine wrapping the real number line around this circle,
with positive numbers corresponding to a counterclockwise
wrapping and negative numbers corresponding to a
clockwise wrapping, as shown below.
5
The Unit Circle
• As the real number line wraps around the unit circle,
each real number t corresponds to a point (x, y) on the
circle.
• For example, the real number 0 corresponds to the
point (1, 0).
• Moreover, because the unit circle has a circumference
of 2, the real number 2 also corresponds to the point
(1, 0).
6
The Unit Circle
• In general, each real number t also corresponds to a
central angle  (in standard position) whose radian
measure is t.
• With this interpretation of t, the arc length formula
s = r
(with r = 1)
indicates that the real number t is the (directional) length
of the arc intercepted by the angle  , given in radians.
7
The Trigonometric Functions
8
The Unit Circle
Imagine a circle on
the coordinate
plane, with its center
at the origin, and a
radius of 1.
Choose a point on
the circle
somewhere in
quadrant I.
9
The Unit Circle
Connect the origin to
the point, and from
that point drop a
perpendicular to the
x-axis.
This creates a right
triangle with
hypotenuse of 1.
10
The Unit Circle
The length of its legs
are the x- and ycoordinates of the
chosen point.
Applying the definitions
of the trigonometric
ratios to this triangle
gives
y
sin( )   y
1
 is the
angle of
rotation
1
y
x
11
The Unit Circle
The coordinates of the chosen point are the cosine
and sine of the angle .
This provides a way to define functions sin() and
cos() for all real numbers .
y
sin( )   y
1
(x, y)
1
y
x
The other trigonometric functions can be defined
from these.
12
Trigonometric Functions
sin( )  y
 is the
angle of
rotation
(x, y)
1
y
x
13
The Trigonometric Functions
From the preceding discussion, it follows that the coordinates
x and y are two functions of the real variable t.
You can use these coordinates to define the six trigonometric
functions of t.
sine cosecant cosine secant tangent cotangent
These six functions are normally abbreviated sin, csc, cos,
sec, tan, and cot, respectively.
14
The Trigonometric Functions
15
Around the Circle
As that point
moves around
the unit circle into
quadrants II, III,
and IV, the new
definitions of the
trigonometric
functions still
hold.
16
Reference Angles
• The angles whose terminal sides fall in
quadrants II, III, and IV will have values of sine,
cosine and other trig functions which are
identical (except for sign) to the values of angles
in quadrant I.
• The acute angle which produces the same
values is called the reference angle.
17
Reference Angles
• The reference angle is the angle between the
terminal side and the nearest arm of the x-axis.
• The reference angle is the angle, with vertex at
the origin, in the right triangle created by
dropping a perpendicular from the point on the
unit circle to the x-axis.
18
Quadrant II
Original angle
For an angle, , in
quadrant II, the reference
angle is .
In quadrant II,
Reference angle
sin() is positive
cos() is negative
tan() is negative
19
Quadrant III
For an angle, , in
quadrant III, the reference
angle is -.
In quadrant III,
Original angle
sin() is negative
cos() is negative
Reference angle
tan() is positive
20
Quadrant IV
Reference angle
For an angle, , in
quadrant IV, the reference
angle is 2.
In quadrant IV,
sin() is negative
cos() is positive
tan() is negative
Original angle
21
All Star Trig Class
Use the phrase “All Star Trig Class” to
remember the signs of the trig functions in
different quadrants.
Star
Sine is positive
Trig
Tan is positive
All
All functions
are positive
Class
Cos is positive
22
Unit Circle
sin
One of the most useful tools in trigonometry is
the unit circle.
It is a circle, with radius 1 unit, that is on the x-y
coordinate plane.
1
cos
The x-axis corresponds to the cosine function, and
the y-axis corresponds to the sine function.
The angles are measured from the positive x-axis
(standard position) counterclockwise.
In order to create the unit circle, we must use the
special right triangles below:
1
45º
1
60º
30º
30º -60º -90º
The hypotenuse for each triangle is 1 unit.
45º
45º -45º -90º
23
Unit Circle
You first need to find the lengths of the other sides of
each right triangle...
1
60º
1
2
30º
3
2
45º
1
2
2
45º
2
2
24
Unit Circle
Usefulness of Knowing Trigonometric Functions of
Special Angles: 30o, 45o, 60o
The trigonometric function values derived from knowing the
side ratios of the 30-60-90 and 45-45-90 triangles are
“exact” numbers, not decimal approximations as could be
obtained from using a calculator
You will often be asked to find exact trig function values for
angles other than 30o, 45o and 60o angles that are
somehow related to trig function values of these angles
25
Now, use the corresponding triangle to find the coordinates on the unit circle...
(0, 1)
sin
What are the
coordinates of
This cooresponds
point?
to (costhis
30,sin
30)
 3 1 
 , 
(Use
2 2 your
(cos
30, sin 30)
30-60-90
(–1, 0)
30º
3
2
1
2
triangle)
cos
(1, 0)
(0, –1)
Now, use the corresponding triangle to find the coordinates on the unit circle...
(0, 1)
sin
What are the
(Use your
coordinates of
 245-45-90
this point?
 , 2 
 (cos45, sin 45)
triangle)
 2 2 
 3 1 
 , 
 2 2 
2
2
(–1, 0)
45º
2
2
(cos 30, sin 30)
cos
(1, 0)
(0, –1)
Now, use the
corresponding triangle to
find the coordinates on
the unit circle...
What are the
of
(0, coordinates
1) sin
this point?
 2 2 
 ,  (cos45,
 2 2 
3
(–1, 0)
(Use your
130-60-90
3
triangle)
(cos60, sin 60)
,
2 2
60 ̊
1/2
sin 45)
 3 1 
 , 
 2 2 
2
(cos 30, sin 30)
cos
(1, 0)
(0, –1)
Use this same technique to complete the unit circle.
(0, 1)
sin
1 3
,
2 2
(cos60, sin 60)
 2 2 
 ,  (cos45,
 2 2 
sin 45)
 3 1 
 , 
 2 2 
(cos 30, sin 30)
(–1, 0)
cos
(1, 0)
(0, –1)
Unit Circle
π
2
(0, 1)
(-1, 0)
π
0 (1, 0)
3π (0, -1)
2
30
Unit Circle
(0, 1)
3 1
(
, )
2 2 5π
π 3 1
( , )
6 2 2
6
30°
30°
(-1, 0)
(1, 0)
30°
30°
7π
6 3
1
(
, )
2
2
11π
3 1
( , )
2
2
6
(0, -1)
31
Unit Circle
(0, 1)
2 2
(
,
)
2 2 3π
π 2 2
( ,
)
4 2 2
4
45°
45°
(-1, 0)
(1, 0)
45°
45°
2
2
(
,)
2
2
5π
4
(0, -1)
7π
2
2
( ,)
2
2 32
4
Unit Circle
60°
(-1, 0)
60°
(1, 0)
60°
60°
1
3
( , )
2 2
4π
3
π 1 3
(
,
)
3 2 2
(0, 1)
1 3 2π
( ,
)
2 2
3
(0, -1)
1
3
5π ( , - )
2 2
3
33
Unit Circle
34
Function Values of Special Angles
Are used a lot in Trigonomet ry and can quickly
be determined from memorized triangles .

sin  cos  tan  cot  sec  csc 
30
1
2
3
2
45
2
2
2
2
60
3
2
1
2
3
3
3
1
2 3
3
1
3
2
3
3
2
2
2
2 3
3
This chart can also be quickly completed by
memorizing the first column and using identities .
35
36
The Trigonometric Functions
A circle with center at (0, 0) and radius 1 is called a unit circle.
The equation of this circle would be
x  y 1
2
2
(0,1)
(-1,0)
(1,0)
(0,-1)
So points on this circle must satisfy this equation.
37
The Trigonometric Functions
Let's pick a point on the circle. We'll choose a point where the
x is 1/2. If the x is 1/2, what is the y value?
x=
2
2
x  y 1
1/2
You can see there


(0,1)  1 3 
2
,
are two y values.


1
2 2 
2
They can be found
   y 1
by putting 1/2 into
2
(-1,0)
(1,0)
the equation for x
3
and solving for y.
2
y 
4
3
y
2
1
3
 ,

2

2

(0,-1)
We'll look at a
larger version
of this and
make a right
triangle.
38
The Trigonometric Functions
We know all of the sides of this
triangle. The bottom leg is just
the x value of the point, the
other leg is just the y value (0,1)
and the hypotenuse is
always 1 because it
is a radius of
the circle.
1
(-1,0)
1 3
 ,

2 2 


3
2

sin  
(1,0)
1
2
tan  
(0,-1)
cos
3
2  3
1
2
1
21
 1 2
3
2  3
1
2
Notice the sine is just the y value of the unit circle point and
the cosine is just the x value.
39
So if I want a trig function for  whose terminal side contains a
point on the unit circle, the y value is the sine, the x value is
the cosine and y/x is the tangent.
 2 2


,
 2 2 


(-1,0)
(0,1)

1
3
 ,

2 2 


sin  
(1,0)
tan 
(0,-1)
1
3
 ,

2
2 


cos
2
2
2

2
2
 2  1
2

2
We divide the unit circle into various pieces and learn the point
values so we can then from memory find trig functions.
Here is the unit circle divided into 8 pieces. Can you figure
out how many degrees are in each division?
These are
0,1
easy to
 2 2
 2 2
90°
memorize




,
 2 2 
 2 , 2 

 135°


since they
45°
all have the
2
sin 225  2
same value
with
180°
45°
 1,0
0° 1,0 different
signs
depending
225°
on the
 2
2
315°


2
2




,
 2 , 2 
quadrant.
270°



2
2 
0,1


We can label this all the way around with how many degrees an angle would be
and the point on the unit circle that corresponds with the terminal side of the
angle. We could then find any of the trig functions.
Can you figure out what these angles would be in radians?
0,1
 2 2


 2 , 2 


 1,0
7
sin

4
2
2
135°
3

180°
2
 45°
4
4
5
4225° 3
2
2
2



 2 , 2 


 2 2


 2 , 2 


90°

270°
0° 1,0
7
4 315° 
0,1
2
2


,

 2
2 

The circle is 2 all the way around so half way is . The upper
half is divided into 4 pieces so each piece is /4.
Here is the unit circle divided into 12 pieces. Can you figure
out how many degrees are in each division?
You'll need
 1 3  0,1
1 3



,
to know
 , 
 2 2 


cos 330  23


2 2 
90°
these too
120°
60°
 3 1
but you
 , 
 3 1
 2 2  150°
 ,  can see


 2 2

30°
the pattern.
180°
30°
 1,0
0° 1,0
210°
 3 1
330° 3 , 1 


,

 2 2
 2 2
240°




270° 300°
sin 240  
3
2
 1
3
  , 
 2 2 


0,1
1
3
 , 
2 2 


We can again label the points on the circle and the sine is the y
value, the cosine is the x value and the tangent is y over x.
Can you figure out what the angles would be in radians?
 1 3
 , 
 2 2 


0,1
1 3
 , 
2 2 


120° 90°

60°
 3 1
2
 , 
 3 1

2
 2 2  150°
 , 
5

3


 2 2
3

30°
6


180°
30°
6
 1,0
7
0° 1,0
11
6
210°4
6
5

 3 1
3 3
330° 3 , 1 


,

 2 2
 2 2
240° 2
3




270° 300°
 1
3
  , 
 2 2 


0,1
1
3
 , 
2 2 


It is still  halfway around the circle and the upper half is divided
into 6 pieces so each piece is /6.
The Trigonometric Functions
In the definitions of the
trigonometric functions, note
that the tangent and secant
are not defined when x = 0.
For instance, because t =  /2
corresponds to (x, y) = (0, 1),
1
sec 𝜃 =
𝑥
𝜋 1
sec = = 𝑢𝑛𝑑𝑒𝑓.
2 0
𝑦
tan 𝜃 =
𝑥
𝜋 1
tan = = 𝑢𝑛𝑑𝑒𝑓.
2 0
(0,1)
𝜋
2
(-1,0)
(1,0)
it follows that tan( /2) and
sec( /2) are undefined.
Similarly, the cotangent and
cosecant are not defined
when y = 0. For instance,
because t = 0 corresponds to
(x, y) = (1, 0), cot 0 and csc 0
are undefined.
(0,-1)
45
You should
know this.
This is a
great
reference
because
you can
figure out
the trig
functions of
all these
angles
quickly.
1
3
 ,

2
2 

The Trigonometric Functions
Let t be a real number and let (x, y) be the point on the unit
circle corresponding to t. The 6 trigonometric functions are:
sine (sin); cosine (cos); tangent (tan); cosecant (csc); secant
(sec); cotangent (cot)
sin t = y
csc t = 1/y
(y ≠ 0) (undef. at 0 & 𝜋)
cos t =x
sec t = 1/x
(x ≠ 0)(undef. at
tan t = y/x
(x ≠ 0)(undef. at
𝜋
2
&
𝜋
2
&
3𝜋
)
2
cot = x/y
3𝜋
)
2
(y ≠ 0) (undef. at 0 & 𝜋)
47
Example: Evaluating Circular Functions
Solution
3π
3π
3π
Evaluate sin
and tan
.
, cos
2
2
2
3π
An angle of
2 radians intersects the
unit circle at the point (0, -1)
3𝜋
sin 𝜃 = 𝑦 so sin
= −1
2
3𝜋
cos 𝜃 = 𝑥 so cos
=0
2
𝑦
3𝜋
tan 𝜃 = so tan
= 𝑢𝑛𝑑𝑒𝑓.
𝑥
2
48
Evaluating Circular Functions
Example
(a) Find the exact values of
(b) Find the exact value of
7π
cos
4
and
 5π 
tan   
 3 
7π
sin
4
.
.
49
Evaluating Circular Functions
Solution
cos
(a) From the figure
7π
2
7π
2

, sin  
.
4
2
4
2
(b) The angle -5𝜋/3 radians is
coterminal with an angle of
𝜋/3 radians. From the
figure
3
 5π 
π 2
tan     tan   
 3
 3 
 3 1
2
1
3
 ,

2

2


50
YOUR TURN:

Find the point (x, y) on the unit circle that
corresponds to the real number t

3
5

4
5
3
 2
1 3
,
2 2
2 2
−
,
2 2
1
3
,−
2
2
(1, 0)
51
Your Turn – Evaluating Trigonometric Functions
Evaluate the six trigonometric functions at each real
number.
a.
b.
c. t =  d.
Solution:
For each t-value, begin by finding the corresponding point
(x, y) on the unit circle. Then use the definitions of
trigonometric functions.
a.
corresponds to the point (x, y) =
.
52
Solution
cont’d
53
Solution
cont’d
54
Solution
b.
corresponds to the point (x, y) =
cont’d
.
55
Solution
cont’d
56
Solution
cont’d
c. t =  corresponds to the point (x, y) = (–1, 0).
sin  = y = 0
cos  = x = –1
tan  =
=
=0
57
Solution
cont’d
(–1, 0)
is undefined.
is undefined.
58
Solution
cont’d
d. Moving clockwise around the unit circle, it follows that
corresponds to the point (x, y) =
59
Solution
cont’d
60
YOUR TURN:
EVALUATE THE SIX TRIGONOMETRIC
FUNCTIONS AT EACH REAL NUMBER, T
7
4


3
4

3
61
Solution:
7𝜋
4
7𝜋
2
sin
=−
4
2
7𝜋
2
cos
=
4
2
7𝜋
tan
= −1
4
7𝜋
csc
=− 2
4
7𝜋
sec
= 2
4
7𝜋
cot
= −1
4
62
Solution:
𝜋
−
3
𝜋
3
sin − = −
3
2
𝜋
2 3
csc − = −
3
3
𝜋 1
cos − =
3 2
𝜋
sec − = 2
3
𝜋
tan − = − 3
3
𝜋
3
cot − = −
3
3
63
Solution:
4𝜋
−
3
4𝜋
3
sin −
=
3
2
4𝜋 2 3
csc −
=
3
3
4𝜋
1
cos −
=−
3
2
4𝜋
sec −
= −2
3
4𝜋
tan −
=− 3
3
4𝜋
3
cot −
=−
3
3
64
Domain and Period of Sine and Cosine
65
Domain of Sine and Cosine
Let’s think about the function f() = sin 
What is the domain? (remember domain means the “legal” things you
can put in for  ).
You can put in anything you want so the
domain is all real numbers.
What is the range? (remember range means what you get out of the
function).
The range is: -1  sin   1
(0, 1)
Let’s look at the unit circle to
answer that. What is the
lowest and highest value
you’d ever get for sine?
(sine is the y value so what
is the lowest and highest y
value?)
(1, 0)
(-1, 0)
(0, -1)
66
Domain of Sine and Cosine
Let’s think about the function f() = cos 
What is the domain? (remember domain means the “legal” things you
can put in for  ).
You can put in anything you want so the
domain is all real numbers.
What is the range? (remember range means what you get out of the
function).
The range is: -1  cos   1
(0, 1)
Let’s look at the unit circle to
answer that. What is the
lowest and highest value
you’d ever get for cosine?
(cosine is the x value so
what is the lowest and
highest x value?)
(-1, 0)
(1, 0)
(0, -1)
67
Domain of Sine and Cosine
The domain of the sine and cosine functions is the set of all
real numbers. To determine the range of these two
functions, consider the unit circle shown in Figure 1.19.
sin = 𝑦
cos = 𝑥
Figure 1.19
68
Domain of Sine and Cosine
Range: Sin and Cos
By definition, sin t = y and cos t = x. Because (x, y) is on
the unit circle, you know that –1  y  1 and –1  x  1. So,
the values of sine and cosine also range between –1 and 1.
–1 
y
1
–1  sin t  1
–1 
and
x
1
–1  cos t  1
Range: Sine and Cosine Functions
69
Period of Sine and Cosine
Adding 2 to each value of t in the interval [0, 2 ] results in
a revolution around the unit circle, as shown below.
70
Period of Sine and Cosine
The values of sin(t + 2) and cos(t + 2) correspond to
those of sin t and cos t.
Similar results can be obtained for repeated revolutions
(positive or negative) on the unit circle. This leads to the
general result
sin(t + 2 n) = sin t
and
cos(t + 2 n) = cos t
for any integer n and real number t. Functions that behave
in such a repetitive (or cyclic) manner are called periodic.
71
Period of Sine and Cosine
Because 0 and 2𝜋 are in the same position on
unit circle (2𝜋 completes a second revolution
around a unit circle) the following statements are
true:
sin(t + 2𝜋n) = sin t
cos(t + 2𝜋n) = cos t
72
Period of Sine and Cosine
 3
What about the
period of tangent?
undef
3
1
1
Let's label
the unit
circle with
values of
the tangent.
(Remember
this is just
y/x)
3
3
3

3
0
0
3
3
We see that they repeat
every  so the tangent’s
period is .

1
1
3
 ,

2

2


3
undef
tan( + ) = tan 
3
3
1
 3
73
Reciprocal Functions have the same Period.
PERIODIC PROPERTIES
sin( + 2) = sin 
cosec( + 2) = cosec 
cos( + 2) = cos 
sec( + 2) = sec 
tan( + ) = tan 
cot( + ) = cot 
9
tan
 1
4
This would have the

same value as tan
4
(you can count around on unit circle
or subtract the period twice.)
74
Example – Evaluating Sine and Cosine
a. Because
, you have
Subtract multiples of 2𝜋.
b. Because
, you have
Add multiples of 2𝜋.
75
YOUR TURN:
Evaluate the trigonometric function using its period as an
aid
cos5
9
sin
4
 8 
cos 
 3 
19
sin
6
76
Solution:
cos 5𝜋
Subtract out full periods of 2𝜋.
5𝜋 – 2(2𝜋) = 5𝜋 - 4𝜋 = 𝜋
So, cos 5𝜋 = cos 𝜋 = −1
cos 5𝜋 = −1
77
Solution:
9
sin
4
Subtract out full periods of 2𝜋.
9𝜋
9𝜋 8𝜋 𝜋
− 2𝜋 =
−
=
4
4
4
4
9𝜋
𝜋
2
sin
= sin =
4
4
2
9𝜋
2
sin
=
4
2
78
Solution:
8𝜋
cos −
3
Add in full periods of 2𝜋.
8𝜋
8𝜋 12𝜋 4𝜋
−
+ 4𝜋 = −
+
=
3
3
3
3
8𝜋
4𝜋
1
cos −
= cos
=−
3
3
2
8𝜋
1
cos −
=−
3
2
79
Even and Odd Circular Functions
Now let’s look at the unit circle to compare trig functions of
positive vs. negative angles.
What is cos

3
?
 
What is cos   ?
 3
1
2
1
2
Recall that if we put a negative
in the function and get the
original back (𝑓 −𝑥 = 𝑓(𝑥)) it
is an even function. Therefore,
cosine is an even function and
cos    cos
1
3
 ,

2

2


80
Even and Odd Circular Functions
What is sin

3
?
3
2
   3
What is sin    ?
2
 3
Recall that if we put a negative
in the function and get the
negative of the function back
(𝑓 −𝑥 = −𝑓(𝑥)) it is an odd
function. Therefore, sine is an
odd function and
sin      sin 
1
3
 ,

2
2 

81
Even and Odd Circular Functions
What is tan

3
?
 
What is tan    ?
 3
3
 3
Again, if we put a negative
in the function and get the
negative of the function
back (𝑓 −𝑥 = −𝑓(𝑥)) it is
an odd function. Therefore,
tangent is an odd function
and
tan      tan 
1
3
 ,

2
2 

82
Even and Odd Circular Functions
We know that a function f is even when f(–t) = f(t), and is
odd when f(–t) = –f(t).
If a function is even, its reciprocal function will be also. If
a function is odd its reciprocal will be also.
83
Example:
sin  60  what in terms of a positive angle?
 sin 60
 2
sec  
 3

  what in terms of a positive angle?

2
sec
3
84
Your Turn:
Use the value of the trigonometric function to
evaluate the indicated functions:
sin(-t) = 3/8
1)
sin t
−
3
8
2)
csc t
−
8
3
85
Domain and Period of Sine and Cosine
When evaluating a trigonometric function with a calculator,
you need to set the calculator to the desired mode of
measurement (degree or radian).
Most calculators do not have keys for the cosecant, secant,
and cotangent functions. To evaluate these functions, you
can use the
key with their respective reciprocal
functions: sine, cosine, and tangent.
86
Domain and Period of Sine and Cosine
For instance, to evaluate csc( /8), use the fact that
and enter the following keystroke sequence in radian
mode.
Display 2.6131259
87
Example 3 – Using a Calculator
Function
Mode
Calculator Keystrokes
Display
88
Applications of Circular Functions
The phase F of the moon is given by
1
F (t )  (1 cos t)
2
where t is called the phase angle. F(t) gives the
fraction of the moon’s face illuminated by the sun.
89
Modeling the Phases of the Moon
Evaluate and interpret.
(a) F (0)
π
(b) F
 2  (c) F (π)
 
 3π 
(d) F 

2
 
1
F (t )  (1 cos t)
2
90
Solution:
a)
b)
1
F (t )  (1 cos t)
2
1
1
F (0)  (1 cos0)  (11)  0
2
2
π 1
1
1
F    (1 cos π )  [1 0] 
2
2
2
2 2
new moon
first quarter
c) F (π)  1 (1 cos π)  1 [1 (1)] 1
full moon
d) F  3π   1 (1 cos 3π )  1 [1 0]  1
 2  2
2
2
2
 
last quarter
2
2
91
Practice Exercises
1. Find the value of the sec 360 ̊ without using a calculator.
2. Find the exact value of the tan 420 ̊ .
 5 
3. Find the exact value of sin   6  .
4. Find the tan 270 ̊ without using a calculator.
5. Find the exact value of the csc
 7 


 3 
.
6. Find the exact value of the cot (-225 ̊ ).
7. Find the exact value of the sin
8. Find the exact value of the cos
 13 


 4 
 11 


 6 
.
.
9. Find the value of the cos(-  ) without using a calculator.
10. Find the exact value of the sec 315 ̊ .
92
Key For The Practice Exercises
1. sec 360 ̊ = 1
2. tan 420 ̊ =
3. sin
 5 


6


3
=

1
2
4. tan 270 ̊ is undefined
5. csc  
7 

3 
2
=
3

2 3
3
6. cot (-225 ̊) = -1
7. sin
8. cos
 13 


 4 
 11 


 6 
=

=
1
2

2
2
3
2
9. cos(-  ) = -1
10. sec 315 ̊ =
2
93
Assignment
Pg. 274 -276: #1 – 67 odd, 75 – 79 odd
94