Transcript 7-3_7-4 Teacher Notes

7-3 Sine and Cosine (and Tangent) Functions
7-4 Evaluating Sine and Cosine
• 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
• Use a calculator to evaluate trigonometric functions
sin is an abbreviation for sine
cos is an abbreviation for cosine
tan is an abbreviation for tangent
csc is an abbreviation for cosecant
Sec is an abbreviation for secant
Cot is an abbreviation for cotangent
Which of the following represents r in the figure below?
(Click on the blue.)
y
P(x,y)
r
y

0
x y r
2
2
2
Close. The Pythagorean Theorem
would be a good beginning but you
will still need to “get r alone.”
x
x
r x y
2
CORRECT!
2
r
xy
2
You’re kidding right? (xy)/2
represents the area of the triangle!
Which of the following represents sin  in the figure below?
(Click on the blue.)
y
P(x,y)
r
y

0
x
sin  
r
Sorry. Does
SohCahToa ring a
bell? x/r represents
cos.
x
x
y
sin   , x  0
x
Sorry. Does Some
Old Hippy Caught
Another Hippy
Tripping on Acid
sound familiar? y/x
represents tan.
sin  
y
r
CORRECT! Well done.
Which of the following represents cos  in the figure below?
(Click on the blue.)
y
P(x,y)
r
y

0
x
cos  
r
CORRECT! Yeah!
x
x
y
cos  
r
Oops! Try something else.
cos  
y
,x  0
x
Sorry. Wrong ratio.
Which of the following represents tan  in the figure below?
(Click on the blue.)
y
P(x,y)
r
y

0
x
x
x
r
y
tan  , x  0
x
y
tan 
r
tan  
CORRECT! Yeah!
Try again.
Try again.
In your notes, please copy down the general ratios but keep in mind
that for a unit circle r = 1.
y
P(x,y)
x2  y 2  r 2
r
y
r  x2  y2

0
x
x
Note that csc, sec, and cot are
reciprocals of sin, cos, and tan. Also
note that tan and sec are undefined
when x = 0 and csc and cot are
undefined when y = 0.
In General
y
r
x
cos  
r
sin  
The Unit Circle
y
x2  y 2  r 2
r x y
2
2
P(x,y)

r
0
x
A few key points to write in your notebook:
•
P(x,y) can lie in any quadrant.
•
Since the hypotenuse r, represents distance, the value of r is always
positive.
•
The equation x2 + y2 = r2 represents the equation of a circle with its center at
the origin and a radius of length r. Hence, the equation of a unit circle is
written x2 + y2 = 1.
•
The trigonometric ratios still apply no matter what quadrant, but you will
need to pay attention to the +/– sign of each.
Example: If the terminal ray of an angle  in
standard position passes through (–3, 2), find
sin  and cos .
x  3
y2
r  ( 3 ) 2  2 2
r  13
sin  
y
2
2  13  2 13



 
r
13
13
13  13 
cos  
x 3
 3 13  3 13




r
13
13
13 13
You try this one in your notebook: If the
terminal ray of an angle  in standard position
passes through (–3, –4), find sin  and cos .
Check Answer
4
5
3
cos   
5
sin   
(–3,2)
r
2
–3

Example: If  is a fourth-quadrant angle and sin  = –5/13, find cos .
sin  
5 y

13 r
y  5
x
r  13
–5
13
x 2  ( 5 )2  132
x 2  25  169
x 2  144
x  144
x  12
Since  is in quadrant IV, the coordinate signs
will be (+x, –y), therefore x = +12.
cos  
x 12

r 13
Example: If  is a second quadrant angle and cos  = –7/25, find sin .
Check Answer
sin  
24
25
Determine the signs of sin  , cos  , and tan  according to quadrant.
Quadrant II is completed for you. Repeat the process for quadrants I, III,
and IV. Hint: r is always positive; look at the red P coordinate to determine
the sign of x and y.
y
y 
   (pos)
r 
x 
cos      (neg)
r 
x
y 
tan     (neg)
x 
P(–x,y)
y
Quadrant II
sin  
r
0
P(x,y)
r
y
0
r
P(–x, –y)
x
0
y
x
Quadrant III
y
sin    
r
x
cos    
r
y
tan   
x
x
0
r
P(x, –y)
Quadrant I
y
sin    
r
x
cos    
r
y
tan   
x
Quadrant IV
y
sin    
r
x
cos    
r
y
tan   
x
Check your answers according to the chart below:
•All are positive in I.
y
•Only sine is positive in II.
•Only tangent is positive in III.
•Only cosine is positive in IV.
Sine
All
x
Tangent
Cosine
A handy pneumonic to help you remember! Write it in your notes!
y
Students
All
x
Take
Calculus
Let  be an angle in standard position. The reference angle  associated with 
is the acute angle formed by the terminal side of  and the x-axis.
y
y
P(–x,y)
P(x,y)
r


r
 
x
0
x
0
  180  
   
 
y
0
3. Determine the sign by
noting the quadrant.

x
0

r
r
P(–x, –y)
2. Find the reference
angle.
y

    180
  
1. If necessary, find a
coterminal angle
between 0 and 360 
or 0 and 2π.
x
4. Evaluate and apply the
sign.
P(x, –y)
  360  
  2  
Example: Find the reference angle for  = 135.
Since 135 is in quadrant II :
  180  
  180  135
  45
You try it: Find the reference angle for  = 5/3.
Check Answer

You try it: Find the reference angle for  = 870.

3
Check Answer
  30
Give each of the following in terms of the cosine of a reference angle:
Example: cos 160
The angle =160 is in Quadrant II; cosine is negative in Quadrant II (refer back
to All Students Take Calculus pneumonic). The reference angle in Quadrant II is
as follows: =180 –  or =180 – 160 = 20. Therefore: cos 160 = –cos 20
You try some:
•cos 182
Check Answer
 cos 2
•cos (–100)
Check Answer
 cos 80
•cos 365
Check Answer
cos 5
Try some sine problems now: Give each of the following in terms of the sine
of a reference angle:
•sin 170
Check Answer
sin 10
•sin 330
Check Answer
 sin 30
•sin (–15)
Check Answer
 sin 15
•sin 400
Check Answer
sin 40
Can you complete this chart?
2
45
3
1
x  3 , y  1, r  2
y 1
sin 30  
r 2
x
3
cos 30  
r
2
 (degrees)
 (radians)
sin 
cos 
0
0
0
1
30

6
1
2
3
2
45
60

2
3
60
1
90
30
1
1
30
2
45
2
60
1
0
Check your work!!!!!! Write this table in your notes!
 (degrees)
 (radians)
sin 
cos 
0
0
0
1
30

6
1
2
3
2
45

4
2
2
2
2
60

3
3
2
1
2
90

2
1
0
Give the exact value in simplest radical form.
Example: sin 225
Determine the sign: This angle is in Quadrant III where sine is
negative. Find the reference angle for an angle in Quadrant III:  =  – 180 or
 = 225 – 180 = 45. Therefore:
2
sin 225   sin 45  
2
 (degrees)
 (radians)
sin 
cos 
0
0
0
1
30

6
1
2
3
2
45

4
2
2
2
2
60

3
3
2
1
2
90

2
1
0
You try some: Give the exact value in simplest radical form:
2
2
•sin 45
Check Answer
•sin 135
Check Answer
•sin 225
Check Answer
•cos (–30)
Check Answer
3
2
•cos 330
Check Answer
3
2
•sin 7/6
Check Answer
•cos /4
Check Answer
2
2
2
2


1
2
2
2
Homework: Page 279-280, #1, 3, 11, 13, 15, 17