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5.2 Functions of Angles and
Fundamental Identities
• To define the six trigonometric functions, start with an
angle  in standard position. Choose any point P having
coordinates (x,y) on the terminal side as seen in the figure
below.
r  ( x  0) 2  ( y  0) 2 
x2  y2
• Notice that r > 0 since distance is never negative.
Copyright © 2007 Pearson Education, Inc.
Slide 8-1
5.2 The Six Trigonometric Functions
• The six trigonometric functions are sine, cosine, tangent,
cotangent, secant, and cosecant.
Trigonometric Functions
Let (x,y) be a point other than the origin on the terminal side
of an angle  in standard position. The distance from the
point to the origin is r  x 2  y 2 . The six trigonometric
functions of angle  are as follows.
y
x
y
tan   ( x  0)
cos 
sin  
x
r
r
x
r
r
csc   ( y  0) sec   ( x  0) cot   ( y  0)
y
x
y
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Slide 8-2
5.2 Finding Function Values of an Angle
Example The terminal side of angle  (beta) in
standard position goes through (–3,–4). Find the
values of the six trigonometric functions of .
Solution
r  ( 3) 2  ( 4) 2  25  5
4
4
sin  

5
5
5
5
csc  

4
4
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3
3
cos  

5
5
5
5
sec  

3
3
4 4
tan  

3 3
3 3
cot  

4 4
Slide 8-3
5.2 Quadrantal Angles
•
The six trigonometric functions
can be found from any point on
the line. Due to similar triangles,
y y
 ,
r r
so sin = y/r is the same no matter which point is used to
find it.
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Slide 8-4
5.2 Reciprocal Identities
• Since sin = y/r and csc = r/y,
1
1
sin  
and csc 
.
csc
sin 
Similarly, we have the following reciprocal
identities for any angle  that does not lead to a
0 denominator.
1
sin  
csc 
1
csc  
sin 
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1
cos 
sec 
1
sec  
cos
1
tan  
cot 
1
cot  
tan 
Slide 8-5
5.2 Using the Reciprocal Identities
12
Example Find sin if csc = 
.
2
1
sin  
Solution
csc
1
 12
 2
2

12
 2 1  3



2 3
3
3
Copyright © 2007 Pearson Education, Inc.
Slide 8-6
5.2 Signs and Ranges of Function Values
•
•
•
•
In the definitions of the trigonometric functions,
the distance r is never negative, so r > 0.
Choose a point (x,y) in quadrant I, then both x
and y will be positive, so the values of the six
trigonometric functions will be positive in
quadrant I.
A point (x,y) in quadrant II has x < 0 and y > 0.
This makes sine and cosecant positive for
quadrant II angles, while the other four functions
take on negative values.
Similar results can be obtained for the other
quadrants.
Copyright © 2007 Pearson Education, Inc.
Slide 8-7
5.2 Signs and Ranges of Function Values
Example Identify the quadrant (or quadrants) of any angle 
that satisfies sin > 0, tan < 0.
Solution Since sin > 0 in quadrants I and II, while tan < 0
in quadrants II and IV, both conditions are met only in quadrant II.
Copyright © 2007 Pearson Education, Inc.
Slide 8-8
5.2 Signs and Ranges of Function Values
• The figure shows angle  as it increases from 0º to 90º.
• The value y increases as  increases, but never exceeds r, so
y  r. Dividing both sides by r gives
y
 1.
r
• In a similar way, angles in quadrant IV suggests
y
y
 1  , so  1   1.
r
r
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Slide 8-9
5.2 Signs and Ranges of Function Values
y
• Since sin   ,
r
 1  sin   1.
for any angle .
• In a similar way,
 1  cos  1.
• sec and csc are reciprocals of sin and cos,
respectively, making
sec  1 or sec  1,
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csc  1 or csc  1.
Slide 8-10
5.2 Ranges of Trigonometric Functions
For any angle  for which for which the indicated function
exists:
1. –1  sin  1 and –1  cos  1;
2. tan and cot may be equal to any real number;
3. sec  –1 or sec  1 and csc  –1 or csc  1.
Example Decide whether each statement is possible or
impossible.
(a) sin   8
(b) tan = 110.47
(c) sec = .6
Solution
(a) Not possible since 8  1.
(b) Possible since tangent can take on any value.
(c) Not possible since sec  –1 or sec  1.
Copyright © 2007 Pearson Education, Inc.
Slide 8-11
5.2 Pythagorean Identities
• Three new identities from x2 + y2 = r2
– Divide by r2
2
2
x2 y2 r 2
x
y
    1



   
2
2
2
r
r
r
r r
Since cos = x/r and sin = y/r, this result becomes
2
2
2
2
(cos )  (sin  )  1 or sin   cos   1.
– Divide by x2
2
x2 y2 r 2
y r 

 2  2  1     
2
x
x
x
 x  x
2
2
1  tan   sec 
2
– Dividing by y2 leads to cot2 + 1 = csc2.
Copyright © 2007 Pearson Education, Inc.
Slide 8-12
5.2 Pythagorean Identities
Pythagorean Identities
sin   cos   1 1  tan   sec  1  cot   csc 
2
2
2
2
2
2
Example Find sin and cos, if tan = 4/3 and  is in
quadrant III.
Solution
Since  is in quadrant III, sin and cos will
both be negative.
2
2
1  tan   sec 
2
5
3
   sec     cos 
5
3
4
2

1     sec 
3
sin 2  1 cos2 
2
 3
sin 2  1   
 5
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 sin 2  
16
4
 sin  
25
5
Slide 8-13
5.2 Quotient Identities
•
Recall that sin   ry and cos   rx .Consider the
quotient of sin and cos where cos  0.
sin 

cos
•
Similarly
cos
sin 
y
r
x
r
y r y
    tan 
r x x
 cot  , sin   0.
Quotient Identities
sin 
 tan 
cos
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cos
 cot 
sin 
Slide 8-14