Transcript Section 4.3

Chapter 4
Trigonometric
Functions
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All rights reserved
© 2010 Pearson Education, Inc. All rights reserved
1
SECTION 4.3 Some Properties of the Trigonometric Functions
OBJECTIVES
1
2
3
Determine the signs of the trigonometric
functions in each quadrant.
Find a reference angle.
Use basic trigonometric identities.
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SIGNS OF THE TRIGONOMETRIC
FUNCTIONS
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3
The signs of the
trigonometric
functions are
determined by
the quadrant in
the terminal side
falls. This is clear
if we see that
above the x-axis,
y>0, etc.
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EXAMPLE 1
Determining the Quadrant in Which an
Angle Lies
If tan θ > 0 and cos θ < 0 in which quadrant does
lie?
Solution
Because tan θ > 0, θ lies either in quadrant I or
in quadrant III. However, cos θ > 0 for θ in
quadrant I; so θ must lie in
quadrant III.
It doesn’t hurt to sketch this diagram.
R&M suggest “All Students Take Calculus.”
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EXAMPLE 2
Evaluating Trigonometric Functions
3
Given that tan   , and cos  0, find the
2
exact value of sin  and sec  .
Solution
Since tan θ > 0 and cos θ < 0, θ lies in Quadrant
III; both x and y must be negative
y 3 3
tan   

x 2 2
r x y 
2
2
2   3
2
2
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 4  9  13
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EXAMPLE 2
Evaluating Trigonometric Functions
Solution continued
With x  2, y  3, and r  13 we can find
sin  and sec  .
y 3
3 13
sin   

r
13
13
r
13
13
sec   

x 2
2
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DEFINITION OF A REFERENCE ANGLE
Let  be an angle in standard position that is
not a quadrantal angle.
The reference angle for  is the positive
acute angle  formed by the terminal side
of  and the positive or negative x-axis.
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DEFINITION OF A REFERENCE ANGLE
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DEFINITION OF A REFERENCE ANGLE
Again, there’s no reason not to sketch these to verify that
we have the desired reference angle (always positive).
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EXAMPLE 3
Identifying Reference Angles
Find the reference angle  for each angle .
a.  = 250º
b.  =
c.  = 5.75
Solution
a. Because 250º lies in quadrant III,
 =   180º. So,  = 250º  180º = 70º.
b. Because
lies in quadrant II,  = π  .
 = π
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EXAMPLE 3
Identifying Reference Angles
Solution continued
c.  is in radians.
≈ 4.71 and 2π ≈ 6.28; so 
lies in quadrant IV and  = 2π  ;
 = 2π – 5.75 ≈ 6.28 – 5.75 = 0.53.
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PROCEDURE FOR USING REFERENCE
ANGLES TO FIND TRIGONOMETRIC
FUNCTION VALUES
Step 1 If the  > 360º, then find a coterminal
angle for  between 0º and 360º.
Otherwise, go to Step 2.
Step 2 Find the reference angle  for the angle
resulting in Step 1. Write the
trigonometric function of  .
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PROCEDURE FOR USING REFERENCE
ANGLES TO FIND TRIGONOMETRIC
FUNCTION VALUES
Step 3 Choose the correct sign for the
trigonometric function based on the
quadrant in which  lies.
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EXAMPLE 5
Using the Reference Angle to Find Values
of Trigonometric Functions
Find the exact value of each expression.
59
a. tan 330º
b. sec
6
Solution
Step 1 0º < 330º < 360º; find its reference angle
Step 2 330º is in Q IV, its reference angle  is
   360º 330º  30º
3
tan    tan 30º 
3
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EXAMPLE 5
Using the Reference Angle to Find Values
of the Trigonometric Function
Solution continued
Step 3 In Q IV, tan is negative, so
3
tan 330º   tan 30º  
3
b. Step 1
59 11  48 11


 8
6
6
6
59
11
is between 0 and 2π coterminal with
6
6
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EXAMPLE 5
Using the Reference Angle to Find Values
of the Trigonometric Function
Solution continued
11

is in Q IV, its reference angle  is
Step 2
6
11 
   2 

6
6

2 3
sec    sec 
6
3
Step 3 In Q IV, sec > 0, so
59
11
 2 3
sec
 sec
 sec 
6
6
6
3
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EXAMPLE 6
The Flight of a Golf Ball
A golf ball is hit on a level fairway with an initial
velocity of 128 ft/sec and an initial angle of
flight of 30º.
Find its range (the horizontal distance it traveled
before hitting the ground) and its maximum
height.
Solution
Use the height equation h = v0t sin θ − 16t2 with
v0 = 128 and θ = 30º.
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EXAMPLE 6
The Flight of a Golf Ball
Solution continued
h = 128t sin 30° − 16t2
1
Replace sin 30° with
and simplify.
2
h = 64t − 16t2 = 16t(t – 4)
The graph of this equation is a parabola; the
portion of the graph with h(t) > 0 represents the
flight path of the ball.
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EXAMPLE 6
The Flight of a Golf Ball
Solution continued
Find the vertex:
and
h(2) = 64(2) – 16(2)2 = 64;
so the vertex is (2, 64) and
the maximum height is 64 ft.
Since h(4) = 0 the ball is in
flight for 4 seconds.
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EXAMPLE 6
The Flight of a Golf Ball
Solution continued
The range is the distance, d, traveled after 4
seconds:
The ball reaches a maximum height of 64 feet
and has a range of 443 feet.
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BASIC TRIGONOMETRIC IDENTITIES
Quotient Identities
sin t
tan t 
cost
cost
cot t 
sin t
Pythagorean
Identities
cos2 t  sin 2 t  1
Reciprocal Identities
1 tan t  sec t
1
csc t 
sin t
1 cot 2 t  csc2 t
1
sec t 
cost
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2
2
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EXAMPLE 7
Finding the Exact Value of a Trigonometric
Function Using the Pythagorean Identities
1
a. Given sin t  and cost  0, find cost and tan t.
3
b. Given sect  2 and tant  0, find tant.
Solution
a. Use Pythagorean identity involving sin t.
cos 2 t  sin 2 t  1
2
1
cos t     1
3
1
2
cos t  1 
9
2
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EXAMPLE 7
Solution
Finding the Exact Value of a Trigonometric
Function Using the Pythagorean Identities
8
cos t 
9
2
8
2 2
cost  

9
3
2 2
cos t  
cos t  0 is given
3
1
sin t
1
2
3
tan t 



cos t
4
2 2
2 2

3
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Finding the Exact Value of a Trigonometric
Function Using the Pythagorean Identities
EXAMPLE 7
Solution continued
b. Given sect  2 and tant  0, find tant.
Use Pythagorean identity involving sec t.
1  tan t  sec t
2
2
1  tan t   2 
2
2
tan 2 t  3
tan t   3
tan t  3
tan t  0 is given
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