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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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
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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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