2. sec
Mid-Chapter Quiz: Lessons 5-1 through 5-3
and sin
, tan
= , csc
> 0
SOLUTION: to Use the Pythagorean Identity that involves tan
find sec .
Find the value of each expression using the
given information.
1. sin and cos , cot = 4, cos > 0
SOLUTION: Use the Pythagorean Identity that involves cot
find csc .
to
Since csc
is positive and tan θ is negative, sec
must be negative. Therefore,
Since sin
=
positive, sin
and cos are both
and cot
to find cos Use the reciprocal identity
must also be positive. Therefore,
.
to find cos .
Use the quotient identity
to find sin .
Use the quotient identity
2. sec
and sin
, tan
= , csc
> 0
SOLUTION: to Use the Pythagorean Identity that involves tan
find sec .
3. tan
and csc , cos
= , sin
>0
SOLUTION: eSolutions
Manual
by Cognero
Since
csc - Powered
is positive and tan
θ is negative, sec
must be negative. Therefore,
Use the Pythagorean Identity that involves cos
find sin .
to Page 1
Mid-Chapter Quiz: Lessons 5-1 through 5-3
3. tan
and csc , cos
= , sin
Simplify each expression.
>0
4. SOLUTION: Use the Pythagorean Identity that involves cos
find sin .
SOLUTION: to Use the reciprocal identity
to find cos .
5. SOLUTION: Use the quotient identity
to find tan .
6. SOLUTION: Simplify each expression.
4. SOLUTION: eSolutions Manual - Powered by Cognero
Page 2
Mid-Chapter Quiz: Lessons 5-1 through 5-3
8. ANGLE OF DEPRESSION From his apartment
6. SOLUTION: window, Tim can see the top of the bank building
across the street at an angle of elevation of , as
shown below.
a. If a bank employee looks down at Tim's
apartment from the top of the bank, what identity
could be used to conclude that sin = cos '?
b. If Tim looks down at a lower window of the bank
with an angle of depression of 35 , how far below
his apartment is the bank window?
SOLUTION: 7. SOLUTION: a. From the right triangle in the diagram, ' = 180
– (90 + ) = 90 – . So, cos ' = cos (90 –
). Because cosine and sine are cofunctions, sin θ = cos
(90 – ). Therefore, by the Transitive Property,
sin = cos '.
So, the cofunction identity sin = cos(90
can be used to show that sin = cos '.
–
)
b. Draw a diagram to model the situation.
The angle of depression from Tim's window to the
lower window of the bank is equivalent to the angle
of depression from the lower window of the bank up
to Tim's window because they alternate interior
angles of parallel lines.
Because an acute angle measure and the side
adjacent to the angle are known, the tangent function
can be used to find x.
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8. ANGLE
DEPRESSION
From his apartment
window, Tim can see the top of the bank building
across the street at an angle of elevation of , as
Page 3
to Tim's window because they alternate interior
angles of parallel lines.
Check choice D.
Because an acute
angle
measure and
side
Mid-Chapter
Quiz:
Lessons
5-1the
through
5-3
csc ≠
answer is D.
adjacent to the angle are known, the tangent function
can be used to find x.
. Therefore, the correct
Verify each identity.
10. –
= –2 tan
SOLUTION: Therefore, the bank window is about 56 feet below
Tim's window.
9. MULTIPLE CHOICE Which of the following is not equal to csc θ?
A sec (90 – )
B
C
D
11. csc2
2
2
– sin
– cos
2
– cot
= 0
SOLUTION: SOLUTION: Check choice A.
Secant and cosecant are cofunctions, so csc = sec
(90 – ). Therefore, A is not the correct answer.
Check choice B.
The Pythagorean Identity csc
2
= cot
be rewritten as csc = is not the correct answer.
2
+ 1 can 12. sin
. Therefore, B
+ = csc SOLUTION: Check choice C.
The cosecant function is the reciprocal of the sine
function, so csc = . Therefore, C is not the
correct answer.
Check choice D.
csc ≠
answer is D.
. Therefore, the correct
Verify each identity.
10. –
= –2 tan
SOLUTION: 13. = sec
– tan
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Page 4
Mid-Chapter Quiz: Lessons 5-1 through 5-3
13. = sec
Find all solutions of each equation on the
interval [0, 2π).
– tan
16. 4 sec
SOLUTION: + 2
= sec
SOLUTION: On the interval [0, 2 ), sec
and = = –
when
= .
14. 17. 2 tan
SOLUTION: + 4 = tan + 5
SOLUTION: On the interval [0, 2 ), tan
= 15. = +
18. 4cos2
SOLUTION: .
+ 2 = 3
SOLUTION: On the interval [0, 2π), cos
and = Find all solutions of each equation on the
interval [0, 2π).
16. 4 sec
+ 2
= 1 when = and
= and cos =
=
when = when
= and
.
= sec
19. cos
SOLUTION: – 1 = sin
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On the interval [0, 2 ), sec
Page 5
= –
when
= and = and cos =
when
= and
Therefore, on the interval [0, 2 ),
Mid-Chapter
Quiz: Lessons 5-1 through 5-3
= .
19. cos
= 0 or = .
20. MULTIPLE CHOICE Which of the following is
– 1 = sin
2
the solution set for cos tan − sin
F n, where n is an integer
SOLUTION: G
= 0?
+ n , where n is an integer
H + 2n , where n is an integer
J n , where n is an integer
SOLUTION: Solve cos
2
tan − sin
= 0.
Check for extraneous solutions.
On the interval [0, 2 ), sin = 0 when = 0 or = and sin = 1 when = . Check for
extraneous solutions.
Therefore, on the interval [0, 2 ),
= 0 or = .
20. MULTIPLE CHOICE Which of the following is
2
the solution set for cos tan − sin
F n, where n is an integer
G
= 0?
+ n , where n is an integer
H + 2n , where n is an integer
J n , where n is an integer
SOLUTION: Solve cos
tan − sin
2
= 0.
Because tan
is undefined, the solution = is extraneous. So, the solutions on the interval (– , )
have the general form = 0 + 2nπ or = 2n ,
which can be combined to = n , where n is an
integer.
Solve each equation for all values of θ.
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21. 3 sin2 θ + 6 = 2 sin2 θ + 7
On the interval [0, 2 ), sin
= 0 when = 0 or SOLUTION: Page 6
extraneous. So, the solutions on the interval (– , )
have the general form = 0 + 2nπ or = 2n ,
which can be combined
to = n 5-1
, where
n is an 5-3
Mid-Chapter
Quiz: Lessons
through
integer.
Solve each equation for all values of θ.
2
and sin = 1 when = . Therefore, the
solutions can be written as
= + n , where n is
an integer.
22. sin
+ cos = 0
2
21. 3 sin θ + 6 = 2 sin θ + 7
SOLUTION: SOLUTION: On the interval [0, 2 ), sin
= –1 when
= and sin = 1 when = . Therefore, the
solutions can be written as
= + n , where n is
an integer.
when
On the interval [0, 2 ), cos θ =
22. sin
= or
+ cos = 0
= SOLUTION: and cos = when
= or = . Check for extraneous solutions.
when
On the interval [0, 2 ), cos θ =
= and cos = when
= = or
or = . Check for extraneous solutions.
So, the only valid solutions are
= and = Therefore, the solutions on the interval (–∞, ∞) are + n ; where n is an integer.
23. sec
SOLUTION: Because 1
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So, the only valid solutions are
+ tan = 0
= and = Therefore, the solutions on the interval (–∞, ∞) are 24. 3 – 3 cos2
0, sec + tan = 0 has no solutions.
= 1 + sin
SOLUTION: 2
Page 7
So, the only valid solutions are
= and = Therefore, the solutions on the interval (–∞, ∞) are Mid-Chapter
Quiz:
n is Lessons
an integer. 5-1 through 5-3
+ n ; where
23. sec
24. 3 – 3 cos2
+ tan = 0
SOLUTION: Because 1
24. 3 – 3 cos2
Because 1
0, sec + tan = 0 has no solutions.
= 1 + sin
2
SOLUTION: 0, sec + tan = 0 has no solutions.
= 1 + sin
2
when θ =
On the interval [0, 2π), sin θ =
SOLUTION: sin θ =
when θ =
or θ =
o
. Check for extra
solutions.
On the interval [0, 2π), sin θ =
sin θ =
when θ =
or θ =
when θ =
o
. Check for extra
solutions.
Therefore, the solutions on the interval (–
,
) are
+ n , where n is an integer.
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25. PROJECTILE MOTION The distance d thatPage
a 8
kickball travels in feet with an acceleration of 32 feet
per second squared is given by d =
,
Therefore, the solutions on the interval (–
,
) are
Mid-Chapter
Quiz:
Lessons
n is an
integer. 5-1 through 5-3
+ n , where
25. PROJECTILE MOTION The distance d that a
kickball travels in feet with an acceleration of 32 feet
per second squared is given by d =
Therefore, the possible launch angles are 30.8 and
59.2 .
26. FERRIS WHEEL The height h of a rider in feet on
a Ferris wheel at t seconds is shown below.
,
where v0 is the object’s initial speed and is the angle at which the object is launched. If the ball was
kicked with an initial speed of 82 feet per second and
travels 185 feet, find the possible launch angle(s) of
the ball.
SOLUTION: Find
when d = 185 and v0 = 82.
a. If the Ferris wheel begins at t = 0, what is the
initial height of a rider?
b. When will the rider first reach the maximum
height of 145 feet?
SOLUTION: a. Find h when t = 0.
The range of the inverse sine function is restricted to
acute angles of on the interval [–90 , 90 ].
Because we are finding the inverse of 2θ instead of
θ, we need to consider angles on the interval [–2
(90 ), 2(90 )] or [–180 , 180 ]. Use your
calculator to find the acute angle and the reference
angle relationship sin (180 – ) = sin to find the
obtuse angle.
Therefore, the initial height of the rider is 5 feet.
b. Let h = 145 and solve for t.
Therefore, the possible launch angles are 30.8 and
59.2 .
26. FERRIS WHEEL The height h of a rider in feet on
a Ferris wheel at t seconds is shown below.
Therefore, the rider will first reach the maximum
height of 145 feet at 25 seconds.
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a. If the Ferris wheel begins at t = 0, what is the
initial height of a rider?
Page 9