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Practice Problems for Final Exam
#1 Find the exact value of the expressions below without using a calculator.
A.
⎛π ⎞
⎛π ⎞
cos ⎜ ⎟ ⋅ sin ⎜ ⎟
⎝3⎠
⎝3⎠
B.
⎛
tan −1 ⎜
⎝
⎛
cos −1 ⎜
⎝
1 ⎞
⎟
3⎠
2⎞
⎟
2 ⎠
#2 Identify the amplitude, period, and phase shift of the function below.
⎛π
π2 ⎞
y = 8sin ⎜ x − ⎟
6 ⎠
⎝3
#3 Find the period and the range of the function below.
y = −3csc ( 2 x ) − 4
#4 Discuss why we restrict the domain of tan x to −
discuss the difference between sin
−1
π
( x ) and csc ( x ) .
2
≤x≤
π
2
in order to define tan −1 ( x ) . Then,
#5 In meteorology, the vertical distance from the ground to the base of the clouds is called the ceiling.
To measure the ceiling, weather watchers directed a spotlight vertically overhead then measured the angle
of elevation from a point 200 meters away from the spotlight to the point where the beam of the spotlight
hits the cloud cover as 71.5º.
#6 Prove both identities below.
A.
tan ( x )
= sin ( x )
sec ( x )
B. sin ( x ) cot ( x ) = cos ( x )
#7 Use identities to simplify the expressions below without a calculator.
A.
⎛ π ⎞
⎛ π ⎞
B. 2sin ⎜ − ⎟ cos ⎜ − ⎟
⎝ 12 ⎠
⎝ 12 ⎠
2 cos 2 ( 22.5° ) − 1
#8 Prove both identities below.
A.
(1 + cot θ )
2
− 2 cot θ =
1
(1 − cos θ )(1 + cos θ )
π⎞
⎛
B. cos ⎜ x − ⎟ = cos x tan x.
2⎠
⎝
#9 Prove both identities below.
A.
sin 4t
= cos3 t sin t − sin 3 t cos t
4
#10 Solve
B. ( sin α − cos α ) = 1 − sin 2α
2
ABC with an interior angle β = 10.5° at vertex B and side lengths a = 6.8 and c = 2.4 .
#11 Find the magnitude and direction of b =
3,1 .
#12 Let z1 = −3 + 3i , z2 = −2 − 2i , and z3 = −8 + 8 3 ⋅ i .
A.
Calculate z1 ⋅ z2 .
B. Calculate
4
z3
#13 Recall the formulas for converting polar to rectangular and vice versa.
#14 Determine the standard form of the conic below. Identify any foci.
9 x 2 − 18 x + 4 y 2 + 24 y = 11
∞
#15 Evaluate
∑
k =1
⎡ ⎛ 2 ⎞ k −1 ⎤
⎢10 ⎜ ⎟ ⎥ .
⎣⎢ ⎝ 5 ⎠ ⎦⎥
#16 Find a single rectangular equation to describe the static graph generated by the parametric equations
below.
x = 7 sin t
y = 7 cos t
#17 Find all the solutions to sin x = 2 2 .
⎛θ ⎞
#18 Suppose 0 ≤ θ ≤ 2π . Find the values of θ ∋ 2sin 2 ⎜ ⎟ = cos (θ ) .
⎝2⎠
SOLUTIONS
3
⎛π ⎞
⎛π ⎞ 1 3
=
#1 A) cos ⎜ ⎟ ⋅ sin ⎜ ⎟ = ⋅
4
⎝3⎠
⎝3⎠ 2 2
#1B)
⎛
tan −1 ⎜
⎝
⎛
cos −1 ⎜
⎝
1 ⎞ π
⎟
3⎠ 6 π π π 4 4 2
= = ÷ = ⋅ = =
2⎞ π 6 4 6 π 6 3
⎟
2 ⎠ 4
⎛π
π2 ⎞
⎡π ⎛
π ⎞⎤
If y = 8sin ⎜ x − ⎟ , then y = 8sin ⎢ ⎜ x − ⎟ ⎥ . Hence, the amplitude is 8; the period is 6 as
6 ⎠
2 ⎠⎦
⎣3⎝
⎝3
calculated below, and the phase shift is π 2 units to the right.
#2
period =
#3
2π
π
3
= 2π ÷ = 2π ⋅ = 6
3
π 3
π
The period is π as calculated below. The range of csc x is ( −∞, −1) ∪ (1, ∞ ) . Dilating csc x by
−3 , changes the range to ( −∞, −3) ∪ ( 3, ∞ ) . Translating −3csc x down 4 units changes the range to
( −∞, −7 ) ∪ ( −1, ∞ ) .
period =
2π
=π
2
#4
The graph of tangent fails the horizontal line test, meaning tangent is not a one-to-one function.
Interchanging all the x and y -coordinates will create a relation where some x-values correspond to more
than one y-value, which means that the interchange will not be a function. For instance, tangent passes
through both of the points, ( 0, 0 ) and (π , 0 ) . Interchanging these coordinates gives ( 0, 0 ) and ( 0, π ) ,
but this relation has two y-values for the one x-value of zero. To alleviate this problem, we simply restrict
π
π
the domain of tangent to the interval − < x < . Over this interval, the graph of tangent passes the
2
2
horizontal line test.
The difference between sin −1 ( x ) and csc ( x ) lies in the fact that sin −1 ( x ) is a function that
π
π
≤ x ≤ while csc ( x ) simply reciprocates all
2
2
⎛π ⎞ 1
⎛1⎞ π
the range values of sin ( x ) . Specifically, consider the case where sin ⎜ ⎟ = . We say sin −1 ⎜ ⎟ =
⎝6⎠ 2
⎝2⎠ 6
⎛π ⎞ 1
and csc ⎜ ⎟ = .
⎝6⎠ 2
interchanges domain and range values of sin ( x ) over −
#5 Note that 71.5º is not a convenient value, so Mr. Simpson might tolerate the use of a calculator.
ceiling
200 m
200 m × 2.98868 ≈ ceiling
tan ( 71.5° ) =
597.7 m ≈ ceiling
spotlight beam
71.5°
200 m
#6A) Apply the Fundamental Identity for Tangent and Reciprocal Identity for Secant as below.
tan ( x )
sin ( x )
⋅ cos ( x ) = sin ( x )
= tan ( x ) ÷ sec ( x ) =
sec ( x )
cos ( x )
#6B)
Apply the Fundamental Identity of Cotangent as below.
sin ( x ) cot ( x ) = sin ( x )
cos ( x )
sin ( x )
= cos ( x )
#7A & B) Apply the double angle identity as below.
2
2
π
π
π
π
1
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
2sin ⎜ − ⎟ cos ⎜ − ⎟ = sin ⎜ 2 × − ⎟ = sin ⎜ − ⎟ = −
12 ⎠
2
⎝ 12 ⎠
⎝ 12 ⎠
⎝
⎝ 6⎠
2 cos 2 ( 22.5° ) − 1 = cos ( 2 × 22.5° ) = cos ( 45° ) =
#8A) Verify the identity, (1 + cot θ ) − 2 cot θ =
1
2
(1 − cos θ )(1 + cos θ )
. Squaring the binomial obtains the
following.
1 + 2 cot θ + cot 2 θ − 2 cot θ =
1
(1 − cos θ )(1 + cos θ )
Adding like terms, the left-hand simplifies as below.
1 + cot 2 θ =
1
(1 − cos θ )(1 + cos θ )
The Pythagorean Identity, gives us the following.
csc 2 θ =
1
(1 − cos θ )(1 + cos θ )
The Reciprocal Identity changes the left-hand side as below.
1
1
=
2
sin θ (1 − cos θ )(1 + cos θ )
The Pythagorean Identity then gives us the following.
1
1
=
2
1 − cos θ (1 − cos θ )(1 + cos θ )
Finally, using the property a 2 − b 2 = ( a − b )( a + b ) , we obtain the identity.
1
=
1
(1 − cos θ )(1 + cos θ ) (1 − cos θ )(1 + cos θ )
π⎞
⎛
#8B) Verify the identity, cos ⎜ x − ⎟ = cos x tan x. By the Difference Identity of Cosine, we have:
2⎠
⎝
⎛π ⎞
⎛π ⎞
cos ( x ) cos ⎜ ⎟ + sin ( x ) sin ⎜ ⎟ = cos x tan x.
⎝2⎠
⎝2⎠
⎛π ⎞
⎛π ⎞
Evaluating cos ⎜ ⎟ and sin ⎜ ⎟ yields the following.
⎝2⎠
⎝2⎠
cos ( x ) ⋅ 0 + sin ( x ) ⋅1 = cos x tan x
sin ( x ) = cos x tan x
Multiplying by a fortuitous form of “1” yields the next step.
cos x
⋅ sin x = cos x tan x
cos x
sin x
= cos x tan x
cos x ⋅
cos x
The Fundamental Identity completes the verification.
cos x ⋅ tan x = cos x tan x
#9A) Verify the identity,
sin 4t
= cos3 t sin t − sin 3 t cos t. Note, 4t = 2 ⋅ 2t . By substitution, we have:
4
sin ( 2 ⋅ 2t )
= cos3 t sin t − sin 3 t cos t.
4
By the Double-Angle Identity, we obtain the following on the left-hand side of the identity.
2sin ( 2t ) cos ( 2t )
= cos3 t sin t − sin 3 t cos t
4
sin ( 2t ) cos ( 2t )
= cos3 t sin t − sin 3 t cos t
2
By the Double-Angle Identities, we change the left-hand side as below.
2sin t ⋅ cos t ( cos 2 t − sin 2 t )
= cos3 t sin t − sin 3 t cos t
2
sin t ⋅ cos t ( cos 2 t − sin 2 t ) = cos3 t sin t − sin 3 t cos t
Distribution yields the identity.
cos3 t ⋅ sin t − sin 3 t ⋅ cos t = cos3 t sin t − sin 3 t cos t
#9B) Verify the identity, ( sin α − cos α ) = 1 − sin 2α . Expanding the binomial, we change the left-hand
side as below.
2
sin 2 α − 2 cos α ⋅ sin α + cos 2 α = 1 − sin 2α
Apply the commutative property, to obtain:
sin 2 α + cos 2 α − 2 sin α ⋅ cos α = 1 − sin 2α .
Apply the Pythagorean Identity, to acquire:
1 − 2sin α ⋅ cos α = 1 − sin 2α .
Thus, by the Double-Angle Identity, we complete the verification as below.
1 − sin 2α = 1 − sin 2α
#10 Use Law of Cosines to calculate the missing side, b .
b 2 = a 2 + c 2 − 2ac cos β
b 2 = 6.82 + 2.42 − 2 ( 6.8 )( 2.4 ) cos (10.5° )
b 2 = 19.9
b ≈ 4.46
Use Law of Sines to calculate a missing angle, keeping in mind that α is opposite the longest side of the
triangle.
sin (10.5° ) sin (α ')
=
4.46
6.8
0.277848 ≈ sin (α ')
sin −1 ( 0.277848 ) ≈ α '
16.1° ≈ α '
α ≈ 180° − 16.1° = 163.9°
The angle opposite side a must be obtuse from the construction of the triangle from the given information.
The last angle is trivial.
#11 Denote magnitude of b as b and calculate as below.
b =
( 3)
2
+ (1)
2
b = 3 +1
b =2
Denote the direction of b as θ and use the definition of the trigonometric functions to calculate.
sin (θ ) =
θ=
π
6
1
2
#12A) Calculate directly as below or convert to trigonometric form and use the formula
r1r2 ( cos (θ1 + θ 2 ) + i sin (θ1 + θ 2 ) ) .
z1 ⋅ z2 = ( −3 + 3i )( −2 − 2i )
z1 ⋅ z2 = −3 (1 − i )( −2 − 2i )
z1 ⋅ z2 = 6 (1 − i )(1 + i )
z1 ⋅ z2 = 6 (1 − i 2 )
z1 ⋅ z2 = 6 (1 + 1)
z1 ⋅ z2 = 12
#12B) Use the formula
n
1
⎡ ⎛ θ + 360°k ⎞
⎛ θ + 360°k ⎞ ⎤
z = r n ⎢cos ⎜
⎟ + i sin ⎜
⎟ ⎥ for z = r ( cos θ + i sin θ ) as below.
n
n
⎠
⎝
⎠⎦
⎣ ⎝
If z3 = −8 + 8 3 ⋅ i , then r = 16 and θ = 120° .
4
1
⎡ ⎛ 120° + 360° ( 0 ) ⎞
⎡ 3 1 ⎤
⎛ 120° + 360° ( 0 ) ⎞ ⎤
z = 16 4 ⎢cos ⎜
+ i⎥ = 3 + i
⎟ + i sin ⎜
⎟⎥ = 2 ⎢
4
4
2
2 ⎦
⎢⎣ ⎝
⎥
⎠
⎝
⎠⎦
⎣
4
⎡ ⎛ 120° + 360° (1) ⎞
⎡ 1
⎛ 120° + 360° (1) ⎞ ⎤
3 ⎤
z = 16 ⎢cos ⎜
i ⎥ = −1 + 3 ⋅ i
⎟ + i sin ⎜
⎟⎥ = 2 ⎢− +
4
4
⎠
⎝
⎠ ⎦⎥
⎣ 2 2 ⎦
⎣⎢ ⎝
4
4
1
4
1
⎡ ⎛ 120° + 360° ( 2 ) ⎞
⎡ 3 1 ⎤
⎛ 120° + 360° ( 2 ) ⎞ ⎤
− i⎥ = − 3 − i
z = 16 4 ⎢cos ⎜
⎟⎥ = 2 ⎢−
⎟ + i sin ⎜
4
4
⎢⎣ ⎝
⎠ ⎥⎦
⎠
⎝
⎣ 2 2 ⎦
1
⎡ ⎛ 120° + 360° ( 3) ⎞
⎡1
⎛ 120° + 360° ( 3) ⎞ ⎤
3 ⎤
z = 16 4 ⎢cos ⎜
i⎥ = 1− 3 ⋅i
⎟ + i sin ⎜
⎟⎥ = 2 ⎢ −
4
4
2
2
⎢⎣ ⎝
⎥
⎠
⎝
⎠⎦
⎣
⎦
#13 Recall the following.
y
⇒ y = r sin (θ )
r
x
cos (θ ) = ⇒ x = r cos (θ )
r
2
2
x + y = r2
sin (θ ) =
#14 Complete the square as follows.
9 x 2 − 18 x + 4 y 2 + 24 y = 11
9 ( x 2 − 2 x ) + 4 ( y 2 + 6 y ) = 11
2
2
⎛ 2
⎛2⎞ ⎞ ⎛ 2
⎛6⎞ ⎞
9 ⎜ x − 2 x + ⎜ ⎟ ⎟ + 4 ⎜ y + 6 y + ⎜ ⎟ ⎟ = 11 + 9 ⋅1 + 4 ⋅ 9
⎜
⎝ 2 ⎠ ⎟⎠ ⎜⎝
⎝ 2 ⎠ ⎟⎠
⎝
9 ( x 2 − 2 x + 1) + 4 ( y 2 + 6 y + 9 ) = 11 + 9 + 36
9 ( x − 1) + 4 ( y + 3) = 56
2
2
9 ( x − 1) 4 ( y + 3)
56
+
=
56
56
56
2
( x − 1)
56 9
2
2
( y + 3)
+
14
2
=1
Identify the conic as an ellipse with a vertical major axis. Identify the center as ( h, k ) = (1, −3) . Calculate
the distance from the center to the foci as below.
2
c 2 = 14 −
(
56 9
)
2
c 2 = 14 − 56 9
70
c2 =
9
70
c=
3
⎛
70 ⎞
Hence, the foci are ⎜⎜1, −3 ±
⎟.
3 ⎟⎠
⎝
2
#15 Note that < 1 .
5
∞
∑
k =1
⎡ ⎛ 2 ⎞k −1 ⎤
a1
10
10
5 50
=
= = 10 ⋅ = .
⎢10 ⎜ ⎟ ⎥ =
3 3
⎢⎣ ⎝ 5 ⎠ ⎥⎦ 1 − r 1 − 2 3
5 5
#16 Square both sides as follows.
x = 7 sin t
y = 7 cos t
⇓
x 2 = 49sin 2 t
y 2 = 49 cos 2 t
Now, add the two equations and use the Pythagorean Identity.
x 2 + y 2 = 49sin 2 t + 49 cos 2 t
x 2 + y 2 = 49 ( sin 2 t + cos 2 t )
x 2 + y 2 = 49 (1)
x2 + y 2 = 72
#17 Find all the solutions to sin x = 2 2 . Note that sin ( x ) equals
2 2 when x = 45° and when
x = 135° . To find all the solutions, worry about all the co-terminal angles that occur with full rotations
from either of these angles. Hence, x = 45° + 360°k or x = 135° + 360°k where k ∈ .
#18 We solve the conditional identity as below.
⎛θ ⎞
2sin 2 ⎜ ⎟ = cos (θ )
⎝2⎠
We can simplify the left-hand side using the Pythagorean Identity. This reduces the equation to cosine
expressions only.
⎛
⎛θ ⎞⎞
2 ⎜1 − cos 2 ⎜ ⎟ ⎟ = cos (θ )
⎝ 2 ⎠⎠
⎝
We can rewrite θ as 2 ⋅
θ
2
to see that the Double-Angle Identity can be applied as follows.
⎛
⎛θ ⎞⎞
⎛ θ⎞
2 ⎜1 − cos 2 ⎜ ⎟ ⎟ = cos ⎜ 2 ⋅ ⎟
⎝ 2 ⎠⎠
⎝ 2⎠
⎝
⎛θ ⎞
⎛θ ⎞
2 − 2 cos 2 ⎜ ⎟ = 2 cos 2 ⎜ ⎟ − 1
⎝2⎠
⎝2⎠
Now, we solve as below.
⎛θ ⎞
⎛θ ⎞
2 − 2 cos 2 ⎜ ⎟ = 2 cos 2 ⎜ ⎟ − 1
⎝2⎠
⎝2⎠
⎛θ ⎞
⎛θ ⎞
3 − 2 cos 2 ⎜ ⎟ = 2 cos 2 ⎜ ⎟
⎝2⎠
⎝2⎠
⎛θ ⎞
3 = 4 cos 2 ⎜ ⎟
⎝2⎠
⎛θ ⎞ 3
cos 2 ⎜ ⎟ =
⎝2⎠ 4
3
⎛θ ⎞
cos ⎜ ⎟ = ±
2
⎝2⎠
θ π 5π 7π 11π
= ,
,
,
,…
2 6 6 6
6
Multiplying by 2 on both sides, we find the solutions, but keep in mind the restrictions, namely,
0 ≤ θ ≤ 2π .
θ=
π 5π
3
,
3