Upload
nguyenkien
View
231
Download
1
Embed Size (px)
Citation preview
Trigonometry(Chapter6)–SampleTest#2
First, a couple of things to help out:
Page 1 of 16
Trigonometry (Chapter 6) – Sample Test #2
More Formulas (memorize these):
Law of Sines:
sin sin sin
Law of Cosines:
2 cos
2 cos
2 cos
Area of a Triangle:
12 sin
12 sin
12 sin
12
Find the area of the triangle having the given measurements. Round to the nearest square unit.
1) C = 120°, a = 4 yards, b = 5 yards
sin ∙ 4 ∙ 5 ∙ sin 120° yards2
2)
tan 38°51
51 ∙ tan 38° 39.8456meters
tan 50°51
51 ∙ tan 50° 60.7794meters
60.7794 39.8456 ~ . meters
Page 2 of 16
Trigonometry (Chapter 6) – Sample Test #2
Solve the triangle. Round lengths to the nearest tenth and angle measures to the nearest
degree.
When you are given the lengths of two sides and the measure of the angle between them:
Use the Law of Cosines to determine the third side length,
Use the Law of Sines to determine the measure of one of the two unknown angles, and
Subtract the two known angle measures from 180° to get the measure of the last unknown angle.
2 cos
2 cos Law of Cosines
2 cos
Note: in problems with a lot of calculations, it is a good idea to keep more accuracy than you
are required to provide in the final answer, and round your answer at the end. This avoids the
compounding of rounding errors throughout your calculations.
3) a = 6, c = 12, B = 124°
6 12 2 6 12 cos 124° ~16.14075~ .
sin sin
⇒
6sin
16.14075sin 124°
⇒sin 0.3082
⇒ ∠ °
∠ 180° 124° 18° °
4,2
4,2
4,2
Step 1:
A negative radius is confusing, let’s make it positive.
Recall that changing sign of requires you to either
add or subtract radians (180°) from .
Step 2:
At an angle of , move a distance of 4.
Step 3: Answer: PointA
Page 3 of 16
Trigonometry (Chapter 6) – Sample Test #2
Alternatively, convert the polar coordinates , to rectangular coordinates as follows:
4,34
cos 4 cos34
4 ∙√22
2√2~ 2.8
sin 4 sin34
4 ∙√22
2√2~ 2.8
So, the rectangular coordinates are: . , .
Polar coordinates of a point are given. Find the rectangular coordinates of the point. Give exact answer.
6) (‐3, 270°)
First, note that 3, 270° 3, 270° 180° 3, 90°
3, 90°
cos 3 cos 90° 3 ∙ 0 0
sin 3 sin 90° 3 ∙ 1 3
So, the rectangular coordinates are: ,
Step 1:
The angle is at 225°.
Step 2:
The radius 4 means we move in
the direction of the angle a distance
of 4.
Step 3:
Answer: Shownasagreenpointonthegraph.
Recall that we can change the sign of , and either add or subtract
180° from , and get a new representation for the same point!
Page 4 of 16
Trigonometry (Chapter 6) – Sample Test #2
7) (3, ‐135°)
3, 135°
cos 3 cos 135° 3 ∙√22
3√22
sin 3 sin 135° 3 ∙√22
3√22
So, the rectangular coordinates are: √ , √
8) (5, )
5,
cos 5 cos23
5 ∙12
52
sin 5 sin23
5 ∙√32
5√32
So, the rectangular coordinates are: , √
The rectangular coordinates of a point are given. Find polar coordinates of the point. Express in radians.
9) (8,‐8)
8 8 8√2
tan88
tan 1 in 4 74
So, the polar coordinates are: √ ,
Page 5 of 16
Trigonometry (Chapter 6) – Sample Test #2
10) (3√3, 3)
3√3 3 6
tan3
3√3tan
√33
in 1 6
So, the polar coordinates are: ,
11) (0, √7)
0 √7 √7
tan√70
undefined at:32
So, the polar coordinates are: √ , or √ ,
Convert the rectangular equation to a polar equation that expresses r in terms of .
12) 1
Since sin , we make that substitution and solve for .
sin 1
or
13) 25
Substitute cos and sin .
cos sin 25
cos sin 25 (recall that: cos sin 1)
25
(note that we take the positive root of only)
Page 6 of 16
Trigonometry (Chapter 6) – Sample Test #2
Convert the polar equation to a rectangular equation.
14) r=3
Substitute
3
15) r=6csc
6sin
sin 6
Substitute y rsin
Find the absolute value of the complex number.
17) 8 2
8 2 √68 √
4
The graph has a constant radius, no
matter what the value of the angle.
It is also obviously a circle of radius
2. The polar equation, then, is:
See how much simpler this is than
the rectangular equation:
Page 7 of 16
Trigonometry (Chapter 6) – Sample Test #2
Write the complex number in polar form. Express the argument in radians.
18) 3√3 3
The process of putting a complex number in polar form is very similar to converting a set
of rectangular coordinates to polar coordinates. So, if this process seems familiar, that’s
because it is.
3√3 3 6
tan3
3√3tan
√33
in 3 76
So, the polar form is:
Note: 6cis 6 ∙ because cos sin . However, the student may not
be required to know this at this point in the course.
19) 3 3
3 3 3√2
tan33
tan 1 in 4 74
So, the polar form is: √ √
Note: 3√2cis 3√2 ∙ because cos sin . However, the student
may not be required to know this at this point in the course.
Write the complex number in rectangular form. Give exact answer.
20) 3 cos 120° sin 120°
3cis120° 3cis 120° 180° 3cis300°
We want this in the form
cos 3cos 300° 3 ∙
sin 3sin 300° 3 ∙ √ √
So, the rectangular form is: √
Page 8 of 16
Trigonometry (Chapter 6) – Sample Test #2
21) 9 cos sin
We want this in the form
cos 9cos 9 ∙ 1 9
sin 9sin 9 ∙ 0 0
So, the rectangular form is:
Find the product of the complex numbers. Leave answer in polar form.
22) √3 cos sin shorthand is: √3cis
√6 cos sin shorthand is: √6cis
To multiply two numbers in polar form, multiply the ‐values and add the angles.
∙ √3 ∙ √6 ∙ cis74
94
3√2cis 4 √
Note: multiplication may be easier to understand in exponential form, since exponents are added when values with the same base are multiplied:
√3 ∙ ∙ √6 ∙ √3 ∙ √6 ∙ 3√2 3√2 3√2
23) 4
6 6
Relating to :
0 4 4and tan40
undefinedat: 2
So, the polar form is: 4 cos sin 4cis
Relating to :
6 6 6√2and tan66
tan 1 in 2 34
So, the polar form is: 6√2 cos
sin
6√2cis
Then, multiply:
∙ 4cis2
∙ 6√2cis34
24√2cis2
34
√
This problem is probably best approached by converting each number to polar form and then multiplying.
Page 9 of 16
Trigonometry (Chapter 6) – Sample Test #2
Find the quotient of the complex numbers. Leave answer in polar form.
24) √3 cos sin
√6 cos sin
To divide two numbers in polar form, divide the ‐values and subtract the angles.
∙√3
√6∙ cis
74
94
1
√2cis
2√22cis
22
√
Note: division may be easier to understand in exponential form, since exponents are subtracted when values with the same base are divided:
√3 ∙
√6 ∙
1
√2
√22
√22
√22
Use DeMoivre's Theorem to find the indicated power of the complex number. Write the answer in rectangular form.
25) 3 cos15° sin15°
To take a power of two numbers in polar form, take the power of the ‐value and multiply the angle by the exponent. (This is the essence of DeMoivre’s Theorem.)
3 cos15° sin15° 3 ∙ cis 4 ∙ 15° 81cis 60°
81 cos60° sin60° 8112
√32
√
Note: Consider this in exponential form. Exponents are multiplied when a value with an exponent is taken to a power:
3cis12
3 ∙ 3 ∙ ∙ 81 ∙
26) √3
√3 1 2and tan1
√3tan
√33
in 2 56
Then, √3 2cis 2 cis 6 ∙
64cis5 64 ∙ cos 5 sin 5
Page 10 of 16
Trigonometry (Chapter 6) – Sample Test #2
Find all the complex roots. Write the answer in the indicated form.
27) The complex cube roots of 8 (rectangular form).
We must use DeMoivre’s Theorem for roots to solve this problem.
Let cos sin . Then, has distinct complex ‐th roots that occupy positions
equidistant from each other on a circle of radius √ . Let’s call the roots: , , … ,
Then, these roots can be calculated as follows:
√ ∙ cos2
sin2
Given this, let’s find the cube roots of .
First, since , we have 8 and 0.
Then, √8 0 8; √ 2
And, tan 0°; 0°
The incremental angle for successive roots is: 360° 3roots 120°.
Then create a root development chart like this (answers are in green):
Cube roots of : √ √ ° incremental °
Angle ( ) √ ∙ √ ∙ ∙
0 0°
1 0° 120° 120° √
2 120° 120° 240° √
Notice that if we add another 120°, we get 360°, which is equivalent to our first angle, 0° because360° 360° 0°. This is a good thing to check. The “next angle” will always be equivalent to the first angle! If it isn’t, go back and check your work.
Roots fit on a circle: Notice that, since all of the roots of
have the same magnitude, and their angles that are 120° apart from each other, that they occupy equidistant positions
on a circle with center 0, 0 and radius √8 2.
More information about using DeMoivre’s Theorem for Roots,
including a more complex example, can be found in the
Trigonometry Handbook on www.mathguy.com.
Page 11 of 16
Trigonometry (Chapter 6) – Sample Test #2
⟨ 4, 21⟩ ⟨ 2,7⟩
⟨ 2, 14⟩
‖ ‖ 2 14
√200 √100 ∙ √2 √
Subtracting is the same as adding .
To get – , simply change the sign of each
element of . If you find it easier to add
than to subtract, you may want to adopt
this approach to subtracting vectors.
⟨ 12, 2⟩ ⟨ 6, 7⟩ ⟨ 12 6, 2 7⟩
⟨ , ⟩
Let v be the vector from initial point to terminal point . Write v in terms of and .
An alternative notation for a vector in the form is ⟨ , ⟩. Using this alternative notation makes many vector operations much easier to work with.
28) 5, 5 ; 1, 5
We can calculate as the difference between the two given points.
1, 5
5, 5
⟨ 6, 0⟩ So, in terms of and ,
Find the specified vector or scalar.
29) u = ‐2i ‐ 7j and v = ‐4i ‐ 21j; Find ‖ ‖.
30) u = ‐12i ‐ 2j, v = 6i + 7j; Find u ‐ v.
To add or subtract vectors, simply
line them up vertically and perform
the required operation:
Find the unit vector that has the same direction as the vector v.
A unit vector has magnitude 1. To get a unit vector in the same direction as the original vector, divide the vector by its magnitude.
31) v = 3i ‐ 4j
The unit vector is: ‖ ‖
Page 12 of 16
Trigonometry (Chapter 6) – Sample Test #2
Write the vector v in terms of i and j whose magnitude v and direction angle are given. Give exact answer.
32)‖ ‖ = 7, θ = 225°
The unit vector in the direction θ 225° is:
⟨cos 225°, sin 225°⟩ ⟨√22,
√22⟩
√22
√22
Multiply this by ‖ ‖ to get :
7√22
√22
√ √
Solve the problem.
33) The magnitude and direction of two forces acting on an object are 60 pounds, N40°E, and 70 pounds, N40°W, respectively. Find the magnitude, to the nearest hundredth of a pound, and the direction angle, to the nearest tenth of a degree, of the resultant force.
This problem requires the addition of two vectors. The approach I prefer is:
1) Convert each vector into itsi and j components, call them and ,
2) Add the resulting and values for the two vectors, and
3) Convert the sum to its polar form.
Keep additional accuracy throughout and round at the end. This will prevent error
compounding and will preserve the required accuracy of your final solutions.
Step 1: Convert each vector into itsi and j components
Let be a force of 60 lbs. at bearing: N40°E
From the diagram at right,
θ 90° 40° 50°
60 cos 50° 38.5673
60 sin 50° 45.9627
Let be a force of 70 lbs. at bearing: N40°W
From the diagram at right,
φ 90° 40° 50°
70 cos 50° 44.9951
70 sin 50° 53.6231
Page 13 of 16
Trigonometry (Chapter 6) – Sample Test #2
⟨5, 6⟩
∙ ⟨ 3, 12⟩
⟨ 5, 3⟩
⟨0, 3⟩
∙ 0 ∙ 3 3 ∙ 12 0 36
⟨38.5673, 45.9627⟩ ⟨ 44.9951, 53.6231⟩
⟨ 6.4278, 99.5858⟩
Step 2: Add the results for the two vectors
Step 3: Convert the sum to its polar form
Direction Angle tan .
.. °
Magnitude 6.4278 99.5858 . lbs.
34) One rope pulls a barge directly east with a force of 79 newtons, and another rope pulls the barge directly north with a force of 87 newtons. Find the magnitude of the resultant force acting on the barge.
The process of adding two vectors whose headings are north, east,
west or south (NEWS) is very similar to converting a set of
rectangular coordinates to polar coordinates. So, if this process
seems familiar, that’s because it is.
79 87 .
This problem does not require it, but let’s calculate the direction angle anyway.
tan8779
47.8°
Use the given vectors to find the specified scalar.
35) u = ‐5i + 3j, v = 5i ‐ 6j,w = ‐3i + 12j; Find u ∙ w + v ∙ w.
The alternate notation for vectors comes in especially handy in doing these types of
problems. Also, note that: (u ∙ w) + (v ∙ w) = (u + v) ∙ w. Let’s calculate (u + v) ∙ w.
Using the distributive property for dot
products results in an easier problem
with fewer calculations.
Page 14 of 16
Trigonometry (Chapter 6) – Sample Test #2
cos ∙
‖ ‖‖ ‖ √ ∙√ √
⟨1, 1⟩ ∙ ⟨4, 5⟩
1 ∙ 4 1 ∙ 5 1
‖ ‖ 1 1 √2
‖ ‖ 4 5 √41
cos1
√82. °
∙ ⟨ 11, 6⟩
∙ 4 ∙ 11 4 ∙ 6 44 24
⟨ 4, 4⟩
Clearly, 2
Therefore, the vectors are parallel.
⟨3, 4⟩ ⟨6, 8⟩
It is clearly easier to check whether one
vector is a multiple of the other than to
use the dot product method. The
student may use either, unless
instructed to use a particular method.
To determine if two vectors are parallel using the
dot product, we check to see if:
∙ ‖ ‖‖ ‖
Therefore, the vectors are parallel.
⟨3, 4⟩ ∙ ⟨6, 8⟩
32 50 18
‖ ‖ 4 5 3
‖ ‖ 8 10 6
‖ ‖‖ ‖ 5 ∙ 10 50
36) u = ‐4i + 4j, v = ‐11i ‐ 6j; Find u ∙ v.
The alternate notation for vectors comes in especially handy in calculating a dot product.
Find the angle between the given vectors. Round to the nearest tenth of a degree.
37) u = i ‐ j, v = 4i + 5j
cos ∙
‖ ‖‖ ‖
0° 180°
Use the dot product to determine whether the vectors are parallel, orthogonal, or neither.
If the vectors are parallel, one is a multiple of the other; also ∙ ‖ ‖‖ ‖ .
If the vectors are perpendicular, their dot product is zero.
38) v = 3i + 4j, w = 6i + 8j
Page 15 of 16
Trigonometry (Chapter 6) – Sample Test #2
39) v = 4i + 3j, w = 3i ‐ 4j
Calculate the dot product.
⟨4,3⟩ ∙ ⟨3, 4⟩
4 ∙ 3 3 ∙ 4 12 12 0
Therefore, the vectors are orthogonal.
Page 16 of 16