solutions for trigonometry by gelfand & saul
TRANSCRIPT
Solutions for Trigonometry by Gelfand & Saul
Git commit 43c184a. Published June 19, 2020.
This work is licensed under a Creative Commons“Attribution-ShareAlike 4.0 International” license.
Introduction
Trigonometry by Gelfand and Saul is often recommended as a precalculus textfor self-study. However, those who are learning without the help of a teachercan struggle with the lack of solutions to exercises in the text. A partial set ofsolutions for Trigonometry (odd numbered exercises only) has been published byJohn Beach1. It is hoped that this document will eventually contain a completeset of solutions. Contributions are welcome. These can take the form of pullrequests or issues submitted to the project’s GitHub repository2.
Chapter 0: Trigonometry
Page 8
1. Statement I applies:
c2 = a2 + b2 = 102 + 242 = 100 + 576 = 676
c =√
676 = 26
2. Statement I applies:
a2 + b2 = c2
a2 + 92 = 412
a2 + 81 = 1681
a2 = 1600
a =√
1600 = 40
1https://jbeach50.weebly.com/gelfand–saul-trig-solutions.html2https://github.com/philip-healy/gelfand-trigonometry-solutions
1
3. 52 + 122 = 25 + 144 = 169 = 132. By Statement II, a right triangle existswith legs of length 5 and 12, and hypotenuse of length 13.
4. Statement I applies:
a2 + b2 = c2
a2 + 12 = 32
a2 + 1 = 9
a2 = 8
a =√
8 =√
4√
2 = 2√
2
5. Statement I applies, where a = b:
a2 + a2 = c2
a2 + a2 = 12
2a2 = 1
a2 =1
2
a =
√1
2=
√1√2
=1√2
6. From the diagram at the bottom of Page 11, we can see the shorter leg ishalf the length of the hypotenuse. So in this instance the shorter leg haslength 1/2. We can use Statement 1 to find the length of the longer leg:
a2 + b2 = c2
a2 +
(1
2
)2
= 12
a2 +1
4= 1
a2 =3
4
a =
√3
4=
√3√4
=
√3
2
7. For any point Y , we can draw a triangle with sides AY , BY and AB. Leta be the length of side AY , b be the length of side BY and c be the lengthof side AB. According to Statement II, the subset of these triangles wherea2 + b2 = c2 are right triangles with legs of length a and b and hypotenusec. Let X be the subset of Y that are vertices of these right triangles. Thisset of points describes a circle with its centre at the midpoint of AB, andradius AB/2.
8.
2
Page 9
1. 62 + 82 = 36 + 64 = 100 = 102. By Statement II on Page 7 (converse of thePythagorean Theorem), this is a right triangle.
2. 10-24-26 (Exercise 1), 9-40-41 (Exercise 2), 5-12-13 (Exercise 3)
3. Using the Pythagorean Theorem:
c2 = a2 + b2 = 82 + 152 = 64 + 225 = 289
c =√
289 = 17
4. The first column in the table increases by 3, the second increases by 4 andthe third increases by 5. Continuing to add rows yields triangles 12-16-20,15-20-25 and 18-24-30.
5. Shortest side with length 10: 10-24-26. Shortest side with length 15: 15-36-39.
6. Multiplying all sides by the common denominator (5), we get a similar tri-angle with sides 15/5 = 3, 20/5 = 4 and 5. We know that this is a righttriangle from the table in Question 4.
7. To find a similar triangle with shorter leg 1, divide all sides by 3, resulting insides 1-4/3-5/3. To find a similar triangle with longer leg 1, divide all sidesby 4, resulting in sides 3/4-1-5/4.
8. To find a similar triangle with hypotenuse 1, divide all sides by 13, resultingin sides 5/13-12/13-1. To find a similar triangle with shorter leg 1, divideall sides by 5, resulting in sides 1-12/5-13/5. To find a similar triangle withlonger leg 1, divide all sides by 12, resulting in sides 5/12-1-13/12.
9. To formula for the area of a triangle is 12bh where b is the length of the base
and h is the height. For right triangles, finding the area is easy: one leg is thebase and the other leg is the height. For other triangles, finding the heightis more difficult: we need to find the length of the altitude drawn from thebase. The triangles with sides 5-12-13 and 9-12-15 are both right triangles:see Exercise 3 on Page 8 and Exercise 4 on Page 9. The triangle with sides13-14-15 is not a right triangle. We can confirm this using Statement I:a2 + b2 = 132 + 142 = 365, c2 = 152 = 225, a2 + b2 6= c2. However, if we jointhe 5-12-13 and 9-12-15 triangles using their equal legs, the resulting trianglehas the dimensions we are looking for: 13-14-15. The base of this combinedtriangle has length 5 + 9 = 14. We also know the length of the altitude fromthe base of the combined triangle: 12. So, the area of the 13-14-15 triangleis 1
2 · 14 · 12 = 84 units squared.
10. (a)
(b)
3
Page 11
1. 1√2
(see the solution for Question 5 on page 8).
Challenge: 1√2
=√22 (multiplying above and below by
√2).√
2 is given to
9 decimal places in the diagram on the top of page 11: 1.4141213562373.Dividing this decimal representation by 2 (using long division if necessary)yields a figure of 0.707060678.
2. c2 = a2 + b2 = 32 + 32 = 9 + 9 = 18. c =√
18 =√
9√
2 = 3√
2.
3. The hypotenuse of a 30◦ right triangle is double the length of the shorter leg.In this instance the hypotenuse is 10 units long. We can use the PythagoreanTheorem to find the length of the longer leg:
a2 + b2 = c2
a2 + 52 = 102
a2 + 25 = 100
a2 = 75
a =√
75 =√
25√
3 = 5√
3
4. We can solve these by finding similar triangles to the 30◦ right triangle withsides 1-
√3-2, or the 45◦ right triangle with sides 1-1-
√2.
(a) x =√
3, y = 2
(b) x =√
3, y = 2√
3
(c) x = 1/2, y =√
3/2
(d) x = 4√
3, y = 8
(e) x = y = 2√
2
(f) x = 5, y = 5√
2
Page 14 (Examples)
1. Why didn’t we need to compare 32 with 22 + 42, or 22 with 32 + 42?The obtuse angle will always be opposite the longest side.
2. This conclusion is incorrect. Why?From the footnote at the beginning of Chapter 0: “Given three arbitrarylengths. . . they form a triangle if and only if the sum of any two of them isgreater than the third.” In this case 1 + 2 = 3 which is equal to (not greaterthan) the third side.
4
Page 14 (Exercise)
1. (a) c2 = 82 = 64. a2 + b2 = 62 + 72 = 36 + 49 = 85. c2 < a2 + b2, so thetriangle is acute.
(b) c2 = 102 = 100. a2 + b2 = 62 + 82 = 36 + 64 = 100. c2 = a2 + b2, sothe triangle is a right triangle.
(c) a and b are the same as in question b), but c is smaller, so the triangleis acute.
(d) a and b are the same as in question b), but c is larger, so the triangleis obtuse.
(e) c2 = 122 = 144. a2 + b2 = 52 + 122 = 25 + 144 = 169. c2 < a2 + b2, sothe triangle is acute.
(f) c2 = 142 = 196. a2 + b2 = 169, as above. c2 > a2 + b2, so the triangleis obtuse.
(g) Here a and b are the same as above, but c is larger. So, this triangle isalso obtuse.
Chapter 1: Trigonometric Ratios in a Triangle
Page 23
1. (a) sinα = 5/13
(b) sinα = 4/5
(c) sinα = 5/13
(d) c =√
62 + 82 =√
100 = 10. sinα = 8/10.
(e) sinα = 3/5
(f) sinα = 12/13
(g) sinα = 3/5
(h) c =√
72 + 32 =√
58. sinα = 7/√
58.
2. (a) sinβ = 12/13
(b) sinβ = 3/5
(c) sinβ = 12/13
(d) sinβ = 6/10
(e) sinβ = 4/5
(f) sinβ = 5/13
(g) sinβ = 4/5
(h) sinβ = 3/√
58
5
3. The example 30-60-90 triangle given on page 11 has sides 1,√
3, 2. Let βrepresent the 60◦angle. The opposite leg b has length
√3. The hypotenuse
c has length 2. So, sinβ = b/c =√
3/2 ≈ 1.732/2 = 0.866.Crossing off the numbers listed:��0.1 ��0.2 ��0.3 ��0.4 ��0.5 ��0.6 ��0.7 ��0.8 0.9
Page 25
1. The Altitude-on-Hypotenuse Theorem tells us that when an altitude is drawnto the hypotenuse of a right triangle, the two triangles formed are similarto the given triangle and to each other. Therefore, the triangles with sidesa-b-c, a-p-d and d-b-q are similar, and the ratio for sinα appears in all ofthem:
(a) b/c
(b) d/a
(c) q/b
2. (a) sinα = h/b
(b) Multiplying both sides of formula above by b: h = b sinα
(c) Substituting b sinα for h, the formula for the area of ABC can berewritten as: bc sinα/2.
(d) sinβ = h/a. Rewriting this in terms of h: h = a sinβ. Substitutingthis for h in the area formula: ac sinβ/2.
(e) Let h2 represent the altitude from A to BC. sinβ = h2/c. Rewritingin terms of h2, we get h2 = c sinβ.
3. (a) Expressing h in terms of sinα and b:
sinα =h
bh = b sinα
Expressing h in terms of sinβ and a:
sinβ =h
ah = a sinβ
(b) Both expresssions are equal to h:
a sinβ = h = b sinα
(c) Expressing h2 in terms of sinβ and c:
sinβ =h2c
h2 = c sinβ
6
Expressing h2 in terms of sin γ and b:
sin γ =h2b
h2 = b sin γ
Both expressions are equal to h2:
b sinα = h2 = c sin γ
(d) i. We can rewrite the result from part (b) so that the expressions oneach side are fractions with sine denominators:
a sinβ = b sinα
a sinβ
sinα sinβ=
b sinα
sinα sinβ
a
sinα=
b
sinβ
ii. We can rewrite the result from part (c) similarly:
c sinβ = b sin γ
c sinβ
sinβ sin γ=
b sin γ
sinβ sin γ
c
sin γ=
b
sinβ
We can derive the Law of Sines by combining results i. and ii. using thecommon expression b/ sinβ:
a
sinα=
b
sinβ=
c
sin γ
Page 26
1. (a) cosα = 12/13. cosβ = 5/13.
(b) cosα = 3/5. cosβ = 4/5.
(c) cosα = 12/13. cosβ = 5/13.
(d) cosα = 6/10. cosβ = 8/10.
(e) cosα = 4/5. cosβ = 3/5.
(f) cosα = 5/13. cosβ = 12/13.
(g) cosα = 4/5. cosβ = 3/5.
(h) cosα = 3/√
58. cosβ = 7/√
58.
2. (a) c =√
82 + 62 =√
64 + 36 =√
100 = 10. cosα = 8/10. cosβ = 6/10.
7
(b) c =√
52 + 122 =√
25 + 155 =√
169 = 13. cosα = 12/13. cosβ =5/13.
(c) Scaling up the 1-√
3-2 30◦ triangle gives us a value of 20 units for thelength of c. Next, we will use the Pythagorean Theorem to find thelength of the longer leg:
a2 + b2 = c2
102 + b2 = 202
b2 = 400− 100 = 300
b =√
300 =√
100√
3 = 10√
3
We can now find cosα and cosβ:
cosα =10√
3
20=
√3
2
cosβ =10
20=
1
2
(d) The triangle is congruent to the one above, so the solution is the same.
(e) Consider the 45◦ right triangle with legs of length 1 and hypotenuse√2. cosα = cosβ = 1/
√2.
(f) c =√
32 + 42 =√
9 + 16 =√
25 = 5. cosα = 3/5. cosβ = 4/5.
(g) b = x√
3. cosα = x√
3/2x =√
3/2. cosβ = x/2x = 1/2.
3. The Altitude-on-Hypotenuse Theorem tells us that when an altitude is drawnto the hypotenuse of a right triangle, the two triangles formed are similarto the given triangle and to each other. Therefore, the triangles with sidesa-b-c, a-p-d and d-b-q are similar, and the ratio for cosα appears in all ofthem:
(a) a/c
(b) p/a
(c) d/b
Page 28
1. In this instance, α = 29◦, β = 61◦, and α + β = 90◦. According to thetheorem above, if α+ β = 90◦, then sinα = cosβ.
2. x = 90− 35 = 55◦
3. If α+ β = 90◦, then β = 90◦ − α. According to the theorem above, sinα =cosβ. Substituting (90− α) for β: sinα = cos (90− α).
8
Page 29
First, we need to find the length of the hypotenuse: c =√
32 + 42 =√
9 + 16 =√25 = 5.
1. sin2 α =(45
)2= 16
25
2. sin2 β =(35
)2= 9
25
3. cos2 α =(35
)2= 9
25 (same as sin2 β)
4. cos2 β =(45
)2= 16
25 (same as sin2 α)
5. sin2 α+ cos2 α = 1625 + 9
25 = 2525 = 1
6. sin2 α+ cos2 β = 1625 + 16
25 = 3225
7. cos2 α+ sin2 β = 925 + 9
25 = 1825
Page 30
1. sin2 α+ cos2 α =
(4
5
)2
+
(3
5
)2
=16
25+
9
25=
25
25= 1
2. It’s not an error. According to the corollary of the Pythagoream Theorem,this a right triangle: a2 + b2 = 32 + 42 = 9 + 16 = 25 = c2.
3. sin2 β + cos2 β =
(3
5
)2
+
(4
5
)2
=9
25+
16
25=
25
25= 1
4. cos2 α+ sin2 α = 1
cos2 α = 1− sin2 α = 1−(
5
13
)2
= 1− 25
169=
144
169
cosα =
√144
169=
12
13
5. cos2 α+ sin2 α = 1
cos2 α = 1− sin2 α = 1−(
5
7
)2
= 1− 25
49=
24
49
cosα =
√24
49=
√4√
6√49
=2√
6
7
9
6. We will follow the proof at the bottom of Page 29:
sin2α+ sin2 β =(ac
)2+
(b
c
)2
=a2
c2+b2
c2
=a2 + b2
c2
=a2 + b2
a2 + b2
= 1
7. Again, we will follow the proof at the bottom of Page 29:
cos2α+ cos2 β =
(b
c
)2
+(ac
)2=b2
c2+a2
c2
=a2 + b2
c2
=a2 + b2
a2 + b2
= 1
Page 31
1.
angle x sinx cosx
30◦ 12
√32
45◦ 1√2
1√2
60◦√32
12
α 45
35
β 35
45
2. cos 30◦ =√32 = sin 60◦
3. sin2 30◦ + cos2 30◦ =
(1
2
)2
+
(√3
2
)2
=1
4+
3
4= 1
10
4. We can observe from the table that sinx increases with the size of an acuteangle (sin 30◦ < sin 45◦ < sin 60◦), while cosx decreases with the size of anacute angle. You can compare the fractions or convert to decimal make sure.We know that sinα = 4
5 . We also know that α is an acute angle.Is it larger or smaller than 30◦? Larger, 4
5 >12 so sinα > sin 30◦.
Than 45◦? Larger, 45 >
1√2
so sinα > sin 45◦.
Than 60◦? Smaller, 45 <
√32 so sinα < sin 60◦.
Page 33 (First)
1. As the angle α get smaller, the ratio of the opposite side to the hypotenuseapproaches 0.
2. Recall from the theorem on page 28 that if α + β = 90◦, then sinα = cosβand cosα = sinβ. So, if sin 90◦ = 1, then cos 0◦ = 1.
3. sin2 0◦ + cos2 0◦ = 02 + 12 = 0 + 1 = 1
4. sin2 90◦ + cos2 90◦ = 12 + 02 = 1 + 0 = 1
5. Our friend is mistaken; the sine of an angle can never be greater than 1.
Page 33 (Second)
1.
sin 0◦ + cos 0◦ 0 + 1 1
sin 30◦ + cos 30◦ 12 +
√32 1.366 (approx.)
sin 45◦ + cos 45◦ 1√2
+ 1√2
1.414 (approx.)
sin 60◦ + cos 60◦√32 + 1
2 1.366 (approx.)
sinα+ cosα, where αis the smaller. . .
35 + 4
5 1.4
sinα+ cosα, where αis the larger. . .
45 + 3
5 1.4
2. If sinα = 1, then cosα = 0 and sinα + cosα = 1. If cosα = 1, thensinα = 0 and sinα + cosα = 1. Otherwise, sinα < 1 and cosα < 1, sosinα+ cosα < 2.
3. First we will expand and simplify (sinα+ cosα)2:
(sinα+ cosα)2 = sin2 α+ 2 sinα cosα+ cos2 α
= (sin2 α+ cos2 α) + 2 sinα cosα
= 1 + 2 sinα cosα
11
We know that 0 ≤ sinα ≤ 1 and 0 ≤ cosα ≤ 1 because α is acute. So2 sinα cosα is the product of three nonnegative numbers, and is itself anonnegative number. A nonnegative number added to 1 results in a number≥ 1. Therefore, 1+2 sinα cosα ≥ 1. The square root of a number ≥ 1 is itself≥ 1. Therefore,
√1 + 2 sinα cosα ≥ 1. Rewriting the expression on the left:√
1 + 2 sinα cosα =√
(sinα+ cosα)2 = sinα+ cosα. So, sinα+ cosα ≥ 1.
4. sin 45◦ + cos 45◦ =1√2
+1√2
=2√2
=2√
2√2√
2=
2√
2
2=√
2
5. You should notice that the values for sinα+cosα increases with larger alphawhen 0◦ ≤ α < 45◦, reaches a maximum value when α = 45◦, then decreaseswith larger α when 45◦ < α ≤ 90◦.
Page 35
1.
(sin 0◦)(cos 0◦) 0 · 1 0
(sin 30◦)(cos 30◦) 12 ·√32 0.433 (approx.)
(sin 45◦)(cos 45◦) 1√2· 1√
20.5
(sin 60◦)(cos 60◦)√32 ·
12 0.433 (approx.)
(sinα)(cosα), where αis the smaller. . .
35 ·
45 0.48
(sinα)(cosα), where αis the larger. . .
45 ·
35 0.48
How large can the product (sinα)(cosα) get? We can see from the table thatthe maximum value of the product appears to be when α = 45◦.
Page 37
1. cosα = 3/5, cosβ = 4/5, sinα = 4/5, sinβ = 3/5, tanα = 4/3, tanβ = 3/4,cotα = 3/4, cotβ = 4/3.
2. We can show that this assumption is correct using the corollary of thePythagorean Theorem: a2 + b2 = 32 + 42 = 25 = c2.
3. cosα = a/c, cosβ = b/c, sinα = b/c, sinβ = a/c, tanα = b/a, tanβ = a/b,cotα = a/b, cotβ = b/a.
4. c =√
122 + 52 =√
169 = 13. cosα = 12/13. cosβ = 5/13. cotα = 12/5.cotβ = 5/12.
12
5. First, we will use the Pythagorean Theorem to find the length of the longerleg:
a2 + b2 = c2
a2 + 72 = 252
a2 + 49 = 625
a2 = 576
a = 24
We can now find the numerical values that were asked for: cosα = 24/25,cosβ = 7/25, cotα = 24/7, cotβ = 7/24.
6.a
c= sinα = cosβ
b
c= cosα = sinβ
a
b= tanα = cotβ
b
a= cotα = tanβ
7. First, we will use the Pythagorean Theorem to find the length of the otherleg:
a2 + b2 = c2
a2 + 32 = 52
a2 + 9 = 25
a2 = 16
a = 4
We can now find the numerical values that were asked for: cosα = 4/5,cotα = 4/3.
8. If tanα = 1, then a/b = 1, implying that a = b and α = 45◦. cosα =cos 45◦ = 1/
√2. cotα = 1/1 = 1.
9. tan 45◦ = 1/1 = 1.
10. tan 30◦ = 1/√
3 ≈ 0.57735.
11. tan 45◦ + sin 30◦ = 1 + 12 = 3
2 . We don’t need a calculator because bothnumbers are rational.
13
Chapter 2: Relations among Trigonometric Ra-tios
Page 43
1. cosα =
√1−
(8
17
)2
=
√1− 64
289=
√225
289=
15
17
tanα =8171517
=8
15
cotα =15
8
2. Let the length of the adjacent leg a be 37 and the length of the hypotenuse
be 1 (see the first triangle diagram on page 44).
sinα =√
1− a2 =
√1−
(3
7
)2
=
√1− 9
49=
√40
49=
√4√
10√49
=2√
10
7
tanα =
√1− a2a
=2√10737
=2√
10
3
cotα =a√
1− a2=
3
2√
10
3. sinα =√
1− b2, tanα =
√1− b2b
, cotα =b√
1− b2
4. sinα =d√
1 + d2, tanα =
1√1 + d2
, cotα =1
d
5.
sinα cosα tanα cotα
sinα a√
1− a2 a√1− a2
√1− a2a
cosα√
1− a2 a
√1− a2a
a√1− a2
tanαa√
1 + a21√
1 + a2a
1
a
cotα1√
1 + a2a√
1 + a21
aa
14
Page 45 (First)
1. Given in text
2. sin2 45◦ =
(1√2
)2
=1
2
3.
sinα cosα tanα cotα
sinα sinα√
1− sin2 αa√
1− sin2 α
√1− sin2 α
sinα
cosα√
1− cos2 α cosα
√1− cos2 α
cosα
cosα√1− cos2 α
tanαtanα√
1 + tan2 α
1√1 + tan2 α
tanα1
tanα
cotα1√
1 + cot2 α
a√1 + cot2 α
1
cotαcotα
Page 45 (Second)
1. tanα =a
b= cotβ
2. cotα =b
a= tanβ
3. secα =c
a= cscβ
4. cscα =c
b= secβ
Page 47
1. (a) sin2 30◦ + cos2 30◦ =
(1
2
)2
+
(√3
2
)2
=1
4+
3
4= 1
(b) sin2 45◦ + cos2 45◦ =
(1√2
)2
+
(1√2
)2
=1
2+
1
2= 1
(c) sin2 60◦ + cos2 60◦ =
(√3
2
)2
+
(1
2
)2
=3
4+
1
4= 1
15
2. sin2 α+ cos2 α = 1(√5
4
)2
+ cos2 α = 1
cos2 α = 1−
(√5
4
)2
= 1− 5
16=
11
16
cosα =
√11
16=
√11
4
3. sin2 α+ cos2 α = 1
sin2 α+
(2
3
)2
= 1
sin2 α = 1− 4
9=
5
9
sinα =
√5
9=
√5
3
4.sinα
cosα= tanα =
1√3
sin2 α
cos2 α=
1
3
sin2 α
1− sin2 α=
1
3
3 sin2 α = 1− sin2 α
4 sin2 α = 1
sin2 α =1
4
sinα =
√1
4=
1
2
5. (a) cotx sinx =
(1
tanx
)sinx =
sinx
tanx=
sinxsin xcos x
=sinx cosx
sinx= cosx
(b)tanx
sinx=
sin xcos x
sinx=
sin xcos x ·
1sin x
sinx · 1sin x
=sin x
sin x cos x
1=
sinx
sinx cosx=
1
cosx
(c) cos2 α−sin2 α = cos2 α−(1−cos2 α) = cos2 α−1+cos2 α = 2 cos2 α−1
(d) This one is tricky. You might need to try a few different approaches(squaring above and below, multiplying above and below by cosα sinα).Eventually it becomes clear that you need to multiply above and belowby (1 − cosα) and find a way to cancel out the sinα factor in the
16
numerator:
sinα
1 + cosα=
sinα(1− cosα)
(1 + cosα)(1− cosα)=
sinα(1− cosα)
1− cosα+ cosα− cos2 α
=sinα(1− cosα)
1− cos2 α=
sinα(1− cosα)
1− (1− sin2 α)=
sinα(1− cosα)
1− 1 + sin2 α
=sinα(1− cosα)
sin2 α=
1− cosα
sinα
(e)sin2 α+ 2 cos2 α− 1
cot2 α=
1− cos2 α+ 2 cos2 α− 1
cot2 α=
cos2 α(cosαsinα
)2=
cos2 αcos2 αsin2 α
=cos2 α sin2 α
cos2 α
= sin2 α
(f) cos2 α =cos2 α
1=
cos2 α
cos2 α+ sin2 α
=cos2 αcos2 α
cos2 α+sinα
cos2 α
=1
cos2 αcos2 α + sin2 α
cos2 α
=1
1 + tan2 α
(g) sin2 α =sin2 α
1=
cos2 α
cos2 α+ sin2 α
=sin2 αsin2 α
cos2 α+sinα
sin2 α
=1
cos2 αsin2 α
+ sin2 αsin2 α
=1
cot2 α+ 1
(h)1− cosα
1 + cosα=
(1− cosα)(1 + cosα)
(1 + cosα)(1 + cosα)=
1 + cosα− cosα− cos2 α
(1 + cosα)2
=1− cos2 α
(1 + cosα)2=
sin2 α
(1 + cosα)2
=
(sinα
1 + cosα
)2
(i) The key to solving this one is the formula for factoring a difference ofcubes: a3 − b3 = (a− b)(a2 + ab+ b2).
sin3 α− cos3 α
sinα− cosα=
(sinα− cosα)(sin2 α+ sinα cosα+ cos2 α)
sinα− cosα
= sin2 α+ sinα cosα+ cos2 α
= 1 + sinα cosα
17
6. (a) We can rewrite the LHS to show that sin4 α− cos4 α = cos2 α− sin2 α:
sin4 α− cos4 α = (sin2 α+ cos2 α)(sin2 α− cos2 α) = 1(sin2 α− cos2 α)
= sin2 α− cos2 α
Answer: There are no angles α for which sin4 α−cos4 α > cos2 α−sin2 αbecause the expressions on either side of the inequality are equivalent.
(b) sin4 α−cos4 α >= cos2 α−sin2 α for all angles α because the expressionson either side of the inequality are equivalent.
7. If we rewrite 2 sinα cosα as a fraction, we can divide above and below bycosα to convert the numerator and denominator into expressions in termsof tanα:
2 sinα cosα =2 sinα cosα
1=
2 sinα cosα
sin2 α+ cos2 α
=2 sinα cosα
cos2 αsin2 α+cos2 α
cos2 α
=2 sinαcosα
sin2 αcos2 α + cos2 α
cos2 α
=2 tanα
tan2 α+ 1
Now we can plug in the given value for tanα to find the value of 2 sinα cosαin this instance:
2 sinα cosα =2 tanα
tan2 α+ 1=
2( 25 )
( 25 )2 + 1
=45
425 + 1
=45
425 + 25
25
=20252925
=20
29
8. First, we will rewrite the expresssion cos2 α− sin2 α in terms of tanα:
cos2 α− sin2 α =cos2 α− sin2 α
1=
cos2 α− sin2 α
cos2 α+ sin2 α=
cos2 α−sin2 αcos2 α
cos2 α+sin2 αcos2 α
=1− tan2 α
1 + tan2 α
(a) To find the numerical value of cos2 α − sin2 α when tanα = 25 we can
substitute 25 for tanα in the formula above:
cos2 α− sin2 α =1− tan2 α
1 + tan2 α=
1−(25
)21 +
(25
)2 =1− 4
25
1 + 425
=21252925
=21
29
(b) Substituting r for tanα in the formula above:
cos2 α− sin2 α =1− r2
1 + r2
18
9. First, we will rewrite the expresssion in terms of tanα:
sinα− 2 cosα
cosα− 3 sinα=
sinα−2 cosαcosα
cosα−3 sinαcosα
=sinαcosα −
2 cosαcosα
cosαcosα −
3 sinαcosα
=tanα− 2
1− 3 tanα
Next, we substitute 25 for tanα:
tanα− 2
1− 3 tanα=
25 − 2
1− 3(25
) =25 −
105
55 −
65
=− 8
5
− 15
= 8
10. First, we will rewrite the expresssion in terms of tanα:
a sinα+ b cosα
c cosα+ d sinα=
a sinαcosα + b cosα
cosαc cosαcosα + d cosα
cosα
=a tanα+ b
c+ d tanα
Next, we substitute 25 for tanα and simplify:
a tanα+ b
c+ d tanα=a(25
)+ b
(55
)c(55
)+ d
(25
) =2a+5b
55c+2d
5
=2a+ 5b
5c+ 2d
Now we can see why the problem included the restriction that 5c+ 2d 6= 0;the value of the expression is undefined if the denominator is zero. The sumof two rational numbers is a rational number. Therefore the numerator anddenominator in the expression are both rational numbers. The quotient oftwo rational numbers is a rational number. Therefore, the entire expressionevaluates to a rational number for arbitratrary rational values of a, b, c andd.
11. We can expand and simplify the expression:
(sinα+ cosα)2 + (sinα− cosα)2
= sin2 α+ 2 sinα cosα+ cos2 α+ sin2 α− 2 sinα cosα+ cos2 α
= 2 sin2 α+ 2 cos2 α
= 2(sin2 α+ cos2 α)
= 2(1)
= 2
As the expression evaluates to a constant, it is as large as possible for allvalues of α.
Page 49
1. Rewriting any instances of secα or cscα on either side of the identities:
19
(a) tanα cscα = secα
tanα1
sinα=
1
cosα
tanα
sinα=
1
cosα
(b) cotα cscα = secα
cotα1
cosα=
1
sinα
cotα
cosα=
1
sinα
(c)1
secαcscα = cotα
11
cosα
· 1
sinα= cotα
cosα1
sinα= cotα
cosα
sinα= cotα
(d) tan2 α = (secα+ 1)(secα− 1)
tan2 α = sec2 α− 1
tan2 α =1
cos2 α− 1
(e) csc2 α = 1 + cot2 α
1
sin2 α= 1 + cot2 α
2. Rewriting any instances of sinα or cosα on either side of the identities, andeliminating fractions:
(a)tanα
sinα=
1
cosα
tanα1
sinα= secα
tanα cscα = secα
(b)1
sinαcosα = cotα
cosα
sinα= cotα
cotα = cotα
(c) tan2 α+ 1 =1
cos2 α
tan2 α+ 1 = sec2 α
20
(d)1
sin2 α= 1 + cot2 α
csc2 α = 1 + cot2 α
Page 50
1. First, we find the value of a2 + b2:
a2 + b2 = (cos2 α− sin2 α)2 + (2 sinα cosα)2
= cos4 α− 2 cos2 α sin2 α+ sin4 α+ 4 sin2 α cos2 α
= cos4 α+ 2 cos2 α sin2 α+ sin4 α
= (cos2 α+ sin2 α)2
= (1)2
= 1
According to the lemma on Page 50, as a2 + b2 = 1, an angle θ exists suchthat a = cos θ and b = sin θ.
2. First, we find the value of a2 + b2:
a2 + b2 =
(√1 + cosα
2
)2
+
(√1− cosα
2
)2
=1 + cosα
2+
1− cosα
2
=1 + cosα+ 1− cosα
2
=2
2= 1
3. First, we will rewrite a and b to eliminate the cube exponents:
a = 4 cos3 α− 3 cosα
= 4 cosα cos2 α− 3 cosα
= 4 cosα(1− sin2 α)− 3 cosα
= 4 cosα− 4 sin2 α cosα− 3 cosα
= cosα− 4 sin2 α cosα
b = 3 sinα− 4 sin3 α
= 3 sinα− 4 sinα sin2 α
= 3 sinα− 4 sinα(1− cos2 α)
= − sinα+ 4 sinα cos2 α
21
Next, we will expand a2 and b2:
a2 = (cosα− 4 sin2 α cosα)2
= cos2 α− 8 sin2 α cos2 α+ 16 sin4 α cos2 α
b2 = (− sinα+ 4 sinα cos2 α)2
= sin2 α− 8 sin2 α cos2 α+ 16 sin2 α cos4 α
Next, we add the expressions for a2 and b2 and simplify to 1:
a2 + b2 = cos2 α− 8 sin2 α cos2 α+ 16 sin4 α cos2 α+ sin2 α− 8 sin2 α cos2 α+
16 sin2 α cos4 α
= cos2 α+ sin2 α− 16 sin2 α cos2 α+ 16 sin4 α cos2 α+ 16 sin2 α cos4 α
= cos2 α+ sin2 α+ 16 sin2 α cos2 α(−1 + sin2 α+ cos2 α)
= 1 + 16 sin2 α cos2 α(0)
= 1
According to the lemma on Page 50, as a2 + b2 = 1, an angle θ exists suchthat a = cos θ and b = sin θ.
4. First, we find the value of a2 + b2:
a2 + b2 =
(1− t2
1 + t2
)2
+
(2t
1 + t2
)2
=(1− t2)2
(1 + t2)2+
(2t)2
(1 + t2)2
=(1− t2)2 + (2t)2
(1 + t2)2
=1− 2t2 + t4 + 4t2
(1 + t2)(1 + t2)
=(1 + t2)(1 + t2)
(1 + t2)(1 + t2)
= 1
According to the lemma on Page 50, as a2 + b2 = 1, an angle θ exists suchthat a = cos θ and b = sin θ.
5. We expand (p2 − q2)2 + (2pq)2 and use the fact that p2 + q2 = 1 to simplify
22
to 1:
(p2 − q2)2 + (2pq)2 = p4 − 2p2q2 + q4 + 4p2q2
= p4 + 2p2q2 + q4
= (p2 + q2)2
= (1)2
= 1
This is similar to Exercise 1 above.
Page 51
1. sinα < 1 when α is acute, therefore 1−sinα > 0 when α is acute. 1−sinα =0 when sinα = 1, i.e., α = 90◦.
2. cosα < 1 when α is acute, therefore 1−cosα > 0 when α is acute. 1−cosα =0 when cosα = 1, i.e., α = 0◦.
3. Statement a) is always true. Statements b) and c) both include the case thatsin2 α+ cos2 α = 1, which is always true.
4. Let x be the maximum cost of the items in a supermarket. In SupermarketA, x ≤ $1. In Supermarket B, x < $1. In Supermarket C, x ≤ $1. InSupermarket D, x > $1. We can see that Supermarkets A and C are offeringthe same terms.
5. Inequality a) is correct. For b) to be correct, an angle α would have toexist such that sinα + cosα = 2. We know that this is not the case. Whenα = 90◦, sinα = 1 and cosα = 0. When α = 0◦, sinα = 0 and cosα = 1.When 0◦ < α < 90◦, sinα < 1 and cosα < 1. In all cases, sinα+ cosα < 2.
6. The largest possible value of sinα is 1, and occurs when α = 90◦. The largestpossible value of cosα is 1, and occurs when α = 0◦. See Page 32.
Page 52
1. sin 30◦ = 0.5, sin 45◦ = 0.707, sin 60◦ = 0.866.
2. By using the tan button to calculate tan 60◦, and the sqrt button to calcu-late√
3, Betty can compare the results: both are 1.732.
3. Press tan, then enter the angle degree measure, then press 1/x
4.
23
in radical or rational form
α sinα cosα tanα cotα
30◦1
2
√3
2
1√3
√3
45◦1√2
1√2
1 1
60◦√
3
2
1
2
√3
1√3
in decimal form, from calculator
α sinα cosα tanα cotα
30◦ 0.5 0.866 0.577 1.732
45◦ 0.707 0.707 1 1
60◦ 0.866 0.5 1.732 0.577
Page 53
1. The sine of the larger angle is 4/5 = .8. We can use the inverse sine functionto find the angle: arcsin .8 = 53.1301◦. The sum of the three angles in thetriangles is: arcsin .6 + arcsin .8 + 90◦ = 36.8699◦ + 53.1301◦ + 90◦ = 180◦.
2. (a) arcsin 1 = 90◦
(b) arccos 0.7071067811865 = 45◦
3. arccos 0.8 = 36.8699◦
4. arcsin 0.6 = 36.8699◦
5. Half of sin 30◦ (0.25) seems like a reasonable estimate. The actual value is0.2588.
6.
7.
8.
9.
10.
24
11.
12.
13.
14.
Page 55
1.
2.
3.
4.
Page 56
1.
2.
3.
4.
Page 59
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
25
Page 62
1. The degree measure of a semicircle is 180◦. The degree measure of a quartercircle is 90◦.
2. The measure of arc cut off by one side of regular pentagon inscribed in acircle is 360◦/5 = 72◦. For a regular hexagon: 360◦/6 = 60◦. For a regularoctagon: 360◦/8 = 45◦.
26