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Last edited 11/13/14
7.1 Solutions to Exercises
1. Dividing both sides by 2, we have sin π = β1
2. Since sin π is negative only in quadrants III
and IV, using our knowledge of special angles, π =7π
6 or π =
11π
6.
3. Dividing both sides by 2, cos π =1
2. Using our knowledge of quadrants, this occurs in
quadrants I and IV. In quadrant I, π =π
3; in quadrant IV, π =
5π
3.
5. Start by dividing both sides by 2 to get sin (π
4π₯) =
1
2. We know that sin π =
1
2 for π =
π
6+
2ππ and π =5π
6+ ππ for any integer π. Therefore,
π
4π₯ =
π
6+ 2ππ and
π
4π₯ =
5π
6+ 2ππ. Solving
the first equation by multiplying both sides by 4
π (the reciprocal of
π
4) and distributing, we get
π₯ =4
6+ 8π, or π₯ =
2
3+ 8π. The second equation is solved in exactly the same way to arrive at
π₯ =10
3+ 8π.
7. Divide both sides by 2 to arrive at cos 2π‘ = ββ3
2. Since cos π = β
β3
2 when π =
5π
6+ 2ππ and
when π =7π
6+ 2ππ. Thus, 2π‘ =
5π
6+ 2ππ and 2π‘ =
7π
6+ 2ππ. Solving these equations for π‘
results in π‘ =5π
12+ ππ and π‘ =
7π
12+ ππ.
9. Divide both sides by 3; then, cos (π
5π₯) =
2
3. Since
2
3 is not the cosine of any special angle we
know, we must first determine the angles in the interval [0, 2π) that have a cosine of 2
3. Your
calculator will calculate cosβ1 (2
3) as approximately 0.8411. But remember that, by definition,
cosβ1 π will always have a value in the interval [0, π] -- and that there will be another angle in
(π, 2π) that has the same cosine value. In this case, 0.8411 is in quadrant I, so the other angle
must be in quadrant IV: 2π β 0.8411 β 5.4421. Therefore, π
5π₯ = 0.8411 + 2ππ and
π
5π₯ =
5.442 + 2ππ. Multiplying both sides of both equations by 5
π gives us π₯ = 1.3387 + 10π and
π₯ = 8.6612 + 10π.
Last edited 11/13/14
11. Divide both sides by 7: sin 3π‘ = β2
7. We need to know the values of π that give us sin π =
β2
7. Your calculator provides one answer: sinβ1 (β
2
7) β β0.2898. However, sinβ1 π has a
range of [βπ
2,
π
2], which only covers quadrants I and IV. There is another angle in the interval
(π
2,
3π
2) with the same sine value; in this case, in quadrant III: π + 0.2898 β 3.4314. Therefore,
3π‘ = β0.2898 + 2ππ and 3π‘ = 3.4314 + 2ππ. Dividing both sides of both equations by 3 gives
us π‘ = 1.1438 +2π
3π and π‘ = β0.0966 +
2π
3π.
13. Resist the urge to divide both sides by cos π₯ -- although you can do this, you then have to
separately consider the case where cos π₯ = 0. Instead, regroup all expressions onto one side of
the equation:
10 sin π₯ cos π₯ β 6 cos π₯ = 0
Now factor cos π₯:
cos π₯ (10 sin π₯ β 6) = 0
So either cos π₯ = 0 or 10 sin π₯ β 6 = 0. On the interval [0, 2π), cos π₯ = 0 at π₯ =π
2 and π₯ =
3π
2,
which provides us with two solutions. If 10 sin π₯ β 6 = 0, then 10 sin π₯ = 6 and sin π₯ =6
10.
Using a calculator or computer to calculate sinβ1 6
10 gives us approximately 0.644, which is in
quadrant I. We know there is another value for π₯ in the interval [0, 2π): in quadrant II at
π β 0.644 β 2.498. Our solutions are π
2,
3π
2, 0.644 and 2.498.
15. Add 9 to both sides to get csc 2π₯ = 9. If we rewrite this as 1
sin 2π₯= 9, we have 9 sin 2π₯ = 1
and sin 2π₯ =1
9. sin π =
1
9 at π β 0.1113 (the value from a calculator) and π β 3.0303 (using the
reference angle in quadrant II). Therefore, 2π₯ = 0.1113 + 2ππ and 2π₯ = 3.0303 + 2ππ.
Solving these equations gives us π₯ = 0.056 + ππ and π₯ = 1.515 + ππ for integral π. We choose
π = 0 and π = 1 for both equations to get four values: 0.056, 1.515, 3.198 and 4.657; these are
the only values that lie in the interval [0, 2π).
Last edited 11/13/14
17. Factoring sin π₯, we get sin π₯ (sec π₯ β 2) = 0. Therefore, either sin π₯ = 0 or sec π₯ β 2 = 0.
On the interval [0, 2π), sin π₯ = 0 at π₯ = 0 and π₯ = π, so these are our first two answers.
If sec π₯ β 2 = 0, then sec π₯ = 2 and 1
cos π₯= 2. This leads us to 2 cos π₯ = 1 and cos π₯ =
1
2.
Recognizing this as a well-known angle, we conclude that (again, on the interval [0, 2π)), π₯ =π
3
and π₯ =5π
3.
19. If sin2(π₯) =1
4, then sin π₯ = Β±
1
2. On the interval [0, 2π), this occurs at π₯ =
π
6, π₯ =
5π
6,
π₯ =7π
6 and π₯ =
11π
6.
21. If sec2 π₯ = 7, then sec π₯ = Β±β7, 1
cos π₯= Β±β7, and cos π₯ = Β±
1
β7= Β±
β7
7.
Using a calculator for cosβ1 (β7
7), we get π₯ β 1.183. There is another angle on the interval
[0, 2π) whose cosine is 1
7, in quadrant IV: π₯ = 2π β 1.183 β 5.1. The two angles where
cos π₯ = ββ7
7 must lie in quadrants III and IV at π₯ = π β 1.183 β 1.959 and π₯ = π + 1.183 β
4.325.
23. This is quadratic in sin π€: think of it as 2π₯2 + 3π₯ + 1 = 0, where π₯ = sin π€. This is simple
enough to factor:
2π₯2 + 3π₯ + 1 = (2π₯ + 1)(π₯ + 1) = 0
This means that either 2π₯ + 1 = 0 and π₯ = β1
2, or π₯ + 1 = 0 and π₯ = β1. Therefore, either
sin π€ = β1
2 or sin π€ = β1. We know these special angles: these occur on the interval [0, 2π)
when π€ =7π
6 or π€ =
11π
6 (for sin π€ = β
1
2) or when π€ =
3π
2 (for sin π€ = β1).
25. If we subtract 1 from both sides, we can see that this is quadratic in cos π‘:
2(cos2 π‘ + cos π‘ β 1 = 0
Last edited 11/13/14
If we let π₯ = cos π‘, we have:
2π₯2 + π₯ β 1 = (2π₯ β 1)(π₯ + 1) = 0
Either 2π₯ β 1 = 0 and π₯ =1
2, or π₯ + 1 = 0 and π₯ = β1. Therefore, cos π‘ =
1
2 or cos π‘ = β1. On
the interval [0, 2π), these are true when π‘ =π
3 or π‘ =
5π
3 (for cos π‘ =
1
2) or when π‘ = π (for
cos π‘ = β1).
27. If we rearrange the equation, it is quadratic in cos π₯:
4 cos2 π₯ β 15 cos π₯ β 4 = 0
If we let π’ = cos π₯, we can write this as:
4π’2 β 15π’ β 4 = 0
This factors as:
(4π’ + 1)(π’ β 4) = 0
Therefore, either 4π’ + 1 = 0 and π’ = β1
4, or π’ β 4 = 0 and π’ = 4. Substituting back, we have:
cos π₯ = β1
4
We reject the other possibility that cos π₯ = 4 since cos π₯ is always in the interval [β1, 1].
Your calculator will tell you that cosβ1 (β1
4) β 1.823. This is in quadrant II, and the cosine is
negative, so the other value must lie in quadrant III. The reference angle is π β 1.823, so the
other angle is at π + (π β 1.823) = 2π β 1.823 β 4.460.
29. If we substitute 1 β cos2 π‘ for sin2 π‘, we can see that this is quadratic in cos π‘:
12 sin2 π‘ + cos π‘ β 6 = 12(1 β cos2 π‘) + cos π‘ β 6 = β12 cos2 π‘ + cos π‘ + 6 = 0
Last edited 11/13/14
Setting π’ = cos π‘:
β12π’2 + π’ + 6 = (β4π’ + 3)(3π’ + 2) = 0
This leads us to β4π’ + 3 = 0 or 3π’ + 2 = 0, so either π’ =3
4 or π’ = β
2
3.
Substituting back, cos π‘ =3
4 gives us (via a calculator) π‘ β 0.7227. This is in quadrant I, so the
corresponding angle must lie in quadrant IV at π‘ = 2π β 0.7227 β 5.5605.
Similarly, cos π‘ = β2
3 gives us π‘ β 2.3005. This is in quadrant II; the corresponding angle with
the same cosine value must be in quadrant III at π‘ = 2π β 2.3005 β 3.9827.
31. Substitute 1 β sin2 π for cos2 π:
1 β sin2 π = β6 sin π
β sin2 π + 6 sin π + 1 = 0
This is quadratic in sin π, so set π’ = sin π and we have:
βπ’2 + 6π’ + 1 = 0
This does not factor easily, but the quadratic equation gives us:
π’ =β6 Β± β36 β 4(β1)(1)
2(β1)=
β6 Β± β40
β2=
β6 Β± 2β10
β2= 3 Β± β10
Thus, π’ β 6.1623 and π’ β β0.1623. Substituting back, we have sin π = β0.1623. We reject
sin π = 6.1623 since sin π is always between -1 and 1. Using a calculator to calculate
sinβ1(β0.1623), we get π β β.1630. Unfortunately, this is not in the required interval [0, 2π),
so we add 2π to get π β 6.1202. This is in quadrant IV; the corresponding angle with the same
sine value must be in quadrant III at π + 0.1630 β 3.3046.
33. If we immediately substitute π£ = tan π₯, we can write:
Last edited 11/13/14
π£3 = 3π£
π£3 β 3π£ = 0
π£(π£2 β 3) = 0
Thus, either π£ = 0 or π£2 β 3 = 0, meaning π£2 = 3 and π£ = Β±β3.
Substituting back, tan π₯ = 0 at π₯ = 0 and π₯ = π. Similarly, tan π₯ = β3 at π₯ =π
3 and π₯ =
4π
3 and
tan π₯ = ββ3 at π₯ =2π
3 and π₯ =
5π
3.
35. Substitute π£ = tan π₯ so that:
π£5 = π£
π£5 β π£ = 0
π£(π£4 β 1) = 0
Either π£ = 0 or π£4 β 1 = 0 and π£4 = 1 and π£ = Β±1. Substituting back, tan π₯ = 0 at π₯ = 0 and
π₯ = π. Similarly, tan π₯ = 1 at π₯ =π
4 and π₯ =
5π
4. Finally, tan π₯ = β1 for π₯ =
3π
4 and π₯ =
7π
4.
37. The structure of the equation is not immediately apparent.
Substitute π’ = sin π₯ and π£ = cos π₯, and we have:
4π’π£ + 2π’ β 2π£ β 1 = 0
The structure is now reminiscent of the result of multiplying two binomials in different variables.
For example, (π₯ + 1)(π¦ + 1) = π₯π¦ + π₯ + π¦ + 1. In fact, our equation factors as:
(2π’ β 1)(2π£ + 1) = 0
Last edited 11/13/14
Therefore, either 2π’ β 1 = 0 (and π’ =1
2) or 2π£ + 1 = 0 (and π£ = β
1
2). Substituting back,
sin π₯ =1
2 or cos π₯ = β
1
2. This leads to π₯ =
π
6, π₯ =
5π
6 (for sin π₯ =
1
2) and π₯ =
2π
3, π₯ =
4π
3 (for
cos π₯ = β1
2
39. Rewrite tan π₯ as sin π₯
cos π₯ to give:
sin π₯
cos π₯β 3 sin π₯ = 0
Using a common denominator of cos π₯, we have:
sin π₯
cos π₯β
3 sin π₯ cos π₯
cos π₯= 0
and
sin π₯ β 3 sin π₯ cos π₯
cos π₯= 0
sin π₯ β 3 sin π₯ cos π₯ = 0
sin π₯ (1 β 3 cos π₯) = 0
Therefore, either sin π₯ = 0 or 1 β 3 cos π₯ = 0, which means cos π₯ =1
3.
For sin π₯ = 0, we have π₯ = 0 and π₯ = π on the interval [0, 2π). For cos π₯ =1
3, we need
cosβ1 (1
3), which a calculator will indicate is approximately 1.231. This is in quadrant I, so the
corresponding angle with a cosine of 1
3 is in quadrant IV at 2π β 1.231 β 5.052.
41. Rewrite both tan π‘ and sec π‘ in terms of sin π‘ and cos π‘:
2sin2 π‘
cos2 π‘= 3
1
cos π‘
Last edited 11/13/14
We can multiply both sides by cos2 π‘:
2 sin2 π‘ = 3 cos π‘
Now, substitute 1 β cos2 π‘ for sin2 π‘ to yield:
2(1 β cos2 π‘) β 3 cos π‘ = 0
This is beginning to look quadratic in cos π‘. Distributing and rearranging, we get:
β2 cos2 π‘ β 3 cos π‘ + 2 = 0
Substitute π’ = cos π‘:
β2π’2 β 3π’ + 2 = 0
(β2π’ + 1)(π’ + 2) = 0
Therefore, either β2π’ + 1 = 0 (and π’ =1
2) or π’ + 2 = 0 (and π’ = β2). Since π’ = cos π‘ will
never have a value of -2, we reject the second solution.
cos π‘ =1
2 at π‘ =
π
3 and π‘ =
5π
3 on the interval [0, 2π).
7.2 Solutions to Exercises
1. sin(75Β°) = sin(45Β°+30Β°) = sin(45Β°)cos(30Β°)+cos(45Β°)sin(30Β°) = β2
2β
β3
2+
β2
2β
1
2=
β6+β2
4
3. cos(165Β°) = cos(120Β° + 45Β°) = cos(120Β°)cos(45Β°)-sin(120Β°)sin(45Β°) = β(β6+β2)
4
5. cos (7π
12) = cos (
π
4+
π
3) = cos (
π
4) cos (
π
3) β sin (
π
4) sin (
π
3) =
β2ββ6
4
Last edited 11/13/14
7. sin (5π
12) = sin (
π
6+
π
4) = sin (
π
6) cos (
π
4) + cos (
π
6) sin (
π
4) =
β6+β2
4
9. sin (π₯ +11π
6) = sin(π₯) cos (
11π
6) + cos(π₯) sin (
11π
6) =
β3
2sin(π₯) β
1
2cos(π₯)
11. cos (π₯ β5π
6) = cos(π₯) cos (
5π
6) + sin(π₯) sin (
5π
6) = β
β3
2cos(π₯) +
1
2sin(π₯)
13. csc (π
2β π‘) =
1
sin(π
2βπ‘)
=1
sin(π
2) cos(π‘)βcos(
π
2) sin(π‘)
=1
cos(π‘)= sec(π‘)
15. cot (π
2β π₯) =
cos(π
2) cos(π₯)+sin(
π
2)sin (π₯)
sin(π
2) cos(π‘)βcos(
π
2) sin(π‘)
=sin(π₯)
cos(π₯)= tan(π₯)
17. 16 sin(16π₯) sin(11π₯) = 16 β1
2(cos(16π₯ β 11π₯) β cos(16π₯ + 11π₯)) = 8 cos(5π₯) β
8 cos(27π₯)
19. 2 sin(5π₯) cos(3π₯) = sin(5π₯ + 3π₯) + sin(5π₯ β 3π₯) = sin(8π₯) + sin(2π₯)
21. cos(6π‘) + cos(4π‘) = 2 cos (6π‘+4π‘
2) cos (
6π‘β4π‘
2) = 2 cos(5π‘) cos(π‘)
23. sin(3π₯) + sin(7π₯) = 2 sin (3π₯+7π₯
2) cos (
3π₯β7π₯
2) = 2 sin(5π₯) cos (β2π₯)
25. We know that sin(π) =2
3 and cos(π) = β
1
4 and that the angles are in quadrant II. We can
find cos (π) and sin (π) using the Pythagorean identity sin2(π) + cos2(π) = 1, or by using the
known values of sin(π) and cos(π) to draw right triangles. Using the latter method: we know
two sides of both a right triangle including angle a and a right triangle including angle b. The
triangle including angle a has a hypotenuse of 3 and an opposite side of 2. We may use the
pythagorean theorem to find the side adjacent to angle a. Using the same method we may find
the side opposite to b.
For the triangle containing angle a:
Adjacent = β32 β 22 = β5
However, this side lies in quadrant II, so it will be ββ5.
Last edited 11/13/14
For the triangle containing angle b:
Opposite = β42 β 12 = β15
In quadrant II, y is positive, so we do not need to change the sign.
From this, we know: sin(π) =2
3, cos(π) = β
β5
3, cos(π) = β
1
4, sin(π) =
β15
4.
a. sin(π + π) = sin(π) cos(π) + cos(π) sin(π) =β2β5β3
12
b. cos(π β π) = cos(π) cos(π) + sin(π) sin(π) =β5+2β15
12
27. sin(3π₯) cos(6π₯) β cos(3π₯) sin(6π₯) = β0.9
sin(3π₯ β 6π₯) = β0.9
sin(β3π₯) = β0.9
βsin(3π₯) = β0.9
3π₯ = sinβ1(0.9) + 2ππ or 3π₯ = π β sinβ1(0.9) + 2ππ, where k is an integer
π₯ =sinβ1(0.9)+2ππ
3 or π₯ =
πβsinβ1(0.9)+2ππ
3
π₯ β 0.373 +2π
3π or π₯ β 0.674 +
2π
3π, where k is an integer
29. cos(2π₯) cos(π₯) + sin(2π₯) sin(π₯) = 1
cos(2π₯ β π₯) = 1
π₯ = 0 + 2ππ, where k is an integer
31. cos(5π₯) = βcos (2π₯)
cos(5π₯) + cos(2π₯) = 0
2 cos (5π₯+2π₯
2) cos (
5π₯β2π₯
2) = 0
cos (7π₯
2) = 0 or cos (
3π₯
2) = 0
7π₯
2=
π
2+ ππ or
3π₯
2=
π
2+ ππ, where k is an integer
π₯ =π+2ππ
7 or π₯ =
π+2ππ
3
Last edited 11/13/14
33. cos(6π) β cos(2π) = sin(4π)
β2 sin (6π+2π
2) sin (
6πβ2π
2) = sin(4π)
β2 sin(4π) sin(2π) β sin(4π) = 0
sin(4π) (β2 sin(2π) β 1) = 0
sin(4π) = 0 or β2 sin(2π) β 1 = 0
sin(4π) = 0 or sin(2π) = β1
2
4π = ππ or 2π =7π
6+ 2ππ or
11π
6+ 2ππ
π =ππ
4 or
7π
12+ ππ or
11π
12+ ππ
35. π΄ = β42 + 62 = 2β13
cos(π) =2
β13, sin(π) = β
3
β13
Since sin(C) is negative but cos(C) is positive, we know that C is in quadrant IV.
πΆ = sinβ1 (β3
β13)
Therefore the expression can be written as 2β13 sin (π₯ + sinβ1 (β3
β13)) or approximately
2β13 sin(π₯ β 0.9828).
37. π΄ = β52 + 22 = β29
cos(πΆ) =5
β29, sin(πΆ) =
2
β29
Since both sin(C) and cos(C) are positive, we know that C is in quadrant I.
πΆ = sinβ1 (2
β29)
Therefore the expression can be written as β29 sin (3π₯ + sinβ1 (2
β29)) or approximately
β29 sin(3π₯ + 0.3805).
39. This will be easier to solve if we combine the 2 trig terms into one sinusoidal function of the
form π΄sin(π΅π₯ + πΆ).
Last edited 11/13/14
π΄ = β52 + 32 = β34, cos(πΆ) = β5
β34, sin(πΆ) =
3
β34
C is in quadrant II, so πΆ = π β sinβ1 (3
β34)
Then:
β34 sin (π₯ + π β sinβ1 (3
β34)) = 1
β sin (π₯ β sinβ1 (3
β34)) =
1
β34 (Since sin(π₯ + π) = βsin (π₯))
π₯ β sinβ1 (3
β34) = sinβ1 (β
1
β34) or, to get the second solution π₯ β sinβ1 (
3
β34) = π β
sinβ1 (β1
β34)
π₯ β 0.3681 or π₯ β 3.8544 are the first two solutions.
41. This will be easier to solve if we combine the 2 trig terms into one sinusoidal function of the
form π΄sin(π΅π₯ + πΆ).
π΄ = β52 + 32 = β34, cos(πΆ) =3
β34, sin(πΆ) = β
5
β34
C is in quadrant IV, so πΆ = sinβ1 (β5
β34).
Then:
β34 sin (2π₯ + sinβ1 (β5
β34)) = 3
sin (2π₯ + sinβ1 (β5
β34)) =
3
β34
2π₯ + sinβ1 (β5
β34) = sinβ1 (
3
β34) and 2π₯ + sinβ1 (β
5
β34) = π β sinβ1 (
3
β34)
π₯ =
sinβ1(3
β34)βsinβ1(β
5
β34)
2 or
πβsinβ1(3
β34)βsinβ1(β
5
β34)
2.
43. sin(7π‘)+sin (5π‘)
cos(7π‘)+cos(5π‘)=
2 sin(7π‘+5π‘
2) cos(
7π‘β5π‘
2)
2 cos(7π‘+5π‘
2) cos(
7π‘β5π‘
2)
=2 sin(6π‘)cos (π‘)
2 cos(6π‘)cos (π‘) = tan(6π‘)
Last edited 11/13/14
45. tan (π
4β π‘) =
sin(π
4βπ‘)
cos(π
4βπ‘)
=sin(
π
4) cos(π‘)βcos(
π
4) sin(π‘)
cos(π
4) cos(π‘)+sin(
π
4) sin(π‘)
=β2
2(cos(π‘)βsin(π‘))
β2
2(cos(π‘)+sin(π‘))
=(cos(π‘))(1β
sin(π‘)
cos(π‘))
(cos(π‘))(1+sin(π‘)
cos(π‘))
=1βtan(π‘)
1+tan(π‘)
47. cos(π+π)
cos(πβπ)=
cos(π) cos(π)βsin(π) sin(π)
cos(π) cos(π)+sin(π) sin(π)
=cos(π) cos(π)(1β
sin(π) sin(π)
cos(π) cos(π))
cos(π) cos(π)(1+sin(π) sin(π)
cos(π) cos(π))
=1βtan(π) tan(π)
1+tan(π) tan(π)
49. Using the Product-to-Sum identity:
2 sin(π + π) sin(π β π) = 2 (1
2) cos((π + π) β (π β π)) β cos((π + π) + (π β π))
= cos(2π) β cos(2π)
51. cos(π+π)
cos(π)cos (π)=
cos(π) cos(π)βsin(π) sin(π)
cos(π) cos(π)
=cos(π) cos(π)
cos(π) cos(π)β
sin(π) sin(π)
cos(π) cos(π)
= 1 β tan(π) tan(π)
Last edited 11/13/14
7.3 Solutions to Exercises
1. a. sin(2π₯) = 2 sin π₯ cos π₯
To find cos π₯:
sin2 π₯ + cos2 π₯ = 1
cos2 π₯ = 1 β1
64=
63
64
cos π₯ = Β±β63
64 Note that we need the positive root since we are told π₯ is in quadrant 1.
cos π₯ =3β7
8
So: sin(2π₯) = 2 (1
8) (
3β7
8) =
3β7
32
b. cos(2π₯) = 2 cos2 π₯ β 1
= (2 (63
64)) β 1 =
63β32
32=
31
32
c. tan(2π₯) =sin(2π₯)
cos(2π₯)=
3β7
32
31
32β =
3β7
31
3. cos2 π₯ β sin2 π₯ = cos(2π₯), so cos2(28Β°) β sin2(28Β°) = cos(56Β°)
5. 1 β 2 sin2(π₯) = cos(2π₯), so 1 β 2sin2(17Β°) = cos(2 (17Β°)) = cos(34Β°)
7. cos2(9π₯) β sin2(9π₯) = cos(2(9π₯)) = cos(18π₯)
9. 4 sin(8π₯) cos(8π₯) = 2 (2sin(8π₯) cos(8π₯)) = 2 sin(16π₯)
11. 6 sin(2π‘) + 9 sin π‘ = 6 β 2 sin π‘ cos π‘ + 9 sin π‘ = 3 sin π‘ (4 cos π‘ + 3), so we can solve
3 sin π‘ (4 cos π‘ + 3) = 0:
sin π‘ = 0 or cos π‘ = β3/4
π‘ = 0, π or π‘ β 2.4186, 3.8643.
13. 9cos (2π) = 9 cos2 π β 4
9(cos2 π- sin2π)
= 9 cos2 π β 4
Last edited 11/13/14
9 sin2 π β 4 = 0
(3 sin π β 2)(3 sin π + 2) = 0
sin π =2
3,-
2
3
π = sinβ1 2
3, sinβ1 β2
3
π β 0.7297, 2.4119, 3.8713, 5.5535
15. sin(2π‘) = cos π‘
2 sin π‘ cos π‘ = cos π‘
2 sin π‘ cos π‘ β cos π‘ = 0
cos π‘ (2 sin π‘ β 1) = 0
cos π‘ = 0 or 2 sin π‘ β 1 = 0
If cos π‘ = 0, then π‘ =π
2 or
3π
2. If 2 sin π‘ β 1 = 0, then sin π‘ =
1
2, so π‘ =
π
6 or
5π
6. So π‘ =
π
2,
3π
2,
π
6
or 5π
6.
17. cos(6π₯) β cos(3π₯) = 0
2 cos2(3π₯) β 1 β cos(3π₯) = 0
2 cos2(3π₯) β cos(3π₯) β 1 = 0
(2 cos(3π₯) + 1)(cos(3π₯) β 1) = 0
cos(3π₯) = β1
2 or 1
Since we need solutions for π₯ in the interval [0, 2π), we will look for all solutions for 3π₯ in the
interval [0, 6π). If cos(3π₯) = β1
2, then there are two possible sets of solutions. First, 3π₯ =
2π
3+ 2ππ where π = 0, 1, or 2, so π₯ =
2π
9+
2ππ
3 where π = 0, 1, or 2. Second, 3π₯ =
4π
3+
2ππ where π = 0, 1, or 2, so π₯ =4π
9+
2ππ
3 where π = 0, 1, or 2 If cos(3π₯) = 1, then 3π₯ =
2ππ where π = 0, 1, or 2, so π₯ =2ππ
3 where π = 0, 1, or 2.
19. cos2(5π₯) =cos(10π₯)+1
2 because cos2 π₯ =
cos(2π₯)+1
2 (power reduction identity)
Last edited 11/13/14
21. sin4(8π₯) = sin2(8π₯) β sin2(8π₯)
= (1βcos(16π₯))
2 .
(1βcos(16π₯))
2 (because power reduction identity sin
2 x =
(1βcos(2π₯))
2
= 1β2 cos(16π₯)
4+
cos2(16π₯)
4
= 1β2 cos(16π₯)
4+
cos(32π₯)+1
8 because cos2 π₯ =
cos(2π₯)+1
2 (power reduction identity)
= 1
4 β
cos (16π₯)
2 +
cos (32π₯)
8 +
1
8
23. cos2 π₯ sin4 π₯
= cos2 π₯ β sin2 π₯ β sin2 π₯
= 1+cos(2π₯)
2β
1βcos (2π₯)
2β
1βcos(2π₯))
2
= 1βcos2(2π₯)
4β
1βcos(2π₯)
2
= 1β
cos(4π₯)+1
2
4β
1βcos (2π₯)
2
=1βcos(4π₯)
8β
1βcos (2π₯)
2
=1βcos(2π₯)βcos(4π₯)+cos(2π₯)cos (4π₯)
16
25. Since csc π₯ = 7 and π₯ is in quadrant 2, sin π₯ =1
7 (reciprocal of cosecant) and cos π₯ = β
4β3
7
(Pythagorean identity).
a. sinπ₯
2= β
(1βcos (π₯))
2 =β7+4β3
14 (Note that the answer is positive because π₯ is in quadrant 2, so
π₯
2
is in quadrant 1.)
b. cos(π₯
2) =β
(cos π₯+1)
2 =β7β4β3
14 (Note that the answer is positive because π₯ is in quadrant 2, so
π₯
2
is in quadrant 1.)
c. tan (π₯
2)=
sin(π₯
2)
cos(π₯
2)
= β7+4β3
7β4β3= β (7+4β3)
2
(7β4β3)(7+4β3)= β(7+4β3)
2
1= 7 + 4β3
27. (sin π‘ β cos π‘)2 = 1 β sin(2π‘)
Left side: (sin2 π‘ β 2 sin π‘ cos π‘ + cos2 π‘)
Last edited 11/13/14
= 1 β 2 sin π‘ cos π‘ (because (sin2 π‘ + cos2 π‘ = 1)
= 1 β sin(2π‘), the right side (because sin(2π‘) = 2 sin π‘ cos π‘)
29. sin(2π₯) =2tan (π₯)
1+tan2(π₯)
The right side:
2sin (π₯)
cos(π₯)
1+sin2 π₯
cos2 π₯
βcos2(π₯)
cos2(π₯)=
2 sin(π₯) cos(π₯)
cos2(π₯)+sin2(π₯) = 2 sin π₯ cos π₯ = sin(2π₯), the left side.
31. cot π₯ β tan π₯ = 2 cot(2π₯)
The left side: cos (π₯)
sin (π₯) β
sin (π₯)
cos (π₯)
= cos2 π₯βsin2 π₯
sin(π₯) cos(π₯) =
cos(2π₯)sin(2π₯)
2
=2 cot(2π₯)
33. cos(2πΌ) =1βtan2(πΌ)
1+tan2(πΌ)
The left side: 1βtan2(πΌ)
1+tan2(πΌ) =
cos2 πΌβsin2 πΌ
cos2 πΌcos2 πΌ+sin2 πΌ
cos2 πΌ
=cos(2πΌ)
cos2 πΌ1
cos2 πΌ
= cos(2πΌ)
35. sin(3π₯) = 3sin(x) cos2 π₯ β sin3 π₯
Left side: sin(π₯ + 2π₯) = sin(π₯) cos(2π₯) + cos(π₯) sin(2π₯) addition rule.
= sin(π₯) (cos2(π₯) β sin2(π₯)) + cos(π₯) (2 sin(π₯) cos(π₯))
= cos2(π₯) sin(x) β sin3(π₯) + 2 cos2(π₯) sin(π₯)
= 3 cos2(π₯) sin(x) β sin3(π₯)
7.4 Solutions to Exercises
1. By analysis, the function has a period of 12 units. The frequency is 1/12 Hz. The average of
the y-values from 0 β€ x < 12 is -1, and since the terms repeat identically there is no change in the
midline over time. Therefore the midline is y = f(x) = -1. The high point (y = 2) and low point (y
= -4) are both 3 units away from the midline. Therefore, amplitude = 3 units. The function also
starts at a minimum, which means that its phase must be shifted by one quarter of a cycle, or 3
Last edited 11/13/14
units, to the right. Therefore, phase shift = 3.
Now insert known values into the function:
π¦ = π΄sin ((2π
period) (π₯ β (phase shift))) + midline
π¦ = 3sin (2π
12(π₯ β 3)) β 1
This can be reduced to π¦ = 3sin (π
6(π₯ β 3)) β 1
Alternatively, had we chosen to use the cosine function: π¦ = β3 cos (π
6π₯) β 1.
3. By analysis of the function, we determine:
A = amplitude = 8 units
Solving 2π
period= 6π, we get: period = 1/3 seconds
Frequency = 3 Hz
5. In this problem, it is assumed that population increases linearly. Using the starting average as
well as the given rate, the average population is then y(x) = 650 + (160/12)x = 650 + (40/3)x ,
where x is measured as the number of months since January.
Based on the problem statement, we know that the period of the function must be twelve months
with an amplitude of 19. Since the function starts at a low-point, we can model it with a cosine
function since -[cos(0)] = -1
Since the period is twelve months, the factor inside the cosine operator is equal to 2π
12=
π
6. Thus,
the cosine function is β19 cos (π
6π₯).
Therefore, our equation is: π¦ = π(π₯) = 650 +40
3π₯ β 19 cos (
π
6π₯)
7. By analysis of the problem statement, the amplitude of the sinusoidal component is 33 units
with a period of 12 months. Since the sinusoidal component starts at a minimum, its phase must
be shifted by one quarter of a cycle, or 3 months, to the right.
π¦ = π(π₯) = 33 sin (π
6(π₯ β 3))
Last edited 11/13/14
Using the starting average as well as the given rate, the average population is then:
π¦ = π(π₯) = 900(1.07)π₯
π(π₯) + π(π₯) = 33 sin (π
6(π₯ β 3)) + 900(1.07)π₯
Alternatively, if we had used the cosine function, weβd get:
β(π₯) = β33 cos (π
6π₯) + 900(1.07)π₯
9. The frequency is 18Hz, therefore period is 1/18 seconds. Starting amplitude is 10 cm. Since
the amplitude decreases with time, the sinusoidal component must be multiplied by an
exponential function. In this case, the amplitude decreases by 15% every second, so each new
amplitude is 85% of the prior amplitude. Therefore, our equation is π¦ = π(π₯) = 10cos (36ππ₯) β
(0.85)π₯.
11. The initial amplitude is 17 cm. Frequency is 14 Hz, therefore period is 1/14 seconds.
For this spring system, we will assume an exponential model with a sinusoidal factor.
The general equation looks something like this: π·(π‘) = π΄(π )π‘ β cos (π΅π‘) where A is amplitude, R
determines how quickly the oscillation decays, and B determines how quickly the system
oscillates. Since π·(0) = 17, we know π΄ = 17. Also, π΅ =2π
period= 28.
We know π·(3) = 13, so, plugging in:
13 = 17(π )3 cos(28π β 3)
13 = 17(π )3 β 1
π 3 =13
17
π = β13
17
3
β 0.9145
Thus, the solution is π·(π‘) = 17(0.9145)π‘ cos(28ππ‘).
Last edited 11/13/14
13. By analysis:
(a) must have constant amplitude with exponential growth, therefore the correct graph is IV.
(b) must have constant amplitude with linear growth, therefore the correct graph is III.
15. Since the period of this function is 4, and values of a sine function are on its midline at the
endpoints and center of the period, π(0) and π(2) are both points on the midline. Weβll start by
looking at our function at these points:
At π(0), plugging into the general form of the equation, 6 = ππ0 + πsin(0), so π = 6.
At π(2): 96 = 6π2 + πsin(π), so π = 4.
At π(1): 29 = 6(4)1 + πsin (π
2). Solving gives π = 5.
This gives a solution of π¦ = 6 β 4π₯ + 5 sin (π
2π₯).
17. Since the period of this function is 4, and values of a sine function are on its midline at the
endpoints and center of the period, π(0) and π(2) are both points on the midline.
At π(0), plugging in gives 7 = πsin(0) + π + π β 0, so π = 7.
At π(2): 11 = πsin(π) + 7 + 2π. Since sin(π) = 0, we get π = 2.
At π(1): 6 = πsin (π
2) + 7 + 2 β 1. Simplifying, π = β3.
This gives an equation of π¦ = β3sin (π
2π₯) + 2π₯ + 7.
19. Since the first two places cos(π) = 0 are when π =π
2 or
3π
2, which for cos (
π
2π₯) occur when
π₯ = 1 or π₯ = 3, weβll start by looking at the function at these points:
At π(1), plugging in gives 3 = ππ1 cos (π
2) + π. Since cos (
π
2) = 0, π = 3. (Note that looking
at π(3) would give the same result.)
At π(0): 11 = ππ0 cos(0) + 3. Simplifying, we see π = 8.
At π(2): 1 = 8π2 cos(π) + 3. Since cos(π) = β1, it follows that π =1
2:
1 = β8π2 + 3
β8π2 = β2
π2 =1
4
Last edited 11/13/14
π = Β±1
2, but since we require exponential expressions to have a positive number as the base,
π =1
2. Therefore, the final equation is: π¦ = 8 (
1
2)
π₯
cos (π
2π₯) + 3.
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