0

Kunci Jawaban dan Pembahasan Soal || Buku Calculus 9th Purcell Chapter 0 - 0.7 Number 1 – 31

Pembahasan Soal Buku Calculus 9th Purcell Chapter 0 Section 0.7

CHAPTER 0 PRELIMINARIES

SECTION 0.7 Trigonometric Functions


Calculus 9th Purcell Chapter 0 - 0.7

Problem Set 0.7, Number 1 – 31.
  1. Convert the following degree measures to radians (leave π in your answer).
    2. (a)
    30°
    (b)
    45°
    (c)
    –60°
    (d)
    240°
    (e)
    –370°
    (f)
    10°
    💬 SOLUTION:
    (a)
     
    30
    (
    π
    )
    =
    π
    180
    6
    (b)
     
    45
    (
    π
    )
    =
    π
    180
    4
    (c)
     
    –60
    (
    π
    )
    =
    –
    π
    180
    3
    (d)
     
    240
    (
    π
    )
    =
    4π
    180
    3
    (e)
     
    –370
    (
    π
    )
    =
    –
    37π
    180
    180
    (f)
     
    10
    (
    π
    )
    =
    π
    180
    18

  2. Convert the following radian measures to degrees.
    3. (a)
    (7/6) π
    (b)
    (3/4) π
    (c)
    –(1/3) π
    (d)
    (4/3) π
    (e)
    –(35/18) π
    (f)
    (3/18) π
    💬 SOLUTION:
    (a)
     
    7
    π
    (
    180
    )
    =
    210°
    6
    π
    (b)
     
    3
    π
    (
    180
    )
    =
    135°
    4
    π
    (c)
     
    –
    1
    π
    (
    180
    )
    =
    –60°
    3
    π
    (d)
     
    4
    π
    (
    180
    )
    =
    240°
    3
    π
    (e)
     
    –
    35
    π
    (
    180
    )
    =
    –350°
    18
    π
    (f)
     
    3
    π
    (
    180
    )
    =
    30°
    18
    π

  3. Convert the following degree measures to radians (1° = π/180 ≈ 1.7453 × 10–2 radian).
    4. (a)
    33.3°
    (b)
    46°
    (c)
    –66.6°
    (d)
    240.11°
    (e)
    –369°
    (f)
    11°
    💬 SOLUTION:
    (a)
     
    33.3
    (
    π
    )
    ≈
    0.5812
    180
    (b)
     
    46
    (
    π
    )
    ≈
    0.8029
    180
    (c)
     
    –66.6
    (
    π
    )
    ≈
    –1.1624
    180
    (d)
     
    240.11
    (
    π
    )
    ≈
    4.1907
    180
    (e)
     
    –369
    (
    π
    )
    ≈
    –6.4403
    180
    (f)
     
    11
    (
    π
    )
    ≈
    0.1920
    180

  4. Convert the following radian measures to degrees (1 radian = 180/π ≈ 57.296 degrees).
    5. (a)
    3.141
    (b)
    6.28
    (c)
    5.00
    (d)
    0.001
    (e)
    –0.1
    (f)
    36.0
    💬 SOLUTION:
    (a)
     
    3.141
    (
    180
    )
    ≈
    180°
    π
    (b)
     
    6.28
    (
    180
    )
    ≈
    359.8°
    π
    (c)
     
    5.00
    (
    180
    )
    ≈
    286.5°
    π
    (d)
     
    0.001
    (
    180
    )
    ≈
    0.057°
    π
    (e)
     
    –0.1
    (
    180
    )
    ≈
    –5.73°
    π
    (f)
     
    36.0
    (
    180
    )
    ≈
    2062.6°
    π

  5. Calculate (be sure that your calculator is in radian or degree mode as neede).
    6. (a)
     
     
    56.4 tan 34.2°
    sin 34.1°
    (b)
     
     
    5.34 tan 21.3°
    sin 3.1° + cot 23.5°
    (c)
    tan 0.452
    (d)
    sin (–0.361)
    💬 SOLUTION:
    (a)
     
    56.4 tan 34.2°
    ≈
    68.37
    sin 34.1°
    (b)
     
    5.34 tan 21.3°
    ≈
    0.8845
    sin 3.1° + cot 23.5°
    (c)
    tan (0.452) ≈ 0.4855
    (d)
    sin (–0.361) ≈ –0.3532

  6. Calculate.
    7. (a)
     
     
    234.1 sin 1.56
    cos 0.34
    (b)
    sin2 2.51 + √(cos 0.51)
    💬 SOLUTION:
    (a)
     
     
    234.1 sin (1.56)
    ≈
    248.3
    cos (0.34)
    (b)
     
     
    sin2 2.51 + √(cos 0.51) ≈ 1.2828

  7. Calculate.
    8. (a)
     
     
    56.3 tan 34.2°
    sin 56.1°
    (b)
     
     
    (
    sin 35°
    )3
    sin 26° + cos 26°
    💬 SOLUTION:
    (a)
     
     
    56.3 tan 34.2°
    ≈
    46.097
    sin 56.1°
    (b)
     
     
    (
    sin 35°
    )3
    ≈
    0.0789
    sin 26° + cos 26°

  8. Verify the values of sin t and cos t in the table used to construct Figure 6.

    9. Calculus 9th Purcell Chapter 0 - 0.7

    💬 SOLUTION:
    Referring to Figure 2, it is clear that sin 0 = 0 and cos 0 = 1. If the angle is π/6, then the triangle in the figure below is equilateral. Thus, |PQ| = 1/2 |OP| = 1/2. This implies that sin π/6 = 1/2. By the Pythagorean Identity, cos2 (π/6) = 1 – sin2 (π/6) = 1 – (1/2)2 = 3/4.
    Thus,

    Calculus 9th Purcell Chapter 0 - 0.7
    cos (π/6) = √3/2. The results sin (π/4) = cos (π/4) = √2/2 were derived in the text. If the angle is π/3 then the triangle in the figure below is equilateral. Thus cos (π/3) = 1/2 and by the Pythagorean Identity, sin (π/3) = √3/2.

    Calculus 9th Purcell Chapter 0 - 0.7
    Referring to Figure 2, it is clear that sin (π/2) = 1 and cos (π/2) = 0. The rest of the values are obtained using the same kind of reasoning in the second quadrant.

    Semoga Bermanfaat 😁

  9. Evaluate without using a calculator.
    10. (a)
    tan π/6
    (b)
    sec π
    (c)
    sec 3π/4
    (d)
    cot π/4
    (e)
    sec 3π/4
    (f)
    tan (– π/4)
    💬 SOLUTION:
    (a)
    tan (π/6) = [sin (π/6)]/[cos(π/6)] = √3/3
    (b)
    sec (π) = 1/cos (π) = –1
    (c)
    sec (3π/4) = 1/cos (3π/4) = –√2
    (d)
    csc (π/2) = 1/sin (π/2) = 1
    (e)
    cot (π/4) = [cos (π/4)]/[sin (π/4)] = 1
    (f)
    tan (–π/4) = [sin (–π/4)]/[cos (–π/4)] = –1

  10. Evaluate without using a calculator.
    11. (a)
    tan π/3
    (b)
    sec π/3
    (c)
    cot π/3
    (d)
    csc π/4
    (e)
    tan (– π/6)
    (f)
    cos (– π/3)
    💬 SOLUTION:
    (a)
    tan (π/3) = [sin (π/3)]/[cos (π/3)] = √3
    (b)
    sec (π/3) = 1/cos (π/3) = 2
    (c)
    cot (π/3) = [cos (π/3)]/[sin (π/3)] = √3/3
    (d)
    csc (π/4) = 1/sin (π/4) = √2
    (e)
    tan (–π/6) = [sin (–π/6)]/cos (–π/6) = –(√3/3)
    (f)
    cos (–π/3) = 1/2

  11. Verify that the following are identities (See Example6).
    12. (a)
    (1 + sin z)(1 – sin z) = 1/sec2 z
    (b)
    (sec t – 1)(sec t + 1) = tan2 t
    (c)
    sec t – sin t tan t = cos t
    (d)
    (sec2 t – 1)/(sec2 t) = sin2 t
    💬 SOLUTION:
    (a)
    (1 + sin z)(1 – sin z) = 1 – sin2 z
     
    = cos2 z
     
    = 1/sec2 z
    (b)
    (sec t – 1)(sec t + 1) = sec2 t – 1
     
    = tan2 t
    (c)
    sec t – sin t tan t = [1/cos t]/[sin2 t/cos t]
     
    = [1 – sin2 t]/[cos t]
     
    = [cos2 t]/[cos t]
     
    = cos t
    (d)
    [sec2 t – 1]/[sec2 t] = [tan2 t]/[sec2 t]
     
    = [(sin2 t)/(cos2 t)]/[(1)/(cos2 t)]
     
    = sin2 t

  12. Verify that the following are identities (See Example6).
    13. (a)
    sin2 v + [(1)/(sec2 v)] = 1
    (b)
    cos 3t = 4 cos3 t – 3 cos t
    Hint: Use a double-angle identity.
    (c)
    sin 4x = 8 sin x cos3 x – 4 sin x cos x
    Hint: Use a double-angle identity twice.
    (d)
    (1 + cos θ)(1 – cos θ) = sin2 θ
    💬 SOLUTION:
    (a)
    sin2 v + [1/sec2 v]
     
    = sin2 v + cos2 v
     
    = 1
    (b)
    cos 3t = cos (2t + t) = cos 2t cos t – sin 2t sin t
     
    = (2 cos2 t – 1) cos t – 2 sin2 t cos t
     
    = 2 cos3 t – cos t – 2(1 – cos2 t) cos t
     
    = 2 cos3 t – cos t – 2 cos t + 2 cos3 t
     
    = 4 cos3 t – 3 cos t
    (c)
    sin 4x = sin [2(2x)] = 2 sin 2x cos 2x
     
    = 2(2 sin x cos x)(2 cos2 x – 1)
     
    = 2(4 sin x cos3 x – 2 sin x cos x)
     
    = 8 sin x cos3 x – 4 sin x cos x
    (d)
    (1 + cos θ)(1 – cos θ) = 1 – cos2 θ = sin2 θ

  13. Verify the following are identities.
    14. (a)
    [(sin u)/(scs u)] + [(cos u)/(sec u)] = 1
    (b)
    (1 – cos2 x)(1 + cot2 x) = 1
    (c)
    sin t (csc t – sin t) = cos2 t
    (d)
    (1 – csc2 t)/(csc2 t) = (–1)/(sec2 t)
    💬 SOLUTION:
    (a)
    [(sin u)/(csc u)] + [(cos u)/(sec u)]
     
    = sin2 u + cos2 u
     
    = 1
    (b)
    (1 – cos2 x)(1 + cot2 x)
     
    = (sin2 x)(csc2 x)
     
    = sin2 x [1/sin2 x]
     
    = 1
    (c)
    sin t (csc t – sin t)
     
    = sin t [(1/sin t) – sin t]
     
    = 1 – sin2 t
     
    = cos2 t
    (d)
    [1 – csc2 t]/[csc2 t]
     
    = – [cot2 t]/[csc2 t]
     
    = – [(cos2 t)/(sin2 t)]/[(1)/(sin2 t)]
     
    = – cos2 t
     
    = – 1/sec2 t

  14. Sketch the graphs of the following on [–π, 2π].
    15. (a)
    y = sin 2x
    (b)
    y = 2 sin t
    (c)
    y = cos (x – π/4)
    (d)
    y = sec t
    💬 SOLUTION:
    (a)
    y = sin 2x

    Calculus 9th Purcell Chapter 0 - 0.7

    (b)
    y = 2 sin t

    Calculus 9th Purcell Chapter 0 - 0.7

    (c)
    y = cos (x – (π/4))

    Calculus 9th Purcell Chapter 0 - 0.7

    (d)
    y = sec t

    Calculus 9th Purcell Chapter 0 - 0.7

  15. Sketch the graphs of the following on [–π, 2π].
    16. (a)
    y = csc t
    (b)
    y = 2 cos t
    (c)
    y = cos 3t
    (d)
    y = cos (t + π/3)
    💬 SOLUTION:
    (a)
    y = csc t

    Calculus 9th Purcell Chapter 0 - 0.7

    (b)
    y = 2 cos t

    Calculus 9th Purcell Chapter 0 - 0.7

    (c)
    y = cos 3t

    Calculus 9th Purcell Chapter 0 - 0.7

    (d)
    y = cos (t + (π/3))

    Calculus 9th Purcell Chapter 0 - 0.7

    Semoga Bermanfaat 😁

    Determine the period, amplitude, and shifts (both horizontal and vertical) and draw a graph over the interval –5 ≤ x ≤ 5 for the functions listed in Problems 16–23.
  16. y = 3 cos (x/2)
    17. 💬 SOLUTION:
    y = 3 cos (x/2)
    Period = 4π, amplitude = 3

    Calculus 9th Purcell Chapter 0 - 0.7

  17. y = 2 sin 2x
    18. 💬 SOLUTION:
    y = 2 sin 2x
    Period = π, amplitude = 2

    Calculus 9th Purcell Chapter 0 - 0.7

  18. y = tan x
    19. 💬 SOLUTION:
    y = tan x
    Period = π

    Calculus 9th Purcell Chapter 0 - 0.7

  19. y = 2 + (1/6) cot 2x
    20. 💬 SOLUTION:
    y = 2 + (1/6) cot (2x)
    Period = π/2, shift: 2 units up

    Calculus 9th Purcell Chapter 0 - 0.7

  20. y = 3 + sec (x – π)
    21. 💬 SOLUTION:
    y = 3 + sec (x – π)
    Period = 2π, shift: 3 units up, π units right

    Calculus 9th Purcell Chapter 0 - 0.7

  21. y = 21 + 7 sin (2x + 3)
    22. 💬 SOLUTION:
    y = 21 + 7 sin (2x + 3)
    Period = π, amplitude = 7, shift: 21 units up, 3/2 units left

    Calculus 9th Purcell Chapter 0 - 0.7

  22. y = 3 cos (x – (π/2)) – 1
    23. 💬 SOLUTION:
    y = 3 cos (x – (π/2)) – 1
    Period = 2π, amplitude = 3, shift: π/2 units right and 1 unit down

    Calculus 9th Purcell Chapter 0 - 0.7

  23. y = tan (2x – (π/3))
    24. 💬 SOLUTION:
    y = tan (2x – (π/3))
    Period = π/2, shift: π/2 units right

    Calculus 9th Purcell Chapter 0 - 0.7

    Semoga Bermanfaat 😁

  24. Which of the following represent the same graph? Check your result analytically using trigonometric identities.
    25. (a)
    y = sin (x + (π/2))
    (b)
    y = cos (x + (π/2))
    (c)
    y = –sin (x + π)
    (d)
    y = cos (x – π)
    (e)
    y = –sin (π – x)
    (f)
    y = cos (x – (π/2))
    (g)
    y = –cos (π – x)
    (h)
    y = sin (x – (π/2))
    💬 SOLUTION:
    (a) and (g)
    y = sin (x + (π/2))
    y = cos x
    y = –cos (π – x)
    (b) and (e)
    y = cos (x + (π/2))
    y = sin (x + π)
    y = –sin (π – x)
    (c) and (f)
    y = cos (x – (π/2))
    y = sin x
    y = –sin (x + π)
    (d) and (h)
    y = sin (x – (π/2))
    y = cos (x + π)
    y = cos (x – π)

  25. Which of the following are odd functions? Even functions? Neither?
    26. (a)
    t sin t
    (b)
    sin2 t
    (c)
    csc t
    (d)
    |sin t|
    (e)
    sin (cos t)
    (f)
    x + sin x
    💬 SOLUTION:
    (a)
    –t sin (–t) = t sin t; even
    (b)
    sin2 (–t) = sin2 t; even
    (c)
    csc (–t) = [1]/[sin (–t)] = –csc t; odd
    (d)
    |sin (–t)| = |–sin t| = |sin t|; even
    (e)
    sin (cos (–t)) = sin (cos t); even
    (f)
    –x + sin (–x) = –x – sin x = –(x + sin x); odd

  26. Which of the following are odd functions? Even functions? Neither?
    27. (a)
    cot t + sin t
    (b)
    sin3 t
    (c)
    sec t
    (d)
    √(sin4 t)
    (e)
    cos (sin t)
    (f)
    x2 + sin x
    💬 SOLUTION:
    (a)
    cot (–t) + sin (–t)
     
    = – cot t – sin t
     
    = – (cot t + sin t); odd
    (b)
    sin3 (–t) = – sin3 t; odd
    (c)
    sec (–t) = [1]/[cos (–t)
     
    = sec t; even
    (d)
    √(sin4 (–t)) = √(sin4 t); even
    (e)
    cos (sin (–t))
     
    = cos (– sin t)
     
    = cos (sin t); even
    (f)
    (–x2) + sin (–x) = x2 – sin x; neither

    Find the exact values in Problems 27–31. Hint: Half-angle identities may be helpful.
  27. cos2 (π/3)
    28. 💬 SOLUTION:
    cos2 (π/3)
    = [cos (π/3)]2
    = (1/2)2
    = 1/4

  28. sin2 (π/6)
    29. 💬 SOLUTION:
    sin2 (π/6)
    = [sin (π/6)]2
    = (1/2)2
    = 1/4

  29. sin3 (π/6)
    30. 💬 SOLUTION:
    sin3 (π/6)
    = [cos (π/6)]3
    = (1/2)3
    = 1/8

  30. cos2 (π/12)
    31. 💬 SOLUTION:
    cos2 (π/12)
    = [1 + cos 2 (π/12)]/[2]
    = [1 + cos (π/6)]/[2]
    = [1 + ((√3)/2)]/[2]
    = [2 + √3]/[4]

  31. sin2 (π/8)
    32. 💬 SOLUTION:
    sin2 (π/8)
    = [1 – cos 2 (π/8)]/[2]
    = [1 – cos (π/4)]/[2]
    = [1 – ((√2)/2)]/[2]
    = [2 – √2]/[4]

Semoga Bermanfaat 😁

Demikian soal serta penjelasan untuk Pembahasan Soal Buku Calculus 9th Purcell Chapter 0 - 0.7 Number 1 – 31. Silahkan untuk berkunjung kembali dikarenakan akan selalu ada update terbaru 😊😄🙏. Silahkan juga untuk memilih dan mendiskusikan di tempat postingan ini di kolom komentar ya supaya semakin bagus diskusi pada setiap postingan. Diperbolehkan request di kolom komentar pada postingan ini tentang rangkuman atau catatan atau soal dan yang lain atau bagian hal yang lainnya, yang sekiranya belum ada di website ini. Terima kasih banyak sebelumnya 👍. Semoga bermanfaat dan berkah untuk kita semua. Amiiinnn 👐👐👐
Jangan lupa untuk 💏 SUBSCRIBE 👪 (Klik lonceng di bawah-kanan layar Anda) dan berikan komentar 💬 atau masukan serta share 👫 postingan ini ke teman-teman untuk berkembangnya https://www.bantalmateri.com/ ini 😀. Terima kasih dan semoga bermanfaat. 😋😆

Ahmad Qolfathiriyus Firdaus

We are bantalmateri.com that utilizes the internet and digital media in delivering material, questions and even the form of discussion. In the current generation, online learning methods (commonly called daring) are considered closer to students who are very integrated and difficult to separate from technology. The emergence of technology has also facilitated the implementation of schools even though students and educators alike have to adapt.

No comments:

Post a Comment