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pablobegins

  • 3 years ago

Really annoying trigonometric graphing, help!

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  1. pablobegins
    • 3 years ago
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  2. pablobegins
    • 3 years ago
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    I need 1,2,5, and 10-24

  3. pablobegins
    • 3 years ago
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    I'm just not grasping the concept.

  4. amistre64
    • 3 years ago
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    f(x) = tan πx/4 g(x) =1/2 sec πx/4 Approximate the interval where f < g. lets set this up then, we know f and g tan (πx/4) < 1/2 sec(πx/4)

  5. amistre64
    • 3 years ago
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    one thing that might help is to rewrite sec, or maybe even tan. into equivalent terms

  6. pablobegins
    • 3 years ago
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    how so?

  7. amistre64
    • 3 years ago
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    well, tan = sin/cos, right? and sec = 1/cos .... that might help us see the resemblence

  8. amistre64
    • 3 years ago
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    \[\frac{sin (πx/4)}{cos (πx/4)} < \frac{1}{2}\frac{1}{cos(πx/4)}\] we can get rid of that 1/2 by multiplying by 2 \[\frac{(2)sin (πx/4)}{cos (πx/4)} < \frac{1(2)}{2}\frac{1}{cos(πx/4)}\] \[\frac{2sin (πx/4)}{cos (πx/4)} < \frac{1}{cos(πx/4)}\] now it looks better

  9. amistre64
    • 3 years ago
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    since the denominators are the same, lets equate numberators

  10. amistre64
    • 3 years ago
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    2sin(pi x/4) < 1

  11. pablobegins
    • 3 years ago
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    so now we graph the equation to find the point?

  12. amistre64
    • 3 years ago
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    id stick to the analysis rather than a graph; the graph can be used to dbl chk the results; but i doubt it will give a definitive result

  13. amistre64
    • 3 years ago
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    sin (t) < 1/2; well, when does sin(t) = 1/2?

  14. pablobegins
    • 3 years ago
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    how would we go about the analysis?

  15. amistre64
    • 3 years ago
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    divide each side by 2 to get: sin(t) < 1/2 im using "y" to help clean up the argument

  16. amistre64
    • 3 years ago
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    "t" that is ... cant type ;)

  17. pablobegins
    • 3 years ago
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    that's not one of the choices though :/ ,

  18. amistre64
    • 3 years ago
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    why would it be? we are only in the middle of it

  19. GT
    • 3 years ago
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    Hang in there and understand the steps.

  20. pablobegins
    • 3 years ago
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    OH! Gotcha. I just have a terrible teacher who goes about everything explaining really fast, and then expecting us to know all this. I'm sorry.

  21. amistre64
    • 3 years ago
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    it would be good to remember the basic angles of trig; i believe this is one of them

  22. amistre64
    • 3 years ago
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    |dw:1323627720794:dw|

  23. amistre64
    • 3 years ago
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    sin(t) = 1/2, when t = 30 degrees, or pi/6

  24. pablobegins
    • 3 years ago
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    is that the special right triangle?

  25. amistre64
    • 3 years ago
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    it is

  26. amistre64
    • 3 years ago
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    i have to remember the interval: -pi/4 to pi/4, that is between 45 and -45 degrees if i recall it correctly

  27. amistre64
    • 3 years ago
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    now i would see about equating pi x/4 and t pi x/4 = pi/6 divide off the pis x/4 = 1/6 and multiply off the 1/4 x = 4/6 , and simplify x = 2/3 so, if i did it right, it should be (-1, 2/3) but if you have questions as to why, it would be good to ask. The numbers tend to be unimportant and it is the process that matters.

  28. pablobegins
    • 3 years ago
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    why the -1?

  29. amistre64
    • 3 years ago
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    its part of our initial interval; and sin(pi *-1/4) is less than 1/2

  30. pablobegins
    • 3 years ago
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    i see. so for number 2 i would just multiply our original interval?

  31. amistre64
    • 3 years ago
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    it would be nice if we could, but no. trig functions are periodic, they do not act like linear functions. so its best to retrace the steps to make sure

  32. pablobegins
    • 3 years ago
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    or would it not change?

  33. amistre64
    • 3 years ago
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    2f just changes how high or low the graph would go; so id guess that its the same intervals. But i would still have to dbl chk the results

  34. pablobegins
    • 3 years ago
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    its not it.

  35. pablobegins
    • 3 years ago
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    like, number 1 was right, but 2f changed it in some way, because now it's not the same

  36. amistre64
    • 3 years ago
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    \[2*\frac{sin (πx/4)}{cos (πx/4)} < 2*\frac{1}{2}\frac{1}{cos(πx/4)}\] \[\frac{2sin (πx/4)}{cos (πx/4)} < \frac{1}{cos(πx/4)}\] \[2sin (πx/4) < 1\] \[sin (πx/4) < 1/2\] it looks to be the same

  37. pablobegins
    • 3 years ago
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    must be a problem with the software. now for number 5,i cant tell the graphs apart

  38. amistre64
    • 3 years ago
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    which one is number 5? this thing is hard to follow

  39. pablobegins
    • 3 years ago
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    graph for y=csc (x)

  40. amistre64
    • 3 years ago
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    well, I know csc is the humpbacks of sine|dw:1323628829369:dw|

  41. amistre64
    • 3 years ago
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    the closest one I see is A

  42. amistre64
    • 3 years ago
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    we can draw the sin(x) in A and see that csc rides the humpbacks

  43. pablobegins
    • 3 years ago
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    yeah, me too! thanks so much

  44. pablobegins
    • 3 years ago
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    number 10?

  45. amistre64
    • 3 years ago
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    id ditch the graph... see that sqrt(3)? that comes about in the 30-60-90 triangle

  46. amistre64
    • 3 years ago
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    and since cot = cos/sin; this is negative when either cos is negative OR sin is negative; but not both. that occurs in q2 and q4

  47. pablobegins
    • 3 years ago
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    so we would relate the degrees with a sin fuction?

  48. amistre64
    • 3 years ago
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    |dw:1323629102467:dw|

  49. amistre64
    • 3 years ago
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    almost; lets draw the 30-60-90 tri again and see what cot comes from

  50. amistre64
    • 3 years ago
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    |dw:1323629160526:dw| the basic angle is then cot(30) do you agree? 30 = pi/6

  51. pablobegins
    • 3 years ago
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    yeah, i see where you're coming from

  52. amistre64
    • 3 years ago
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    q2: 180-30 = 150 degrees q4: 360 - 30 = 330 degrees

  53. amistre64
    • 3 years ago
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    hmmm, 5pi/6 and 11pi/6 seem to fit in there somewhere. but i do get these mixed aroung at times :)

  54. amistre64
    • 3 years ago
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    i see the interval now; it one full revolution forward and one full revolution backwards

  55. pablobegins
    • 3 years ago
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    ignore that

  56. pablobegins
    • 3 years ago
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    it sent a previous message

  57. pablobegins
    • 3 years ago
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    would the pi/6 be negative? since were dealing with cot?

  58. amistre64
    • 3 years ago
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    |dw:1323629494595:dw|

  59. amistre64
    • 3 years ago
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    -pi/6, 5pi/6, -7pi/6, 11pi/6 is my best assessment

  60. amistre64
    • 3 years ago
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    thats option 4 right?

  61. pablobegins
    • 3 years ago
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    for 11, is the period 16/

  62. pablobegins
    • 3 years ago
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    YES, YOU WERE RIGHT ONCE AGAIN! :)

  63. amistre64
    • 3 years ago
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    tan and cot have a normal period of pi; this speeds that period up; by a factor of 8.; so id say its a period of 1/8

  64. amistre64
    • 3 years ago
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    that doesnt work out right ....

  65. pablobegins
    • 3 years ago
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    if we divide it by 2pi, you get 16

  66. amistre64
    • 3 years ago
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    that does sound familiar :)

  67. amistre64
    • 3 years ago
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    but the 2pi is wrong

  68. amistre64
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    divide by pi; since this is a pi period 2pi is for sin and cos

  69. pablobegins
    • 3 years ago
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    so it would be 1/8?

  70. amistre64
    • 3 years ago
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    pi/(pi/8) = pi * 8/pi = 8

  71. amistre64
    • 3 years ago
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    normal period/w

  72. pablobegins
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    and the normal period is 8??

  73. amistre64
    • 3 years ago
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    look again

  74. amistre64
    • 3 years ago
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    tan/cot has a normal pf pi everything else is 2pi

  75. amistre64
    • 3 years ago
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    pi/(w) = periodicity

  76. amistre64
    • 3 years ago
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    since w = pi/8 .... we plug it in

  77. pablobegins
    • 3 years ago
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    we end up with 1/8.

  78. amistre64
    • 3 years ago
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    \[\frac{pi}{pi/8}=\frac{pi*8}{pi}=8\]

  79. pablobegins
    • 3 years ago
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    oh, i was solving it incorrectly. can you explain 12 to me please? i know it's similar to 10, but i want to grasp the concept.

  80. amistre64
    • 3 years ago
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    since csc rides the humpbacks of sin; lets equate this to an angle of sin. csc = 1/sin is a good thing to remember

  81. amistre64
    • 3 years ago
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    \[csc(x)=\frac{2\sqrt{3}}{3}\] \[\frac{1}{sin(x)}=\frac{2\sqrt{3}}{3}\] \[\frac{sin(x)}{1}=\frac{3}{2\sqrt{3}}\] and draw a triangle to help visualise it

  82. amistre64
    • 3 years ago
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    |dw:1323630627950:dw|

  83. amistre64
    • 3 years ago
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    this doesnt seem to match any of our basic triangles does it?

  84. amistre64
    • 3 years ago
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    maybe .... divide off sqrt(3) 2sqrt(3)/sqrt(3) = 2 3/sqrt(3) = 3sqrt(3)/3 = sqrt(3) its our 30-60-90 in disguise

  85. amistre64
    • 3 years ago
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    |dw:1323630843700:dw|

  86. pablobegins
    • 3 years ago
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    'disguise' lol

  87. pablobegins
    • 3 years ago
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    so we're just rearranging the equation so the triangle makes sense, in a way

  88. amistre64
    • 3 years ago
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    \[csc(60) = \frac{2}{\sqrt{3}}\] \[csc(60) = (\frac{2}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}})\] \[csc(60) = \frac{2\sqrt{3}}{3}\] so we are good :)

  89. amistre64
    • 3 years ago
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    60 = pi/3 .... so we need to adapt that to the interval

  90. amistre64
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    |dw:1323631158475:dw|

  91. amistre64
    • 3 years ago
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    |dw:1323631248664:dw|

  92. amistre64
    • 3 years ago
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    thatss what, option 7?

  93. pablobegins
    • 3 years ago
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    seven is incorrect :-(nt and so is 3

  94. amistre64
    • 3 years ago
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    trig takes alot of mental gymnastics :)

  95. amistre64
    • 3 years ago
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    hmmm

  96. amistre64
    • 3 years ago
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    i think I know where I went astray at

  97. amistre64
    • 3 years ago
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    csc is only positive in q1 and q2, i confused it with a tan ....

  98. amistre64
    • 3 years ago
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    |dw:1323631597639:dw|

  99. amistre64
    • 3 years ago
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    -5,-4,1,2 ---------, might be better 3

  100. pablobegins
    • 3 years ago
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    what choice is that?

  101. amistre64
    • 3 years ago
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    option 4, since that aint in the choices

  102. amistre64
    • 3 years ago
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    youre right tho, these ARE really annoying ;)

  103. pablobegins
    • 3 years ago
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    finally correct! this thing penalizes me for every wrong answer too... now for the period of number 13, is 7?

  104. amistre64
    • 3 years ago
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    the value attached to x is out "w" and sec is a 2pi normal \[\frac{2pi}{-7pi/2}\] \[\frac{2pi*2}{-7pi}\] \[\frac{2*2}{-7}=-4/7\] id go with option 1

  105. pablobegins
    • 3 years ago
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    ow do we find the frequency?

  106. amistre64
    • 3 years ago
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    frequency is how fast it moves along .... id have to look it up :)

  107. amistre64
    • 3 years ago
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    http://regentsprep.org/Regents/math/algtrig/ATT7/graphvocab.htm

  108. amistre64
    • 3 years ago
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    freq = 1/period

  109. pablobegins
    • 3 years ago
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    initial phase?

  110. amistre64
    • 3 years ago
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    i dont recall ever doing an initial phase ...

  111. amistre64
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    is that a phase shift perhaps?

  112. pablobegins
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    it says initial phase, it might be referring to the same thing

  113. amistre64
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    i cant make any sense of the online stuff. I doubt that it is the same thing then

  114. amistre64
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    initial phase is when t=0 ....

  115. amistre64
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    but i got no idea at the moment what the means

  116. pablobegins
    • 3 years ago
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    what about the period of 1 7? i got 10/9, but that's apparently wrong

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