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brazellnicole25 Group Title

Can anyone help me with a trig problem. I need to turn y1=-4sin(t/2)+3cos(t/2) into the form y2=Asin(Bt+C)

  • one year ago
  • one year ago

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  1. brazellnicole25 Group Title
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    I would like to know how to obtain the answer step by step. I don't want to just know the answer. I would like to understand how it was obtained.

    • one year ago
  2. asnaseer Group Title
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    do you know how to expand sin(Bt+C)?

    • one year ago
  3. asnaseer Group Title
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    using the angle sum formulae?

    • one year ago
  4. asnaseer Group Title
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    @brazellnicole25 I can only help you if you respond to my questions.

    • one year ago
  5. brazellnicole25 Group Title
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    I will repond. i know i'm suppost to use the transformation identity to make the first equation the other one. and c is suppost to be the smallest positive ange.

    • one year ago
  6. asnaseer Group Title
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    first expand sin(Bt+C) using the angle sum formulae - do you know how to do that?

    • one year ago
  7. asnaseer Group Title
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    i.e do you know how to express sin(x+y) in terms of sin(x) and cos(x)?

    • one year ago
  8. asnaseer Group Title
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    and sin(y) and cos(y)

    • one year ago
  9. brazellnicole25 Group Title
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    yes i have done that before

    • one year ago
  10. asnaseer Group Title
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    ok, good, so please first write sin(Bt+C) in its expanded form

    • one year ago
  11. brazellnicole25 Group Title
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    okay is it sin(a+-B)=sinacosb+-cos(a)sin(b)

    • one year ago
  12. asnaseer Group Title
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    yes - that is the correct formulae to use here

    • one year ago
  13. brazellnicole25 Group Title
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    okay.

    • one year ago
  14. asnaseer Group Title
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    so sin(Bt+C)=?

    • one year ago
  15. brazellnicole25 Group Title
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    does it equal the equation that I wrote before? I am stuck here for some reason

    • one year ago
  16. asnaseer Group Title
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    ok, let me take you through this one step-by-step and then hopefully you will be able to do others that are similar to this one

    • one year ago
  17. brazellnicole25 Group Title
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    This is where I'm having my block for this problem for some reason. I've used the above before but I dont' know why i'm having so much trouble. I'm sorry I'm really trying

    • one year ago
  18. asnaseer Group Title
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    no problem - lets work through it together. we start by first expanding sin(Bt+C) to get: sin(Bt+C) = sin(Bt)cos(C) + cos(Bt)sin(C) agreed?

    • one year ago
  19. brazellnicole25 Group Title
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    yes

    • one year ago
  20. asnaseer Group Title
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    good, next we multiply this by A to get: Asin(Bt+C) = Asin(Bt)cos(C) + Acos(Bt)sin(C) agreed?

    • one year ago
  21. brazellnicole25 Group Title
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    yes because we have to do the same to all of them (distribute) okay i'm following

    • one year ago
  22. asnaseer Group Title
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    great, now lets rearrange this to get it into a similar form to the original equation. we therefore get: Asin(Bt+C) = Acos(C) * sin(Bt) + Asin(C) * cos(Bt)

    • one year ago
  23. asnaseer Group Title
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    compare this to the original equation which is: -4*sin(t/2) + 3*cos(t/2)

    • one year ago
  24. asnaseer Group Title
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    can you now infer what the value of B must be?

    • one year ago
  25. asnaseer Group Title
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    compare the term sin(Bt) and sin(t/2) and also cos(Bt) and cos(t/2)

    • one year ago
  26. brazellnicole25 Group Title
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    b must be t is that right?

    • one year ago
  27. brazellnicole25 Group Title
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    t/2

    • one year ago
  28. asnaseer Group Title
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    not quite - if you compare these terms then we see that: Bt = t/2 therefore B = ?

    • one year ago
  29. asnaseer Group Title
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    divide both sides by 't'

    • one year ago
  30. sirm3d Group Title
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    if you write \(t/2\) as \((1/2)t\) you should be able to infer the value of \(B\)

    • one year ago
  31. brazellnicole25 Group Title
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    2

    • one year ago
  32. asnaseer Group Title
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    we have B times 't' is equal to (1/2) times 't' so what must B equal?

    • one year ago
  33. brazellnicole25 Group Title
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    1/2

    • one year ago
  34. asnaseer Group Title
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    perfect!

    • one year ago
  35. asnaseer Group Title
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    ok, so now we can use this to get: -4*sin(t/2) + 3*cos(t/2) = Acos(C) * sin(t/2) + Asin(C) * cos(t/2)

    • one year ago
  36. asnaseer Group Title
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    now compare the terms that are multiplying the sin(t/2) terms and the cos(t/2) terms on both sides of the equals sign

    • one year ago
  37. asnaseer Group Title
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    can you see that this means that: -4 = Acos(C) 3 = Asin(C)

    • one year ago
  38. brazellnicole25 Group Title
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    yes I can see that. The right side of the equal sign it just streched out. yeah that makes sense

    • one year ago
  39. asnaseer Group Title
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    good, next are you aware of the identity:\[\sin^2(x)+\cos^2(x)=1\]

    • one year ago
  40. brazellnicole25 Group Title
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    yes and i know that can me manipulated in a couple of ways. that is one of the pythagorean identies

    • one year ago
  41. asnaseer Group Title
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    correct, so if we use this identity, we can square both sides of: -4 = Acos(C) 3 = Asin(C) to get:\[16 = A^2\cos^2(C)\]\[9=A^2\sin^2(C)\]and then add them together to get:\[16+9=A^2\cos^2C)+A^2\sin^2(C)\]

    • one year ago
  42. asnaseer Group Title
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    therefore:\[25=A^2(\cos^2(C)+\sin^2(C))=A^2(1)=A^2\]does that make sense?

    • one year ago
  43. brazellnicole25 Group Title
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    yes it does your just combining the two to make one equation.

    • one year ago
  44. asnaseer Group Title
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    correct - now we take square roots of both sides to get:\[A=\sqrt{25}=5\]so now we have found A=5 and B=1/2 which just leaves C to find

    • one year ago
  45. brazellnicole25 Group Title
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    yeah that's right

    • one year ago
  46. asnaseer Group Title
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    we can find C using either one of the two earlier equations we got to: -4 = Acos(C) 3 = Asin(C) lets use the second one to get: 3 = Asin(C) therefore: sin(C) = 3/A = 3/5 therefore: C = arcsin(3/5) = 36.87 degrees (approximately)

    • one year ago
  47. asnaseer Group Title
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    make sense?

    • one year ago
  48. brazellnicole25 Group Title
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    yes that does. you solved for c.

    • one year ago
  49. brazellnicole25 Group Title
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    wow i made that harder than it had to be for sure.

    • one year ago
  50. asnaseer Group Title
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    so we can finally write: y2 = 5sin(t/2 + 36.87) or, if you prefer: y2 = 5sin(t/2 + arcsin(3/5))

    • one year ago
  51. asnaseer Group Title
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    I hope you understood all the steps

    • one year ago
  52. brazellnicole25 Group Title
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    yes I did. Thank you so much. I really appriciate it. I was looking back at some of these problems that I missed on a previous test and it was bothering me so I wanted to figure them out. Just so I would know. I don't want to be behind in my calculus class next semester.

    • one year ago
  53. sirm3d Group Title
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    hmmm, i think C should be 180-36.87.

    • one year ago
  54. brazellnicole25 Group Title
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    I thank you so much for your time.

    • one year ago
  55. asnaseer Group Title
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    yw :) @sirm3d - let me check what you are saying...

    • one year ago
  56. asnaseer Group Title
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    both are correct @sirm3d since:\[\sin(x)=\sin(180-x)\]

    • one year ago
  57. asnaseer Group Title
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    actually no

    • one year ago
  58. sirm3d Group Title
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    \[y_2=5 \sin (\frac{1}{2}t +143.17)\] because \(-4=5 \cos C\) abd there's no way to reconcile the sign if C is in quadrant I

    • one year ago
  59. asnaseer Group Title
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    I believe you are right @sirm3d since: -4 = Acos(C)

    • one year ago
  60. asnaseer Group Title
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    which implies cos(C) must be negative thank you for correcting me. :)

    • one year ago
  61. asnaseer Group Title
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    @brazellnicole25 - to avoid this type of mistake, what we should have done is gone from these two equations: -4 = Acos(C) 3 = Asin(C) to this one by diving the two: tan(C) = -3/4 that guarantees that we get the correct quadrant.

    • one year ago
  62. sirm3d Group Title
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    dividing the two equations should yield an equation in tangent.\[\frac{3}{-4}=\frac{\sin C}{\cos C}=\tan C\]

    • one year ago
  63. asnaseer Group Title
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    exactly

    • one year ago
  64. asnaseer Group Title
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    I hope that is also clear to you @brazellnicole25

    • one year ago
  65. brazellnicole25 Group Title
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    okay yeah so you combined the two right? and then solved for C. is that correct

    • one year ago
  66. asnaseer Group Title
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    yes because the angle C has to satisfy BOTH equations: -4 = Acos(C) 3 = Asin(C)

    • one year ago
  67. asnaseer Group Title
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    so we just divided the second one by the first one to get:\[\frac{3}{-4}=\frac{A\sin(C)}{A\cos(C)}\]which lead to: tan(C) = -3/4

    • one year ago
  68. asnaseer Group Title
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    so the correct solution is: y2 = 5sin(t/2 - arctan(3/4))

    • one year ago
  69. asnaseer Group Title
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    sorry: y2 = 5sin(t/2 + arctan(-3/4))

    • one year ago
  70. brazellnicole25 Group Title
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    ok I got you. So can I can I ask you something about this if you have time?

    • one year ago
  71. asnaseer Group Title
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    if its related to this question then sure

    • one year ago
  72. brazellnicole25 Group Title
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    y1 and y2 would be the same equations right? I mean if I was to graph them?

    • one year ago
  73. asnaseer Group Title
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    yes

    • one year ago
  74. brazellnicole25 Group Title
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    yes it is about this one. sorry i ment to put that first.

    • one year ago
  75. brazellnicole25 Group Title
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    okay thank you so much. I really appriciate it. and your time.

    • one year ago
  76. brazellnicole25 Group Title
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    I haven't ever used this site is there anything I can do to help you get a beter rating or anything? I don't know alot about this site.

    • one year ago
  77. asnaseer Group Title
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    I am glad I was able to help - I do this because I enjoy teaching those who are willing to learn, not for any rating. But, in general, the usual workflow is: 1. You post a question. 2. One or more people come to your question to help you out. 3. You select the one that best answered your question by clicking on the blue Best Response button next to any one of their replies. 4. You close the question by clicking on the Close button then the whole process starts again.

    • one year ago
  78. brazellnicole25 Group Title
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    could i ask you another question about this?

    • one year ago
  79. asnaseer Group Title
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    BTW: You can use wolframalpha to see the graphs as follows: y1: http://www.wolframalpha.com/input/?i=y%3D-4sin%28t%2F2%29%2B3cos%28t%2F2%29 y2: http://www.wolframalpha.com/input/?i=y+%3D+5sin%28t%2F2+%2B+2.4981%29

    • one year ago
  80. brazellnicole25 Group Title
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    would the amplitude of this be 5 , the period be 2pie (symbol i mean) and the phanse shift be .... sorry i'm thinking about that for a min.

    • one year ago
  81. brazellnicole25 Group Title
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    okay thanks for showing me the graphs. I got the same graphs on my calculator at on these you gave me. Thank you so much for that.

    • one year ago
  82. asnaseer Group Title
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    you can learn more about phase shift and amplitudes here: http://www.purplemath.com/modules/grphtrig.htm

    • one year ago
  83. asnaseer Group Title
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    in general:\[Y=A\sin(wt+\phi)\]means: A=amplitude \(w=2*\pi/P\) P= period \(\phi\)=phase shift

    • one year ago
  84. brazellnicole25 Group Title
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    okay sorry to be so full of questions. I'm just rying to fully understand it. I hope i'm not being to much trouble. I really do appriciate it though.

    • one year ago
  85. asnaseer Group Title
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    not at all - I am glad you find the subject interesting :)

    • one year ago
  86. brazellnicole25 Group Title
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    and apply things I've already learned in class. That's why i'm asking so much.

    • one year ago
  87. brazellnicole25 Group Title
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    I do love math and I like to do it more when I understand it more.

    • one year ago
  88. asnaseer Group Title
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    it is the same with any subject - if you really enjoy a subject then you will naturally get better at it.

    • one year ago
  89. asnaseer Group Title
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    a thirst for knowledge is always a good sign - I am sure you will do really well and achieve all that you want to achieve. :)

    • one year ago
  90. brazellnicole25 Group Title
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    Thank you so much. I really appriciate it. Your very smart though. do you teach math somewhere? Not that I mean to be to nosey. Your good at it though.

    • one year ago
  91. asnaseer Group Title
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    again, you are more than welcome. And no I am not an actual maths teacher, I work as a fulltime software engineer but always enjoyed maths since high school, so I try and keep up with the subject by /trying/ to teach others. :)

    • one year ago
  92. asnaseer Group Title
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    I have also learnt a LOT of things from other on this site - @sirm3d being one of them. :)

    • one year ago
  93. asnaseer Group Title
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    *others

    • one year ago
  94. sirm3d Group Title
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    ^^ thanks, @asnaseer i teach math so i actually get to see the ins and outs of most mathematics.

    • one year ago
  95. asnaseer Group Title
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    what age group do you teach @sirm3d ?

    • one year ago
  96. brazellnicole25 Group Title
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    well you are good I mean great at it. I like how you take the time to help others and make sure the person your helping will interact with you. if they really want to learn then they would have no problem answering you.

    • one year ago
  97. sirm3d Group Title
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    college math. up to engineering mathematics, @asnaseer

    • one year ago
  98. brazellnicole25 Group Title
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    I like sirm3d how you helped us to.

    • one year ago
  99. asnaseer Group Title
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    thats good to know @sirm3d - I'll keep you in mind if I run into any problems that I get stuck in. ok folks - I have to go now as it is 1:40am where I am and I need to catch some Zzzz's.... :)

    • one year ago
  100. sirm3d Group Title
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    i love math, and i find joy seeing students learn and appreciate math.

    • one year ago
  101. brazellnicole25 Group Title
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    you were very polite in giving your input. I have gone to get help before and some people are just so rude. but your so nice about helping.

    • one year ago
  102. brazellnicole25 Group Title
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    okay asnaseer I hope you have a good night. thanks alot.

    • one year ago
  103. asnaseer Group Title
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    night night... :)

    • one year ago
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