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brazellnicole25

  • 2 years ago

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)

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  1. brazellnicole25
    • 2 years ago
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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.

  2. asnaseer
    • 2 years ago
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    do you know how to expand sin(Bt+C)?

  3. asnaseer
    • 2 years ago
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    using the angle sum formulae?

  4. asnaseer
    • 2 years ago
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    @brazellnicole25 I can only help you if you respond to my questions.

  5. brazellnicole25
    • 2 years ago
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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.

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

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

  8. asnaseer
    • 2 years ago
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    and sin(y) and cos(y)

  9. brazellnicole25
    • 2 years ago
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    yes i have done that before

  10. asnaseer
    • 2 years ago
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    ok, good, so please first write sin(Bt+C) in its expanded form

  11. brazellnicole25
    • 2 years ago
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    okay is it sin(a+-B)=sinacosb+-cos(a)sin(b)

  12. asnaseer
    • 2 years ago
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    yes - that is the correct formulae to use here

  13. brazellnicole25
    • 2 years ago
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    okay.

  14. asnaseer
    • 2 years ago
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    so sin(Bt+C)=?

  15. brazellnicole25
    • 2 years ago
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    does it equal the equation that I wrote before? I am stuck here for some reason

  16. asnaseer
    • 2 years ago
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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

  17. brazellnicole25
    • 2 years ago
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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

  18. asnaseer
    • 2 years ago
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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?

  19. brazellnicole25
    • 2 years ago
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    yes

  20. asnaseer
    • 2 years ago
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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?

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

  22. asnaseer
    • 2 years ago
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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)

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

  24. asnaseer
    • 2 years ago
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    can you now infer what the value of B must be?

  25. asnaseer
    • 2 years ago
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    compare the term sin(Bt) and sin(t/2) and also cos(Bt) and cos(t/2)

  26. brazellnicole25
    • 2 years ago
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    b must be t is that right?

  27. brazellnicole25
    • 2 years ago
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    t/2

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

  29. asnaseer
    • 2 years ago
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    divide both sides by 't'

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

  31. brazellnicole25
    • 2 years ago
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    2

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

  33. brazellnicole25
    • 2 years ago
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    1/2

  34. asnaseer
    • 2 years ago
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    perfect!

  35. asnaseer
    • 2 years ago
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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)

  36. asnaseer
    • 2 years ago
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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

  37. asnaseer
    • 2 years ago
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    can you see that this means that: -4 = Acos(C) 3 = Asin(C)

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

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

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

  41. asnaseer
    • 2 years ago
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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:\[16+9=A^2\cos^2C)+A^2\sin^2(C)\]

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

  43. brazellnicole25
    • 2 years ago
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    yes it does your just combining the two to make one equation.

  44. asnaseer
    • 2 years ago
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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

  45. brazellnicole25
    • 2 years ago
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    yeah that's right

  46. asnaseer
    • 2 years ago
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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)

  47. asnaseer
    • 2 years ago
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    make sense?

  48. brazellnicole25
    • 2 years ago
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    yes that does. you solved for c.

  49. brazellnicole25
    • 2 years ago
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    wow i made that harder than it had to be for sure.

  50. asnaseer
    • 2 years ago
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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))

  51. asnaseer
    • 2 years ago
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    I hope you understood all the steps

  52. brazellnicole25
    • 2 years ago
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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.

  53. sirm3d
    • 2 years ago
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    hmmm, i think C should be 180-36.87.

  54. brazellnicole25
    • 2 years ago
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    I thank you so much for your time.

  55. asnaseer
    • 2 years ago
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    yw :) @sirm3d - let me check what you are saying...

  56. asnaseer
    • 2 years ago
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    both are correct @sirm3d since:\[\sin(x)=\sin(180-x)\]

  57. asnaseer
    • 2 years ago
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    actually no

  58. sirm3d
    • 2 years ago
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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

  59. asnaseer
    • 2 years ago
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    I believe you are right @sirm3d since: -4 = Acos(C)

  60. asnaseer
    • 2 years ago
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    which implies cos(C) must be negative thank you for correcting me. :)

  61. asnaseer
    • 2 years ago
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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.

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

  63. asnaseer
    • 2 years ago
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    exactly

  64. asnaseer
    • 2 years ago
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    I hope that is also clear to you @brazellnicole25

  65. brazellnicole25
    • 2 years ago
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    okay yeah so you combined the two right? and then solved for C. is that correct

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

  67. asnaseer
    • 2 years ago
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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

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

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

  70. brazellnicole25
    • 2 years ago
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    ok I got you. So can I can I ask you something about this if you have time?

  71. asnaseer
    • 2 years ago
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    if its related to this question then sure

  72. brazellnicole25
    • 2 years ago
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    y1 and y2 would be the same equations right? I mean if I was to graph them?

  73. asnaseer
    • 2 years ago
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    yes

  74. brazellnicole25
    • 2 years ago
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    yes it is about this one. sorry i ment to put that first.

  75. brazellnicole25
    • 2 years ago
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    okay thank you so much. I really appriciate it. and your time.

  76. brazellnicole25
    • 2 years ago
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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.

  77. asnaseer
    • 2 years ago
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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.

  78. brazellnicole25
    • 2 years ago
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    could i ask you another question about this?

  79. asnaseer
    • 2 years ago
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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

  80. brazellnicole25
    • 2 years ago
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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.

  81. brazellnicole25
    • 2 years ago
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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.

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

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

  84. brazellnicole25
    • 2 years ago
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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.

  85. asnaseer
    • 2 years ago
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    not at all - I am glad you find the subject interesting :)

  86. brazellnicole25
    • 2 years ago
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    and apply things I've already learned in class. That's why i'm asking so much.

  87. brazellnicole25
    • 2 years ago
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    I do love math and I like to do it more when I understand it more.

  88. asnaseer
    • 2 years ago
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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.

  89. asnaseer
    • 2 years ago
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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. :)

  90. brazellnicole25
    • 2 years ago
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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.

  91. asnaseer
    • 2 years ago
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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. :)

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

  93. asnaseer
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    *others

  94. sirm3d
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    ^^ thanks, @asnaseer i teach math so i actually get to see the ins and outs of most mathematics.

  95. asnaseer
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    what age group do you teach @sirm3d ?

  96. brazellnicole25
    • 2 years ago
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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.

  97. sirm3d
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    college math. up to engineering mathematics, @asnaseer

  98. brazellnicole25
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    I like sirm3d how you helped us to.

  99. asnaseer
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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.... :)

  100. sirm3d
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    i love math, and i find joy seeing students learn and appreciate math.

  101. brazellnicole25
    • 2 years ago
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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.

  102. brazellnicole25
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    okay asnaseer I hope you have a good night. thanks alot.

  103. asnaseer
    • 2 years ago
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    night night... :)

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