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MrCarey

  • 3 years ago

sin(arcsin(-11π/13) I'd really just like an explanation of this one.

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  1. MrCarey
    • 3 years ago
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    I don't know why I'm forgetting how to do this, but it's been awhile and my final is coming up and I want a better understanding of it.

  2. godorovg
    • 3 years ago
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    do you know the init Circle?

  3. MrCarey
    • 3 years ago
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    Yep

  4. godorovg
    • 3 years ago
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    when you at look the unit circle what quard would Sin be in?

  5. MrCarey
    • 3 years ago
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    I see that it goes -11π, so I believe it's going clockwise around the unit circle and I think it gets to Q3.

  6. godorovg
    • 3 years ago
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    yes correct, so in the problem what does that tell you?

  7. MrCarey
    • 3 years ago
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    I don't know...hahaha, rephrase that maybe?

  8. godorovg
    • 3 years ago
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    from looking at this problem and taking the information you already have what would you say the answer is? sin(arcsin(-11π/13)

  9. godorovg
    • 3 years ago
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    do you understand arcsin and what that means??

  10. MrCarey
    • 3 years ago
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    crap, I wrote it down wrong. It was arcsin(sin(-11π/13))

  11. MrCarey
    • 3 years ago
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    I have a pretty good understanding of how to use it. I'm rusty, though.

  12. godorovg
    • 3 years ago
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    okay, let us take this and break it down on what you know, you start I will fill in if needed.

  13. MrCarey
    • 3 years ago
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    I believe sin(ø) = arcsin (-11π/13) in this situation.

  14. godorovg
    • 3 years ago
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    you are right, but how did you do that?

  15. MrCarey
    • 3 years ago
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    I just kind of have that memorized. I guess I never really grasped that...

  16. godorovg
    • 3 years ago
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    okay, now I have a stupid what is the question asking for??? I am a little lost with the question here does it ask to solve or what?

  17. MrCarey
    • 3 years ago
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    Find the exact value of the given expression. Simplify your answer (rationalize the denom)

  18. MrCarey
    • 3 years ago
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    I also have that the answer is -2π/13

  19. MrCarey
    • 3 years ago
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    I don't really understand where the 2π comes from.

  20. godorovg
    • 3 years ago
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    because recall the unit circle has a total of 360 degrees right?

  21. MrCarey
    • 3 years ago
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    Yep and 2π is 360º

  22. godorovg
    • 3 years ago
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    2/pie is a point on the Unit the circle in which the angle is 360.. you beat to that one good job

  23. MrCarey
    • 3 years ago
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    So, my picture shows the line -11π/13 and the distance between that line and the 180º line is 2π/13.

  24. MrCarey
    • 3 years ago
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    Having a hard time understanding why it's 2π and not just π.

  25. MrCarey
    • 3 years ago
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    Probably just overthinking it.

  26. godorovg
    • 3 years ago
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    what I am questioning here the negative here in this case?

  27. godorovg
    • 3 years ago
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    because you start with a negative you will most likely end that way..

  28. godorovg
    • 3 years ago
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    2pie is at the 360 degrees which has points 0,0 so how can that be a negative result help here?

  29. MrCarey
    • 3 years ago
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    One sec, gonna post a photo of what is giving me a headache.

  30. godorovg
    • 3 years ago
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    I think we are both getting the same headache on this. This problem is going down..

  31. MrCarey
    • 3 years ago
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    Hahaha, I'm tellin' you. I haven't had any problems with any other questions, but this one is just destroying my brain.

  32. godorovg
    • 3 years ago
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    yeah, the same thing only that my brain is dead because I am stuying Vectors myself.

  33. godorovg
    • 3 years ago
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    what I am missing here????

  34. lgbasallote
    • 3 years ago
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    is the question still \[\sin (arcsin (-11 \pi/3)\]

  35. MrCarey
    • 3 years ago
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    arcsin(sin(-11π/3))

  36. godorovg
    • 3 years ago
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    yes sir, please help me I am getting a major headache here

  37. MrCarey
    • 3 years ago
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    Messed up the on the original post

  38. lgbasallote
    • 3 years ago
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    here's a hint.. arcsin( sin x) = x sin (arcsin x) = x cos( arc cos x) = x arc cos (cos x) = x tan (arctan x) = x arctan (tan x) = x does that help?

  39. MrCarey
    • 3 years ago
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  40. MrCarey
    • 3 years ago
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    There is the picture that I was trying to describe.

  41. MrCarey
    • 3 years ago
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    I wish that helped, but I still can't figure out why it is -2π/13

  42. lgbasallote
    • 3 years ago
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    is that included in the question? or is that your solytion?

  43. MrCarey
    • 3 years ago
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    Solution. It's an answer key and I'm trying to decipher it.

  44. MrCarey
    • 3 years ago
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    I know it is -2π/13 because of the range, but I don't know where the 2π came from.

  45. lgbasallote
    • 3 years ago
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    use the hint i gave you...here's a demo \[\sin (\sin^{-1} ( \frac{\pi}{2} ) ) = \frac{\pi}{2}\] \[\sin^{-1} (\sin ( \frac{2 \pi}{5} ) ) = \frac{2 \pi}{5}\] getting it now?

  46. lgbasallote
    • 3 years ago
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    \[\sin^{-1} (\sin ( \frac{-11\pi}{3} ) ) = -\frac{11\pi}{3}\] then simplify that you get \[-\frac{2\pi}{3}\] make sense?

  47. godorovg
    • 3 years ago
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    lgbasallote what did I miss here? I mean I thought for sure I was going the right way what did I do in my steps that went wrong here? Help me to understand..

  48. MrCarey
    • 3 years ago
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    How'd you simplify to that?

  49. lgbasallote
    • 3 years ago
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    i think i wrote the wrong thingies lol...

  50. lgbasallote
    • 3 years ago
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    but that's the concept

  51. MrCarey
    • 3 years ago
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    Yeah, I knew what you were saying with the others. It's this deal here. I don't understand how it is simplifying from 11π to 2π.

  52. MrCarey
    • 3 years ago
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    It's like it just magically disappears.

  53. lgbasallote
    • 3 years ago
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    reference angles

  54. MrCarey
    • 3 years ago
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    Heh, I feel dumb for not knowing still.

  55. godorovg
    • 3 years ago
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    Mrcarey I am so sorry I took the long way here.

  56. MrCarey
    • 3 years ago
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    Haha, no problem at all. I've been at it for so long, it didn't even feel that long.

  57. godorovg
    • 3 years ago
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    ha ha yeah better next time right? HA HA..

  58. MrCarey
    • 3 years ago
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    Exactly! Thanks for trying at least. Good luck with the vector calc. I'll be THERE soon enough.

  59. godorovg
    • 3 years ago
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    I am Physics Engineering major myelf

  60. MrCarey
    • 3 years ago
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    Mech E. for me. Math math math math and more math!

  61. godorovg
    • 3 years ago
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    you got this now

  62. MrCarey
    • 3 years ago
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    Yeah, I feel better about it. Got one more day of review in class, so I'll probably bring it up with him anyway. Thanks again!

  63. godorovg
    • 3 years ago
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    do you another problem you need help on? Just asking here

  64. MrCarey
    • 3 years ago
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    Nope, this one seemed to be the only one I kept coming back to. This stupid thing made everything else seem easy!

  65. godorovg
    • 3 years ago
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    yeah, I only hope for me it is just one problem that will allow everything to be easy after that. Fat chance ha ha

  66. MrCarey
    • 3 years ago
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    Hahahaha, no way. I've peeked at that stuff and needed to take a nap after.

  67. godorovg
    • 3 years ago
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    did that one already.. I took a 5hrs nap

  68. godorovg
    • 3 years ago
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    I am doing Calc1 andPhysics 1 with Calc

  69. MrCarey
    • 3 years ago
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    I'll be doing that in the Fall. Well, no Physics yet, couldn't get in the class. General Chem 1 and Calc 1. Gonna be exciting.

  70. godorovg
    • 3 years ago
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    what school?

  71. MrCarey
    • 3 years ago
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    Doing community for now, so I can transfer to University of Washington Seattle.

  72. MrCarey
    • 3 years ago
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    You?

  73. godorovg
    • 3 years ago
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    cool I live Iin Portland Ore myself..

  74. godorovg
    • 3 years ago
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    I will start Sept 24th at Jr College than transfer to Tarleton ST university in TX in Spring 2013

  75. MrCarey
    • 3 years ago
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    Awesome, TX should be pretty fun. Definitely a change of scenery.

  76. godorovg
    • 3 years ago
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    I am studying math and Physics by watching videos and taking notes on my own

  77. MrCarey
    • 3 years ago
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    Khan Academy?

  78. godorovg
    • 3 years ago
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    yeah StephenVille is very small town though

  79. godorovg
    • 3 years ago
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    who is Khan A?

  80. MrCarey
    • 3 years ago
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    www.khanacademy.com MIT grad who made his own site that teaches every kind of math and most sciences.

  81. MrCarey
    • 3 years ago
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    Free site for learning. I've used it for awhile now. Especially when I get bad professors.

  82. godorovg
    • 3 years ago
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    funny you just posted MIT, Because I am using thier open courses on line for the video in math and Physics

  83. MrCarey
    • 3 years ago
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    Haha, that works, too! You'd probably like his teaching style. He is very helpful and has tons of videos. I always recommend the site.

  84. godorovg
    • 3 years ago
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    thanks will check it out, hey you have a great night and if you need me to help me message me

  85. godorovg
    • 3 years ago
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    you

  86. MrCarey
    • 3 years ago
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    Sounds like a plan! Have a good night!

  87. godorovg
    • 3 years ago
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    you too.

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