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lilsis76

  • 2 years ago

Given CSC(s)= 13/12, where cos(s)<0 & tan(t)=-sqrt3 where 3pi/2 < t < 2pi. Find exact value of cos(s-t)

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  1. lilsis76
    • 2 years ago
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    @phi or @jim_thompson5910 or @zepdrix can anyone of you explain this to me please. thank you so much

  2. zepdrix
    • 2 years ago
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    Hmm, do you happen to know what the answer is? I came up with something, but I'd hate to explain this to you if I'm wrong lol.

  3. lilsis76
    • 2 years ago
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    lol its okay, and no i dont have the answer. i have to correct these problems inorder to retake the test. and i have to turn this in tomorrow morning. Do you mind helping me with these problems, so far this is 1 out of 6 problems. and they are all different.

  4. zepdrix
    • 2 years ago
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    I doubt I'll have time :c Important exam tomorrow. But just post your questions with the little @ tag and I'll try to take a look if I get a moment. c: As for this problem.... Hmm, lemme see if I'm on the right track here. cos(s) < 0, so we're in the 2nd or 3rd quadrant.\[\csc(s)=\frac{13}{12} \quad \rightarrow \quad \frac{1}{\sin(s)}=\frac{13}{12} \quad \rightarrow \quad \sin(s)=\frac{12}{13}\]

  5. zepdrix
    • 2 years ago
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    Sine positive, cosine negative, so we're in the 2nd quadrant.

  6. zepdrix
    • 2 years ago
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    |dw:1355197009318:dw|

  7. zepdrix
    • 2 years ago
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    They told us that t lies in the 4th quadrant. And they gave us a special angle for t. \[\tan(t)=-\sqrt3\]

  8. lilsis76
    • 2 years ago
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    how was the cosine negative?

  9. zepdrix
    • 2 years ago
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    cos(s) < 0 was in the problems description. Less than zero means it's negative :)

  10. zepdrix
    • 2 years ago
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    Lots of little details to this problem, it's easy to miss stuff :D heh

  11. zepdrix
    • 2 years ago
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    So hmm, let's try to solve for t.

  12. lilsis76
    • 2 years ago
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    okay

  13. zepdrix
    • 2 years ago
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    If it was in the first quadrant it would be... lemme try to remember.. it's the larger angle, pi/3 i guess. Since we're in the 4th quadrant we gotta think a lil bit.. it's ummm

  14. lilsis76
    • 2 years ago
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    pi/3 is the 60 degree

  15. zepdrix
    • 2 years ago
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    So in the 4th quadrant is it, 5pi/3 maybe?

  16. lilsis76
    • 2 years ago
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    haha im sorry im superly bad at this, im retaking this class next fall but im going to give it my best shot to retake a test and on the final :)

  17. zepdrix
    • 2 years ago
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    aw :D

  18. lilsis76
    • 2 years ago
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    yes its 5pi/3

  19. lilsis76
    • 2 years ago
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    i checked by drawing haha

  20. zepdrix
    • 2 years ago
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    ah good :)

  21. zepdrix
    • 2 years ago
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    |dw:1355197419619:dw|

  22. lilsis76
    • 2 years ago
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    by doing it the same way? but wouldnt we get....hm...the same 12 and 13?

  23. zepdrix
    • 2 years ago
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    It's not straight across even though it might look that way based on how I drew it. 12/13 is not a special angle. :D I don't think so at least :3

  24. zepdrix
    • 2 years ago
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    \[\large \cos(s-t)\]So what I did from here is... Apply the Difference Formula for Cosine. Remember the Sum and Difference formulas?

  25. zepdrix
    • 2 years ago
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    \[\large \cos(s-t)=\cos(s)\cos(t)+\sin(s)\sin(t)\]

  26. zepdrix
    • 2 years ago
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    This problem is a bit of a doozy.. Once you get to this point though, it's not too bad. Just a bunch of plugging in angles. We know what t is, so we can plug that value into cos(t) and sin(t). We already know what sin(s) is. And by finding the missing side of that triangle above, we can easily determine cos(s).

  27. zepdrix
    • 2 years ago
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    This is a pretty rough problem :D I don't remember doing anything like this in trig lol

  28. lilsis76
    • 2 years ago
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    haha, dang...okay hmmm.. okay so i just plug in right? . lol

  29. zepdrix
    • 2 years ago
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    So looking at our triangle we drew on the coordinate plane, can you determine cos(s)?

  30. lilsis76
    • 2 years ago
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    well the cos s. would be the path. the. that equaled 5, right?

  31. zepdrix
    • 2 years ago
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    Cosine would be Adjacent over Hypotenuse, soh cah toa :D CAH! C=A/H

  32. zepdrix
    • 2 years ago
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    If this had been the UNIT circle, then yes you would be correct :)

  33. lilsis76
    • 2 years ago
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    lol ya, so 5 right? wait.....lol hold on

  34. zepdrix
    • 2 years ago
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    Look at the triangle. The Hypotenuse is ummm 13 I think.

  35. lilsis76
    • 2 years ago
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    i got 12/13=.923

  36. zepdrix
    • 2 years ago
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    woops you calculated sin again :O

  37. zepdrix
    • 2 years ago
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    12 is the side OPPOSITE s.

  38. lilsis76
    • 2 years ago
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    what/! haha hold on

  39. lilsis76
    • 2 years ago
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    okay i got 13/12 = 1.0833

  40. zepdrix
    • 2 years ago
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    5 is the side ADJACENT to s. 13 is the HYPOTENUSE :D silly sis ^^

  41. zepdrix
    • 2 years ago
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    Keep in mind we're going to leave it as a FRACTION, they want an EXACT answer.

  42. lilsis76
    • 2 years ago
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    okay

  43. zepdrix
    • 2 years ago
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    sin(s)=12/13, we'll be leaving it that way. We want to find cos(s), it should be similar, but it will relate 2 other sides together. not the 12 and 13.

  44. lilsis76
    • 2 years ago
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    so 5/13 = .3846

  45. zepdrix
    • 2 years ago
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    Ok cool c:

  46. zepdrix
    • 2 years ago
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    \[\large \sin(s)=\frac{12}{13}, \quad \cos(s)=-\frac{5}{13}\] \[\large \cos(t) \quad \rightarrow \quad \cos\left(\frac{5\pi}{3}\right)=?\]\[\large \sin(t) \quad \rightarrow \quad \sin\left(\frac{5\pi}{3}\right)=?\]

  47. zepdrix
    • 2 years ago
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    Ah crap I just noticed, the cosine should be a NEGATIVE value, remember why?

  48. lilsis76
    • 2 years ago
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    because its in the 3rd quad?

  49. lilsis76
    • 2 years ago
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    AH! I means 2nd

  50. zepdrix
    • 2 years ago
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    lol ya c: because of that.

  51. zepdrix
    • 2 years ago
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    now you have to think back to your special angles. What do you get for sin(t) and cos(t)? :o

  52. lilsis76
    • 2 years ago
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    crud, i forgot the special angles :/ okay um...sin t would be.... - 1/2 ?

  53. zepdrix
    • 2 years ago
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    Ummm let's think. This is an angle where the y value is larger. So the sine is going to be the bigger one, I think it's -sqrt3/2.

  54. lilsis76
    • 2 years ago
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    oh man :? im so bad at this haha your right

  55. zepdrix
    • 2 years ago
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    lol :D

  56. lilsis76
    • 2 years ago
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    the cos is going to be 1/2 then

  57. zepdrix
    • 2 years ago
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    K sounds good :D positive right? since we're in the 4th.

  58. lilsis76
    • 2 years ago
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    yes. but how did we get from 2nd to fourth? haha i just noticed that

  59. zepdrix
    • 2 years ago
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    So your problem should give you something like this,\[\large \cos(s-t)=\cos(s)\cos(t)+\sin(s)\sin(t)\] \[\large \rightarrow \quad \left(-\frac{5}{13}\right)\cdot\left(\frac{1}{2}\right)+\left(\frac{12}{13}\right)\cdot\left(-\frac{\sqrt3}{2}\right)\]

  60. zepdrix
    • 2 years ago
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    Oh S was in 2nd :) t was in 4th

  61. zepdrix
    • 2 years ago
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    They're making us jump around and use 2 different angles.

  62. lilsis76
    • 2 years ago
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    WAIT! i see it, tan t is - sqrt 3, so that means -sin/+ cos haha okay 4th Q

  63. zepdrix
    • 2 years ago
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    Oh also, they told us that 3pi/2 < t < 2pi These are the boundaries on t, remember where this region lies? :D

  64. zepdrix
    • 2 years ago
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    -sqrt3 isn't quite enough information. Because tan t could be +sin/-cos right? :) Which is second quad.

  65. lilsis76
    • 2 years ago
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    oh...thats true

  66. lilsis76
    • 2 years ago
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    okay so now i solve that equation?? or do we gotta get another anser?

  67. zepdrix
    • 2 years ago
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    See the big ugly fractions I plugged in? We just need to simplify that down a tiny bit, NO DECIMAL, and we're finished.

  68. lilsis76
    • 2 years ago
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    haha okay hold on let me try it

  69. lilsis76
    • 2 years ago
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    i got - 5/12, haha and i cant do the other one

  70. lilsis76
    • 2 years ago
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    OH! -5/26 sorry haha

  71. zepdrix
    • 2 years ago
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    It won't simply very far :) We're just combining fractions. Giving you something like this,\[\huge \cos(s-t)\quad =\quad \frac{-5-12\sqrt3}{26}\]

  72. zepdrix
    • 2 years ago
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    Sounds like you're on the right track though :D

  73. lilsis76
    • 2 years ago
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    dang. haha thank you so much

  74. lilsis76
    • 2 years ago
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    just let me know when you have to get off okay :)

  75. zepdrix
    • 2 years ago
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    I'm not positive that's correct. It's a rather burdensome problem. Hopefully we didn't make any silly mistakes in there. :D

  76. zepdrix
    • 2 years ago
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    Post your new questions in NEW threads. That way, just in case I can't get to them, someone else won't have to scroll AAAALL the way down this thread to find your question. :D

  77. lilsis76
    • 2 years ago
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    okay :)

  78. phi
    • 2 years ago
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    fyi, you got the exact answer \[ \cos(s-t)= \frac{-(5+12\sqrt{3})}{26} \]

  79. phi
    • 2 years ago
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    btw, it is helpful to remember that if the lengths of a triangle are 3,,4,5 or 5,12,13 they are right triangles (it saves time solving for the 3rd side if you know this) also, if all 3 legs are multiplied by the same number (example: 10,24,26 (5,12,13 all multiplied by 2), then you still have a right triangle.

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