Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

lilsis76

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

  • one year ago
  • one year ago

  • This Question is Closed
  1. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    @phi or @jim_thompson5910 or @zepdrix can anyone of you explain this to me please. thank you so much

    • one year ago
  2. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • one year ago
  3. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • one year ago
  4. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    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.\[\csc(s)=\frac{13}{12} \quad \rightarrow \quad \frac{1}{\sin(s)}=\frac{13}{12} \quad \rightarrow \quad \sin(s)=\frac{12}{13}\]

    • one year ago
  5. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    Sine positive, cosine negative, so we're in the 2nd quadrant.

    • one year ago
  6. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1355197009318:dw|

    • one year ago
  7. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  8. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    how was the cosine negative?

    • one year ago
  9. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    cos(s) < 0 was in the problems description. Less than zero means it's negative :)

    • one year ago
  10. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    Lots of little details to this problem, it's easy to miss stuff :D heh

    • one year ago
  11. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    So hmm, let's try to solve for t.

    • one year ago
  12. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    okay

    • one year ago
  13. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • one year ago
  14. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    pi/3 is the 60 degree

    • one year ago
  15. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    So in the 4th quadrant is it, 5pi/3 maybe?

    • one year ago
  16. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    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 :)

    • one year ago
  17. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    aw :D

    • one year ago
  18. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    yes its 5pi/3

    • one year ago
  19. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    i checked by drawing haha

    • one year ago
  20. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    ah good :)

    • one year ago
  21. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1355197419619:dw|

    • one year ago
  22. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    by doing it the same way? but wouldnt we get....hm...the same 12 and 13?

    • one year ago
  23. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • one year ago
  24. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  25. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\large \cos(s-t)=\cos(s)\cos(t)+\sin(s)\sin(t)\]

    • one year ago
  26. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    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).

    • one year ago
  27. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    This is a pretty rough problem :D I don't remember doing anything like this in trig lol

    • one year ago
  28. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    haha, dang...okay hmmm.. okay so i just plug in right? . lol

    • one year ago
  29. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    So looking at our triangle we drew on the coordinate plane, can you determine cos(s)?

    • one year ago
  30. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    well the cos s. would be the path. the. that equaled 5, right?

    • one year ago
  31. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    Cosine would be Adjacent over Hypotenuse, soh cah toa :D CAH! C=A/H

    • one year ago
  32. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    If this had been the UNIT circle, then yes you would be correct :)

    • one year ago
  33. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    lol ya, so 5 right? wait.....lol hold on

    • one year ago
  34. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    Look at the triangle. The Hypotenuse is ummm 13 I think.

    • one year ago
  35. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    i got 12/13=.923

    • one year ago
  36. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    woops you calculated sin again :O

    • one year ago
  37. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    12 is the side OPPOSITE s.

    • one year ago
  38. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    what/! haha hold on

    • one year ago
  39. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    okay i got 13/12 = 1.0833

    • one year ago
  40. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    5 is the side ADJACENT to s. 13 is the HYPOTENUSE :D silly sis ^^

    • one year ago
  41. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    Keep in mind we're going to leave it as a FRACTION, they want an EXACT answer.

    • one year ago
  42. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    okay

    • one year ago
  43. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • one year ago
  44. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    so 5/13 = .3846

    • one year ago
  45. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    Ok cool c:

    • one year ago
  46. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\large \sin(s)=\frac{12}{13}, \quad \cos(s)=-\frac{5}{13}\] \[\large \cos(t) \quad \rightarrow \quad \cos\left(\frac{5\pi}{3}\right)=?\]\[\large \sin(t) \quad \rightarrow \quad \sin\left(\frac{5\pi}{3}\right)=?\]

    • one year ago
  47. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    Ah crap I just noticed, the cosine should be a NEGATIVE value, remember why?

    • one year ago
  48. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    because its in the 3rd quad?

    • one year ago
  49. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    AH! I means 2nd

    • one year ago
  50. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    lol ya c: because of that.

    • one year ago
  51. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    now you have to think back to your special angles. What do you get for sin(t) and cos(t)? :o

    • one year ago
  52. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  53. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • one year ago
  54. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    oh man :? im so bad at this haha your right

    • one year ago
  55. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    lol :D

    • one year ago
  56. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    the cos is going to be 1/2 then

    • one year ago
  57. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    K sounds good :D positive right? since we're in the 4th.

    • one year ago
  58. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    yes. but how did we get from 2nd to fourth? haha i just noticed that

    • one year ago
  59. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  60. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh S was in 2nd :) t was in 4th

    • one year ago
  61. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    They're making us jump around and use 2 different angles.

    • one year ago
  62. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    WAIT! i see it, tan t is - sqrt 3, so that means -sin/+ cos haha okay 4th Q

    • one year ago
  63. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  64. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  65. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    oh...thats true

    • one year ago
  66. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    okay so now i solve that equation?? or do we gotta get another anser?

    • one year ago
  67. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  68. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    haha okay hold on let me try it

    • one year ago
  69. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    i got - 5/12, haha and i cant do the other one

    • one year ago
  70. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    OH! -5/26 sorry haha

    • one year ago
  71. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  72. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    Sounds like you're on the right track though :D

    • one year ago
  73. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    dang. haha thank you so much

    • one year ago
  74. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    just let me know when you have to get off okay :)

    • one year ago
  75. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  76. zepdrix
    Best Response
    You've already chosen the best response.
    Medals 1

    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

    • one year ago
  77. lilsis76
    Best Response
    You've already chosen the best response.
    Medals 0

    okay :)

    • one year ago
  78. phi
    Best Response
    You've already chosen the best response.
    Medals 1

    fyi, you got the exact answer \[ \cos(s-t)= \frac{-(5+12\sqrt{3})}{26} \]

    • one year ago
  79. phi
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.