## lilsis76 Group Title 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

1. lilsis76 Group Title

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

2. zepdrix Group Title

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 Group Title

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 Group Title

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 Group Title

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

6. zepdrix Group Title

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7. zepdrix Group Title

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

8. lilsis76 Group Title

how was the cosine negative?

9. zepdrix Group Title

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

10. zepdrix Group Title

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

11. zepdrix Group Title

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

12. lilsis76 Group Title

okay

13. zepdrix Group Title

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 Group Title

pi/3 is the 60 degree

15. zepdrix Group Title

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

16. lilsis76 Group Title

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 Group Title

aw :D

18. lilsis76 Group Title

yes its 5pi/3

19. lilsis76 Group Title

i checked by drawing haha

20. zepdrix Group Title

ah good :)

21. zepdrix Group Title

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22. lilsis76 Group Title

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

23. zepdrix Group Title

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 Group Title

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

25. zepdrix Group Title

$\large \cos(s-t)=\cos(s)\cos(t)+\sin(s)\sin(t)$

26. zepdrix Group Title

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 Group Title

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

28. lilsis76 Group Title

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

29. zepdrix Group Title

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

30. lilsis76 Group Title

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

31. zepdrix Group Title

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

32. zepdrix Group Title

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

33. lilsis76 Group Title

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

34. zepdrix Group Title

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

35. lilsis76 Group Title

i got 12/13=.923

36. zepdrix Group Title

woops you calculated sin again :O

37. zepdrix Group Title

12 is the side OPPOSITE s.

38. lilsis76 Group Title

what/! haha hold on

39. lilsis76 Group Title

okay i got 13/12 = 1.0833

40. zepdrix Group Title

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

41. zepdrix Group Title

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

42. lilsis76 Group Title

okay

43. zepdrix Group Title

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 Group Title

so 5/13 = .3846

45. zepdrix Group Title

Ok cool c:

46. zepdrix Group Title

$\large \sin(s)=\frac{12}{13}, \quad \cos(s)=-\frac{5}{13}$ $\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 Group Title

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

48. lilsis76 Group Title

because its in the 3rd quad?

49. lilsis76 Group Title

AH! I means 2nd

50. zepdrix Group Title

lol ya c: because of that.

51. zepdrix Group Title

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

52. lilsis76 Group Title

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

53. zepdrix Group Title

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 Group Title

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

55. zepdrix Group Title

lol :D

56. lilsis76 Group Title

the cos is going to be 1/2 then

57. zepdrix Group Title

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

58. lilsis76 Group Title

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

59. zepdrix Group Title

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 Group Title

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

61. zepdrix Group Title

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

62. lilsis76 Group Title

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

63. zepdrix Group Title

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

64. zepdrix Group Title

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

65. lilsis76 Group Title

oh...thats true

66. lilsis76 Group Title

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

67. zepdrix Group Title

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 Group Title

haha okay hold on let me try it

69. lilsis76 Group Title

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

70. lilsis76 Group Title

OH! -5/26 sorry haha

71. zepdrix Group Title

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 Group Title

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

73. lilsis76 Group Title

dang. haha thank you so much

74. lilsis76 Group Title

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

75. zepdrix Group Title

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 Group Title

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 Group Title

okay :)

78. phi Group Title

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

79. phi Group Title

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.