## 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)

1. lilsis76

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

2. zepdrix

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

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

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

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

6. zepdrix

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7. zepdrix

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

8. lilsis76

how was the cosine negative?

9. zepdrix

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

10. zepdrix

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

11. zepdrix

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

12. lilsis76

okay

13. zepdrix

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

pi/3 is the 60 degree

15. zepdrix

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

16. lilsis76

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

aw :D

18. lilsis76

yes its 5pi/3

19. lilsis76

i checked by drawing haha

20. zepdrix

ah good :)

21. zepdrix

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22. lilsis76

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

23. zepdrix

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

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

25. zepdrix

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

26. zepdrix

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

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

28. lilsis76

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

29. zepdrix

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

30. lilsis76

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

31. zepdrix

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

32. zepdrix

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

33. lilsis76

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

34. zepdrix

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

35. lilsis76

i got 12/13=.923

36. zepdrix

woops you calculated sin again :O

37. zepdrix

12 is the side OPPOSITE s.

38. lilsis76

what/! haha hold on

39. lilsis76

okay i got 13/12 = 1.0833

40. zepdrix

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

41. zepdrix

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

42. lilsis76

okay

43. zepdrix

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

so 5/13 = .3846

45. zepdrix

Ok cool c:

46. zepdrix

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

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

48. lilsis76

because its in the 3rd quad?

49. lilsis76

AH! I means 2nd

50. zepdrix

lol ya c: because of that.

51. zepdrix

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

52. lilsis76

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

53. zepdrix

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

55. zepdrix

lol :D

56. lilsis76

the cos is going to be 1/2 then

57. zepdrix

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

58. lilsis76

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

59. zepdrix

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

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

61. zepdrix

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

62. lilsis76

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

63. zepdrix

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

64. zepdrix

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

65. lilsis76

oh...thats true

66. lilsis76

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

67. zepdrix

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

haha okay hold on let me try it

69. lilsis76

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

70. lilsis76

OH! -5/26 sorry haha

71. zepdrix

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

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

73. lilsis76

dang. haha thank you so much

74. lilsis76

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

75. zepdrix

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

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

okay :)

78. phi

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

79. phi

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.