lilsis76
  • 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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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lilsis76
  • lilsis76
@phi or @jim_thompson5910 or @zepdrix can anyone of you explain this to me please. thank you so much
zepdrix
  • 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.
lilsis76
  • 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.

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More answers

zepdrix
  • 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.\[\csc(s)=\frac{13}{12} \quad \rightarrow \quad \frac{1}{\sin(s)}=\frac{13}{12} \quad \rightarrow \quad \sin(s)=\frac{12}{13}\]
zepdrix
  • zepdrix
Sine positive, cosine negative, so we're in the 2nd quadrant.
zepdrix
  • zepdrix
|dw:1355197009318:dw|
zepdrix
  • zepdrix
They told us that t lies in the 4th quadrant. And they gave us a special angle for t. \[\tan(t)=-\sqrt3\]
lilsis76
  • lilsis76
how was the cosine negative?
zepdrix
  • zepdrix
cos(s) < 0 was in the problems description. Less than zero means it's negative :)
zepdrix
  • zepdrix
Lots of little details to this problem, it's easy to miss stuff :D heh
zepdrix
  • zepdrix
So hmm, let's try to solve for t.
lilsis76
  • lilsis76
okay
zepdrix
  • 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
lilsis76
  • lilsis76
pi/3 is the 60 degree
zepdrix
  • zepdrix
So in the 4th quadrant is it, 5pi/3 maybe?
lilsis76
  • 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 :)
zepdrix
  • zepdrix
aw :D
lilsis76
  • lilsis76
yes its 5pi/3
lilsis76
  • lilsis76
i checked by drawing haha
zepdrix
  • zepdrix
ah good :)
zepdrix
  • zepdrix
|dw:1355197419619:dw|
lilsis76
  • lilsis76
by doing it the same way? but wouldnt we get....hm...the same 12 and 13?
zepdrix
  • 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
zepdrix
  • zepdrix
\[\large \cos(s-t)\]So what I did from here is... Apply the Difference Formula for Cosine. Remember the Sum and Difference formulas?
zepdrix
  • zepdrix
\[\large \cos(s-t)=\cos(s)\cos(t)+\sin(s)\sin(t)\]
zepdrix
  • 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).
zepdrix
  • zepdrix
This is a pretty rough problem :D I don't remember doing anything like this in trig lol
lilsis76
  • lilsis76
haha, dang...okay hmmm.. okay so i just plug in right? . lol
zepdrix
  • zepdrix
So looking at our triangle we drew on the coordinate plane, can you determine cos(s)?
lilsis76
  • lilsis76
well the cos s. would be the path. the. that equaled 5, right?
zepdrix
  • zepdrix
Cosine would be Adjacent over Hypotenuse, soh cah toa :D CAH! C=A/H
zepdrix
  • zepdrix
If this had been the UNIT circle, then yes you would be correct :)
lilsis76
  • lilsis76
lol ya, so 5 right? wait.....lol hold on
zepdrix
  • zepdrix
Look at the triangle. The Hypotenuse is ummm 13 I think.
lilsis76
  • lilsis76
i got 12/13=.923
zepdrix
  • zepdrix
woops you calculated sin again :O
zepdrix
  • zepdrix
12 is the side OPPOSITE s.
lilsis76
  • lilsis76
what/! haha hold on
lilsis76
  • lilsis76
okay i got 13/12 = 1.0833
zepdrix
  • zepdrix
5 is the side ADJACENT to s. 13 is the HYPOTENUSE :D silly sis ^^
zepdrix
  • zepdrix
Keep in mind we're going to leave it as a FRACTION, they want an EXACT answer.
lilsis76
  • lilsis76
okay
zepdrix
  • 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.
lilsis76
  • lilsis76
so 5/13 = .3846
zepdrix
  • zepdrix
Ok cool c:
zepdrix
  • zepdrix
\[\large \sin(s)=\frac{12}{13}, \quad \cos(s)=-\frac{5}{13}\] \[\large \cos(t) \quad \rightarrow \quad \cos\left(\frac{5\pi}{3}\right)=?\]\[\large \sin(t) \quad \rightarrow \quad \sin\left(\frac{5\pi}{3}\right)=?\]
zepdrix
  • zepdrix
Ah crap I just noticed, the cosine should be a NEGATIVE value, remember why?
lilsis76
  • lilsis76
because its in the 3rd quad?
lilsis76
  • lilsis76
AH! I means 2nd
zepdrix
  • zepdrix
lol ya c: because of that.
zepdrix
  • zepdrix
now you have to think back to your special angles. What do you get for sin(t) and cos(t)? :o
lilsis76
  • lilsis76
crud, i forgot the special angles :/ okay um...sin t would be.... - 1/2 ?
zepdrix
  • 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.
lilsis76
  • lilsis76
oh man :? im so bad at this haha your right
zepdrix
  • zepdrix
lol :D
lilsis76
  • lilsis76
the cos is going to be 1/2 then
zepdrix
  • zepdrix
K sounds good :D positive right? since we're in the 4th.
lilsis76
  • lilsis76
yes. but how did we get from 2nd to fourth? haha i just noticed that
zepdrix
  • zepdrix
So your problem should give you something like this,\[\large \cos(s-t)=\cos(s)\cos(t)+\sin(s)\sin(t)\] \[\large \rightarrow \quad \left(-\frac{5}{13}\right)\cdot\left(\frac{1}{2}\right)+\left(\frac{12}{13}\right)\cdot\left(-\frac{\sqrt3}{2}\right)\]
zepdrix
  • zepdrix
Oh S was in 2nd :) t was in 4th
zepdrix
  • zepdrix
They're making us jump around and use 2 different angles.
lilsis76
  • lilsis76
WAIT! i see it, tan t is - sqrt 3, so that means -sin/+ cos haha okay 4th Q
zepdrix
  • zepdrix
Oh also, they told us that 3pi/2 < t < 2pi These are the boundaries on t, remember where this region lies? :D
zepdrix
  • zepdrix
-sqrt3 isn't quite enough information. Because tan t could be +sin/-cos right? :) Which is second quad.
lilsis76
  • lilsis76
oh...thats true
lilsis76
  • lilsis76
okay so now i solve that equation?? or do we gotta get another anser?
zepdrix
  • 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.
lilsis76
  • lilsis76
haha okay hold on let me try it
lilsis76
  • lilsis76
i got - 5/12, haha and i cant do the other one
lilsis76
  • lilsis76
OH! -5/26 sorry haha
zepdrix
  • zepdrix
It won't simply very far :) We're just combining fractions. Giving you something like this,\[\huge \cos(s-t)\quad =\quad \frac{-5-12\sqrt3}{26}\]
zepdrix
  • zepdrix
Sounds like you're on the right track though :D
lilsis76
  • lilsis76
dang. haha thank you so much
lilsis76
  • lilsis76
just let me know when you have to get off okay :)
zepdrix
  • 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
zepdrix
  • 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
lilsis76
  • lilsis76
okay :)
phi
  • phi
fyi, you got the exact answer \[ \cos(s-t)= \frac{-(5+12\sqrt{3})}{26} \]
phi
  • 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.

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