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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)
@phi or @jim_thompson5910 or @zepdrix can anyone of you explain this to me please. thank you so much
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.
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.
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.\[\csc(s)=\frac{13}{12} \quad \rightarrow \quad \frac{1}{\sin(s)}=\frac{13}{12} \quad \rightarrow \quad \sin(s)=\frac{12}{13}\]
Sine positive, cosine negative, so we're in the 2nd quadrant.
They told us that t lies in the 4th quadrant. And they gave us a special angle for t. \[\tan(t)=-\sqrt3\]
how was the cosine negative?
cos(s) < 0 was in the problems description. Less than zero means it's negative :)
Lots of little details to this problem, it's easy to miss stuff :D heh
So hmm, let's try to solve for t.
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
So in the 4th quadrant is it, 5pi/3 maybe?
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 :)
i checked by drawing haha
by doing it the same way? but wouldnt we get....hm...the same 12 and 13?
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
\[\large \cos(s-t)\]So what I did from here is... Apply the Difference Formula for Cosine. Remember the Sum and Difference formulas?
\[\large \cos(s-t)=\cos(s)\cos(t)+\sin(s)\sin(t)\]
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).
This is a pretty rough problem :D I don't remember doing anything like this in trig lol
haha, dang...okay hmmm.. okay so i just plug in right? . lol
So looking at our triangle we drew on the coordinate plane, can you determine cos(s)?
well the cos s. would be the path. the. that equaled 5, right?
Cosine would be Adjacent over Hypotenuse, soh cah toa :D CAH! C=A/H
If this had been the UNIT circle, then yes you would be correct :)
lol ya, so 5 right? wait.....lol hold on
Look at the triangle. The Hypotenuse is ummm 13 I think.
woops you calculated sin again :O
12 is the side OPPOSITE s.
okay i got 13/12 = 1.0833
5 is the side ADJACENT to s. 13 is the HYPOTENUSE :D silly sis ^^
Keep in mind we're going to leave it as a FRACTION, they want an EXACT answer.
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.
\[\large \sin(s)=\frac{12}{13}, \quad \cos(s)=-\frac{5}{13}\] \[\large \cos(t) \quad \rightarrow \quad \cos\left(\frac{5\pi}{3}\right)=?\]\[\large \sin(t) \quad \rightarrow \quad \sin\left(\frac{5\pi}{3}\right)=?\]
Ah crap I just noticed, the cosine should be a NEGATIVE value, remember why?
because its in the 3rd quad?
lol ya c: because of that.
now you have to think back to your special angles. What do you get for sin(t) and cos(t)? :o
crud, i forgot the special angles :/ okay um...sin t would be.... - 1/2 ?
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.
oh man :? im so bad at this haha your right
the cos is going to be 1/2 then
K sounds good :D positive right? since we're in the 4th.
yes. but how did we get from 2nd to fourth? haha i just noticed that
So your problem should give you something like this,\[\large \cos(s-t)=\cos(s)\cos(t)+\sin(s)\sin(t)\] \[\large \rightarrow \quad \left(-\frac{5}{13}\right)\cdot\left(\frac{1}{2}\right)+\left(\frac{12}{13}\right)\cdot\left(-\frac{\sqrt3}{2}\right)\]
Oh S was in 2nd :) t was in 4th
They're making us jump around and use 2 different angles.
WAIT! i see it, tan t is - sqrt 3, so that means -sin/+ cos haha okay 4th Q
Oh also, they told us that 3pi/2 < t < 2pi These are the boundaries on t, remember where this region lies? :D
-sqrt3 isn't quite enough information. Because tan t could be +sin/-cos right? :) Which is second quad.
okay so now i solve that equation?? or do we gotta get another anser?
See the big ugly fractions I plugged in? We just need to simplify that down a tiny bit, NO DECIMAL, and we're finished.
haha okay hold on let me try it
i got - 5/12, haha and i cant do the other one
It won't simply very far :) We're just combining fractions. Giving you something like this,\[\huge \cos(s-t)\quad =\quad \frac{-5-12\sqrt3}{26}\]
Sounds like you're on the right track though :D
dang. haha thank you so much
just let me know when you have to get off okay :)
I'm not positive that's correct. It's a rather burdensome problem. Hopefully we didn't make any silly mistakes in there. :D
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
fyi, you got the exact answer \[ \cos(s-t)= \frac{-(5+12\sqrt{3})}{26} \]
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.