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

- lilsis76

- jamiebookeater

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- lilsis76

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

- 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

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

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

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

- zepdrix

|dw:1355197009318:dw|

- zepdrix

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

- lilsis76

how was the cosine negative?

- zepdrix

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

- zepdrix

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

- zepdrix

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

- lilsis76

okay

- 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

pi/3 is the 60 degree

- zepdrix

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

- 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

aw :D

- lilsis76

yes its 5pi/3

- lilsis76

i checked by drawing haha

- zepdrix

ah good :)

- zepdrix

|dw:1355197419619:dw|

- lilsis76

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

- 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

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

- zepdrix

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

- 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

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

- lilsis76

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

- zepdrix

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

- lilsis76

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

- zepdrix

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

- zepdrix

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

- lilsis76

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

- zepdrix

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

- lilsis76

i got 12/13=.923

- zepdrix

woops you calculated sin again :O

- zepdrix

12 is the side OPPOSITE s.

- lilsis76

what/! haha hold on

- lilsis76

okay i got 13/12 = 1.0833

- zepdrix

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

- zepdrix

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

- lilsis76

okay

- 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

so 5/13 = .3846

- zepdrix

Ok cool c:

- 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

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

- lilsis76

because its in the 3rd quad?

- lilsis76

AH! I means 2nd

- zepdrix

lol ya c: because of that.

- zepdrix

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

- lilsis76

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

- 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

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

- zepdrix

lol :D

- lilsis76

the cos is going to be 1/2 then

- zepdrix

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

- lilsis76

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

- 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

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

- zepdrix

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

- lilsis76

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

- zepdrix

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

- zepdrix

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

- lilsis76

oh...thats true

- lilsis76

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

- 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

haha okay hold on let me try it

- lilsis76

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

- lilsis76

OH! -5/26 sorry haha

- 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

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

- lilsis76

dang. haha thank you so much

- lilsis76

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

- 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

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

okay :)

- phi

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

- 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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