Question

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

Find exact value of cos(s-t)

Answer 1

@phi or @jim_thompson5910 or @zepdrix

can anyone of you explain this to me please. thank you so much

Answer 2

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.

Answer 3

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.

Answer 4

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.\[\csc(s)=\frac{13}{12} \quad \rightarrow \quad \frac{1}{\sin(s)}=\frac{13}{12} \quad \rightarrow \quad \sin(s)=\frac{12}{13}\]

Answer 5

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

Answer 6

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

Answer 7

how was the cosine negative?

Answer 8

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

Answer 9

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

Answer 10

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