How Do You Determine CscØ with CosØ and TanØ?

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arildno said:
BTW, the positive square root of 3/4 may be written as follows:
[tex]\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{\sqrt{4}}=\frac{\sqrt{3}}{2}[/tex]
So then the answer would be -1/ [tex]\sqrt{\frac{3}{2}[/tex]?

But again, something seems wrong.
 
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Why don't you think it is right?
It is; however, why are you dissatisfied with it?
(It is important when trying to learn maths to express your own ideas/doubts, otherwise other persons won't be able to show you the right way in a manner you understand/are satisfied with)
 
Poweranimals said:
So then the answer would be -1/ [tex]\sqrt{\frac{3}{2}[/tex]?
Certainly, this can be simplified as:
[tex]csc\phi=\frac{-1}{\frac{\sqrt{3}}{2}}=-\frac{2}{\sqrt{3}}[/tex]
 
Wouldn't it be -2 square root of 3/3?
 
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Isn't that what I wrote?
Besides, I just saw that you wrote in post 31 that
"So then the answer would be [tex]\frac{-1}{\sqrt{\frac{3}{2}}}[/tex] ?
This is wrong; it should be:

[tex]\frac{-1}{\frac{\sqrt{3}}{2}}[/tex]
 
Poweranimals said:
Wouldn't it be -2 square root of 3/3?
What do you mean by this??
The answer is, again:
[tex]csc\phi=-\frac{2}{\sqrt{3}}[/tex]
EDIT:
Yes, you can write this as:
[tex]csc\phi=-\frac{2}{\sqrt{3}}=-\frac{2\sqrt{3}}{3}[/tex]
 
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Yeah, I meant to write something else, but I kept screwing up the code. Anyway, what you have looks different from what I had, so I got confused.
 
arildno said:
What do you mean by this??
The answer is, again:
[tex]csc\phi=-\frac{2}{\sqrt{3}}[/tex]
EDIT:
Yes, you can write this as:
[tex]csc\phi=-\frac{2}{\sqrt{3}}==-\frac{2\sqrt{3}}{3}[/tex]
Well, judging by the multiple choice answers, the options are:

a) 2, b) -2, c) 2 and sqare root of 3/3, and d) -2 and the square root of 3/3. Meh, I could just guess, but I wouldn't really learn anything; and I've got a test on it tonight.
 
As you see, my edit shows that alternative d) is what you should choose! :smile:

I'm sorry, I don't understand your wording of c) and d)
They're unclear!

I believe it is d) you're after