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Poweranimals
Sep20-04, 05:13 AM
Okay, I'm confused as to how you get the cscØ. The only information I have available is the cosØ, which is 1/2 and the tanØ, which is listed as negative.

arildno
Sep20-04, 05:20 AM
What is the relation between cscØ and sinØ?

Poweranimals
Sep20-04, 05:26 AM
Well, my book says that cscØ = (fraction: 1/sinØ).

devious_
Sep20-04, 05:39 AM
And what's the relation between tanØ, sinØ and cosØ?

HallsofIvy
Sep20-04, 07:21 AM
Since you are given cos(φ) you should be able to find sin(&phi) (that's a famous identity)- except that sin can be positive or negative even when cos is positive. That's why you need the tan. If tan is positive what can you say about the signs of sin and cos? If tan is negative?

relinquished™
Sep20-04, 07:30 AM
Poweranimals, you need to see where the tangent, cosine and sine functions ' signs are in the four quadrants, and then just remember what the tangent, sine and cosine are :smile: You know the definition of cosecant earlier.. That's the best advice (so far) I can give you instead of spoiling the answer.

Poweranimals
Sep20-04, 12:51 PM
I'd kind of rather you spoil it. ;) For the life of me, I don't understand all of this sine, cosine, tangent, ect.

arildno
Sep20-04, 01:12 PM
You should be able to answer the following:
1. How can tanØ be expressed by sinØ and cosØ?
2. Given that cosØ=1/2 and tanØ negative, what must the sign of sinØ be?
3. Given that you know the sign of sinØ, what must the sign of cscØ be?
4. What identity relates the squares of sinØ and cosØ?
5. Given the answers to the above, what value must sinØ have?
6. Hence, what value must cscØ have?

Poweranimals
Sep20-04, 01:28 PM
I think #1 is where I'm having trouble. tanØ = (fraction: sinØ/cosØ), but I'm still not sure how I would get sinØ. But I'm pretty sure that since tanØ is negative, that either sinØ or cosØ would be negative too, but since cosØ is 1/2, it would be positive, so sinØ must be negative, but I'm still not sure how I'd solve for it.

arildno
Sep20-04, 01:35 PM
Do you agree that you now have answered questions 1&2?
What's then the answer to 3?

Poweranimals
Sep20-04, 01:43 PM
The sign of cscØ must be negative.

arildno
Sep20-04, 01:53 PM
OK, so now we have 4,5,6 left!
What you now need is an equation by which you may calculate sinØ!
4. is crucial here; there exists a famous equation relating the squares of sinØ and cosØ.
(This is sufficient for your purpose; since you know cosØ (1/2) you also know the square of cosØ, and hence, by this equation, what the square of sinØ must be)
Try to think of such an equation!

Poweranimals
Sep20-04, 01:57 PM
sin² +cos² = 1 ?

Edits: So if cos = 1/2, then sin would have to be 1/2 as well.

arildno
Sep20-04, 02:05 PM
You're right about the identity!
But:
If cosØ=1/2, what is then the square of cosØ?
And, what must therefore the square of sinØ be?

Poweranimals
Sep20-04, 02:06 PM
Wait, so if csc = 1 / 1/2, then that would be the same as 1 * 2, which would equal 2.. and since the sign of the csc is negative, then the final answer should be -2, right?

arildno
Sep20-04, 02:08 PM
Read my last post; sinØ is not equal to 1/2!

Poweranimals
Sep20-04, 02:15 PM
You're right about the identity!
But:
If cosØ=1/2, what is then the square of cosØ?
And, what must therefore the square of sinØ be?
The square root of 1/2 turns up 0.707106781

arildno
Sep20-04, 02:17 PM
It was NOT the square root of cosØ (=1/2) I asked for, but \cos^{2}\phi!!
(That is, the square)
This is the term appearing in your equation (you should be able to write it as a fraction)

Poweranimals
Sep20-04, 02:28 PM
Well, if csc = sin² +cos² = 1, then wouldn't that mean that sin² and cos² both equal 1/2? So I did the square root thing, because the 1/2 is actually the sin or cos squared. I think.

arildno
Sep20-04, 02:32 PM
You have:
cos^{2}\phi+\sin^{2}\phi=1
Right?
Now:
cos^{2}\phi=\cos\phi*\cos\phi=\frac{1}{2}*\frac{1} {2}=\frac{1}{4}
Right?
Now, try to use this information to determine \sin^{2}\phi

Poweranimals
Sep20-04, 02:35 PM
So would I be correct in assuming that:

sin^{2}\phi=\sin\phi*\sin\phi=\frac{1}{2}*\frac{1} {2}=\frac{1}{4}

?

arildno
Sep20-04, 02:43 PM
Why?
Your equation now reads, by substituting for the square of cos:
\frac{1}{4}+\sin^{2}\phi=1

Poweranimals
Sep20-04, 02:52 PM
So then:

sin^{2}\phi=\sin\phi*\sin\phi=\frac{3}{8}*\frac{1} {2}=\frac{3}{4}

?

arildno
Sep20-04, 03:12 PM
It is true that \sin^{2}\phi=\frac{3}{4}
but do you have any understanding whatsoever of mathematics??
Where do you get 1/2 from, and where do you get 3/8 from??
Are they equal??
Do their product equal 3/4??

Poweranimals
Sep20-04, 03:22 PM
Well, I originally thought to get 3/4, you'd have to multiply two of the same number. But I wasn't sure how to get the square root of 3/4 as a fraction. The decimal is 0.866025403, but I'm not sure how that translates to decimal other than \frac{3}{4}/2

Edit: and that fraction isn't even right either, since it calculates out to 3/4 before even multiplying it by itself.

arildno
Sep20-04, 03:34 PM
Well we're agreed that we have:
\frac{1}{4}+\sin^{2}\phi=1
Now move the 1/4 over to the right-hand side:
\sin^{2}\phi=1-\frac{1}{4}
But this can be simplified to:
\sin^{2}\phi=\frac{3}{4}
Given from before that \sin\phi is a negative number, what number does the last equation tell you that \sin\phi must be?

Poweranimals
Sep20-04, 03:45 PM
-0.866025403?

arildno
Sep20-04, 03:47 PM
Now, I don't like decimal numbers as such, because they are so uninformative.
So, if you could explain:
1) How did you get that particular number? (it looks right)
2) How can you now find cscØ?

Poweranimals
Sep20-04, 03:55 PM
I did the square root of 3/4.

But anyway, I guess if csc = 1/sin, which is apparently 1/-0.866025403, then that equals -1.54700593.. but something doesn't seem right.

arildno
Sep20-04, 03:56 PM
BTW, the positive square root of 3/4 may be written as follows:
\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{\sqrt{4}}=\frac {\sqrt{3}}{2}

Poweranimals
Sep20-04, 03:59 PM
BTW, the positive square root of 3/4 may be written as follows:
\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{\sqrt{4}}=\frac {\sqrt{3}}{2}
So then the answer would be -1/ \sqrt{\frac{3}{2}?

But again, something seems wrong.

arildno
Sep20-04, 03:59 PM
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)

arildno
Sep20-04, 04:01 PM
So then the answer would be -1/ \sqrt{\frac{3}{2}?
Certainly, this can be simplified as:
csc\phi=\frac{-1}{\frac{\sqrt{3}}{2}}=-\frac{2}{\sqrt{3}}

Poweranimals
Sep20-04, 04:03 PM
Wouldn't it be -2 square root of 3/3?

arildno
Sep20-04, 04:08 PM
Isn't that what I wrote?
Besides, I just saw that you wrote in post 31 that
"So then the answer would be \frac{-1}{\sqrt{\frac{3}{2}}} ?
This is wrong; it should be:

\frac{-1}{\frac{\sqrt{3}}{2}}

arildno
Sep20-04, 04:09 PM
Wouldn't it be -2 square root of 3/3?
What do you mean by this??
The answer is, again:
csc\phi=-\frac{2}{\sqrt{3}}
EDIT:
Yes, you can write this as:
csc\phi=-\frac{2}{\sqrt{3}}=-\frac{2\sqrt{3}}{3}

Poweranimals
Sep20-04, 04:10 PM
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
Sep20-04, 04:12 PM
I think we're finally landed together here..
At least if you read my previous post..

Poweranimals
Sep20-04, 04:13 PM
What do you mean by this??
The answer is, again:
csc\phi=-\frac{2}{\sqrt{3}}
EDIT:
Yes, you can write this as:
csc\phi=-\frac{2}{\sqrt{3}}==-\frac{2\sqrt{3}}{3}

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.

arildno
Sep20-04, 04:15 PM
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