Finding CscØ: CosØ & TanØ Info

  • Thread starter Poweranimals
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In summary: I let you think about that!In summary, the conversation discusses the relationship between cscØ, sinØ, cosØ, and tanØ. It is determined that cscØ = 1/sinØ, and that the square of cosØ can be used to find the square of sinØ. Through algebraic manipulation, it is determined that sinØ must be equal to -0.866025403, and therefore cscØ = -2/sqrt(3).
  • #1
Poweranimals
68
0
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.
 
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  • #2
What is the relation between cscØ and sinØ?
 
  • #3
Well, my book says that cscØ = (fraction: 1/sinØ).
 
  • #4
And what's the relation between tanØ, sinØ and cosØ?
 
  • #5
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?
 
  • #6
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.
 
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  • #7
I'd kind of rather you spoil it. ;) For the life of me, I don't understand all of this sine, cosine, tangent, ect.
 
  • #8
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?
 
  • #9
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.
 
  • #10
Do you agree that you now have answered questions 1&2?
What's then the answer to 3?
 
  • #11
The sign of cscØ must be negative.
 
  • #12
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!
 
  • #13
sin² +cos² = 1 ?

Edits: So if cos = 1/2, then sin would have to be 1/2 as well.
 
  • #14
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?
 
  • #15
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?
 
  • #16
Read my last post; sinØ is not equal to 1/2!
 
  • #17
arildno said:
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
 
  • #18
It was NOT the square root of cosØ (=1/2) I asked for, but [tex]\cos^{2}\phi[/tex]!
(That is, the square)
This is the term appearing in your equation (you should be able to write it as a fraction)
 
  • #19
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.
 
  • #20
You have:
[tex]cos^{2}\phi+\sin^{2}\phi=1[/tex]
Right?
Now:
[tex]cos^{2}\phi=\cos\phi*\cos\phi=\frac{1}{2}*\frac{1}{2}=\frac{1}{4}[/tex]
Right?
Now, try to use this information to determine [tex]\sin^{2}\phi[/tex]
 
  • #21
So would I be correct in assuming that:

[tex]sin^{2}\phi=\sin\phi*\sin\phi=\frac{1}{2}*\frac{1}{2}=\frac{1}{4}[/tex]

?
 
  • #22
Why?
Your equation now reads, by substituting for the square of cos:
[tex]\frac{1}{4}+\sin^{2}\phi=1[/tex]
 
  • #23
So then:

[tex]sin^{2}\phi=\sin\phi*\sin\phi=\frac{3}{8}*\frac{1}{2}=\frac{3}{4}[/tex]

?
 
  • #24
It is true that [tex]\sin^{2}\phi=\frac{3}{4}[/tex]
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??
 
  • #25
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 [tex]\frac{3}{4}[/tex]/2

Edit: and that fraction isn't even right either, since it calculates out to 3/4 before even multiplying it by itself.
 
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  • #26
Well we're agreed that we have:
[tex]\frac{1}{4}+\sin^{2}\phi=1[/tex]
Now move the 1/4 over to the right-hand side:
[tex]\sin^{2}\phi=1-\frac{1}{4}[/tex]
But this can be simplified to:
[tex]\sin^{2}\phi=\frac{3}{4}[/tex]
Given from before that [tex]\sin\phi[/tex] is a negative number, what number does the last equation tell you that [tex]\sin\phi[/tex] must be?
 
  • #27
-0.866025403?
 
  • #28
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?
 
  • #29
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.
 
  • #30
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]
 
  • #31
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.
 
  • #32
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)
 
  • #33
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]
 
  • #34
Wouldn't it be -2 square root of 3/3?
 
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  • #35
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]
 

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