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Finding the cscØ?

  1. Sep 20, 2004 #1
    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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  3. Sep 20, 2004 #2

    arildno

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    What is the relation between cscØ and sinØ?
     
  4. Sep 20, 2004 #3
    Well, my book says that cscØ = (fraction: 1/sinØ).
     
  5. Sep 20, 2004 #4
    And what's the relation between tanØ, sinØ and cosØ?
     
  6. Sep 20, 2004 #5

    HallsofIvy

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    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?
     
  7. Sep 20, 2004 #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.
     
    Last edited: Sep 20, 2004
  8. Sep 20, 2004 #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.
     
  9. Sep 20, 2004 #8

    arildno

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    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?
     
  10. Sep 20, 2004 #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.
     
  11. Sep 20, 2004 #10

    arildno

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    Do you agree that you now have answered questions 1&2?
    What's then the answer to 3?
     
  12. Sep 20, 2004 #11
    The sign of cscØ must be negative.
     
  13. Sep 20, 2004 #12

    arildno

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    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!
     
  14. Sep 20, 2004 #13
    sin² +cos² = 1 ?

    Edits: So if cos = 1/2, then sin would have to be 1/2 as well.
     
  15. Sep 20, 2004 #14

    arildno

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    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?
     
  16. Sep 20, 2004 #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?
     
  17. Sep 20, 2004 #16

    arildno

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    Read my last post; sinØ is not equal to 1/2!
     
  18. Sep 20, 2004 #17
    The square root of 1/2 turns up 0.707106781
     
  19. Sep 20, 2004 #18

    arildno

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    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)
     
  20. Sep 20, 2004 #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.
     
  21. Sep 20, 2004 #20

    arildno

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    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]
     
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