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Write the trigonometric expression as an algebraic expression in u.

  1. Oct 4, 2011 #1
    1. The problem statement, all variables and given/known data
    csc(cos^-1 u)


    2. Relevant equations
    Fundamental identities properties


    3. The attempt at a solution
    The book is incredibly vague on how to do this. I tried multiplying csc by (cosθ/cosθ) to get cotθ*secθ. I'm not sure if this is the right thing to do, or what to do afterward if it is.
     
    Last edited: Oct 4, 2011
  2. jcsd
  3. Oct 4, 2011 #2
    Try drawing a right triangle and seeing what cos-1u is equal to.
     
  4. Oct 4, 2011 #3
    How am I supposed to find cos^-1 from drawing a right triangle?
     
  5. Oct 4, 2011 #4
    Think of it this way, "the cosecant of the angle who's cosine is u."

    If the cosine is u, what does that say about two sides of the triangle? Remember the cosine is a ratio of two sides, and in our case it is u. Express u as a ratio (think simple) and you have two sides of the triangle.

    You can find the third side via pythagorean, and you have all sides of the triangle. Cosecant is just a different a ratio of two different sides, and you have them all!
     
  6. Oct 4, 2011 #5
    Okay, the first two things you said make sense. I am not getting how I am supposed to get the third side, as I'm not sure what values I am supposed to input into the Pythagorean theorem. I'm frustrated because I've been trying to work this problem for hours, and I'm still lost...
     
  7. Oct 4, 2011 #6
    The angle whose cosine is u. Draw a right triangle, lable one of the acute angles as theta or whatever.

    You know that the cosine of this angle is u. You know that cosine is adjacent over hypotenuse. So, you know what the adjacent side and hypotenuse are.

    If I told you to express 3 or 11, or 92 as the simplest possible ratio, what would you say?

    Do the same for u, and you have adjacent side over hypotenuse. If you have those two sides, that is exactly what you plug into pythagorean theorem.
     
    Last edited: Oct 4, 2011
  8. Oct 4, 2011 #7
    I've still got you here.

    Wait. How do I know this by knowing the definition of cosine? I'm not given any values...

    I don't understand how I am supposed to find a 2nd side from one of the sides...
     
  9. Oct 4, 2011 #8
    I understand your confusion, but I don't want to nudge you too far because these things stick better when you have the "eureka" moment.

    Now, the first thing I want to clear up for you is that we aren't going to end up with a number, we will end up with an identical expression.

    For example, if I say "x + 1" and then ask what x is, well, that doesn't make sense. I can add whatever x I want to 1!

    If I ask you to factor the equation "x^2 +5x + 6", you don't care what x is because your goal is to get another equation which is the same as the original for any x.

    Similarly, I can take the arccos of whatever I want (within the function's range) and take the cosecant of that.

    U is any number.

    So, you are given that the cosine of this angle is U.

    You are given two sides. A trig function is a ratio. Just because you see one digit, doesn't mean you are only given one side. It is a ratio of two sides. If it's just a whole number, that just means the ratio is equal to that whole number.


    Do you agree that 6 is the same as 6/1?
    Do you agree that 3 = 3/1?
    What about that 16 = 16/1?
    22 = 22/1?

    Do you agree that any number divided by 1 is itself? Why wouldn't that apply for U?

    U is any number! So, U = U/1, for any concievable value of U.

    So, on your triangle, label an acute angle theta. It's cosine is U, OR U/1, because they are the same for any U.

    Given that you know the definition of cosine is adjacent over hypotenuse, and that your cosine is U/1, doesn't that mean that the adjacent side is U, and the hypotenuse is 1?

    If the adjacent side is U, and the hypotenuse is 1, you know two values, 1, and U (which is ANY number!).

    [itex]a^{2}+b^{2} = c^{2}[/itex]

    Just rearranging that equation:

    [itex]a^{2} = c^{2} - b^{2}[/itex]

    So, if we know b to be U, which is ANY number, and we know c to be 1, what is an expression for a, the missing side?
     
    Last edited: Oct 4, 2011
  10. Oct 4, 2011 #9

    HallsofIvy

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    [tex]cos(\theta)= \frac{opp.side}{hypotenuse}= \frac{a}{1}[/tex]
     
  11. Oct 4, 2011 #10
    I've got it now! Thank you so much!

    Your method is sooo much easier than the book is making it out to be.
     
    Last edited: Oct 4, 2011
  12. Oct 4, 2011 #11
    Glad to hear it. So what's your final answer?
     
  13. Oct 4, 2011 #12
    1/√(1-u^2) :smile:
     
  14. Oct 4, 2011 #13
    Bingo! You've just proven an identity.
     
  15. Feb 20, 2012 #14
    I know this is an old thread, but I just had a similar question

    write tan(cos-1 x) as an expression without trig or inverse trig functions.
    And the explanation here was very helpful.

    Thank you
    Jesper
     
  16. Feb 21, 2012 #15

    eumyang

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    Homework Helper

    (It's not good form to resurrect an old thread. It would have been better to start a new thread, I think.)

    When you say "cos-1 x," think of it as "an angle whose cosine is x" (or x/1). You now have two sides of a right triangle. Find the third, and use it to find the tangent.
     
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