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Trigonometric question

  1. Dec 2, 2011 #1
    1. The problem statement, all variables and given/known data
    Calculate tan(arcsin(1/7))


    2. Relevant equations
    arcsinv = x => sinx = v
    sinx = (opposite/hypotenuse) (in a right-angled triangle)
    tanx = (opposite/adjacent)

    3. The attempt at a solution
    Well I realize that arcsin(1/7) is an angle. Let's call it v. I also know that arcsin(1/7) => sin(v) = 1/7.

    So i try to picture myself a triangle like this:
    math_problem.png
    After this I'm lost. I'd like to use tanx = o/a, but I only know the value of o. I'm not sure how these kind of problems are solved. Any ideas?
     
  2. jcsd
  3. Dec 2, 2011 #2

    BruceW

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    technically, you only have the shape of the triangle, so it would be more correct to label the sides as 7a and a (where a is some undetermined constant).

    From here you can find the other side using a well known theorem about triangles. (I'm sure you know it, maybe you've forgotten). And then once you have the other side, you can find tan.
     
  4. Dec 2, 2011 #3
    @BruceW

    Thanks man. I knew I could just hit arcsin(1/7) in on my calculator to find the angle and then from there find the adjacent side, but we're not allowed to use the calculator.

    But then I realized you meant Pythagoras theorem, which I ofcourse didnt think about...

    Thanks again! I think I solved it (got no key/solutions so cant check it). I got the answer:
    tan(arcsin(1/7)) = tan(v) = 1(4sqrt(3))
     
  5. Dec 2, 2011 #4

    dextercioby

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    [tex] \tan\arcsin x = \frac{\sin \arcsin x}{\cos \arcsin x} = \frac{x}{\sqrt{1-x^2}} [/tex]
     
  6. Dec 2, 2011 #5

    BruceW

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    Yes, Pythagoras, that's the one. And you meant 1/(4sqrt(3)), right?

    And P.S. dextercioby has written the same method, but all in one step.
     
  7. Dec 2, 2011 #6
    Alright. Yes i meant 1/(4sqrt(3)).
     
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