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Write the trigonometric expression as an algebraic expression

  1. Mar 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Not long ago I had a similar problem, which I was able to solve after reading this thread, but for this question I'm stuck and I could use a small hint.

    attachment.php?attachmentid=45020&stc=1&d=1331609688.png

    2. Relevant equations



    3. The attempt at a solution

    Thanks
     

    Attached Files:

  2. jcsd
  3. Mar 12, 2012 #2
    I would first solve for [itex]arctan(v)[/itex] in terms of [itex]\alpha[/itex]. See if you can do the rest!
     
  4. Mar 12, 2012 #3

    SammyS

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    Label the triangle differently. Label it so that the tangent of α is v .

    [itex]\displaystyle \tan(\alpha)=\frac{v}{1}\,.[/itex]
     
  5. Mar 12, 2012 #4
    >>I would first solve for arctan(v) in terms of α

    >>Label the triangle differently. Label it so that the tangent of α is v .

    hmm, I think that is what I have, tan (α) = v = sqrt..., I'm still missing something.
     
  6. Mar 12, 2012 #5
    Okay, so you have [itex]tan(\alpha) = v[/itex]. What is [itex]arctan(v)[/itex]? Then, working from the inside out, what is [itex]sin(arctan(v))[/itex]?
     
  7. Mar 13, 2012 #6
    That is the part I'm missing, how do I find the exact value of atan(v). I know what the result is, but I dont know how to get to it.
     
  8. Mar 13, 2012 #7
    Look in your original picture. v is equal to four quantities, which one would give you a nice value for [itex]arctan(v)[/itex]?
     
  9. Mar 13, 2012 #8
    I appreciate the help and the hints, but at the moment I'm blank. I'm sure the answer is staring right at me, and it is pretty straight forward - but I just can't see it. I will think more about this during the day and hopefully tonight when I get back to it, the answer jumps right out of the picture.
     
  10. Mar 13, 2012 #9
    ah - I think I got it, will try later. I need to find the inverse of v = sqrt()/x.
     
  11. Mar 13, 2012 #10
    A little closer, but not there yet.
    attachment.php?attachmentid=45034&stc=1&d=1331656005.png
     

    Attached Files:

  12. Mar 13, 2012 #11
    Sorry for the late reply, I took a nap!

    What is [itex]arctan(tan(x))[/itex] equal to?

    Can you see which value to plug in now?

    Then you need to calculate sine of that number you get, which is also listed in the picture as a number not involving trig functions or trig inverses.
     
  13. Mar 13, 2012 #12
    I have not been able to figure this out by myself, I kept going in circles. Finally when I was about to give up I Bing'ed it and found this explanation http://mathforum.org/library/drmath/view/53946.html - and now I can see how simple it is, and I can see my main problem was the labeling as was pointed out in one of the answers. Had I labeled the figure correct, then I might have been able to figure it out on my own.
     
  14. Mar 13, 2012 #13
    Yeah, I've had plenty of those moments, and then it just clicks. Just so we all know you understand it correctly, what answer did you get?
     
  15. Mar 14, 2012 #14
    This is how I solved it. It is exactly the same steps as on the link. I read the explanation, closed the browser and solved it myself by first drawing a new trigangle with the 'correct' labels, and then it pretty much fell in place.

    I think the trick is to label the triangle with the sides as v/1 = v for the inner trig function, and that is where I made the mistake.... I think

    attachment.php?attachmentid=45076&stc=1&d=1331740423.png

    Again, thank you for your help.

    Edit: I just noticed that the tanθ does not provide any additional information.
     

    Attached Files:

    Last edited: Mar 14, 2012
  16. Mar 14, 2012 #15
    That looks good to me based off of the labels of the diagram! Just wanted to make sure you figured out the problem correctly!
     
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