1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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.


    2. Relevant equations

    3. The attempt at a solution


    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    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.

    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


    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!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook