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: 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:
    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


    User Avatar
    Homework Helper

    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

    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


    User Avatar
    Science Advisor
    Homework Helper

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


    User Avatar
    Homework Helper

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