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Can you derive a trigonometric function from its inverse dx?

  1. Jan 28, 2016 #1
    1. The problem statement, all variables and given/known data
    Arbitrary derivative of inverse trigonometric function:
    (sin-1x) = 1/(√1 - x2)

    2. Relevant equations
    f-1(f(x)) = 1/f`(x)

    3. The attempt at a solution
    So basically I learned about derivatives of trigonometric functions in class, and I thought maybe this would work: deriving the derivative of the regular sine function from its inverse sine, and then integrating that, to express a regular arbitrary trigonometric function in terms of x. Here is my thinking:

    (sin-1x) = 1/(√1-x2)
    f-1(sin x) = 1/(sin x)` = 1/(√1 - x2)

    Which is to say that:

    (sin x)` = (√1 - x2)

    Integrating this, we have

    sin x = ∫(1 - x2)½dx

    I haven't yet figured out how to evaluate this integral, but can someone tell me if my thinking is straight or not?
     
  2. jcsd
  3. Jan 28, 2016 #2

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    It's not really clear what you are trying to achieve, but there clearly are a number of mistakes.
    (sin-1x) = 1/(√1 - x2): you probably meant ##(\sin^{-1})'(x)=\frac{1}{\sqrt{1-x²}}##
    f-1(f(x)) = 1/f`(x): you probably meant ##(f^{-1})'(f(x))=\frac{1}{f'(x)}##
    In both cases there is a derivative sign missing.

    As you started from wrong equations, the final result is also incorrect: ##\sin(x) \neq \int \sqrt{1-x²}dx##.
     
    Last edited: Jan 28, 2016
  4. Jan 28, 2016 #3
    Can I ask anybody to refrain from posting in this topic?
     
    Last edited: Jan 28, 2016
  5. Jan 28, 2016 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Then what was your point in posting it?
     
  6. Jan 28, 2016 #5
    Never mind... I can't change what other people think, and I certainly can't stop people from saying what's on their mind.
     
  7. Jan 28, 2016 #6

    Mark44

    Staff: Mentor

    You could mark it as "Solved" using the new button at the top right of the thread.
     
  8. Jan 28, 2016 #7
  9. Jan 28, 2016 #8

    Mark44

    Staff: Mentor

    You're a member, so you should be able to use it. If you scroll all the way up to the top of the thread, don't you see a button that says "MARK SOLVED"?

    This is a new change, so it could be that you have an old version of our software in your cache. Try logging out, and then logging back in.
     
  10. Jan 28, 2016 #9
    I still can't see it. If you can't find it in the link I posted, then I won't be able to find it, either.

    It's probably a premium feature.
     
  11. Jan 28, 2016 #10

    Mark44

    Staff: Mentor

    Maybe your browser doesn't support it. I don't think we have and "premium features."
     
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