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Sin/cos inverse - derivitives

  1. Nov 7, 2005 #1
    I'm studying for my exam which is in four days and I've searched through my text book (im using an old book because thats the one the school uses) and I cant seem find how to do the derivitives for the sine/cosine inverse functions:



    Could anyone please tell me what these derive to please?
  2. jcsd
  3. Nov 7, 2005 #2
    Use [tex]{y'}_x = \frac{1}{{x'}_y}[/tex]. Take [tex]y=\arcsin x[/tex] and [tex]x=\sin y[/tex].

    - Kamataat
  4. Nov 7, 2005 #3
    sin¯¹x=1/√1-x² and cos¯¹x=-1/√1-x²

    Hope this is what you are looking for.
  5. Nov 7, 2005 #4


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    I'll derive the sine one for you, try to use the process i use to find cosine.


    f(x)=sin(x), g(x)=sin-1(x)


    y=sin(x) implies that x=sin(y) for the inverse

    We know the derivative of sine is cosine, so the equation becomes..


    but we know that


    also using the idea that [itex]cos(sin^{-1}(x))=cos(y)[/itex]


    and from an equation above, x=sin(y), so the final product is:


    Last edited: Nov 7, 2005
  6. Nov 7, 2005 #5
    Ah, I get it now. Thanks everyone!
  7. Nov 7, 2005 #6


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    It's too complicated.

    Differentiate in both sides wrt "x"

    [tex] x=\sin\left(\arcsin x\right) [/tex]


    [tex] x=\cos\left(\arccos x\right) [/tex]

    using for the RHS the chain rule.

  8. Nov 7, 2005 #7
    I think everyone's is too complicated. Here's how I derive it.





    Use a right trianlge to evaluate the cosine of y. So it's four written steps and however many you want to call using the triangle.
  9. Nov 13, 2005 #8
    They are positive(for sin) and negative(for cos) 1/[(1-x^2)^(1/2)]

    Unfortunatelly I had to memorize them...I have no clue why they're that, but they're 1 divided by square root of 1 - x^2 and negative that for cos.
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