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Homework Help: Prove cos(sin^-1 x)= [itex]\sqrt{1-x}[/itex]

  1. May 29, 2012 #1
    1. The problem statement, all variables and given/known data

    cos(sin-1x) = [itex]\sqrt{1-x^2}[/itex]

    2. Relevant equations

    I would assume trigonometrical identities would be used to prove this.
  2. jcsd
  3. May 29, 2012 #2
    Hello nowayjose! :smile:

    Yes, they would...

    Why don't you start by assuming [itex]\theta = sin^{-1}x[/itex], and then draw out a triangle to find a relation between theta and cosine, that you can use...

    PS : your thread title is misleading :uhh:
  4. May 29, 2012 #3
    Thanks for the prompt reply!
    Sorry, and the question's undoubtedly stupid. I've used this method before and haven't happened to used any identities (or so i believe...).

    [itex]\theta = sin^{-1}x[/itex]
    [itex]sin\theta = x[/itex]
    [itex] sin = 1/X [/itex]
    the cosine side must therefore be [itex]\sqrt{1-x^2}[/itex]
    therefore the cosine angle is
    [itex]\sqrt{1-x^2} / 1[/itex]
  5. May 29, 2012 #4


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    Or, different wording of the same idea: [itex]sin^2(\theta)+ cos^2(\theta)= 1[/itex] so that [itex]cos(\theta)= \pm\sqrt{1- sin^2(\theta)}[/itex]. So
    [tex]cos(sin^{-1}(x))= \pm\sqrt{1- sin^2(sin^{-1}(x)}= \pm\sqrt{1- x^2}[/tex]
  6. May 29, 2012 #5


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    What you have written here makes little sense. If [itex]\theta= sin^{-1}(x)[/itex] then, yes, [itex]sin(\theta)= x[/itex], but you cannot write "sin" without some argument. And the "-1" does NOT indicate reciprocal (1/x), it means the inverse function.
  7. May 30, 2012 #6
    consider the attached triangle picture (sorry its sloppy)

    in that case Sin(theta) = x (hypotenuse is 1, opposite is x)

    thus sin^-1(x) = theta.

    For that same theta, using a^2 + b^2 = c^2......

    x^2 + b^2 = 1^2
    b^2 = 1-x^2
    b = sqrt( 1 - x^2)

    and cos(theta) = adj / hyp
    cos(theta) = sqrt( 1 - x^2) / 1
    cos(theta) = sqrt( 1 - x^2)

    sin^-1(x) = theta

    so sub in theta

    cos(sin^-1(x)) = sqrt( 1 - x^2)


    Attached Files:

  8. May 30, 2012 #7
    People sorry for the typo and for not being clear about my thought process.

    Bascially what i meant was:

    [itex]\theta = sin^{-1}x[/itex]

    [itex]sin\theta = x[/itex]

    If the sine angle is X, then the opposite is X and the hypotenuse 1.

    the adjacent side can now be calculated using pythagoras, which gives [itex]\sqrt{1-x^2}[/itex]

    The cosine angle is the quotient of the adjacent and the hypotenuse:

    [itex]\sqrt{1-x^2} / 1[/itex]
  9. May 30, 2012 #8
    Would you care explaining me how you multiplied out what's rooted?

    I know sin(sin^-1) cancel out because you add the indices, so shouldn't that leave sin x..
    Last edited: May 30, 2012
  10. May 30, 2012 #9
    Those are inverse functions, so they don't simplify like exponents do.
    Since sinx and sin-1x are inverses and sin2x = (sinx)2, sin(sin-1x) = x and
    sin2(sin-1x) = (sin(sin-1x))2 = x2
  11. May 30, 2012 #10


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    No, don't think of it that way. Inverse functions are a fancy word of saying "doing the opposite". You have some function, such as [itex]y=\sin(x)[/itex] and you want to make a process to get back to just x, and for this case its inverse will be [itex]\sin^{-1}(y)[/itex].
    Other inverses are, for example, the inverse of [itex]y=x^2[/itex] is [itex]\sqrt{y}[/itex] because [itex]\sqrt{x^2}=x[/itex] (technically it's |x| so that's why we specify domains, in this case [itex]x\geq 0[/itex])
    Another would include [itex]y=\ln(x)[/itex] and [itex]e^y[/itex]

    Also, keep in mind that if you have a function [itex]y=x^n[/itex] and applying its inverse [tex]\sqrt[n]{y}=y^{1/n}[/tex] the reason we get back to x is because [tex]\left(x^n\right)^{1/n}=x^{n\cdot\frac{1}{n}}=x^{\frac{n}{n}}=x[/tex]

    You multiply the indices, not add.
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