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Math Help - inverse function for arcsin

  1. #1
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    inverse function for arcsin

    I have a webworks problem that I am not doing correctly. here it is
    f(x)=arcsin(x/7)+10
    This is what I did. Y=arcsin(x/7)+10
    X=arcsin(y/7)+10
    x-10=arcsin(y/7)
    y/7=arcsin^-1(x-10)
    answer y=7*arcsin^-1(x-10)
    the answer is supposed to be in this form, webworks said this is incorrect. I did the rest of the problems this way but they were tan and cos and I am sure this one must be done differently or I must have skipped or omitted something. Please let me know if you can help me understand how this one is solved.
    Thank you!
    keith stevens
    keithstevens@ctc.net
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  2. #2
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    Quote Originally Posted by kcsteven View Post
    I have a webworks problem that I am not doing correctly. here it is
    f(x)=arcsin(x/7)+10
    This is what I did. Y=arcsin(x/7)+10
    X=arcsin(y/7)+10
    x-10=arcsin(y/7)
    y/7=arcsin^-1(x-10)
    answer y=7*arcsin^-1(x-10)
    the answer is supposed to be in this form, webworks said this is incorrect. I did the rest of the problems this way but they were tan and cos and I am sure this one must be done differently or I must have skipped or omitted something. Please let me know if you can help me understand how this one is solved.
    Thank you!
    keith stevens
    keithstevens@ctc.net
    It is okay but!
    \arcsin^{-1}(x)=\sin x
    Because the inverse of the inverse is the original function.
    So you have,
    y=7\sin (x-10)

    WARNING. If you want to be stirctly correct it is not really the regular sine curve (along the x-axis) this one must be restricted on some domain because the original inverse sine you had was itself restricted.
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  3. #3
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    Hello, Keith!

    Find the inverse of: f(x) \:=\:\arcsin\left(\frac{x}{7}\right)+10

    This is what I did.
    y\:=\:\arcsin\left(\frac{x}{7}\right)+10
    x \:=\:\arcsin\left(\frac{y}{7}\right)+10
    x-10\:=\:\arcsin\left(\frac{y}{7}\right)
    \frac{y}{7}\:=\:\arcsin^{-1}(x-10) . . . ??

    Answer: . y\:=\:7\arcsin^{-1}(x-10)

    You just invented some new notation: "the inverse of the inverse sine" . . .


    You had: . x - 10 \:=\:\arcsin\left(\frac{y}{7}\right)

    Take the sine of both sides:
    . . \sin(x - 10) \:=\:\sin\left[\arcsin\left(\frac{y}{7}\right)\right]

    . . \sin(x-10) \:=\:\frac{y}{7}\quad\Rightarrow\quad 7\sin(x-10)\:=\:y

    Therefore: . f^{-1}(x)\:=\:7\sin(x-10)

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