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Math Help - Just a few doubts

  1. #1
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    Just a few doubts

    Determine if True or False

    A. sin(sin inverse x) = x for all real numbers x
    Not sure...(what is it asking?)sin of sin inverse x(can any explain what this means?)

    B. The graph of y = tan inverse x has two horizontal asymptotes
    True...i am sure

    C. 2cosxsinx - cosx = 0
    Find X between [0,2pi)

    x = pi/2, 3pi/2

    I am also sure about number 3.

    Wanted to check.

    Thanks for any help...
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  2. #2
    Super Member Matt Westwood's Avatar
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    \sin \left({\sin^{-1} x}\right) does not equal x for all x.

    Because \sin^{-1} x is not defined for x < -1 or x > 1.
    Last edited by Matt Westwood; August 31st 2009 at 01:08 PM. Reason: correction
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  3. #3
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    Quote Originally Posted by Matt Westwood View Post
    \sin \left({\sin^{-1} x}\right) does not equal x for all x.

    Because \sin^{-1} x is not defined for x < -1 or x > 1.
    alright, but i do not understand what
    \sin \left({\sin^{-1} x}\right)

    is saying

    Thanks for help though
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  4. #4
    Super Member Matt Westwood's Avatar
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    Okay, let's say you have y = \sin^{-1} x.

    That is, y is the "inverse sine" of x.

    That means, "What value of y do you have to take the sine of to get x?

    That is, it's doing x = \sin y but this time you know what x is and you want to find y.

    So doing \sin \left({\sin^{-1} x}\right) is like saying:

    "We've got this value x, and we find the number it is the sine of. Then we take this number and take its sine."

    You seem to understand what \tan^{-1} x is, you could do question B.

    The trouble is here, is that not all numbers are the sine of another number, because the only possible values for the sine of a number are between -1 and 1.

    So if someone says: "What's the inverse sine of 2?" you'd reply "Don't be silly, it doesn't exist." (Not at this level of mathematics anyway.)
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