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Math Help - Finding asymptotes+Fundamental identities

  1. #1
    Member Chokfull's Avatar
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    Finding asymptotes+Fundamental identities

    Hey, I'm new to calculus, and I can't figure out this problem.

    You start with:

    y = TAN (2 SIN x) -pi < x < pi

    Since: TAN x = (SIN x / COS x),

    you can get

    y= SIN (2 SIN x) / COS (2 SIN x)

    Then, since the asymptote is where the equation is undefined (or equals infinity), we need to find the value of x which causes the denominator to equal 0.

    COS (2 SIN x) = 0

    The cosine of +pi/2 is 0, so

    2 SIN x = +pi/2
    SIN x = +pi/4
    x = + SIN^-1 pi/4

    The problem is that there are two answers, but I can't figure out how to work out the next one, which the book says is

    x = + (pi - SIN^-1[pi/4]

    Help, plz!

    I'd also like a way to type in a pi symbol, if that's possible.
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  2. #2
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    If you have an equation \sin x = C then the possible solutions are x=\arcsin C+2n\pi and x=\pi - \arcsin C+2n\pi
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  3. #3
    Member Chokfull's Avatar
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    Is that just a general rule? My Math book doesn't show it. Also, does it work for just pi - arcsin C? Otherwise I don't see how it works. The 2n pi doesn't seem to fit at all. If it does, then thanks a lot.
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  4. #4
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    It's the general solution to an equation like this. It contains all the solutions to the equation, but for your problem the question makes it clear that we are only interested in the domain x\in [-\pi,\pi], so the only value of n that doesn't take us out from the valid domain is n=0.

    Hence the two solutions are x=\pm\arcsin\frac{\pi}{4} and x=\pm\left(\pi-\arcsin\frac{\pi}{4}\right)

    If you think about it, going 2n\pi radians in a circle always takes us back to where we started.
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  5. #5
    Member Chokfull's Avatar
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    Oh, yeah, I guess I was being pretty dumb about the 2n pi thing. Thanks again.
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