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Math Help - Inverse trig functions and implied domains

  1. #1
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    Inverse trig functions and implied domains

    Hello all,

    I just need to confirm some answers to some questions I have attempted, unfortunately i do not have the solutions to them:

    1.) evalute arctan (2 sin (-pi/3))

    my solution: first i evaluated 2 sin (-pi/3)

    since sin (-pi/3) = - root (3) / 2 and since it is multiplied by 2, we now have arctan (-root 3)

    which again from the triangle i found to equal -pi/3.
    __________________________________________________ _____________________

    2.) arccos( cosec (pi/2)).

    cosec (pi/2) = 1 / sin (pi/2) = 1

    so we have: arccos (1) and cos x = 1
    so x: 2n * pi where n is any integer.
    __________________________________________________ _____________________

    3.) cos (arcsin (-root 3 / 2) + arccos (root 2 / 2))

    the solution is very long-winded so i will cut it short.

    by using the cos (x-y) formula after evaluating the terms in the brackets i got root 3 + 1 / 2 * (root 2) as the final answers.

    __________________________________________________ _____________________

    4.) Find the implied domain and range of h(x) = log (sec (x))

    range: [0, inf)
    domain: all real values excluding integer multiples of the region [pi/2, 3pi/2] where the range is below [1, inf]

    Regards.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Inverse trig functions and implied domains

    1) Right! But if you want to be more general like in 2) you can say: \frac{-\pi}{3}+k\cdot \pi
    2) Right!
    3) Right!
    4)
    Domain: First of all we now that \log(x) is defined if x>0 that means in this case \sec(x)>0 and \cos(x)>0 in the 1th and 4th quadrant, so there are solutions between \frac{3\pi}{2}<x<\frac{5\pi}{2}. So check again, because if domain h: [1,\infty] that would mean for example \log\left[\sec\left(\frac{3\pi}{4}\right)\right] would be defined, which is not.

    EDIT: I see answer 3 is correct, sorry I red it wrong.
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  3. #3
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    Re: Inverse trig functions and implied domains

    Hello Siron,

    First of all, I would like to thank you for helping me.

    Secondly, I'm not exactly sure how to put what I mean of the domain of the above function in mathematical terms. I'm trying to say that there are gaps where the domain is invalid and these gaps are periodic, any tips on how I could go about doing this?
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Inverse trig functions and implied domains

    Can you continue with the hint I gave about the cosinus? Indeed if you look at the graph (you can use wolphram alpha to graph it), this is a periodic function.
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  5. #5
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    Re: Inverse trig functions and implied domains

    unfortunately I can't, i'm still having trouble definining (mathematically) the gaps that are to be excluded from the domain
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  6. #6
    MHF Contributor Siron's Avatar
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    Re: Inverse trig functions and implied domains

    First of all, do you agree the domain of the function h can't be: [1,\infty]? (look at my previous example)
    Like I said \log(x) is only defined if x>0, in this case: \frac{1}{\cos(x)} has to be positive, because the numerator is positive the denominator has also to be positive that means \cos(x)>0 which is true if x lays in the 1th or 4th quadrant. So \frac{3\pi}{2}<x<\frac{5\pi}{2}, but because we can continue on the goniometric circle we may write: \frac{3\pi}{2}+2k\pi<x<\frac{5\pi}{2}+2k\pi.

    So the domain of h: x \in \left]\frac{3\pi}{2},\frac{5\pi}{2}\right[+2k\pi and \frac{3\pi}{2},\frac{5\pi}{2} can not be in the interval because \sec\left(\frac{3\pi}{2}\right) is undefined.

    And your range is fine!
    Last edited by Siron; August 14th 2011 at 06:10 AM.
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