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Math Help - sin (arc cos)

  1. #1
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    sin (arc cos)

    Hello,

    How would I evaluate the following:

    sin (arc cos (-5/13))

    would I put it into the calculator this way:

    sin (cos^-1(-5/13)) ????????

    -----therefor------

    How would I evaluate the following:

    sin (arc cos 5/13) <---- noticed there are no parenthesis around the 5/13 like in the first problem. Is this because of the negative? Please help on how I would put them into the calculator? Thanks so very much!

    IS IT CORRECT TO GET THE SAME ANSWER FOR BOTH???? 0.923???? ODD WHY IS THIS?
    Last edited by Justardms; June 28th 2005 at 08:18 AM. Reason: mistype
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  2. #2
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    Yes, it is correct, becoase sin(x) positive in first two quadrants.
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  3. #3
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    You can do it without a calculator.

    The special right-angle triangle is 5-12-13.

    sin(arccos( something ) ) is saying:

    find the opp \over hyp of the angle of adj \over hyp

    The answer is 12/13.
    Last edited by paultwang; June 28th 2005 at 02:45 PM.
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  4. #4
    vms
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    This always works

    You can use the formula

    arc cos(-x) = Pi - arc cos(x) whenever x is in [-1, 1]
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  5. #5
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    I don't know if you are taught to use calculators for this exercise. Or, I don't know if the emphasis is on the use of calculators.
    I believe the idea is for you to analyze/study the angle through its properties.

    Anyway, this is how I will answer your question:

    arccos(-5/13) is an angle whose cosine is -5/13.
    cosine is (adjacent side)/(hypotenuse), so,
    -5/13 = adj/hyp
    adj = -5
    hyp = 13 ----hyp is always positive, so adj above has to be negative.

    Now imagine the reference right triangle of this angle. We know the adjacent side and the hypotenuse. We don't know the opposite side, but we can find it by using the Pythagorean theorem:
    (hyp)^2 = (opp)^2 +(adj)^2
    (13)^2 = (opp)^2 +(-5)^2
    169 = (opp)^2 +25
    169 -25 = (opp)^2
    (opp)^2 = 144
    opp = +,-sqrt(144) = +,-12.

    But what is the sign of that opp of this angle whose cos is -5/13?
    Is it positive only? Negative only? Either positive or negative?

    Back to adj = -3.
    Adj, is adjacent side, is also the x-component of hypotenuse.
    The adj = x = -5 is on the negative side of the x-axis. This negative side is to the left of the y-axis. And in this side we find the 2nd and 3rd quadrants.
    In the 2nd quadrant, y is positive. In the 3rd quadrant, y is negative.
    Since y is the opposite side of the angle, then, opp is either positive or negative.
    So, opp = +,-12 really.

    So the reference rigth triangle of the angle whose cosine is -5/13 has these properties:
    adj = -5
    opp = +,-12
    hyp = 13

    You are looking for the sine of the angle. Sine is opp/hyp, so,
    sin(arccos(-5/13)) = (+,-12)/13.

    Meaning, the sine is either 12/13 or -12/13.
    Either 0.923, or -0.923 -----answer.

    --------------------
    I tried on my calculator the sin(arccos -5/13)) and my calculator showed "SYNTAX Error". Meaning, cannot be.
    While with the sin(arccos 5/13)), my calculator showed "0.923076923".

    Maybe even my calculator cannot follow my explanation above---why it should show "+,-0.923076923" :-)
    (No, my calculator was programmed to show only one sign at a time, I suppose.)
    Last edited by ticbol; June 29th 2005 at 12:08 PM.
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  6. #6
    Junior Member theprof's Avatar
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    Quote Originally Posted by Justardms
    Hello,

    How would I evaluate the following:

    sin (arc cos (-5/13))

    IS IT CORRECT TO GET THE SAME ANSWER FOR BOTH???? 0.923???? ODD WHY IS THIS?
    consider that
    <br />
\sin^2{x}+\cos^2{x}=1<br />
    hence
    <br />
\sin{x}=\pm\sqrt{1-\cos^2{x}}<br />
    Sine could be either negative or positive, but pocket calculator and math software like Derive will give you only positive value:
    <br />
\sin{\arccos{\left(-\frac{5}{13}\right)}}=\pm\sqrt{1-\left(-\frac{5}{13}\right)^2}=\pm\frac{12}{13}<br />
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