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Math Help - Help with trig algebra. I can't figure this out

  1. #1
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    Help with trig algebra. I can't figure this out

    Ok, so I am looking at the min and max for the angle x:

    -12cos(x) + (2)12sin(x) = 5

    and

    -12cos(x) + (2)12sin(x) = -5

    which I reduced to:

    (12)(2sin(x) - cos(x)) = 5

    2sin(x) - cos(x) = 5/12

    I'm not sure how to solve for the angle (x) from here.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by mathdude09 View Post
    Ok, so I am looking at the min and max for the angle x:

    -12cos(x) + (2)12sin(x) = 5

    and

    -12cos(x) + (2)12sin(x) = -5

    which I reduced to:

    (12)(2sin(x) - cos(x)) = 5

    2sin(x) - cos(x) = 5/12

    I'm not sure how to solve for the angle (x) from here.
    I'm not sure about what is going on here, but:

    12(2\sin{x}-\cos{x})=5

    2\sqrt{1-\cos^2{x}}-\cos{x}=\frac{5}{12}

    2\sqrt{1-cos^2{x}}=\frac{5}{12}+\cos{x}

    Squaring both sides

    4-4\cos^2{x}=\frac{25}{144}+\frac{5}{6}\cos{x}+\cos^  2{x}


    Now you have a quadratic in cos{x}.

    Can you solve quadratics?
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  3. #3
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    Working with what you gave me I was able to successfully get both angles using the quadratic formula, thanks.

    Just to understand exactly how you manipulated the sin function:

    sin(x) = sqrt(1-cos(x)^2) ???

    Is there a listing of these conversions somewhere?
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by mathdude09 View Post
    Working with what you gave me I was able to successfully get both angles using the quadratic formula, thanks.

    Just to understand exactly how you manipulated the sin function:

    sin(x) = sqrt(1-cos(x)^2) ???

    Is there a listing of these conversions somewhere?
    This was derived from the pythagorean identity IE

    \sin^2{x}+\cos^2{x}=1

    Subtracting...
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  5. #5
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    Oh yeah. Pythagoras, unit circle...it's all coming back now. Thanks
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