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Math Help - tanget converting to cosine

  1. #1
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    tanget converting to cosine

    if tan β = 6 then what is cos β equal to?

    i really don't know to approach this problem. i don't even know where to begin

    please and thankyou to anyone who can help
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  2. #2
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    Quote Originally Posted by luckystar001 View Post
    if tan β = 6 then what is cos β equal to?

    i really don't know to approach this problem. i don't even know where to begin

    please and thankyou to anyone who can help
    0^o<\beta<\frac{{\pi}}{2}

    tan\beta=\frac{6x}{x}

    The hypotenuse of the right-angled triangle is \sqrt{x^2+36x^2}=x\sqrt{37}

    cos\beta=\frac{x}{x\sqrt{37}}=\frac{1}{\sqrt{37}}


    {\pi}<\beta<\frac{3{\pi}}{2}

    cos\beta=-\frac{1}{\sqrt{37}}

    You can also work with the unit circle and let the radius be 1 (hypotenuse=1).
    Last edited by Archie Meade; August 10th 2010 at 10:58 AM. Reason: typo
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    wow you are amazing! i would have never thought of using a right triangle to solve this problem; i was trying to find an answer using algebraic means and i only got as far as tan β = sin β / sin β

    but why is β limited to (pi, 2pi) ?
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  4. #4
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    Quote Originally Posted by luckystar001 View Post
    wow you are amazing! i would have never thought of using a right triangle to solve this problem; i was trying to find an answer using algebraic means and i only got as far as tan β = sin β / sin β

    but why is β limited to (pi, 2pi) ?
    It isn't, but tan\beta gives the slope of a line through the origin

    and if that's positive, then the line is increasing.
    If you draw a unit-circle (radius=1, centred at the origin)
    and draw a line through the origin with slope=6,
    you will see this corresponds to two angles,
    as there are two points on the circumference of the circle.

    If we go beyond 360 degrees, we only wind up where we started!

    When you have a geometric or trigonometric representation of the situation,
    it can be very clear.
    Attached Thumbnails Attached Thumbnails tanget converting to cosine-tan-beta-.jpg  
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  5. #5
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    Quote Originally Posted by luckystar001 View Post

    but why is β limited to (pi, 2pi) ?
    Sorry, I didn't answer that correctly!

    If the angle was in either of the other two quadrants, tan(angle) would be negative.

    I should have written

    0<\beta<\frac{{\pi}}{2}

    and

    {\pi}<\beta<\frac{3{\pi}}{2}
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  6. #6
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    Quote Originally Posted by luckystar001 View Post
    if tan β = 6 then what is cos β equal to?
    1 + \tan^2{\beta} = \sec^2{\beta}

    1 + 6^2 = \sec^2{\beta}

    37 = \sec^2{\beta}

    \displaystyle \frac{1}{37} = \cos^2{\beta}<br />

    \displaystyle \pm \frac{1}{\sqrt{37}} = \cos{\beta}

    (+) if \beta is in quad I , (-) if \beta is in quad III
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  7. #7
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    i think i finally understand this problem!

    A big Thank you to all who helped! you've shown me so many ways to approach this problem!
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