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Thread: trig help / simplifying

  1. #1
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    trig help / simplifying

    Given that $\displaystyle tan^{2}K = 2secK $

    a) find the value of sec k.
    (b) deduce that cosk=√2−1


    $\displaystyle tan^{2}k = 2secK $

    $\displaystyle sec^2k-1 = 2secK $

    $\displaystyle sec^2K-2secK-1 = 0 $

    $\displaystyle secK = 1+ \sqrt{2} $

    b) $\displaystyle \frac{1}{cosk} = 1+\sqrt{2} $

    $\displaystyle cosK = \frac{1}{1 + \sqrt{2}} $

    I don't get the right answer, if I rationalise the denominator by multiplying top and bottom by $\displaystyle 1 - \sqrt{2} $ I still don't get $\displaystyle cosK = \sqrt{2} -1 $
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  2. #2
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    Quote Originally Posted by Tweety View Post
    Given that $\displaystyle tan^{2}K = 2secK $

    a) find the value of sec k.
    (b) deduce that cosk=√2−1


    $\displaystyle tan^{2}k = 2secK $

    $\displaystyle sec^2k-1 = 2secK $

    $\displaystyle sec^2K-2secK-1 = 0 $

    $\displaystyle secK = 1+ \sqrt{2} $

    b) $\displaystyle \frac{1}{cosk} = 1+\sqrt{2} $

    $\displaystyle cosK = \frac{1}{1 + \sqrt{2}} $

    I don't get the right answer, if I rationalise the denominator by multiplying top and bottom by $\displaystyle 1 - \sqrt{2} $ I still don't get $\displaystyle cosK = \sqrt{2} -1 $
    $\displaystyle \frac{1}{1+\sqrt{2}}\cdot\frac{1-\sqrt{2}}{1-\sqrt{2}}=\frac{1-\sqrt{2}}{1-(\sqrt{2})^2}=\frac{1-\sqrt{2}}{1-2}=\frac{1-\sqrt{2}}{-1}=\sqrt{2}-1$
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  3. #3
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    what mistake i have done????????????please help
    $\displaystyle
    $
    $\displaystyle tan^{2}k = 2secK
    $

    $\displaystyle
    $
    $\displaystyle sec^2k-1 = 2secK
    $

    $\displaystyle sec^2K-2secK-1 = 0
    $
    after that if we take $\displaystyle \sec k = x$
    then
    $\displaystyle x^2-2x-1=0$
    $\displaystyle x=1$
    $\displaystyle \sec x=1$









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  4. #4
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    jashansinghal: your mistake is having the color tags inside of the math tags in LaTex...


    Oh, you mean the problem?
    $\displaystyle x^2-2x-1=0$
    $\displaystyle x=1$
    $\displaystyle \sec x=1$
    I don't know why you got x = 1 from the quadratic. You need to use the quadratic formula.
    $\displaystyle x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-1}}{2(1)}$
    $\displaystyle x = \frac{2 \pm \sqrt{4 + 4}}{2}$
    $\displaystyle x = \frac{2 \pm 2\sqrt{2}}{2}$
    $\displaystyle x = 1 \pm \sqrt{2}$

    Reject $\displaystyle x = 1 - \sqrt{2}$ because secent of an angle cannot be in between -1 and 1.
    $\displaystyle x = 1 + \sqrt{2}$
    $\displaystyle \sec k = 1 + \sqrt{2}$


    01
    Last edited by yeongil; Jul 24th 2009 at 07:45 PM.
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  5. #5
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    okkk sorry
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