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Thread: Six trigonometric equations

  1. #1
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    Six trigonometric equations

    Given: cot theta = sqrt2/4 , pi < theta < 3pi/2
    find the exact value of the other five trig functions of theta, leave answers in radical form.

    could someone show me how to do this, when i did it i got:
    sin theta= -2sqrt2/3
    cos theta= -1/3
    tan theta= 2/sqrt2
    cot theta= sqrt2/4
    sec theta= -3
    csc theta= -3sqrt2/4

    is this correct?

    a friend in class said that it wasn't and that i had to start by doing cot theta= cos/sin cos = sqrt2 sin = 4....
    Last edited by FailingTrig; Jul 21st 2010 at 02:40 PM.
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  2. #2
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    image link doesn't work. some you got correct.

    $\displaystyle \cot{t} = \frac{\sqrt{2}}{4}
    $

    noted that $\displaystyle t$ is in quad III ...

    $\displaystyle \tan{t} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$

    using the identity $\displaystyle \csc^2{t} = 1 + \cot^2{t}$ ...

    $\displaystyle \csc{t} = -\frac{3}{2\sqrt{2}}$

    $\displaystyle \sin{t} = -\frac{2\sqrt{2}}{3}$

    $\displaystyle \cos{t} = -\frac{1}{3}$

    $\displaystyle \sec{t} = -3$
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  3. #3
    Senior Member eumyang's Avatar
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    $\displaystyle cot \theta = \frac{x}{y} = \frac{\sqrt{2}}{4}$, and we're given that θ is in quadrant III, so $\displaystyle x = -\sqrt{2}$ and $\displaystyle y = -4$

    Use $\displaystyle x^2 + y^2 = r^2$ to find r:
    $\displaystyle (-\sqrt{2})^2 + (-4)^2 = r^2$
    $\displaystyle 2 + 16 = r^2$
    $\displaystyle 18 = r^2$
    $\displaystyle r = 3\sqrt{2}$

    So...
    $\displaystyle sin \theta = \frac{y}{r} = \frac{-4}{3\sqrt{2}} = ...$
    $\displaystyle cos \theta = \frac{x}{r} = ...$
    ...
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  4. #4
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    Hello, FailingTrig!


    Here's another approach . . .



    Given: .$\displaystyle \cot\theta \,=\, \frac{\sqrt{2}}{4},\;\; \pi < \theta < \frac{3\pi}{2}$

    Find the exact value of the other five trig functions of $\displaystyle \theta$.

    We are given: .$\displaystyle \cot\theta \:=\:\dfrac{\sqrt{2}}{4} \:=\:\dfrac{adj}{opp}$

    That is, $\displaystyle \theta$ is in a right triangle with: .$\displaystyle adj = \sqrt{2},\;opp = 4$

    Pythagorus says: .$\displaystyle hyp^2 \:=\:(\sqrt{2})^2 + 4^2 \:=\:18 \quad\Rightarrow\quad hyp \:=\:3\sqrt{2}$


    Since $\displaystyle \theta$ is in Quadrant 3, only $\displaystyle \tan\theta$ and $\displaystyle \cot\theta$ are positive.



    Therefore:

    . . $\displaystyle \begin{array}{cccccc}
    \sin\theta &=& -\dfrac{4}{3\sqrt{2}} &=& -\dfrac{2\sqrt{2}}{3} \\ \\[-3mm]
    \cos\theta &=& -\dfrac{\sqrt{2}}{3\sqrt{2}} &=& -\dfrac{1}{3} \\ \\[-3mm]
    \tan\theta &=& \dfrac{4}{\sqrt{2}} &=& 2\sqrt{2} \\ \\[-3mm]
    \cot\theta &=& \dfrac{\sqrt{2}}{4} \\ \\[-3mm]
    \sec\theta &=& -3 \\ \\[-3mm]
    \csc\theta &=& -\dfrac{3\sqrt{2}}{4}\end{array}$

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  5. #5
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    Quote Originally Posted by Soroban View Post
    Hello, FailingTrig!


    Here's another approach . . .




    We are given: .$\displaystyle \cot\theta \:=\:\dfrac{\sqrt{2}}{4} \:=\:\dfrac{adj}{opp}$

    That is, $\displaystyle \theta$ is in a right triangle with: .$\displaystyle adj = \sqrt{2},\;opp = 4$

    Pythagorus says: .$\displaystyle hyp^2 \:=\\sqrt{2})^2 + 4^2 \:=\:18 \quad\Rightarrow\quad hyp \:=\:3\sqrt{2}$


    Since $\displaystyle \theta$ is in Quadrant 3, only $\displaystyle \tan\theta$ and $\displaystyle \cot\theta$ are positive.



    Therefore:

    . . $\displaystyle \begin{array}{cccccc}
    \sin\theta &=& -\dfrac{4}{3\sqrt{2}} &=& -\dfrac{2\sqrt{2}}{3} \\ \\[-3mm]
    \cos\theta &=& -\dfrac{\sqrt{2}}{3\sqrt{2}} &=& -\dfrac{1}{3} \\ \\[-3mm]
    \tan\theta &=& \dfrac{4}{\sqrt{2}} &=& 2\sqrt{2} \\ \\[-3mm]
    \cot\theta &=& \dfrac{\sqrt{2}}{4} \\ \\[-3mm]
    \sec\theta &=& -3 \\ \\[-3mm]
    \csc\theta &=& -\dfrac{3\sqrt{2}}{4}\end{array}$

    Did it right then! Awesome, thanks for showing me everyone =)
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