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Math Help - 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; July 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.

    \cot{t} = \frac{\sqrt{2}}{4}<br />

    noted that t is in quad III ...

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

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

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

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

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

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

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

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


    Here's another approach . . .



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

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

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

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

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


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



    Therefore:

    . . \begin{array}{cccccc}<br />
\sin\theta &=& -\dfrac{4}{3\sqrt{2}} &=& -\dfrac{2\sqrt{2}}{3} \\ \\[-3mm]<br />
\cos\theta &=& -\dfrac{\sqrt{2}}{3\sqrt{2}} &=& -\dfrac{1}{3} \\ \\[-3mm]<br />
\tan\theta &=& \dfrac{4}{\sqrt{2}} &=& 2\sqrt{2} \\ \\[-3mm]<br />
\cot\theta &=& \dfrac{\sqrt{2}}{4} \\ \\[-3mm]<br />
\sec\theta &=& -3 \\ \\[-3mm]<br />
\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: . \cot\theta \:=\:\dfrac{\sqrt{2}}{4} \:=\:\dfrac{adj}{opp}

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

    Pythagorus says: . \sqrt{2})^2 + 4^2 \:=\:18 \quad\Rightarrow\quad hyp \:=\:3\sqrt{2}" alt="hyp^2 \:=\\sqrt{2})^2 + 4^2 \:=\:18 \quad\Rightarrow\quad hyp \:=\:3\sqrt{2}" />


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



    Therefore:

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

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