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Math Help - Feel pretty dumb, basic trig question

  1. #1
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    Feel pretty dumb, basic trig question

    If \sin\theta = \frac{1}{3} and \tan\theta < 0 then what is \cos\theta

    Like I said I feel completely dumb here. I know this is so basic I'm just having a mental block with it.

    Multiple choice answers are

    a) \frac{1}{3}

    b) \frac{-3}{\sqrt{10}}

    c) \frac{-1}{2\sqrt{2}}

    d) \frac{-2\sqrt{2}}{3}

    e) \frac{1}{\sqrt{10}}

    Because of the nature of the answers I know I'd have to use the unit triangle somewhere. Because tan is negative and sin is positive the answer is in quadrent 2. In other words sin1/3 = 180 - cos.

    What I don't know though is how we've got the unit triangle to give sin 1/3, my stupid question is how do I get sin 1/3 on the unit triangle?
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  2. #2
    MHF Contributor
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    Re :

    Quote Originally Posted by Peleus View Post
    If \sin\theta = \frac{1}{3} and \tan\theta < 0 then what is \cos\theta

    Like I said I feel completely dumb here. I know this is so basic I'm just having a mental block with it.

    Multiple choice answers are

    a) \frac{1}{3}

    b) \frac{-3}{\sqrt{10}}

    c) \frac{-1}{2\sqrt{2}}

    d) \frac{-2\sqrt{2}}{3}

    e) \frac{1}{\sqrt{10}}

    Because of the nature of the answers I know I'd have to use the unit triangle somewhere. Because tan is negative and sin is positive the answer is in quadrent 2. In other words sin1/3 = 180 - cos.

    What I don't know though is how we've got the unit triangle to give sin 1/3, my stupid question is how do I get sin 1/3 on the unit triangle?

    From the information given , we know that \theta is in the second quadrant . Draw a triangle and find cos\theta
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  3. #3
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    Lexington, MA (USA)
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    Hello, Peleus!

    If \sin\theta = \tfrac{1}{3} and \tan\theta < 0, find \cos\theta.

    . . (a)\;\frac{1}{3}\qquad (b)\;-\frac{3}{\sqrt{10}} \qquad (c)\;-\frac{1}{2\sqrt{2}} \qquad (d)\;-\frac{2\sqrt{2}}{3} \qquad (e)\;\frac{1}{\sqrt{10}}

    Because \tan\theta < 0\text{ and }\sin\theta > 0, the answer is in quadrant 2. . Good!
    I don't use the unit circle . . .


    We have: . \sin\theta \:=\:\frac{1}{3} \:=\:\frac{opp}{hyp}

    So \theta is in a right triangle with: . opp = 1,\;hyp = 3

    Pythagorus says: . opp^2 + adj^2 \:=\:hyp^2

    So we have: . 1^2 + adj^2 \:=\:3^2 \quad\Rightarrow\quad adj^2 \:=\:8 \quad\Rightarrow\quad adj \:=\:\pm2\sqrt{2}

    Then: . \cos\theta \:=\:\frac{adj}{hyp} \:=\:\pm\frac{2\sqrt{2}}{3}


    In quadrant 2, cosine is negative: . {\color{blue}\cos\theta\:=\:-\frac{2\sqrt{2}}{3}}\quad\hdots answer (d)

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  4. #4
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    It's times like this that the answer smacks you in the face and tells you how silly you are.

    Thank you very much and quite obvious now. I guess I just have to not get focused on looking at one method to solve it (unit circle) and look at the big picture.

    Thanks again.
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