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Math Help - Triganomic Ratios

  1. #1
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    Triganomic Ratios

    Find sec(theta) if tan(theta)=sqrt(3)/3, and sin(theta)<0.
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    Quote Originally Posted by Chinnie15 View Post
    Find sec(theta) if tan(theta)=sqrt(3)/3, and sin(theta)<0.
    use the identity ...

    sec^2{t} = 1 + \tan^2{t}
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    Thanks!
    Last edited by Chinnie15; September 27th 2009 at 05:55 PM.
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    Quote Originally Posted by Chinnie15 View Post
    Thanks!

    I'm not sure how to find t? I've been trying all day to learn this stuff, and it's just not clicking.
    \sec^2{t} = 1 + \left(\frac{\sqrt{3}}{3}\right)^2

    solve for \sec{t} ... mind your signs.
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    Ok, I understand the rest now, but what is the sin<theta? What does that have to do with the problem?

    I took the root of both sides and have ended up with sqrt(3)+1/3. Is this correct?
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  6. #6
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    Quote Originally Posted by Chinnie15 View Post
    Ok, I understand the rest now, but what is the sin<theta? What does that have to do with the problem?
    you were given \tan{t} = \frac{\sqrt{3}}{3} ... a (+) value

    you were also told \sin{\theta} < 0 ... a (-) value

    those two pieces of information tell you what quadrant \theta is in ... so you can determine the correct sign for \sec{\theta}
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  7. #7
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    Ok, thanks!

    I have 2/sq(3)=secΘ
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  8. #8
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    Find sec(theta) if tan(theta)=sqrt(3)/3, and sin(theta)<0.

    tan is + and sin is < 0, that is - ; that is in the 3rd quadrant.

    But in the 3rd quadrant only tan and cot are positive, the rest negative.

    , from this we have,
    sec = -(2/3)(sqrt 3)

    the negative sign ( - ) is introduced since it is in the 3rd quadtant
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