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Math Help - Value of function given point?

  1. #1
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    Value of function given point?

    No idea where to even start on this one. I'm asked to find the exact value of \sec \theta for angle \theta in standard position if the point (-3, 2) lies on its terminal side. I'm horrible with standard positions and the terminal side, can anyone help shine some light on this problem?

    Thanks!
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Failbait View Post
    No idea where to even start on this one. I'm asked to find the exact value of \sec \theta for angle \theta in standard position if the point (-3, 2) lies on its terminal side. I'm horrible with standard positions and the terminal side, can anyone help shine some light on this problem?

    Thanks!
    just note that the triangle is in the second quadrant. so \sec \theta is negative.

    you have a right-triangle with an acute angle \theta. opposite to that angle is a side of length 2, and adjacent to that angle is a side of size 3.

    what is the hypotenuse of this triangle?

    when we find the hypotenuse, how do we get \sec \theta ?
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  3. #3
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    (-3,2) is a point in the second quadrant.
    Both the cosine and the secant functions are negative in the second quadrant.
    \begin{array}{l}<br />
 \phi  \equiv \left( {x,y} \right),\;r = \sqrt {x^2  + y^2 }  \\ <br />
 \cos (\phi ) = \frac{x}{r}\quad \& \quad \sec (\phi ) = \frac{1}{{\cos (\phi )}} \\ <br />
 \end{array}
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  4. #4
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    Alright, using the Pythagorean theorem, I've got that the hypotenuse is \sqrt{13}. Given that, would \sec \theta = -(\sqrt{13}/3)?

    Thanks again.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Failbait View Post
    Alright, using the Pythagorean theorem, I've got that the hypotenuse is \sqrt{13}. Given that, would \sec \theta = -(\sqrt{13}/3)?

    Thanks again.
    yes
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