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Thread: [SOLVED] Find Cos (theta), sin(theta), and tan(theta)

  1. #1
    Junior Member moonman's Avatar
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    [SOLVED] Find Cos (theta), sin(theta), and tan(theta)

    The problem states,

    If sec(theta) = 3 with theta in QIV find cos(theta), sin(theta) and tan(theta).
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  2. #2
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    Re:

    $\displaystyle \sec\theta=3$

    $\displaystyle \cos\theta=\frac{1}{3}$

    $\displaystyle \theta$ is in the fourth quadrant so cos is positive .

    $\displaystyle \sin\theta=-\frac{\sqrt{8}}{3}$

    $\displaystyle
    \tan\theta=-\sqrt{8}
    $
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  3. #3
    Junior Member moonman's Avatar
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    Thank you for helping
    Can you explain in a little more detail how you got those answers?
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  4. #4
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    you should know that secant is the reciprocal of cosine , so the fact that

    $\displaystyle \cos{\theta} = \frac{1}{3}$ should be immediate.

    since $\displaystyle \theta$ is in quad IV , you should also know that $\displaystyle \sin{\theta}$ and $\displaystyle \tan{theta}$ are both negative values.

    there are two ways to determine the values of sine and tangent ...

    1) use of reference triangles

    sketch a reference right triangle in quad IV ... since $\displaystyle \cos{\theta} = \frac{1}{3}$, then the adjacent side = 1 and the hypotenuse = 3

    the size of the opposite side = $\displaystyle \sqrt{3^2 - 1^2} = \sqrt{8} = 2\sqrt{2}$

    since the opposite side has a downward direction, its value is $\displaystyle -2\sqrt{2}$.

    $\displaystyle \sin{\theta} = \frac{opp}{hyp} = -\frac{2\sqrt{2}}{3}$

    $\displaystyle \tan{\theta} = \frac{opp}{adj} = -\frac{2\sqrt{2}}{1} = -2\sqrt{2}$

    2) use of Pythagorean identities ...

    $\displaystyle \sin{\theta} = \pm \sqrt{1 - \cos^2{\theta}}$

    $\displaystyle \tan{\theta} = \pm \sqrt{\sec^2{\theta} - 1}$

    and, of course, you have to determine the correct sign from the quadrant info.
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