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Math Help - [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:

     \sec\theta=3

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

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

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

     <br />
\tan\theta=-\sqrt{8}<br />
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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

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

    since \theta is in quad IV , you should also know that \sin{\theta} and \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 \cos{\theta} = \frac{1}{3}, then the adjacent side = 1 and the hypotenuse = 3

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

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

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

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

    2) use of Pythagorean identities ...

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

    \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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