The problem states,

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

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- Feb 5th 2009, 11:12 PMmoonman[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). - Feb 6th 2009, 01:57 AMmathaddictRe:
$\displaystyle \sec\theta=3$

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

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

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

$\displaystyle

\tan\theta=-\sqrt{8}

$ - Feb 6th 2009, 02:55 PMmoonman
Thank you for helping :)

Can you explain in a little more detail how you got those answers? - Feb 6th 2009, 03:19 PMskeeter
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.