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

moonman · #1 $Old$ February 6th, 2009, 12:12 AM

The problem states,

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

mathaddict · #2 $Old$ February 6th, 2009, 02:57 AM

$\sec\theta=3$

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

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

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

$\tan\theta=-\sqrt{8}$

moonman · #3 $Old$ February 6th, 2009, 03:55 PM

Thank you for helping

Can you explain in a little more detail how you got those answers?

skeeter · #4 $Old$ February 6th, 2009, 04:19 PM

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.