# How do you use tantheta=4 to find costheta?

Mar 7, 2018

$\cos \theta = \pm \frac{1}{\sqrt{17}}$

#### Explanation:

You could use the identity ${\sec}^{2} \theta = {\tan}^{2} \theta + 1$ to find that in this case ${\sec}^{2} \theta = 16 + 1 = 17$, so that $\sec \theta = \pm \sqrt{17}$ Since $\sec \theta = \frac{1}{\cos} \theta$, this tells us that

$\cos \theta = \pm \frac{1}{\sqrt{17}}$

Unfortunately the sign of $\cos \theta$ can not be found from the value of $\tan \theta$. Since $\tan \theta$ is positive in this case, the angle $\theta$ may lie in either the first or the third quadrant. If the angle is in the first quadrant, then $\cos \theta = \frac{1}{\sqrt{17}}$. If, on the other hand, it is in the third quadrant, then $\cos \theta = - \frac{1}{\sqrt{17}}$

Mar 7, 2018

$\cos \left(\theta\right) = \frac{1}{\sqrt{17}} \textcolor{w h i t e}{\text{xxx")"or"color(white)("xxx}} \cos \left(\theta\right) = - \frac{1}{\sqrt{17}}$

#### Explanation:

If $\tan \left(\theta\right) = 4$
then the ratio of the opposite side to the adjacent side is $4 : 1$

Two possibilities exist depending upon whether $\theta$ is in Quadrant I or Quandrant III; as indicated below:

In either case the relative length of the hypotenuse is given by the Pythagorean Theorem as
$\textcolor{w h i t e}{\text{XXX}} h = \sqrt{{4}^{2} + {1}^{2}} = \sqrt{17}$

Since
$\textcolor{w h i t e}{\text{XXX")cos(theta)="adjacent"/("hypotenuse}}$
for this case
$\textcolor{w h i t e}{\text{XXX")cos(theta)=1/sqrt(17)color(white)("xx")"or"color(white)("xx}} \frac{- 1}{\sqrt{17}}$