What is the exact value of cos(tan^-1 (2/3))?

Apr 15, 2015

${\tan}^{-} 1 \left(\frac{2}{3}\right)$ is a $t$ in $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$ with $\tan t = \frac{2}{3}$

$t$ is in quadrant 1 or 2, so $\cos t$ is positive.

The problem has become:
$\tan t = \frac{2}{3}$ and $\cos t > 0$ .Find $\cos t$.

Use your favorite method for solving such problems.

It's hard to draw and label a triangle or unit circle here, so I'll use the identities:

$\cos t = \frac{1}{\sec} t$ and ${\tan}^{2} t + 1 = {\sec}^{2} t$

${\sec}^{2} t = {\left(\frac{2}{3}\right)}^{2} + 1 = \frac{4}{9} + 1 = \frac{13}{9}$

So, $\sec t = \frac{\sqrt{13}}{3}$ (Remember $\cos t > 0$, so $\sec t > 0$ as well)

$\cos t = \frac{3}{\sqrt{13}} = \frac{3 \sqrt{13}}{13}$