# How do you find the unit vectors that are parallel to the tangent line to the curve y = 2sinx at the point (pi/6,1)?

Aug 24, 2016

$\pm < \cos \left(\frac{\pi}{6}\right) , \sin \left(\frac{\pi}{6}\right) > = \pm < \frac{1}{2} , \frac{\sqrt{3}}{2} >$

#### Explanation:

y'at$x = \frac{\pi}{6}$ is $2 \cos \left(\frac{\pi}{6}\right) = \sqrt{3}$.

This direction $\theta = \psi$ is given by $\tan \psi = \sqrt{3}$.

Inversely, $\psi = {\tan}^{- 1} \sqrt{3}$ is $\frac{\pi}{6}$. For the opposite direction ,

it is $\pi + \frac{\pi}{6}$.

The unit vector in the direction $\theta = \frac{\pi}{6}$ is

$< \cos \left(\frac{\pi}{6}\right) , \sin \left(\frac{\pi}{6}\right) >$.

For the opposite direction, it is

$< \cos \left(\pi + \frac{\pi}{6}\right) , \sin \left(\pi + \frac{\pi}{6}\right) >$

$= < - \cos \left(\frac{\pi}{6}\right) , - \sin \left(\frac{\pi}{6}\right) >$ .