# Question #4bd79

Sep 27, 2017

Yes. It's probably looking for the answers in exact form. The answers are $x \left(\frac{2 \pi}{3}\right) = 4 \sqrt{3}$, $v \left(\frac{2 \pi}{3}\right) = - 4$, and $a \left(\frac{2 \pi}{3}\right) = - 4 \sqrt{3}$.

#### Explanation:

The reasons are that $\sin \left(\frac{2 \pi}{3}\right) = \frac{\sqrt{3}}{2}$ and $\cos \left(\frac{2 \pi}{3}\right) = - \frac{1}{2}$.

These facts can be derived from drawing a 30-60-90 (degrees) triangle (with 30 degree angle at the top) whose hypotenuse has length 1 and whose base has length 1/2 (because its half the length of an equilateral triangle whose sides all have length 1). (See the figure below)

The Pythagorean Theorem then implies that the height of the triangle is $\sqrt{{1}^{2} - {\left(\frac{1}{2}\right)}^{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$. Now use SOH, CAH, TOA to say that $\cos \left({60}^{\circ}\right) = \frac{1}{2}$ and $\sin \left({60}^{\circ}\right) = \frac{\sqrt{3}}{2}$.

Since $\frac{2 \pi}{3} = {120}^{\circ}$ and $\frac{\pi}{3} = {60}^{\circ}$ are supplementary angles, their cosines are opposites ($\cos \left({120}^{\circ}\right) = - \cos \left({60}^{\circ}\right)$).