The position of two particles on the $x$-axis are $x_1 = \sin t$ and $x_2 = \sin \left( t + \frac{\pi}{3}\right)$
(a) At what time(s) in the interval $[0,2\pi]$ do the particles meet?
(b) What is the farthest apart that the particles every get?
(c) When in the interval $[0,2\pi]$ is the distance between the particles changing the fastest?
I am studying for the AP Calculus BC exam and this is a problem out of the Calculus Problem book. I know the answers: (a) $\frac{\pi}{3}, \, \frac{4\pi}{3}$ (b) 1 (c) $\frac{\pi}{3}, \, \frac{4\pi}{3}$
I am have trouble with (a)
I rewrote $\sin\left(t + \frac{\pi}{3}\right) \Rightarrow \frac{\sqrt{3}}{2}\sin t + \frac{1}{2}\cos t$
I get this which I can't seem to solve for the answer given:
$\left(2 - \sqrt{3}\right)\sin t = \cos t $
For part (b), I would maximized the distance formula. $d(x)=\sqrt{x_1^2 + x_2^2}$
Set the 1st derivative to zero and use the 2nd derivative test to find the max.
for part (c) I would find the maximum of $d^{\prime}(x)$