I'm literally stuck at A. The answer is ($\frac{2\pi}{3},\frac{3}{2}$) but I don't know how to get it, most significantly because I don't know how to find the answer from the points when the equation doesn't have an amplitude coefficient or a period coefficient– the graph has just been shifted. And because the graph has been shifted, the period isn't divided into 2 equal halves anymore– the curve between $0$ and $\frac{4\pi}{3}$ is bigger and taller than the one between $\frac{4\pi}{3}$ and $2\pi$.
What I've tried: I realized that the period would still be $2\pi$ and the amplitude would still be 1, meaning the height of the entire graph would be 2. I understood how to figure out the minimum point as $\frac{5\pi}{3}$, but I don't know how to get A after that. I assumed that because the difference between $\frac{5\pi}{3}$ and $\frac{4\pi}{3}$ is one-half of the difference between $\frac{4\pi}{3}$ and 0, that the size of the curve on the left is twice the size of the curve on the right. But beyond that, I have no clue.
Best Answer
So we have the function $f(x) = \sin(x-k) + c$
And looking at the graph we know 2 things:
$ f(0) = 0 \\ f \left ( \frac{4\pi}{3} \right ) = 0 $
Using the first we have
$$ 0 = \sin(-k) + c = -\sin(k) + c \Rightarrow c = \sin(k) $$
Now using the second and substituting the value we just got for $c$ we have $$ \sin \left ( \frac{4\pi}{3} -k \right ) + \sin(k) = 0 \\ $$
Now recall the identity $$ \sin(\alpha) + \sin(\beta) = 2\sin \left ( \frac{\alpha + \beta}{2} \right ) \cos \left ( \frac{\alpha - \beta}{2} \right ) $$
This leaves us with $$ \begin{align} 2\sin \left ( \frac{2\pi}{3} \right ) \cos \left ( \frac{2\pi}{3} - k \right ) &= 0 \\ \cos \left ( \frac{2\pi}{3} - k \right ) &= 0 \\ \cos^{-1} \left ( \cos \left ( \frac{2\pi}{3} - k \right ) \right ) &= \cos^{-1}(0) \\ \frac{2\pi}{3} - k &= \frac{\pi}{2} \\ k &= \frac{\pi}{6} \end{align} $$
So we have $k = \frac{\pi}{6} \text{ and } c = \frac{1}{2}$