<s>[INST] Find the value of csc$\frac{16\pi}{3}$. Q.  Find the value of csc$\frac{16\pi}{3}$.
A.  This angle is equal to two full revolutions plus $\frac{4\pi}{3}$. I got this by subtracting, $\frac{16\pi}{3} - 4\pi = \frac{16\pi}{3} - \frac{12\pi}{3} = \frac{4\pi}{3}$. The terminal side is in Quadrant III. The reference angle is $\frac{4\pi}{3} - \pi = \frac{\pi}{3}$. The cosecant of $\frac{\pi}{3}$ is $\frac{2\sqrt{3}}{3}$. Using the negative, the answer is $-\frac{2\sqrt{3}}{3}$.
My questions:
is how do one tell how many revolutions it takes, based off the fraction they give?
And how does one find the cosecant of $\frac{\pi}{3}$? On the calculator I use sine and then take the reciprocal of sine, but sine come out to $.018276028$. And I'm not sure on how to get the reciprocal from that or even find the cosecant of that number.
 [/INST] A formula for determining the quadrant containing an angle $\theta$ when $\theta$ is not an integer multiple of $\frac{\pi}{2}$ (the quadrantal angles) is
\begin{equation}
Q\left(\theta\right)=\left(\left\lfloor\frac{2\theta}{\pi} \right\rfloor\mod{4}\right)+1
\end{equation}
For example,
\begin{eqnarray}
Q\left(\frac{16\pi}{3}\right)&=&\left(\left\lfloor\frac{32\pi}{3\pi} \right\rfloor\mod{4}\right)+1\\
&=&(10\mod{4})+1\\
&=&2+1=3
\end{eqnarray}
and 
\begin{eqnarray}
Q\left(-\frac{16\pi}{3}\right)&=&\left(\left\lfloor-\frac{32\pi}{3\pi} \right\rfloor\mod{4}\right)+1\\
&=&(-11\mod{4})+1\\
&=&1+1=2
\end{eqnarray}

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