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.

Just observe that $16=5\cdot 3+1$.
The cosecant of $16\pi/3$ is the reciprocal of the sine; from $16=5\cdot3+1$, we get
$$
\frac{16\pi}{3}=5\pi+\frac{\pi}{3}=\pi+\frac{\pi}{3}+4\pi
$$
you have
$$
\sin\frac{16\pi}{3}=\sin\left(\pi+\frac{\pi}{3}\right)=
-\sin\frac{\pi}{3}=-\frac{\sqrt{3}}{2}
$$
Therefore
$$
\csc\frac{16\pi}{3}=-\frac{2}{\sqrt{3}}=-\frac{2\sqrt{3}}{3}\approx-1.1547
$$


*

*You probably had the calculator set to angles in degrees, rather than radians.

*In a right triangle having one angle of $\pi/3$, the other acute angle is $\pi/6$. Thus the triangle is half of an equilateral triangle, and so $\cos\frac{\pi}{3}=\frac{1}{2}$. With Pythagoras' theorem, $\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$.
