# How do you determine the quadrant in which (9pi)/8 lies?

Dec 17, 2017

#### Explanation:

We can see from the drawring that $\frac{9 \pi}{8}$ is just over $\pi$
As its $1.125 \cdot \pi$ So hence in the 3rd quadrant

As $\pi < \frac{9 \pi}{8} < \frac{3 \pi}{2}$

Dec 17, 2017

I would visualize its rotation in the unit circle to find out that it is in Quadrant III.

#### Explanation:

Visualize a unit circle:

The angle $\frac{9}{8} \pi$ is a counterclockwise rotation of $\frac{9}{8} \pi$ radians from the starting point, which is in case of the unit circle, $\left(1 , 0\right)$.

Since $\frac{9}{8} \pi$ is $\frac{8}{8} \pi + \frac{1}{8} \pi$, simplify out $\frac{8}{8}$ to get $\pi + \frac{1}{8} \pi$.

This tells me, intuitively, to rotate $\pi$ radians, and then another $\frac{1}{8}$ of $\pi$. We first rotate counterclockwise by $\pi$ radians, which is half of a full rotation:

Then, since we have $\pi$ radians left to complete a full rotation, yet our angle only tells us to rotate $\frac{1}{8} \pi$ more, let's slice up the remaining semicircle into $8$ parts:

And rotate our angle accordingly:

It seems that our angle lands on Quadrant III.