Where is the angle for $-{7\pi\over8}$ in a unit circle?

Question

I am having difficulty finding the angle for $-{7\pi\over8}$ in a unit circle. I have found the coordinates for $\frac{\pi}{8}$ and have to use the symmetry to locate the coordinates for $-{7\pi\over8}$. Is the angle just below the $(1,0)$ and symmetry to ${\pi\over8}$? Thank you.

Answer 1

$-\frac{7\pi}{8}=\frac{\pi}{8}-\pi$. So you must rotate your point clockwise through an angle $\pi$. This will take you to a point directly opposite the point you started from.

Answer 2

$\dfrac{-7\pi}8$ is in the third quadrant. So, using $\dfrac {\pi}8$ as your reference angle, make both coordinates negative.

(It is also coterminal with $\dfrac{9\pi}8$, since they differ by $2\pi$.)

Answer 3

You can easily turn $-\frac{7\pi}{8}$ into $-(\frac{8\pi}{8}-\frac{\pi}{8})$ which simplified become $-(\pi - \frac{\pi}{8})$. Negative angle mean you're rotating clockwise , so instead of being in 2nd quadrant it's in $2\pi - \pi + \frac{\pi}{8} = \pi + \frac{\pi}{8}$ (in 3rd quadrant).

Answer 4

A full circle is going $2\pi$ radians in the counter-clockwise direction from $(1,0)$. You are going $3$ circles clock-wise plus one more $\pi$ clock-wise so you are at $-180^\circ=180^\circ$. The coordiates are on the x-axis at $(-1,0)$