# Question #3c386

Feb 18, 2018

Do $\frac{- 11 \pi}{6} = \frac{- 12 \pi}{6} + \frac{\pi}{6} = - 2 \pi + \frac{\pi}{6}$.
See explanation.

#### Explanation:

Good question.
Here is one way of thinking about it.
You see that $- \frac{11}{6}$ is very close to $- \frac{12}{6}$ which would be equal to $- 2$, so I would first try to rewrite as such:

$\frac{- 11 \pi}{6} = \frac{- 12 \pi}{6} + \frac{\pi}{6} = - 2 \pi + \frac{\pi}{6}$

You are asked to compute the sine and cosine of that angle, so this can be written:
$\cos \left(\frac{- 11 \pi}{6}\right) = \cos \left(- 2 \pi + \frac{\pi}{6}\right)$
and
$\sin \left(\frac{- 11 \pi}{6}\right) = \sin \left(- 2 \pi + \frac{\pi}{6}\right)$

Now, picture yourself the trigonometric circle.
Going around a full circle is $2 \pi$ (radian), so if you add $2 \pi$, or any multiple of $2 \pi$ if you're going full-circle many times, to whatever angle, you come back to where you started from (of course, it's the same thing if you subtract $2 \pi$!).

So,
$\cos \left(- 2 \pi + \frac{\pi}{6}\right) = \cos \left(\frac{\pi}{6}\right)$
and
$\sin \left(- 2 \pi + \frac{\pi}{6}\right) = \sin \left(\frac{\pi}{6}\right)$
and these are equal to
$\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$
and
$\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$.

I hope this helps.