## anonymous one year ago Trig/ Pre Cal How do I determine what quadrant this is in? 2 owlbucks! $\tan^-1(\tan(\frac{5\pi}{6}))$ Thank you.

1. anonymous

It was suppose to be $\tan^1(\tan\frac{5\pi}{6})$ I don't know how to make the -1 power so 1 = -1. It is an inverse

2. anonymous

I'm not sure what you mean... You mean where it is on the unit circle?

3. anonymous

Just a number does not have a "quadrant".

4. anonymous

Btw to make an exponent negative, just do tan^{-1} in latex :)

5. anonymous

Yes. For instance, we have a domain of $$-\frac{\pi}{2}< x< \frac{\pi}{2}$$ and range of $$-infinity< x< infinity$$

6. anonymous

How do I find the quadrant that $$\frac{5\pi}{6} lies in?$$

7. anonymous

Easy way is to just look at the unit circle http://upload.wikimedia.org/wikipedia/commons/thumb/4/4c/Unit_circle_angles_color.svg/2000px-Unit_circle_angles_color.svg.png

8. anonymous

I know by looking at the unit circle but lets say we had a strange angle like $$\frac{4\pi}{5}$$

9. anonymous

$\large \frac{\pi}{5} = \frac{180}{5} = 36~degrees$So $\large \frac{4\pi}{5} = 4 \times \frac{\pi}{5} = 4 \times 36 = 144$ Since 144 is above 90 but below 180, it would be in quadrant 2 :) Also if the numerator is lower than the denominator, you know it will be in it either the first or second quadrants

10. anonymous

Thank you!!

11. anonymous

I sent your owlbucks :-)