ashleedean12 2 years ago inverse tan[tan(6pi/7)] i need an exact answer using pi as needed. can someone explain to me how to solve this?

1. mathteacher1729

Sorry I mean $\huge \arctan\left( \tan (6\pi/7) \right)$ or $\huge \tan^{-1}\left( \tan (6\pi/7) \right)$

2. ashleedean12

$\tan^{-1} [\tan (6\pi/7)]$

3. mathteacher1729

What do you know about inverse functions? (they're super important to answering this question).

4. ashleedean12

um i know how to put them in my calculator and that they have restrictions this particular one the answer must be in either quadrant 1 or 4...

5. mathteacher1729

The key is that an inverse function "undoes" a function. arctan ( tan ( stuff ) ) = stuff and tan ( arctan ( stuff ) ) = stuff (assuming domains and ranges are valid) also $$\large \arctan(x) = \tan^{-1}(x)$$ it's just notation. They mean the same exact thing.

6. ashleedean12

so would the answer just be 6pi/7 since the inverse tan would undo tan?

7. phi

yes

8. ashleedean12

thanks (: does that apply to any function that has the inverse of the same function on the outside?

9. mathteacher1729

6pi/7 or -pi/7, they are the same thing. And yes, inverse function ( function ( stuff ) ) = stuff function ( inverse function ( stuff ) ) = stuff Inverse functions and functions "undo" each other. That's glossing over a lot of details, but that's the big idea. :)

10. phi

yes, and vice versa.

11. ashleedean12

what if it was something like $\tan [\tan^{-1}(\pi/10)]$

12. ashleedean12

the simplified version of 6pi/7 would be just pi/7 ?

13. phi

yes, pi/10 you can check with your calculator tan^-1 (pi/10) you are treating pi/10 as a number not as an angle.

14. phi

the simplified version of 6pi/7 would be just pi/7 ? I would not say "simplified" If you mean Is pi/7 the "reference angle" of 6pi/7, that is true

15. ashleedean12

the problem says simplify and use pi as needed..

16. phi

17. phi

unless they want the answer between -pi/2 and pi/2 in which case they want -pi/7 (see math teacher's post)

18. ashleedean12

ok thank you sooo much ! i have test on this tomorrow and i had no clue how to do it! you are a lifesaver!