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## anonymous one year ago Find the exact value of cos^(-1)(cos(17pi/5))?

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1. chrisdbest

Can you put that in equation form using the equation button located on the left of the "Post" button?

2. anonymous

$\cos^{-1}(\cos( 17\pi/5))$

3. chrisdbest

Aye!

4. chrisdbest

Now that's better!

5. chrisdbest

Based on my calculations, $\frac{ 3\pi }{ 5 }$

6. anonymous

@_alex_urena_ what's the $$cos(\pi)?$$

7. anonymous

what do you mean @jdoe0001 ?

8. anonymous

how did you get that @chrisdbest ?

9. anonymous

ohh, just that... if you were to get the $$cos(\pi)$$ what would that give you?

10. chrisdbest

Sure, I'll tell you

11. anonymous

-1? @jdoe0001

12. anonymous

ok... so... what is now the $$cos^{-1}(-1)?$$

13. anonymous

pi

14. anonymous

yeap.... thus... one sec

15. anonymous

$$\bf cos(\pi )={\color{brown}{ -1}}\qquad cos^{-1}({\color{brown}{ -1}})=\pi \\ \quad \\ \textit{thus we could say that }cos^{-1}[cos(\pi )]=\pi \\ \quad \\ \textit{thus, we could also say that }cos^{-1}\left[ cos\left( \cfrac{17\pi }{5} \right) \right]\implies \cfrac{17\pi }{5}$$

16. chrisdbest

What he's trying to say is that when you have an inverse cosine and a cosine, they cancel out

17. anonymous

so in short, $$\bf cos^{-1}[cos(whatever)]=whatever\qquad \\ \quad \\sin^{-1}[sin(whatever)]=whatever \\ \quad \\ tan^{-1}[tan(whatever)]=whatever$$

18. chrisdbest

Lols, that's basically what I said

19. anonymous

so the answer is 17pi/5 ?

20. chrisdbest

It might, I thought it was 3pi/5

21. anonymous

yeap, unless, the inverse function apply, which so far I don't see they do

22. anonymous

inverse functions restrictions I meant, doesn't seem like in this context they do

23. chrisdbest

Because the way you wrote it in equation form, I plugged that into my calculator the same way, and I got 3pi/5

24. anonymous

hmmm well how did you get 3pi/5 @chrisdbest

25. chrisdbest

Because the way you wrote it in equation form, I plugged that into my calculator the same way, and I got 3pi/5

26. anonymous

ok but how would you do without the calculator? or do you have to use one to solve this problem?

27. chrisdbest

I use one to solve the problem.

28. anonymous

hmmm ok thnx

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