- anonymous

Find the exact value of cos^(-1)(cos(17pi/5))?

- schrodinger

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- chrisdbest

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

- anonymous

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

- chrisdbest

Aye!

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## More answers

- chrisdbest

Now that's better!

- chrisdbest

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

- jdoe0001

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

- anonymous

what do you mean @jdoe0001 ?

- anonymous

how did you get that @chrisdbest ?

- jdoe0001

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

- chrisdbest

Sure, I'll tell you

- anonymous

-1? @jdoe0001

- jdoe0001

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

- anonymous

pi

- jdoe0001

yeap.... thus...
one sec

- jdoe0001

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

- chrisdbest

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

- jdoe0001

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

- chrisdbest

Lols, that's basically what I said

- anonymous

so the answer is 17pi/5 ?

- chrisdbest

It might, I thought it was 3pi/5

- jdoe0001

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

- jdoe0001

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

- chrisdbest

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

- anonymous

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

- chrisdbest

- anonymous

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

- chrisdbest

I use one to solve the problem.

- anonymous

hmmm ok thnx

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