August 3rd, 2016, 03:28 PM  #1 
Member Joined: Jul 2016 From: Usa Posts: 59 Thanks: 3  Exact values?
Cos^1 (cos (pi/3)) couldn't it be either 2pi/3 or 4pi/3? Tan^1(1) couldn't it be either 7pi/4 or 3pi/4 ? Can there be more than 1 exact value? 
August 3rd, 2016, 04:38 PM  #2  
Math Team Joined: Jul 2011 From: Texas Posts: 2,924 Thanks: 1521  neither ... $Cos^{1}\bigg[\cos\left(\dfrac{\pi}{3}\right)\bigg] = Cos^{1}\left(\dfrac{1}{2}\right) = \dfrac{\pi}{3}$ Quote:
Quote:
if the problem were stated ... Given that $\tan{x} = 1$, then for $0 \le x < 2\pi$, determine all value(s) of $x$ ... $x = \dfrac{\pi}{4}$ and $x = \dfrac{5\pi}{4}$  
August 3rd, 2016, 05:06 PM  #3  
Member Joined: Jul 2016 From: Usa Posts: 59 Thanks: 3  Quote:
 
August 3rd, 2016, 05:07 PM  #4 
Member Joined: Jul 2016 From: Usa Posts: 59 Thanks: 3 
What does that 1 exponent even represent then?

August 3rd, 2016, 05:15 PM  #5 
Math Team Joined: Jul 2011 From: Texas Posts: 2,924 Thanks: 1521  it's not an exponent ... it is notation for an inverse function the inverse of $f(x)$ is notated $f^{1}(x)$ an inverse swaps x and y values ... in other words, if $f(1)=2$, then $f^{1}(2)=1$ I'm not a big fan of the $1$ notation, which is why I prefer "arc" notation ... $y =\sin{x} \implies x = \arcsin{y}$ 
August 3rd, 2016, 06:41 PM  #6 
Member Joined: Jul 2016 From: Usa Posts: 59 Thanks: 3 
Thanks!

August 3rd, 2016, 06:42 PM  #7 
Member Joined: Jul 2016 From: Usa Posts: 59 Thanks: 3 
I was way over complicating what I had to do.


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