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 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
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Quote:
 Originally Posted by Pete23 Cos^-1 (cos (-pi/3)) couldn't it be either 2pi/3 or 4pi/3?
neither ...

$Cos^{-1}\bigg[\cos\left(-\dfrac{\pi}{3}\right)\bigg] = Cos^{-1}\left(\dfrac{1}{2}\right) = \dfrac{\pi}{3}$

Quote:
 Tan^-1(1) couldn't it be either 7pi/4 or 3pi/4 ?
no ... $Tan^{-1}(1) = \dfrac{\pi}{4}$

Quote:
 Can there be more than 1 exact value?
no, the inverse trig functions yield a single value because they are functions ...

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
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Quote:
 Originally Posted by skeeter neither ... $Cos^{-1}\bigg[\cos\left(-\dfrac{\pi}{3}\right)\bigg] = Cos^{-1}\left(\dfrac{1}{2}\right) = \dfrac{\pi}{3}$ no ... $Tan^{-1}(1) = \dfrac{\pi}{4}$ no, the inverse trig functions yield a single value because they are functions ... 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 $x = \dfrac{5\pi}{4}$
But doesnt cos^-1 (x)= cos theta -(x)?

 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
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Quote:
 Originally Posted by Pete23 What does that -1 exponent even represent then?
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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