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Given that $\cos x = \frac{1}{\sqrt{5}}$ and $\tan x < 0$, what is the exact value of $\cos^{-1} x$?
Since $\sin x = - \frac{2}{\sqrt{5}}$, we can see that $\tan x$ is in fact $-2$. But how do we get the (exact) value of $\cos^{-1} x$?

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According to Wolfram|Alpha, there doesn't seem to be one. What makes you think there should be one? –  joriki Feb 8 '12 at 20:13
    
@I.J.Kennedy: If Joriki is reading your mind correctly, you meant to ask for the value of $x$, not $\cos^{-1} x$, right? –  Hans Lundmark Feb 8 '12 at 20:21
    
I copied this question from a problem book in the library. The book wasn't Olympiad level, but neither was it a book of run-of-the-mill problems. I guess I miscopied the problem but on the other hand, $\cos^{-1} \frac{1}{\sqrt{5}}$ doesn't seem to have a nice answer either. –  I. J. Kennedy Feb 8 '12 at 20:58
    
$\cos^{-1}\frac{1}{\sqrt 5}$ is transcedential. Wolfran alpha give the continued fraction expansion $\frac{2}{1+\frac{4}{3+\frac{16}{5+\frac{36}{...}}}$ –  Angela Richardson Feb 9 '12 at 10:20
    
Are you sure you want $\cos^{-1}(x)$? Or do you need $\frac{1}{\cos(x)}$? –  N. S. Feb 10 '12 at 0:41
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1 Answer

up vote 3 down vote accepted

Since $1/\sqrt5 < 1/2$, $\cos \pi/3 = 1/2$ and since the cosine function is decreasing on $[0.\pi]$, we deduce that if $x>0$, then $x>\pi/3>1$; as the cosine is an even function, this implies that if $x<0$, then $x<-1$.

All in all, if $\cos x = 1/\sqrt5$, then $|x|>1$ and thus $\cos^{-1}x$ does not exist.

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...on the other hand, if one widens his/her horizon to the complex plane, the arccosine of a number with magnitude bigger than $1$ is certainly possible. –  J. M. Feb 10 '12 at 0:14
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$$\arccos \left( \arccos\frac1{\sqrt{5}} \right) = i\;\mathrm{arcosh}\left(\arctan\,2\right)$$ –  J. M. Feb 10 '12 at 0:20
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