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 $\cos^{-1} x$?
If $\cos x = \frac{1}{\sqrt{5}}$, what is $\cos^{-1} x$?
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trigonometry
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0Are you sure you want $\cos^{-1}(x)$? Or do you need $\frac{1}{\cos(x)}$? – 2012-02-10
1 Answers
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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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2$\arccos \left( \arccos\frac1{\sqrt{5}} \right) = i\;\mathrm{arcosh}\left(\arctan\,2\right)$ – 2012-02-10