How do you evaluate $tan (arccos (\frac{2}{3}))$ $tan(arccos(2/3))$ ?

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How do you evaluate $tan (arccos (\frac{2}{3}))$ $tan(arccos(2/3))$ ?

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Answer 1 · 2015-06-10T14:13:41

$tan (arccos (\frac{2}{3})) = \frac{\sqrt{5}}{2}$ $tan(arccos(2/3))=sqrt(5)/2$ .

Explanation:

$α = arccos (\frac{2}{3})$ $alpha=arccos(2/3)$ .
$α$ $alpha$ isn't a known value, but it's about 48,19°.
$tan (α) = \frac{sin α}{cos α}$ $tan(alpha)=sinalpha/cosalpha$
We can say something about $cos α$ $cosalpha$ and $sin α$ $sinalpha$ :
$cos α = \frac{2}{3}$ $cosalpha=2/3$
$sin α = \sqrt{1 - {(cos α)}^{2}}$ $sinalpha=sqrt(1-(cosalpha)^2)$ (for the first fundamental relation*).
So $sin α = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$ $sinalpha=sqrt(1-4/9)=sqrt(5)/3$ .

$tan (α) = \frac{sin α}{cos α} = \frac{\frac{\sqrt{5}}{3}}{\frac{2}{3}} = \frac{\sqrt{5}}{2} .$ $tan(alpha)=sinalpha/cosalpha=(sqrt(5)/3)/(2/3)=sqrt(5)/2.$

So $tan (arccos (\frac{2}{3})) = \frac{\sqrt{5}}{2}$ $tan(arccos(2/3))=sqrt(5)/2$ .

*The first fundamental relation:
${(cos α)}^{2} + {(sin α)}^{2} = 1$ $(cosalpha)^2+(sinalpha)^2=1$
From which we can get $sin α$ $sinalpha$ :
${(sin α)}^{2} = 1 - {(cos α)}^{2}$ $(sinalpha)^2=1-(cosalpha)^2$
$sin α = \pm \sqrt{1 - {(cos α)}^{2}}$ $sinalpha=+-sqrt(1-(cosalpha)^2)$
But in this case we consider only positive values.

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