Question

How do you evaluate `tan(arccos(2/3))`?

Answer 1

`tan(arccos(2/3))=sqrt(5)/2`.

Explanation:

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

`tan(alpha)=sinalpha/cosalpha=(sqrt(5)/3)/(2/3)=sqrt(5)/2.`

So `tan(arccos(2/3))=sqrt(5)/2`.

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*The first fundamental relation:
`(cosalpha)^2+(sinalpha)^2=1`
From which we can get `sinalpha`:
`(sinalpha)^2=1-(cosalpha)^2`
`sinalpha=+-sqrt(1-(cosalpha)^2)`
But in this case we consider only positive values.