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

1 Answer
Jun 10, 2015

Answer:

#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#.


*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.