How do you evaluate $tan[arccos(1/3)]$ ?

Question

8785 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-03T12:06:46

$tan[arccos(1/3)]=2sqrt(2)$

arccos is the reversing of the process of cos to give the angle

$=> theta=[ arccos(1/3) = arccos(("adjacent")/("hypotenuse"))]$

So this is giving us 2 length of sides for a right triangle. From which we can work out the tangent value.
Tony B Tony B

By Pythagoras and using the notation in the diagram.

$c^2=b^2+a^2" " =>" " 3^2=1^2+a^2$

Thus $a=sqrt(8) = sqrt(2xx2^2)=2sqrt(2)$

$tan(theta) = ("opposite")/("adjacent")=a/b = (2sqrt(2))/1$

$tan(theta)=2sqrt(2)$

Thus: $" "tan[arccos(1/3)]=2sqrt(2)$

How do you evaluate tan[arccos(1/3)]tan[arccos(1/3)]?