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

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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

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)#