How do you evaluate #Tan[arccos ( - sqrt ( 3 ) / 2 ) ]#?

1 Answer
Lovecraft
Nov 14, 2015

#tan(arccos(-sqrt(3)/2)) = -(sqrt(3))/3#

Explanation:

#tan(theta) = sin(theta)/cos(theta)#
#sin(theta) = +-sqrt(1 - cos^2(theta))#

During the arccosine range, the sine is always positive so

#tan(arccos(-sqrt(3)/2)) = sqrt(1 - (-sqrt(3)/2)^2)/(-sqrt(3)/2)#
#tan(arccos(-sqrt(3)/2)) = -(2*sqrt(1 - 3/4))/sqrt(3)#
#tan(arccos(-sqrt(3)/2)) = -(2*sqrt(3)*sqrt(1 - 3/4))/3#
#tan(arccos(-sqrt(3)/2)) = -(2*sqrt(3)*sqrt((4 - 3)/4))/3#
#tan(arccos(-sqrt(3)/2)) = -(2*sqrt(3)*sqrt(1/4))/3#
#tan(arccos(-sqrt(3)/2)) = -(2*sqrt(3)*(1/2))/3#
#tan(arccos(-sqrt(3)/2)) = -(sqrt(3))/3#

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