How do you evaluate tan[arccos(−√32)]?

1 Answer
Lovecraft
Nov 14, 2015

tan(arccos(−√32))=−√33

Explanation:

tan(θ)=sin(θ)cos(θ)
sin(θ)=±√1−cos2(θ)

During the arccosine range, the sine is always positive so

tan(arccos(−√32))=√1−(−√32)2−√32
tan(arccos(−√32))=−2⋅√1−34√3
tan(arccos(−√32))=−2⋅√3⋅√1−343
tan(arccos(−√32))=−2⋅√3⋅√4−343
tan(arccos(−√32))=−2⋅√3⋅√143
tan(arccos(−√32))=−2⋅√3⋅(12)3
tan(arccos(−√32))=−√33

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