How do you evaluate #sec^-1( 2/ sqrt3)#?

2 Answers
Nov 25, 2015

#pi/6#

Explanation:

Ask yourself: "secant of what angle gives me #2/sqrt3#?"

Since #sectheta=1/costheta#, an easier question to ask would be, "cosine of what angle gives me #sqrt3/2#?"

We know that #cos(pi/6)=sqrt3/2#, so #sec(pi/6)=2/sqrt3#.

While secant is also positive in quadrant four, and there are an infinite amount of coterminal angles where secant is #2/sqrt3#, the domain of #sec^-1(x)# is restricted from #(0,pi)#, so #pi/6# is the only valid angle.

Nov 26, 2015

#30^@# or #330^@#

Explanation:

#sec^-1(2/sqrt(3))#
#sectheta=2/sqrt(3)#
#1/costheta=2/sqrt(3)#
#costheta=sqrt(3)/2#
#theta=cos^-1(sqrt(3)/2)#
#theta=30^@#

However, since #2/sqrt(3)# is positive, there is more than one answer because according to the CAST rule, #cos# is positive in more than one quadrant:

http://mathonline.wikidot.com/cast-rule

#cos# is positive in quadrants #1# and #4#.

To find the other angle that would also give an answer of #2/sqrt(3)#, subtract #30^@# from #360^@#. This ensures that your principal angle is in quadrant #4# :

#360^@-30^@#
#=330^@#

#:.#, #sec^-1(2/sqrt(3))# is #30^@# or #330^@#.