How do you evaluate #sec((11pi)/6)#?

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How do you evaluate #sec((11pi)/6)#?

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Answer 1 · 2016-10-25T12:34:42

#(2sqrt3)/3#

Explanation:

#sec ((11pi)/6) = 1/(cos ((11pi)/6)#
Trig table and unit circle -->
#cos ((11pi)/6) = cos (-pi/6 + (12pi)/6) = cos (-pi/6 + 2pi) = #
#= cos (-pi/6) = cos (pi/6) = sqrt3/2#
There for:
#sec ((11pi)/6) = 2/sqrt3 = (2sqrt3)/3#

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How do you evaluate #sec((11pi)/6)#?

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