How do you evaluate #sec ((-pi)/3)#?

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Answer 1 · 2016-11-22T15:55:51

#sec(-pi/3) = 2#

When drawn in standard position, negative angles are clockwise while positive angles are counterclockwise.

Hence, #2pi - |"negative angle"| = "positive angle"#. Our positive angle is therefore

#2pi - pi/3 = (5pi)/3#

The question becomes evaluate #sec((5pi)/3)#. We know that #sectheta = 1/costheta#.

#=> 1/cos((5pi)/3)#

#(5pi)/3# has a reference angle of #pi/3#. #cos(pi/3) = 1/2#, and cosine is positive in quadrant IV.

#=> 1/(1/2)#

#=> 2#

Hopefully this helps!