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

1 Answer
Nov 22, 2016

#sec(-pi/3) = 2#

Explanation:

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!