How do I find the value of #sec(-pi)#?

1 Answer
Alan P.
Sep 6, 2015

#sec(-pi) = -1#

Explanation:

Using a unit circle in standard position,
the angle #(-pi)# is a semicircle with its terminal point on the negative X-axis (at #(x,y) = -1,0#)

The length of the hypotenuse will be
#color(white)("XXX")sqrt((-1)^2+0^2) = 1#

and since the definition of #sec# is
#color(white)("XXX")sec = ("hypotenuse")/("x-coordinate")# (of unit circle)

#sec(-pi) = 1/(-1) = -1#

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