How do you find the exact value of #cos^-1 (-1)#?

1 Answer
Aug 24, 2015

for angles in the range #[0,2pi]#
#cos^-1(-1) = pi# (radians)

Explanation:

If #cos^(-1)= theta#
#rArrcolor(white)("XXXXX")cos(theta) = -1#

This means that the adjacent side is equal in magnitude to the hypotenuse but negative.

Within the range #[0,2pi]#
this is only true at #theta=pi (= 180^@)#

For all solutions (unrestricted in range:
#color(white)("XXX")cos^-1(-1)=pi+n2picolor(white)("XXX")AAn in ZZ#