How do you evaluate #sec^-1(-sqrt2)#?

2 Answers
Mar 6, 2018

#color(green)( theta = (3pi) / 4 + 2kpi->k inZZ)#, #color(blue )(theta = (5pi) / 4 + 2kpi->k inZZ)#

Explanation:

#theta = sec^-1 (-sqrt2)#

#sec theta = - sqrt2#

#sec theta# is negative in #II, III# quadrants.

Hence #theta = (3pi) / 4, (5pi) / 4#

Since trigonometric function #sec# is periodic and repeats every #2pi#,

#color(green)( theta = (3pi) / 4 + 2kpi->k inZZ)#, #color(blue )(theta = (5pi) / 4 + 2kpi->k inZZ)#

Mar 6, 2018

#x = (3pi)/4 + 2kpi#
#x = (5pi)/4 + 2kpi#

Explanation:

#sec^-1 (-sqrt2)#
Find arccos x, that #cos x = 1/(sec) = 1/-sqrt2 = - sqrt2/2#.
#cos x = -sqrt2/2#
Trig Table and unit circle give 2 general solutions:
#x = +- (3pi)/4 + 2kpi#
Reminder:
arc #x = - (3pi)/4# is co-terminal to arc #x = (5pi)/4#.
Answers:
#x = (3pi)/4 + 2kpi#
#x = (5pi)/4 + 2kpi#