Question

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

Answer 1

`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)`

Answer 2

`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`