Evaluating Inverse Trig Functions

[CB1101]

For part of my homework, I need to evaluate arcsec(0) and arccos(sqrt(14)). I have no idea where to begin, can someone point me in the right direction?

 

[Ishuda]

For part of my homework, I need to evaluate arcsec(0) and arccos(sqrt(14)). I have no idea where to begin, can someone point me in the right direction?

An arcsec of a number is an angle such that sec(x) equals that number. So what angle x gives sec(x) = 0?

Basically the same for the arccos, what angle x gives cos(x) = $\displaystyle \sqrt{14}$?

However, I think you may have copied those wrong since there is no Real Number x such that sec(x) < 1 nor is there one such that cos(x) >1.

 

[CB1101]

here's a picture of it: http://postimg.org/image/yxnnnngnx/

 

[CB1101]

An arcsec of a number is an angle such that sec(x) equals that number. So what angle x gives sec(x) = 0?

I can't figure out what that number is. It looks like it is undefined to me.

 

[Steven G]

For part of my homework, I need to evaluate arcsec(0) and arccos(sqrt(14)). I have no idea where to begin, can someone point me in the right direction?

When you have no idea where to begin then you look at the definitions!

 

[Ishuda]

here's a picture of it: http://postimg.org/image/yxnnnngnx/

Do you know what a complex number is and how to define the cosine, for example, in terms of exponentials?

We have
$\displaystyle cos(x) = \frac{1}{2} (e^{i x} - e^{-ix}) = \sqrt{14}$
letting u = e^ix and a = $\displaystyle \sqrt{14}$,
this translates to
u² - 2 a u - 1 = 0
or, using the quadratic equation,
u = $\displaystyle a \pm \sqrt{a^2 + 1}$
or,
i x = ln(u)

You can do the same sort of thing for the secant.