# My reasoning: the inverse secx not the reciprocal of inverse cosx?

anniecvc
This came up when I was trying to evaluate to a decimal value an arcsec(something), but on TI-83 there is no inverse secant button. So, I punched 1/(arccos(something)). Naturally, this came up as a domain error.

Then my instructor pointed out arcsesc is not 1/arccosx. Oops.

Thinking about it, the domain of the cos-1x is [-1,1].

The domain the sec-1x is (-∞,-1]U[1,∞).

So although secx =1/cosx is an identity, it could not be the case for the inverses since there is a discrepancy in domain.

Also, the inverse cosine is the function which takes the y value and maps it to the x value, e.g cos-1(1/2) = ∏/3. The inverse secant for this domain value of 1/2 would be...well it wouldn't be since this value is not in the domain of the arcsec! That is an example of why the arcsec is not the multiplicative inverse of arccos.

Is there anything I should add or other pressing points or ways of looking at this?

## Answers and Replies

Homework Helper
Yes. You can also see this by playing around a bit with the definitions.

arcsec(x) is the value y such that sec(y) = x, in other words, arcsec( 1/cos(x) ) = x.

If you look at 1/cos(1 / arccos(x)) you don't get x back. The best you can do is plug arccos(x) into the secant and get sec( arccos(x) ) = 1/cos( arccos(x) ) = 1/x, leading to the identity 1/sec(arccos(x)) = x.

Combining these, you can conclude that
arcsec(1/cos(x)) = 1/sec(arccos(x)).

You can try numerous tricks like taking the secant on both sides,
1/cos(x) = sec(1/sec(arccos(x)),
but because of the pesky 1/...'s you cannot get your (inverse) trig functions to cancel out.

Staff Emeritus
but because of the pesky 1/...'s you cannot get your (inverse) trig functions to cancel out.
##\operatorname{arcsec}(x) = \arccos(1/x)## works quite nicely, as do ##\cos(\operatorname{arcsec}(x)) = \sec(\arccos(x)) = 1/x##.

• 2 people
Gold Member
As others have said, playing around with trig identities can give you interesting simplifications for nested trig functions. For instance;

##\sin^{2}x + \cos^{2}x = 1 \implies \sin x = \sqrt{1 - \cos^{2}x} \implies \sin(\arccos x) = \sqrt{1 - x^2}##.

Try finding an expression not involving a trig function for ##\sec(\arctan x)##, and perhaps other combinations if you're interested, e.g. ##\sin(\arctan x)##.

anniecvc
As others have said, playing around with trig identities can give you interesting simplifications for nested trig functions. For instance;

##\sin^{2}x + \cos^{2}x = 1 \implies \sin x = \sqrt{1 - \cos^{2}x} \implies \sin(\arccos x) = \sqrt{1 - x^2}##.

Try finding an expression not involving a trig function for ##\sec(\arctan x)##, and perhaps other combinations if you're interested, e.g. ##\sin(\arctan x)##.

Question:

Let x=sint => arcsinx = t

sin2t + cos2t = 1

cos2t = 1 - sin2t

cos2t = 1 - x2

cost = + (1-x2)1/2

cos(arcsinx) = (1-x)1/2

But why is it necessarily the positive root of cos2t that we take? I don't believe the "lengths of a side of a right triangle" argument suffices, since x can be a positive or negative number. I want to argue that since we are taking t = arcsinx,
then 0 ≤ t ≤∏, but then the cos t for
∏/2 < t ≤ ∏ is negative, so it's not guaranteed that the root is positive.

anniecvc
##\operatorname{arcsec}(x) = \arccos(1/x)## works quite nicely, as do ##\cos(\operatorname{arcsec}(x)) = \sec(\arccos(x)) = 1/x##.

Hm interesting, I worked it out and it does seem to be the case.

So in the future, I now know what to input into my calculator if I wanted to find an arcsecx. Thank You!

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