Prove identity sec^-1(x) = cos^-1(1/x)

[physicsernaw]

Homework Statement

Find and prove the identity sec^-1(x) in terms of cos^-1(arg) (Note that 1/cos^-1(x) is not equal to sec^-1(x).

Homework Equations

None.

The Attempt at a Solution

sec(sec^-1(x)) = x

1/cos(sec^-1(x)) = x

1/cos(cos^-1(x)) = 1/x

1/cos(cos^-1(1/x)) = 1/1/x = x

cos^-1(1/x) = sec^-1(x).

Is this correct?

 

[Mark44]

Homework Statement

Find and prove the identity sec^-1(x) in terms of cos^-1(arg) (Note that 1/cos^-1(x) is not equal to sec^-1(x).

Homework Equations

None.

The Attempt at a Solution

sec(sec^-1(x)) = x

1/cos(sec^-1(x)) = x

1/cos(cos^-1(x)) = 1/x

How do you justify the previous step?

1/cos(cos^-1(1/x)) = 1/1/x = x

cos^-1(1/x) = sec^-1(x).

Is this correct?

 

[HallsofIvy]

The fundamental problem with try to prove "$sec^{-1}(x)= cos^{-1}(1/x)$" is that it is NOT true! This is NOT an identity. For example, if $x= \pi/4$ $sec^{-1}(x)= \sqrt{2}/2$ while $1/x= 4/\pi$, cos(1/x)= 3.41, approximately.

 

[SammyS]

The fundamental problem with try to prove "$sec^{-1}(x)= cos^{-1}(1/x)$" is that it is NOT true! This is NOT an identity. For example, if $x= \pi/4$ $sec^{-1}(x)= \sqrt{2}/2$ while $1/x= 4/\pi$, cos(1/x)= 3.41, approximately.

$\displaystyle \sec(\pi/4)=\sqrt{2}\$

$\displaystyle \sec^{-1}(\pi/4)\$ is undefined, since $\displaystyle\ \ \pi/4<1\ .$

 

[SammyS]

I would be inclined to write sec^-1(x) as $\displaystyle \ \sec^{-1}\left(\frac{1}{1/x}\right)\,,\$ then use the identity, cos(cos^-1(u) = u , for the denominator.