Can you derive a trigonometric function from its inverse dx?

[Eclair_de_XII]

Homework Statement

Arbitrary derivative of inverse trigonometric function:
(sin^-1x) = 1/(√1 - x²)

Homework Equations

f^-1(f(x)) = 1/f`(x)

The Attempt at a Solution

So basically I learned about derivatives of trigonometric functions in class, and I thought maybe this would work: deriving the derivative of the regular sine function from its inverse sine, and then integrating that, to express a regular arbitrary trigonometric function in terms of x. Here is my thinking:

(sin^-1x) = 1/(√1-x²)
f^-1(sin x) = 1/(sin x)` = 1/(√1 - x²)

Which is to say that:

(sin x)` = (√1 - x²)

Integrating this, we have

sin x = ∫(1 - x²)^½dx

I haven't yet figured out how to evaluate this integral, but can someone tell me if my thinking is straight or not?

 

[Samy_A]

Homework Statement

Arbitrary derivative of inverse trigonometric function:
(sin^-1x) = 1/(√1 - x²)

Homework Equations

f^-1(f(x)) = 1/f`(x)

The Attempt at a Solution

So basically I learned about derivatives of trigonometric functions in class, and I thought maybe this would work: deriving the derivative of the regular sine function from its inverse sine, and then integrating that, to express a regular arbitrary trigonometric function in terms of x. Here is my thinking:

(sin^-1x) = 1/(√1-x²)
f^-1(sin x) = 1/(sin x)` = 1/(√1 - x²)

Which is to say that:

(sin x)` = (√1 - x²)

Integrating this, we have

sin x = ∫(1 - x²)^½dx

I haven't yet figured out how to evaluate this integral, but can someone tell me if my thinking is straight or not?

It's not really clear what you are trying to achieve, but there clearly are a number of mistakes.
(sin^-1x) = 1/(√1 - x²): you probably meant $(\sin^{-1})'(x)=\frac{1}{\sqrt{1-x²}}$
f^-1(f(x)) = 1/f`(x): you probably meant $(f^{-1})'(f(x))=\frac{1}{f'(x)}$
In both cases there is a derivative sign missing.

As you started from wrong equations, the final result is also incorrect: $\sin(x) \neq \int \sqrt{1-x²}dx$.

 

[Eclair_de_XII]

Can I ask anybody to refrain from posting in this topic?

 

[HallsofIvy]

Then what was your point in posting it?

 

[Eclair_de_XII]

Never mind... I can't change what other people think, and I certainly can't stop people from saying what's on their mind.

 

[Mark44]

Never mind... I can't change what other people think, and I certainly can't stop people from saying what's on their mind.

You could mark it as "Solved" using the new button at the top right of the thread.

 

[Eclair_de_XII]

http://i.imgur.com/FSPhDrm.png

I don't see it. Is it a members-only feature?

 

[Mark44]

http://i.imgur.com/FSPhDrm.png

I don't see it. Is it a members-only feature?

You're a member, so you should be able to use it. If you scroll all the way up to the top of the thread, don't you see a button that says "MARK SOLVED"?

This is a new change, so it could be that you have an old version of our software in your cache. Try logging out, and then logging back in.

 

[Eclair_de_XII]

I still can't see it. If you can't find it in the link I posted, then I won't be able to find it, either.

It's probably a premium feature.

 

[Mark44]

Maybe your browser doesn't support it. I don't think we have and "premium features."