# Can you derive a trigonometric function from its inverse dx?

1. Jan 28, 2016

### Eclair_de_XII

1. The problem statement, all variables and given/known data
Arbitrary derivative of inverse trigonometric function:
(sin-1x) = 1/(√1 - x2)

2. Relevant equations
f-1(f(x)) = 1/f(x)

3. 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?

2. Jan 28, 2016

### Samy_A

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: $\sin(x) \neq \int \sqrt{1-x²}dx$.

Last edited: Jan 28, 2016
3. Jan 28, 2016

### Eclair_de_XII

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

Last edited: Jan 28, 2016
4. Jan 28, 2016

### HallsofIvy

Then what was your point in posting it?

5. Jan 28, 2016

### 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.

6. Jan 28, 2016

### Staff: Mentor

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

7. Jan 28, 2016

### Eclair_de_XII

8. Jan 28, 2016

### Staff: Mentor

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.

9. Jan 28, 2016

### 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.