# Help with inverse trig functions

1. Sep 5, 2012

### vysero

Here is my problem:

cot(arcsin(x))

my awnser:

cot= x/(1-x)^1/2

The online program were suppose to use says im wrong but im not sure what I did wrong.

2. Sep 5, 2012

### SammyS

Staff Emeritus
What's the question that you're trying to answer?

The answer you give makes no sense whatsoever.

3. Sep 5, 2012

### Staff: Mentor

Draw a right triangle, with θ being the acute angle in the triangle. Let θ = arcsin(x), and label the sides and hypotenuse according to this definition.

The problem boils down to evaluating cot(θ).

Your answer SHOULD NOT start with "cot =" since cot by itself is meaningless in this context. If you are doing your work on paper, your work should start with cot(arcsin(x)) = ... and end up with what you found, which is actually not too far off.

4. Sep 6, 2012

### HallsofIvy

Staff Emeritus
The simplest way to do this is to imagine a right triangle with one angle $\theta$, "opposite side" to $\theta$ of length x, and hypotenuse of length 1 so that $sin(\theta)= x/1$ and thus $\arcsin(x)= \theta$.

Now, what is the length of the near side and what is $tan(\theta)= tan(arcsin(x))$?

You are missing a square in your answer.