# Trigonometric question

1. Dec 2, 2011

### beborche

1. The problem statement, all variables and given/known data
Calculate tan(arcsin(1/7))

2. Relevant equations
arcsinv = x => sinx = v
sinx = (opposite/hypotenuse) (in a right-angled triangle)

3. The attempt at a solution
Well I realize that arcsin(1/7) is an angle. Let's call it v. I also know that arcsin(1/7) => sin(v) = 1/7.

So i try to picture myself a triangle like this:

After this I'm lost. I'd like to use tanx = o/a, but I only know the value of o. I'm not sure how these kind of problems are solved. Any ideas?

2. Dec 2, 2011

### BruceW

technically, you only have the shape of the triangle, so it would be more correct to label the sides as 7a and a (where a is some undetermined constant).

From here you can find the other side using a well known theorem about triangles. (I'm sure you know it, maybe you've forgotten). And then once you have the other side, you can find tan.

3. Dec 2, 2011

### beborche

@BruceW

Thanks man. I knew I could just hit arcsin(1/7) in on my calculator to find the angle and then from there find the adjacent side, but we're not allowed to use the calculator.

But then I realized you meant Pythagoras theorem, which I ofcourse didnt think about...

Thanks again! I think I solved it (got no key/solutions so cant check it). I got the answer:
tan(arcsin(1/7)) = tan(v) = 1(4sqrt(3))

4. Dec 2, 2011

### dextercioby

$$\tan\arcsin x = \frac{\sin \arcsin x}{\cos \arcsin x} = \frac{x}{\sqrt{1-x^2}}$$

5. Dec 2, 2011

### BruceW

Yes, Pythagoras, that's the one. And you meant 1/(4sqrt(3)), right?

And P.S. dextercioby has written the same method, but all in one step.

6. Dec 2, 2011

### beborche

Alright. Yes i meant 1/(4sqrt(3)).