Calculating tan(arcsin(1/7)) using trigonometric equations

[beborche]

Homework Statement

Calculate tan(arcsin(1/7))

Homework Equations

arcsinv = x => sinx = v
sinx = (opposite/hypotenuse) (in a right-angled triangle)
tanx = (opposite/adjacent)

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?

 

[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.

 

[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 can't check it). I got the answer:
tan(arcsin(1/7)) = tan(v) = 1(4sqrt(3))

 

[dextercioby]

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

 

[BruceW]

@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 can't check it). I got the answer:
tan(arcsin(1/7)) = tan(v) = 1(4sqrt(3))

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.

 

[beborche]

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