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

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Homework Help Overview

The problem involves calculating tan(arcsin(1/7)), which requires understanding the relationships between the sides of a right triangle and trigonometric functions. The original poster attempts to visualize the problem using a triangle but expresses uncertainty about how to proceed without knowing all side lengths.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of using a triangle to represent the problem, with one suggesting to label the sides in terms of an undetermined constant. There is also mention of using the Pythagorean theorem to find missing side lengths, and a participant questions the need for a calculator in this context.

Discussion Status

The discussion is ongoing, with participants exploring different methods to find the adjacent side of the triangle and confirming the use of the Pythagorean theorem. Some guidance has been provided regarding the relationship between the sides of the triangle and trigonometric identities, but no consensus has been reached on a final solution.

Contextual Notes

Participants note that the use of calculators is not permitted, which influences their approach to solving the problem. There is also a reference to a lack of access to solution keys for verification.

beborche
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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:
math_problem.png

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?
 
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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.
 
@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))
 
[tex]\tan\arcsin x = \frac{\sin \arcsin x}{\cos \arcsin x} = \frac{x}{\sqrt{1-x^2}}[/tex]
 
beborche said:
@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.
 
Alright. Yes i meant 1/(4sqrt(3)).
 

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