Angle Help15 = arctan(2/x) - arctan (1/x)

  • Thread starter Thread starter Pawnag3
  • Start date Start date
  • Tags Tags
    Angle
Click For Summary

Homework Help Overview

The problem involves solving the equation 15 = arctan(2/x) - arctan(1/x), which is situated in the context of trigonometric functions and inverse tangent relationships.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to manipulate the equation using the tangent subtraction formula and expresses concern over the lack of real roots in their quadratic result. Some participants suggest taking the tangent of both sides as a potential approach.

Discussion Status

The discussion has evolved with the original poster indicating a possible resolution after realizing a misinterpretation of the equation. However, another participant points out that rearranging the equation leads to a quadratic with no real solutions, suggesting further exploration is needed.

Contextual Notes

There is a noted confusion regarding the equation's constants, with the original poster initially stating "15" instead of "tan 15," which impacts the problem's setup. The discussion reflects on assumptions about the equation's structure and potential solutions.

Pawnag3
Messages
8
Reaction score
0

Homework Statement


Basically, solve for x
15 = arctan(2/x) - arctan (1/x)

Homework Equations


tan (A-B) = (Tan A -Tan B) / (1+Tan A*Tan B)

The Attempt at a Solution


I really tried everything.
My first step was to:
Let y = arctan (2/x)
Therefore, tan y = 2/x
Similarly, u = 1/x
Then, tan (y-u) = (Tan y -Tan u) / (1+Tan u*Tan y) = 15
15 = (2/x-(1/x) / (1+(2/x^2)
15 = (1/x) / (x^2 + 2 / x^2)
15 = (x) / (x^2+2)
15x^2 + 30 - x = 0
Which has no real roots :(
But, with guess and check, it's around 0.65
 
Physics news on Phys.org
You need to post an attempt before we can help you. Start by taking the tangent of both sides.
 
Yeah, sorry. I just misclicked the first time
 
Alright, sorry guys to waste your time, but I believe I figured it out. Thanks for the hint of "tanning" both sides.
Instead of 15, it's supposed to be tan 15.
So that,
x/(x^2+2) = tan 15
x = 0.64
x = 3.08 (approximately)

Thanks for the help!
 
If you rearrange that equation you'll get the quadratic

[tex]x^2-\frac{1}{tan(15)}x+2=0[/tex]

There are no real solutions to this quadratic.
 

Similar threads

  • · Replies 17 ·
Replies
17
Views
4K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 17 ·
Replies
17
Views
3K