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-   -   Angle Help!15 = arctan(2/x) - arctan (1/x) (http://www.physicsforums.com/showthread.php?t=465935)

Pawnag3 Jan22-11 11:22 PM

Angle Help!15 = arctan(2/x) - arctan (1/x)
 
1. The problem statement, all variables and given/known data
Basically, solve for x
15 = arctan(2/x) - arctan (1/x)

2. Relevant equations
tan (A-B) = (Tan A -Tan B) / (1+Tan A*Tan B)

3. 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

rock.freak667 Jan22-11 11:26 PM

Re: Angle Help!15 = arctan(2/x) - arctan (1/x)
 
You need to post an attempt before we can help you. Start by taking the tangent of both sides.

Pawnag3 Jan22-11 11:27 PM

Re: Angle Help!15 = arctan(2/x) - arctan (1/x)
 
Yeah, sorry. I just misclicked the first time

Pawnag3 Jan22-11 11:35 PM

Re: Angle Help!15 = arctan(2/x) - arctan (1/x)
 
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!

Mentallic Jan22-11 11:40 PM

Re: Angle Help!15 = arctan(2/x) - arctan (1/x)
 
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.


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