Problem Regarding Inverse Tangents

  • Thread starter Thread starter studiousStud
  • Start date Start date
  • Tags Tags
    Inverse
AI Thread Summary
To calculate cos(tan-1(d/2x)), one can use a right triangle approach. By letting tan-1(d/2x) equal an angle θ, the opposite side is d/2x, the adjacent side is 1, and the hypotenuse is √(1 + (d/2x)²). This leads to the formula cos(tan-1(d/2x)) = 1/√(d²/(4x²) + 1), which aligns with the result from Wolfram Alpha. The discussion highlights the importance of visualizing trigonometric functions through triangles for better understanding. This method can be applied to similar inverse trigonometric calculations.
studiousStud
Messages
2
Reaction score
0
How do I calculate:

cos(tan-1(d/2x))


This is part of a problem from electric fields an such but it can be regarded as irrelevant
Wolfram Alpha gives an answer of

1/sqrt(d2/(4 x2)+1)


Here's the page:
http://www.wolframalpha.com/input/?i=cos%28tan^-1%28d%2F2x%29%29

I would like to know how, why, and where I can learn more.
Many thanks in advance.
 
Physics news on Phys.org
Ok so to solve something like \cos(\tan^{-1}x) or \sin(\cos^{-1}y) etc. First draw a right triangle and denote one of its angles as \theta. Now if you let \tan^{-1}x=\theta or equivalently, x=\tan(\theta) that means you can now label the opposite side as x, the adjacent side as 1, and thus the hypotenuse will be \sqrt{1+x^2}. Now since we're trying to solve \cos(\tan^{-1}x) this is the same as solving \cos(\theta).
 
WWWWOOOW!
I never thought it like that!
That just stretched my molasses like mind to new limits.
My mind must've glazed over when I saw that inverse.
Thanks for that mind blowing explanation!
OMG OMG OMG OMG OMG OMG
 
mmm … molasses! :-p
 
studiousStud said:
WWWWOOOW!
I never thought it like that!
That just stretched my molasses like mind to new limits.
My mind must've glazed over when I saw that inverse.
Thanks for that mind blowing explanation!
OMG OMG OMG OMG OMG OMG

I'm guessing you're satisfied :wink:
 

Thread 'Family of lines that are at a distance of 5 from the origin'

I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
View full post »

Thread 'Making the denominator of a certain fraction real'

Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
View full post »

Similar threads

How would this equation be simplified?
Replies
4
Views
2K
I What are the applications of inverses of vector functions?
Replies
17
Views
3K
How to deal with INVERSE TANGENT?
Replies
3
Views
3K
Solving equations of the form ##a\sin\theta+b\cos\theta = c##
Replies
2
Views
2K
Somewhat convoluted algebra problem
Replies
1
Views
2K
Computing arctan (inverse tan) function
Replies
10
Views
3K
B Solving Kepler's 1st law as a function of time
Replies
23
Views
4K
Proving a Trigonometric Identity Involving Inverse Functions
Replies
5
Views
2K
I Troubleshooting a difficult integral
Replies
6
Views
2K
A Series expansion of ##(1-cx)^{1/x}##
Replies
3
Views
2K

Hot Threads

Recent Insights

Back
Top