Problem Regarding Inverse Tangents

[studiousStud]

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.

 

[Mentallic]

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).

 

[studiousStud]

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

 

[tiny-tim]

mmm … molasses!

 

[Mentallic]

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