# Arctan of a fraction of two tangents?

1. Jan 10, 2012

### huey910

if:

arctan[tan(f(x))/tan(g(x))]

then,

f(x)/tan(g(x)) ?

Is this correct or does the tan function at the denominator also vanish?

2. Jan 10, 2012

### JJacquelin

No, both are not correct.
Where is no simple formula of this kind.

3. Jan 10, 2012

### Curious3141

Nothing as simple as the OP wants, clearly, but there is an interesting elementary way to "split" the arctangent of a ratio into a sum of arctangents.

$\frac{\tan x + \tan y}{1 - {\tan x}{\tan y}} = \tan{(x+y)}$

Put $A = \tan{x}, B = \tan{y}$

$\frac{A + B}{1 - AB} = \tan{(\arctan{A}+\arctan{B})}$

Finally take arctan of both sides:

$\arctan{(\frac{A + B}{1 - AB})} = \arctan{A}+\arctan{B}$

This is a pretty nifty formula that works both ways: to convert the arctan of a quotient to a sum of arctans, and vice versa.

To see how to do it the first way, let's say we want to convert the expression for $\arctan({\frac{f}{g}})$ into a sum of arctangents.

We start by setting up the simult. eqn. pair:

$A + B = f$ ---eqn 1

$1 - AB = g$ ---eqn 2

Solving those gives:

$A = \frac{f \pm \sqrt{f^2 + 4(g-1)}}{2}$

and $B = \frac{f \mp \sqrt{f^2 + 4(g-1)}}{2}$

giving the result as:

$\arctan({\frac{f}{g}}) = \arctan{(\frac{f + \sqrt{f^2 + 4(g-1)}}{2})} + \arctan{(\frac{f - \sqrt{f^2 + 4(g-1)}}{2})}$

As I said, this is probably not what the OP was looking for (or thinking of), but it's an interesting result I thought bore mentioning.

4. Jan 10, 2012

### DivisionByZro

Very nice Curious, I like the trick. +1

5. Jan 10, 2012

### JJacquelin

If "smart" is the good word, I would say : Smart !

6. Jan 10, 2012

### Curious3141

Thanks to the above 2 posters for the nice compliments.

In fact, this is also a nice way to see this beautiful relationship:

$$\arctan \phi - \arctan \frac{1}{\phi} = \arctan \frac{1}{2}$$

where $\phi$ is the golden ratio 1.618...

EDIT: Sorry, made a sign error in my original post. Micromass and others - please take note.

Last edited: Jan 10, 2012
7. Jan 10, 2012

### micromass

Very nice!!

Yes, math can be beautiful indeed...

8. Jan 10, 2012

### Curious3141

This is the correct relationship (reposted as I don't want anyone to get it wrong on account of my original typo):

$$\arctan \phi - \arctan \frac{1}{\phi} = \arctan \frac{1}{2}$$