# Arctangent Simplification

• Sep 7th 2009, 03:04 PM
Manuel
Arctangent Simplification
Hello!

I am stuck with this challenge:

I want to derive an expression for angle BAC, using angle ABC, distances BC and AB.

Translating the problem to mathematics, It looks like this:
$\displaystyle BAC=tan^{-1} {\lgroup {\gamma \cdot sin(\alpha) \over 1-\gamma \cdot cos(\alpha)} \rgroup}$

Where

$\displaystyle \gamma \equiv {\overline {BC} \over \overline {AB}}$

After some manipulation, trying to simplify the expression in order to get the angle directly (since that sine and cosine are converting an angle to cartesian, an then the arctangent goes backward, to an angle again - seems redundant)

I could only manage to get this:

$\displaystyle BAC=tan^{-1} \lgroup { {2 \cdot \gamma \cdot tan({\alpha \over 2})} \over {(1-\gamma)+(1+\gamma) \cdot tan^2({\alpha \over 2}))}} \rgroup$

I can't simplify more than that - what am I missing? Can it be so ugly?

Greetings, and thanks in advance!

Manuel
• Sep 7th 2009, 03:08 PM
Manuel
http://img216.imageshack.us/img216/5343/trim.pngI forgot to include the image! (Giggle)