1. ## Simplify arctan expression

Simplify:

$\displaystyle 3arctan(2) - arctan(2/11)$

I'd know how to do it if it weren't for the 3 infront of arctan(2).

EDIT: I guess I could do it like this:

$\displaystyle arctan(2) + arctan(2) + arctan(2) - arctan(2/11)$

That seems a bit over the top though. I'm still looking for an easier solution.

Also, I've never encountered three arctan terms before:

$\displaystyle arctan(2) + arctan(3) + arctan(4)$

I'm guessing I need to first simplify $\displaystyle arctan(2) + arctan(3)$, and then use that to get $\displaystyle (arctan(2) + arctan(3)) + arctan(4)$ by using the $\displaystyle tan(u+v)$ rule again. And of course make sure it's in the range of $\displaystyle \pi/2<arctan(2)+arctan(3)+arctan(4)<3\pi/2$

2. Call $\displaystyle \alpha=3\arctan2$ & $\displaystyle \beta=\arctan\frac2{11}$

Some identity could be useful for this.

The answer is $\displaystyle \pi$

3. I already knew the answer and I've found a way to calculate it, so now I'm just looking for an easier solution than the one I proposed.

4. Originally Posted by Spec
Simplify:

$\displaystyle 3arctan(2) - arctan(2/11)$

I'd know how to do it if it weren't for the 3 infront of arctan(2).
I am just curious. Can you please show us how to simplify
arctan(2) -arctan(2/11)
then?

5. $\displaystyle tan(arctan(2)-arctan(\frac{2}{11})) = \frac{tan(arctan(2)) - tan(arctan(\frac{2}{11}))}{1 + tan(arctan(2)) * tan(arctan(\frac{2}{11}))} =$
$\displaystyle \frac{2 - \frac{2}{11}}{1 + 2 * \frac{2}{11}} = \frac{20}{15} = \frac{4}{3}$

Therefore,
$\displaystyle arctan(2) - arctan(\frac{2}{11}) = arctan(\frac{4}{3}) + n\pi$

Note that's there's only one n for which the solution is within the defined range.

6. Originally Posted by Spec
$\displaystyle tan(arctan(2)-arctan(\frac{2}{11})) = \frac{tan(arctan(2)) - tan(arctan(\frac{2}{11}))}{1 + tan(arctan(2)) * tan(arctan(\frac{2}{11}))} =$
$\displaystyle \frac{2 - \frac{2}{11}}{1 + 2 * \frac{2}{11}} = \frac{20}{15} = \frac{4}{3}$

Therefore,
$\displaystyle arctan(2) - arctan(\frac{2}{11}) = arctan(\frac{4}{3}) + n\pi$

Note that's there's only one n for which the solution is within the defined range.
Umm, I see.

Then 3arctan(2) -arctan(2/11)
= [arctan(2) +arctan(2)] +[arctan(2) -arctan(2/11)]
===> -4/3 +4/3
= 0

Hence,
3arctan(2) -arctan(2/11)
= arctan(0)
= 0 or pi