# arctan(1/3)+arctan(1/2)

• Jan 27th 2008, 09:16 PM
JOhkonut
arctan(1/3)+arctan(1/2)
simplify: Tan-1(1/3)+tan-1(1/2)

is there an algebraeic way to do this?
• Jan 27th 2008, 11:14 PM
mr fantastic
Quote:

Originally Posted by JOhkonut
simplify: Tan-1(1/3)+tan-1(1/2)

is there an algebraeic way to do this?

Yes.

Oh, now you want to see a way ......

There are many ways. Here's a simple but brutal one (watch out for falling Grand Pianos):

Let $\displaystyle \tan \alpha = \frac{1}{3}$ and $\displaystyle \tan \beta = \frac{1}{3}$.

Then $\displaystyle \tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = \frac{\frac{1}{3} + \frac{1}{2}}{1 - \left( \frac{1}{3} \right) \left( \frac{1}{2} \right)} = 1$.

Therefore $\displaystyle \alpha + \beta = \tan^{-1} \left( \frac{1}{3}\right) + \tan^{-1} \left( \frac{1}{2}\right) = \tan^{-1} (1) = \frac{\pi}{4}$.

Bet you didn't see that coming ;)
• Jan 27th 2008, 11:41 PM
topher0805
I think the easiest way is to just use the special triangles. Does that not work in this case?
• Jan 27th 2008, 11:45 PM
mr fantastic
Quote:

Originally Posted by topher0805
I think the easiest way is to just use the special triangles. Does that not work in this case?

Well it's your suggestion, sport. So why don't you try it - see if it does work ....
• Jan 27th 2008, 11:49 PM
topher0805
Oh, I see. 2 is never an adjacent side. Disregard my last post.