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

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January 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?
January 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 $\tan \alpha = \frac{1}{3}$ and $\tan \beta = \frac{1}{3}$ .

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

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

Bet you didn't see that coming ;)
January 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?
January 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 ....
January 27th 2008, 11:49 PM
topher0805

Oh, I see. 2 is never an adjacent side. Disregard my last post.