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August 10

Double angle tangent function

Hey guys, I've been puzzling out the following diagram for hours now, and have to admit I'm stumped:

I understand how they get tan α, and I understand how they get 1 / cos α, but how do they get tan β? - 110.20.126.106 (talk) 04:13, 10 August 2014 (UTC)[reply]

It's just applying the simple tan formula in the purple triangle to find the side opposite β, then applying the simple cosine formula (adjacent over hypotenuse) in the top right triangle. If it helps, call the unknown sides "x" and "y" and express these in terms of α and β. Dbfirs 06:41, 10 August 2014 (UTC)[reply]

it's a geometric proof of a tan(α + β) - I'm not sure I'm following how they get tan β in the first place! - Letsbefiends (talk) 09:48, 10 August 2014 (UTC)[reply]

Once you have 1 / cos α, use that and the angle β in the bottom left corner to get tan β / cos α. Then use that and the angle α in the top right to get tan β.--80.109.80.78 (talk) 14:27, 10 August 2014 (UTC)[reply]

In other words, once you have the short side of the pink triangle (tan β / cos α) which is also the hypotenuse of the small triangle in the upper right, you can let x be the adjacent side of the small triangle (with α) and set up the equation cos α = x / (tan β / cos α) then solving for x gives you tan β. El duderino ^(abides) 20:26, 10 August 2014 (UTC)[reply]

...or simply go from (tan β / cos α) and observe that in the white triangle, the vertical side is cos α (tan β / cos α) , or

tan β cos α / cos α and cancel the cosine. - ¡Ouch! (^{hurt me} / _{more pain}) 06:51, 13 August 2014 (UTC)[reply]