Wikipedia:Reference desk/Archives/Mathematics/2014 August 10

= 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 &alpha;, and I understand how they get 1 / cos &alpha;, but how do they get tan &beta;? - 110.20.126.106 (talk) 04:13, 10 August 2014 (UTC)


 * It's just applying the simple tan formula in the purple triangle to find the side opposite &beta;, 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 &alpha; and &beta;.    D b f i r s   06:41, 10 August 2014 (UTC)
 * it's a geometric proof of a tan(&alpha; + &beta;) - I'm not sure I'm following how they get tan &beta; in the first place! - Letsbefiends (talk) 09:48, 10 August 2014 (UTC)
 * Once you have 1 / cos &alpha;, use that and the angle &beta; in the bottom left corner to get tan &beta; / cos &alpha;.  Then use that and the angle &alpha; in the top right to get tan &beta;.--80.109.80.78 (talk) 14:27, 10 August 2014 (UTC)
 * In other words, once you have the short side of the pink triangle (tan &beta; / cos &alpha;) 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 &alpha;) and set up the equation  cos &alpha; = x / (tan &beta; / cos &alpha;)  then solving for x gives you tan &beta;. El duderino (abides) 20:26, 10 August 2014 (UTC)


 * ...or simply go from (tan &beta; / cos &alpha;) and observe that in the white triangle, the vertical side is cos &alpha; (tan &beta; / cos &alpha;), or
 * tan &beta; cos &alpha; / cos &alpha; and cancel the cosine. - ¡Ouch! (hurt me / more pain) 06:51, 13 August 2014 (UTC)