Editing Tangent half-angle formula (section)

=== Geometric proofs ===
[[File:Tan.half.svg|right|400px|thumb|The sides of this rhombus have length&nbsp;1.  The angle between the horizontal line and the shown diagonal is&nbsp;{{math|{{sfrac|1|2}} (''a'' + ''b'')}}.  This is a geometric way to prove the particular tangent half-angle formula that says {{math|tan {{sfrac|1|2}} (''a'' + ''b'') {{=}} (sin ''a'' + sin ''b'') / (cos ''a'' + cos ''b'')}}. The formulae {{math|sin {{sfrac|1|2}}(''a'' + ''b'')}} and {{math|cos {{sfrac|1|2}}(''a'' + ''b'')}} are the ratios of the actual distances to the length of the diagonal.]]
Applying the formulae derived above to the rhombus figure on the right, it is readily shown that

<math display="block">\tan \tfrac12 (a+b) = \frac{\sin \tfrac12 (a + b)}{\cos \tfrac12 (a + b)} = \frac{\sin a + \sin b}{\cos a + \cos b}.</math>

In the unit circle, application of the above shows that <math display="inline">t = \tan \tfrac12 \varphi</math>. By [[similar triangles|similarity of triangles]],

<math display="block">\frac{t}{\sin \varphi} = \frac{1}{1+ \cos \varphi}.</math>

It follows that

<math display="block">t = \frac{\sin \varphi}{1+ \cos \varphi} = \frac{\sin \varphi(1- \cos \varphi)}{(1+ \cos \varphi)(1- \cos \varphi)} = \frac{1- \cos \varphi}{\sin \varphi}.</math>

{{clear}}