The Proof of the Tangent Half-Angle Formula

In this post, we prove the following trigonometric identity:

\displaystyle \tan \frac{\theta}{2} = \frac{\sin\theta}{1 + \cos \theta} = \frac{1 - \cos \theta}{\sin \theta}.


Consider a semi-circle with “center” O and diameter AB and radius equal to 1 unit as shown below.  If we let \angle BOC =\theta, then by the Inscribed Angle Theorem, \angle CAB = \frac{\theta}{2}.

Draw CD perpendicular to OB as shown in the second figure. We can compute for the sine and cosine of \theta which equal to the lengths of CD and OD, respectively. In effect, BD = 1 - \cos \theta and AD = 1 + \cos \theta.

Draw BC. Notice that ADC and CDB are similar triangles, so their corresponding angles are congruent. So, \angle CAD = \angle BCD = \frac{\theta}{2}.

half angle proof without words


Now, we compute for the tangent of  \displaystyle \frac{\theta}{2}.

In triangle ACD,

\displaystyle \tan \frac{\theta}{2} = \frac{\sin \theta}{1 + \cos \theta}.

In triangle BCD,

\displaystyle \tan \frac{\theta}{2}= \frac{1-\cos \theta}{\sin \theta}.


\displaystyle \tan \frac{\theta}{2} = \frac{\sin\theta}{1 + \cos \theta} = \frac{1 - \cos \theta}{\sin \theta}.

And we are done.


The last figure is the proof without words of R. J. Walker.

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