# The Proof of the Tangent Half-Angle Formula

In this post, we prove the following trigonometric identity:

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

Proof

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, $\angle CAD = \angle BCD = \frac{\theta}{2}$.

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 $BCD$,

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

Therefore,

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

And we are done.

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