Half Angle Formulas

Theorem

 * 1) $$\sin\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{2}}$$
 * 2) $$\cos\frac{\theta}{2}=\pm\sqrt{\frac{1+\cos\theta}{2}}$$
 * 3) $$\tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}=\frac{1-\cos\theta}{\sin\theta}$$

Proof
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Since $$\cos\theta \geq -1$$, it follows that $$\cos\theta + 1 \geq 0$$.

We also have $$\sin\theta > 0$$ if $$\theta$$ is in the first or second quadrants and $$\sin\theta < 0$$ if $$\theta$$ is in the third or fourth quadrants

But $$\tan\frac{\theta}{2} \geq 0$$ if $$\theta$$ is in the first or second quadrants (because $$\tan\theta > 0$$ if $$\theta$$ is in the first or third quadrant), and $$\tan\frac{\theta}{2} < 0$$ if $$\theta$$ is in the third or fourth quadrants (because $$\tan\theta < 0$$ if $$\theta$$ is in the second or fourth quadrant)

Thus, $$\tan\frac{\theta}{2}$$ and $$\sin\theta$$ have the same sign, so we can drop the $$\pm$$, and we finally have $$\tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}$$

If, when we had $$\tan\frac{\theta}{2} = \pm\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$$, we had instead multiplied by $$\sqrt{1-\cos\theta}$$, we end up with $$\tan\frac{\theta}{2}=\frac{1-\cos\theta}{\sin\theta}$$

Note
Technically, we should also check the boundaries between the first and fourth quadrants and the second and third quadrants.

If $$\theta = \pi + 2k \pi, k \in \Z$$, then $$\tan\frac{\theta}{2}$$ is undefined, $$\frac{\sin\theta}{1+\cos\theta}$$ is undefined, $$\frac{1-\cos\theta}{\sin\theta}$$ is undefined.

If $$\theta = 2k \pi, k \in \Z$$, then $$\tan\frac{\theta}{2} = 0$$, $$\frac{\sin\theta}{1+\cos\theta} = 0$$, and $$\frac{1-\cos\theta}{\sin\theta}$$ is undefined (although by L'Hôpital's Rule, $$\lim_{\theta \to 0}\frac{1-\cos\theta}{\sin\theta} = 0$$).

Thus, $$\frac{1-\cos\theta}{\sin\theta}$$ is not a perfect formula for $$\tan\frac{\theta}{2}$$.