How many tangent half-angle formulas are there?

Question

\begin{align} \tan \frac{\alpha+\beta} 2 & = \frac{\sin\alpha+\sin\beta}{\cos\alpha + \cos\beta} \tag 1 \\[10pt] \tan \left( \frac \pi 4 \pm \frac \alpha 2 \right) & = \sec\alpha \pm \tan\alpha \tag 2 \\[10pt] \frac{1 + i\tan\frac\alpha2}{1-i\tan\frac\alpha2} & = e^{i\alpha} \tag 3 \\[10pt] \tan\frac\alpha2\cdot\tan\frac\beta2 = \tan\frac\gamma2 & \text{ if and/or only if } \left|\frac{\cos\alpha+\cos\beta}{1+\cos\alpha\cos\beta}\right| = \left|\cos\gamma\right| \tag 4 \end{align} All of these relate tangents of half angles to trigonometric functions of whole angles.

Are there other tangent half-angle formulas essentially different from these? And how does one decide whether they're essentially different? (I'm not sure I would count the case of $(1)$ in which $\alpha=0$ as essentially different. One could make a case that $(1)$ is not essentially different from $(2)$. We're all accustomed to the mapping $(x,y)\mapsto \dfrac{x+y}{1-xy}$ in connection with tangents, but $(4)$ relates tangents to $(x,y)\mapsto\dfrac{x+y}{1+xy}.$)

Answer 1

For $\alpha+\beta+\gamma=\pi$, \begin{align} \tan\tfrac\alpha2\,\tan\tfrac\beta2\tan\tfrac\gamma2 &= \frac{\cos\alpha+\cos\beta+\cos\gamma-1}{\sin\alpha+\sin\beta+\sin\gamma}. \end{align}

Edit

Some more

\begin{align} \tan\tfrac\alpha2\,\tan\tfrac\beta2\tan\tfrac\gamma2 &= \frac{2\sin\alpha\sin\beta\sin\gamma}{(\sin\alpha+\sin\beta+\sin\gamma)^2} \\ &= \frac{\sin2\alpha+\sin2\beta+\sin2\gamma}{2(\sin\alpha+\sin\beta+\sin\gamma)^2} \end{align}

\begin{align} \cot\tfrac\alpha2+\cot\tfrac\beta2+\cot\tfrac\gamma2-(\tan\tfrac\alpha2+\tan\tfrac\beta2+\tan\tfrac\gamma2) &= 2(\cot\alpha+\cot\beta+\cot\gamma) \end{align}

Answer 2

I don't know if the following are essentially different, but

$$\cot(x/2)=\cot(x)+\csc(x)$$

and hence

$$\begin{align} \tan(x/2)&=\frac{1}{\cot(x)+\csc(x)}\\\\ &=\frac{\sin(x)}{\cos(x)+1}\\\\ &=\frac{1-\cos(x)}{\sin(x)}\\\\ &=\csc(x)-\cot(x) \end{align}$$

Answer 3

Here's a fragment of an answer. I found this derivation in G. H. Hardy's Divergent Series. $$ 1 + e^{i\theta} + e^{2i\theta} + e^{3i\theta} + \cdots = \frac 1 {1-e^{i\theta}}, $$ The identity above holds if $|e^{i\theta}|<1$ but the series diverges of $\theta$ is real.

Therefore \begin{align*} & \sin\theta + \sin2\theta+ \sin3\theta + \cdots = \operatorname{Im}\frac 1 {1-e^{i\theta}} = \operatorname{Im} \frac{1- e^{-i\theta}}{2-2\cos\theta} \\[10pt] = {} & \frac{ \sin\theta }{2-2\cos\theta} = \frac 1 2 \cot\frac\theta2. \end{align*} $$ \text{If } \operatorname{Re}\theta>0 \text{ then } \sin\theta + \sin2\theta+ \sin3\theta + \cdots = \frac 1 2 \cot\frac\theta2. $$ $$ \text{or if you like } \frac 1 {2(\sin\theta + \sin2\theta + \sin3\theta + \cdots)} = \tan\frac\theta2. $$ If $\theta$ is real, this doesn't converge in the usual sense, but perhaps it converges in the sense of some "summation method".