<s>[INST] Is this proof of $\tan \frac{x}{2} = \frac{1-\cos x}{\sin x}$ incomplete? So, for any angle $\alpha$ :
$$\cos(2\alpha) = \cos^2\alpha - \sin^2\alpha = \dfrac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha+\sin^2\alpha} = \dfrac{\dfrac{\cos^2\alpha-\sin^2\alpha}{\cos^2\alpha}}{\dfrac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha}}= \dfrac{1-\tan^2\alpha}{1+\tan^2\alpha}$$
Now, $\cos\alpha = \cos\Big(2\cdot\dfrac{\alpha}{2}\Big) = \dfrac{1-\tan^2\dfrac{\alpha}{2}}{1+\tan^2\dfrac{\alpha}{2}}$
Now, using the componendo and dividendo rule, we get :
$$\dfrac{\cos\alpha+1}{\cos\alpha-1} = \dfrac{2}{-2\tan^2\dfrac{\alpha}{2}} = \dfrac{-1}{\tan^2\dfrac{\alpha}{2}} \implies \tan^2\dfrac{\alpha}{2} = \dfrac{1-\cos\alpha}{1+\cos\alpha}$$
$$\implies \tan^2\dfrac{\alpha}{2} = \dfrac{(1-\cos\alpha)(1-\cos\alpha)}{(1+\cos\alpha)(1-\cos\alpha)} = \Big(\dfrac{1-\cos\alpha}{\sin\alpha}\Big)^2$$
$$\implies \Bigg|\tan\Big(\dfrac{\alpha}{2}\Big)\Bigg| = \Bigg|\dfrac{1-\cos\alpha}{\sin\alpha}\Bigg|$$
Now, only if $\mathrm{sign}\Big(\tan\dfrac{\alpha}{2}\Big) = \mathrm{sign}\Big(\dfrac{1-\cos\alpha}{\sin\alpha}\Big)$ is true, we can say that $\tan\dfrac{\alpha}{2} = \dfrac{1-\cos\alpha}{\sin\alpha}$
So, I think that without proving that, the proof will be incomplete but my Math textbook doesn't prove it.
So, is it necessary to prove it? If not, why not?
Thanks!
 [/INST] Your comment is correct. You can only get the final equality by proving that $\tan\left(\frac{\alpha}{2}\right)$ and $\frac{1-\cos \alpha}{\sin\alpha}$ have the same sign.
But this is not complicated to prove. $\tan\left(\frac{\alpha}{2}\right)$ is positive if and only if $\frac{\alpha}{2} \in (k\pi, k\pi +\frac{\pi}{2})$. Like $\sin \alpha$ while $1- \cos \alpha$ is always non negative.
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