tangent of halved angle

The formulae

$$\cos 2\alpha =1-2{\sin }^2\alpha ,$$

$$\cos 2\alpha =2{\cos }^2\alpha -1$$

may be solved for  $\sin \alpha$  and  $\cos \alpha$, respectively.  One gets the equations

$$\sin \alpha =\pm \sqrt{\frac{1-\cos 2\alpha }{2}},\cos \alpha =\pm \sqrt{\frac{1+\cos 2\alpha }{2}},$$

where the signs have to be chosen according to the quadrant where the angle $\alpha$ is.  Changing $\alpha$ to $\frac{x}{2}$ and dividing these equations gives us the formula

$\tan {\displaystyle \frac{x}{2}}=\pm \sqrt{{\displaystyle \frac{1-\cos x}{1+\cos x}}}.$(1)

Also here one must chose the sign according to the quadrant of  ${\displaystyle \frac{x}{2}}$.

We obtain two alternative forms of (1) when we multiply both the numerator andthe denominator of the radicand the first time by  $1-\cos x$  and thesecond time by  $1+\cos x$; note that  $1-{\cos }^{2}x={\sin }^{2}x$:

$\tan {\displaystyle \frac{x}{2}}={\displaystyle \frac{1-\cos x}{\sin x}},$(2)

$\tan {\displaystyle \frac{x}{2}}={\displaystyle \frac{\sin x}{1+\cos x}}$(3)

Here,  $\sin x$  determines the sign of the hand sides;it can be justified that it has always the same sign as $\tan \frac{x}{2}$.