# 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

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

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

 $\displaystyle\tan{\frac{x}{2}}\,=\,\pm\sqrt{\frac{1-\cos{x}}{1+\cos{x}}}.$ (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 and the denominator of the radicand the first time by  $1-\cos{x}$  and the second time by  $1+\cos{x}$; note that  $1-\cos^{2}{x}=\sin^{2}{x}$:

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

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

Title tangent of halved angle TangentOfHalvedAngle 2013-03-22 17:00:32 2013-03-22 17:00:32 pahio (2872) pahio (2872) 9 pahio (2872) Derivation msc 26A09 DerivationOfHalfAngleFormulaeForTangent GoniometricFormulae