In trigonometry, tangent half-angle formulas relate the tangent of half of an angle to trigonometric functions of the entire angle.

The tangent of half an angle is the stereographic projection of the circle through the point at angle ${\textstyle \pi }$ radians onto the line through the angles ${\textstyle \pm {\frac {\pi }{2}}}$. Tangent half-angle formulae include $${\displaystyle {\begin{aligned}\tan {\tfrac {1}{2}}(\eta \pm \theta )&={\frac {\tan {\tfrac {1}{2}}\eta \pm \tan {\tfrac {1}{2}}\theta }{1\mp \tan {\tfrac {1}{2}}\eta \,\tan {\tfrac {1}{2}}\theta }}={\frac {\sin \eta \pm \sin \theta }{\cos \eta +\cos \theta }}=-{\frac {\cos \eta -\cos \theta }{\sin \eta \mp \sin \theta }}\,,\end{aligned}}}$$ with simpler formulae when η is known to be 0, π/2, π, or 3π/2 because sin(η) and cos(η) can be replaced by simple constants.

In the reverse direction, the formulae include $${\displaystyle {\begin{aligned}\sin \alpha &={\frac {2\tan {\tfrac {1}{2}}\alpha }{1+\tan ^{2}{\tfrac {1}{2}}\alpha }}\\[7pt]\cos \alpha &={\frac {1-\tan ^{2}{\tfrac {1}{2}}\alpha }{1+\tan ^{2}{\tfrac {1}{2}}\alpha }}\\[7pt]\tan \alpha &={\frac {2\tan {\tfrac {1}{2}}\alpha }{1-\tan ^{2}{\tfrac {1}{2}}\alpha }}\,.\end{aligned}}}$$

Using the angle addition and subtraction formulae for both the sine and cosine one obtains $${\displaystyle {\begin{aligned}\sin(a+b)+\sin(a-b)&=2\sin a\cos b\\[15mu]\cos(a+b)+\cos(a-b)&=2\cos a\cos b\,.\end{aligned}}}$$

Setting ${\textstyle a={\tfrac {1}{2}}(\eta +\theta )}$ and ${\displaystyle b={\tfrac {1}{2}}(\eta -\theta )}$ and substituting yields $${\displaystyle {\begin{aligned}\sin \eta +\sin \theta =2\sin {\tfrac {1}{2}}(\eta +\theta )\,\cos {\tfrac {1}{2}}(\eta -\theta )\\[15mu]\cos \eta +\cos \theta =2\cos {\tfrac {1}{2}}(\eta +\theta )\,\cos {\tfrac {1}{2}}(\eta -\theta )\,.\end{aligned}}}$$

Dividing the sum of sines by the sum of cosines gives $${\displaystyle {\frac {\sin \eta +\sin \theta }{\cos \eta +\cos \theta }}=\tan {\tfrac {1}{2}}(\eta +\theta )\,.}$$

Also, a similar calculation starting with ${\displaystyle \sin(a+b)-\sin(a-b)}$ and ${\displaystyle \cos(a+b)-\cos(a-b)}$ gives $${\displaystyle -{\frac {\cos \eta -\cos \theta }{\sin \eta -\sin \theta }}=\tan {\tfrac {1}{2}}(\eta +\theta )\,.}$$

Furthermore, using double-angle formulae and the Pythagorean identity ${\textstyle 1+\tan ^{2}{\tfrac {1}{2}}\alpha =1{\big /}\cos ^{2}{\tfrac {1}{2}}\alpha }$ gives $${\displaystyle \sin \alpha =2\sin {\tfrac {1}{2}}\alpha \cos {\tfrac {1}{2}}\alpha ={\frac {2\sin {\tfrac {1}{2}}\alpha \,\cos {\tfrac {1}{2}}\alpha {\Big /}\cos ^{2}{\tfrac {1}{2}}\alpha }{1+\tan ^{2}{\tfrac {1}{2}}\alpha }}={\frac {2\tan {\tfrac {1}{2}}\alpha }{1+\tan ^{2}{\tfrac {1}{2}}\alpha }}}$$ $${\displaystyle \cos \alpha =\cos ^{2}{\tfrac {1}{2}}\alpha -\sin ^{2}{\tfrac {1}{2}}\alpha ={\frac {\left(\cos ^{2}{\tfrac {1}{2}}\alpha -\sin ^{2}{\tfrac {1}{2}}\alpha \right){\Big /}\cos ^{2}{\tfrac {1}{2}}\alpha }{1+\tan ^{2}{\tfrac {1}{2}}\alpha }}={\frac {1-\tan ^{2}{\tfrac {1}{2}}\alpha }{1+\tan ^{2}{\tfrac {1}{2}}\alpha }}\,.}$$ Taking the quotient of the formulae for sine and cosine yields $${\displaystyle \tan \alpha ={\frac {2\tan {\tfrac {1}{2}}\alpha }{1-\tan ^{2}{\tfrac {1}{2}}\alpha }}\,.}$$