Fórmula tangente de meio ângulo

Na trigonometria, as fórmulas de tangente de meio ângulo relacionam a tangente de metade de um ângulo às funções trigonométricas de todo o ângulo. Entre estas estão as seguintes

Trigonometria

História Funções Funções inversas Aprofundamento

Referência

Lista de identidades CORDIC

Teoria euclidiana

Lei dos senos Lei dos cossenos Lei das tangentes Teorema de Pitágoras

Cálculo

Integração trigonométrica Substituição trigonométrica Integrais de funções Diferenciação trigonométrica

${\begin{aligned}\tan \left({\frac {\eta \pm \theta }{2}}\right)&={\frac {\sin \eta \pm \sin \theta }{\cos \eta +\cos \theta }}=-{\frac {\cos \eta -\cos \theta }{\sin \eta \mp \sin \theta }},\\[10pt]\tan \left(\pm {\frac {\theta }{2}}\right)&={\frac {\pm \sin \theta }{1+\cos \theta }}={\frac {\pm \tan \theta }{\sec \theta +1}}={\frac {\pm 1}{\csc \theta +\cot \theta }},&&(\eta =0)\\[10pt]\tan \left(\pm {\frac {\theta }{2}}\right)&={\frac {1-\cos \theta }{\pm \sin \theta }}={\frac {\sec \theta -1}{\pm \tan \theta }}=\pm (\csc \theta -\cot \theta ),&&(\eta =0)\\[10pt]\tan \left({\frac {1}{2}}(\theta \pm {\frac {\pi }{2}})\right)&={\frac {1\pm \sin \theta }{\cos \theta }}=\sec \theta \pm \tan \theta ={\frac {\csc \theta \pm 1}{\cot \theta }},&&(\eta ={\frac {\pi }{2}})\\[10pt]\tan \left({\frac {1}{2}}(\theta \pm {\frac {\pi }{2}})\right)&={\frac {\cos \theta }{1\mp \sin \theta }}={\frac {1}{\sec \theta \mp \tan \theta }}={\frac {\cot \theta }{\csc \theta \mp 1}},&&(\eta ={\frac {\pi }{2}})\\[10pt]{\frac {1-\tan(\theta /2)}{1+\tan(\theta /2)}}&=\pm {\sqrt {\frac {1-\sin \theta }{1+\sin \theta }}}\\[10pt]\tan {\frac {\theta }{2}}&=\pm {\sqrt {\frac {1-\cos \theta }{1+\cos \theta }}}\end{aligned}}$

Destas, podemos derivar identidades que expressam seno, cosseno e tangente como funções de tangentes de semi-ângulos:

${\begin{aligned}\sin \alpha &={\frac {2\tan {\dfrac {\alpha }{2}}}{1+\tan ^{2}{\dfrac {\alpha }{2}}}}\\[7pt]\cos \alpha &={\frac {1-\tan ^{2}{\dfrac {\alpha }{2}}}{1+\tan ^{2}{\dfrac {\alpha }{2}}}}\\[7pt]\tan \alpha &={\frac {2\tan {\dfrac {\alpha }{2}}}{1-\tan ^{2}{\dfrac {\alpha }{2}}}}\end{aligned}}$

Verificação

Provas algébricas

Use fórmulas de ângulo duplo e sin² α + cos² α = 1,

$\sin \alpha =2\sin {\frac {\alpha }{2}}\cos {\frac {\alpha }{2}}={\frac {2\sin {\frac {\alpha }{2}}\cos {\frac {\alpha }{2}}}{\cos ^{2}{\frac {\alpha }{2}}+\sin ^{2}{\frac {\alpha }{2}}}}={\frac {2{\frac {\sin {\frac {\alpha }{2}}}{\cos {\frac {\alpha }{2}}}}{\frac {\cos {\frac {\alpha }{2}}}{\cos {\frac {\alpha }{2}}}}}{{\frac {\cos ^{2}{\frac {\alpha }{2}}}{\cos ^{2}{\frac {\alpha }{2}}}}+{\frac {\sin ^{2}{\frac {\alpha }{2}}}{\cos ^{2}{\frac {\alpha }{2}}}}}}={\frac {2\tan {\frac {\alpha }{2}}}{1+\tan ^{2}{\frac {\alpha }{2}}}}$

$\cos \alpha =\cos ^{2}{\frac {\alpha }{2}}-\sin ^{2}{\frac {\alpha }{2}}={\frac {\cos ^{2}{\frac {\alpha }{2}}-\sin ^{2}{\frac {\alpha }{2}}}{\cos ^{2}{\frac {\alpha }{2}}+\sin ^{2}{\frac {\alpha }{2}}}}={\frac {{\frac {\cos ^{2}{\frac {\alpha }{2}}}{\cos ^{2}{\frac {\alpha }{2}}}}-{\frac {\sin ^{2}{\frac {\alpha }{2}}}{\cos ^{2}{\frac {\alpha }{2}}}}}{{\frac {\cos ^{2}{\frac {\alpha }{2}}}{\cos ^{2}{\frac {\alpha }{2}}}}+{\frac {\sin ^{2}{\frac {\alpha }{2}}}{\cos ^{2}{\frac {\alpha }{2}}}}}}={\frac {1-\tan ^{2}{\frac {\alpha }{2}}}{1+\tan ^{2}{\frac {\alpha }{2}}}}$

tomando o quociente da fórmula para produtos de seno e cosseno

$\tan \alpha ={\frac {2\tan {\frac {\alpha }{2}}}{1-\tan ^{2}{\frac {\alpha }{2}}}}$

Combinando a identidade pitagórica $\cos ^{2}\alpha +\sin ^{2}\alpha =1$ com a fórmula de ângulo duplo para o cosseno, $\cos 2\alpha =\cos ^{2}\alpha -\sin ^{2}\alpha =1-2\sin ^{2}\alpha =2\cos ^{2}\alpha -1$,

reorganizando, e tomando as raízes quadradas produz

$|\sin \alpha |={\sqrt {\frac {1-\cos 2\alpha }{2}}}$ e $|\cos \alpha |={\sqrt {\frac {1+\cos 2\alpha }{2}}}$

que, mediante divisão, dá

$|\tan \alpha |={\frac {\sqrt {1-\cos 2\alpha }}{\sqrt {1+\cos 2\alpha }}}$ = ${\frac {{\sqrt {1-\cos 2\alpha }}{\sqrt {1+\cos 2\alpha }}}{1+\cos 2\alpha }}$ = ${\frac {\sqrt {1-\cos ^{2}2\alpha }}{1+\cos 2\alpha }}$ = ${\frac {|\sin 2\alpha |}{1+\cos 2\alpha }}$

ou alternativamente

$|\tan \alpha |={\frac {\sqrt {1-\cos 2\alpha }}{\sqrt {1+\cos 2\alpha }}}$ = ${\frac {1-\cos 2\alpha }{{\sqrt {1+\cos 2\alpha }}{\sqrt {1-\cos 2\alpha }}}}$ = ${\frac {1-\cos 2\alpha }{\sqrt {1-\cos ^{2}2\alpha }}}$ = ${\frac {1-\cos 2\alpha }{|\sin 2\alpha |}}$.

Além disso, usando as fórmulas de adição e subtração de ângulos para o seno e o cosseno, obtém-se:

$\cos(a+b)=\cos a\cos b-\sin a\sin b$

$\cos(a-b)=\cos a\cos b+\sin a\sin b$

$\sin(a+b)=\sin a\cos b+\cos a\sin b$

$\sin(a-b)=\sin a\cos b-\cos a\sin b$

A adição pareada das quatro fórmulas acima produz:

${\begin{aligned}\sin(a+b)+\sin(a-b)&=\sin a\cos b+\cos a\sin b+\sin a\cos b-\cos a\sin b\\&=2\sin a\cos b\\[3pt]\cos(a+b)+\cos(a-b)&=\cos a\cos b-\sin a\sin b+\cos a\cos b+\sin a\sin b\\&=2\cos a\cos b\end{aligned}}$

Configurando $a={\frac {p+q}{2}}$ e $b={\frac {p-q}{2}}$ e substituindo produzimos:

${\begin{aligned}\sin \left({\frac {p+q}{2}}+{\frac {p-q}{2}}\right)+\sin \left({\frac {p+q}{2}}-{\frac {p-q}{2}}\right)&=\sin(p)+\sin(q)\\&=2\sin \left({\frac {p+q}{2}}\right)\cos \left({\frac {p-q}{2}}\right)\\[6pt]\cos \left({\frac {p+q}{2}}+{\frac {p-q}{2}}\right)+\cos \left({\frac {p+q}{2}}-{\frac {p-q}{2}}\right)&=\cos(p)+\cos(q)\\&=2\cos \left({\frac {p+q}{2}}\right)\cos \left({\frac {p-q}{2}}\right)\end{aligned}}$

Dividindo a soma dos senos pela soma dos cossenos, chega-se a:

${\frac {\sin(p)+\sin(q)}{\cos(p)+\cos(q)}}={\frac {2\sin \left({\frac {p+q}{2}}\right)\cos \left({\frac {p-q}{2}}\right)}{2\cos \left({\frac {p+q}{2}}\right)\cos \left({\frac {p-q}{2}}\right)}}=\tan \left({\frac {p+q}{2}}\right)$