Fórmula tangente de meio ângulo

Predefinição:Trigonometria 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

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

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

$\begin{array}{cc}\sin \alpha & =\frac{2\tan \frac{\alpha }{2}}{1+{\tan }^2\frac{\alpha }{2}}\\ \cos \alpha & =\frac{1-{\tan }^2\frac{\alpha }{2}}{1+{\tan }^2\frac{\alpha }{2}}\\ \tan \alpha & =\frac{2\tan \frac{\alpha }{2}}{1-{\tan }^2\frac{\alpha }{2}}\end{array}$

Verificação

Provas algébricas

Use fórmulas de ângulo duplo e Predefinição:Math,

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

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

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:

$\begin{array}{cc}\sin (a+b)+\sin (a-b)& =\sin a\cos b+\cos a\sin b+\sin a\cos b-\cos a\sin b\\ & =2\sin a\cos b\\ \cos (a+b)+\cos (a-b)& =\cos a\cos b-\sin a\sin b+\cos a\cos b+\sin a\sin b\\ & =2\cos a\cos b\end{array}$

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

$\begin{array}{cc}\sin (\frac{p+q}{2}+\frac{p-q}{2})+\sin (\frac{p+q}{2}-\frac{p-q}{2})& =\sin (p)+\sin (q)\\ & =2\sin \left(\frac{p+q}{2}\right)\cos \left(\frac{p-q}{2}\right)\\ \cos (\frac{p+q}{2}+\frac{p-q}{2})+\cos (\frac{p+q}{2}-\frac{p-q}{2})& =\cos (p)+\cos (q)\\ & =2\cos \left(\frac{p+q}{2}\right)\cos \left(\frac{p-q}{2}\right)\end{array}$

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

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

Predefinição:Referências Predefinição:Esboço-matemática